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NCERT Solutions For Class 11 Chemistry Chapter 12 Organic Chemistry Basic Principles And Techniques Long Question And Answers

Question 1. Depict the bonding in the following compounds In terms of atomic orbitals involved and predict all the bond angles:

1. CD₃CH=CH₂
2. CH₃OCH₃

Answer:

1. The 3 carbon atoms are sp³ , sp² and sp² -hybridised respectively. Therefore the bond angles about these carbons are 109.5°, 120° and 120° respectively corresponding to tetrahedral and trigonal planar geometries.

2. The carbon and oxygen atoms are all sp³ -hybridized. So, the bond angles are nearly 109.5° corresponding to tetrahedral geometry.

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Class 11 Chemistry Chapter 12 Organic Chemistry Long Questions

Question 2. Mention the number of primary (1°), secondary (2°), and tertiary (3°) hydrogen atoms in the following

Answer:

Hydrogen Atoms Given in the table:

Question 3. How many alkyl groups can be derived from the alkane, (CH₃)₂ CHCH₂ CH(CH₃)₂ and why? Write their IUPAC names.
Answer:

Since this hydrocarbon molecule contains 3 types of nonequivalent hydrogen atoms, the removal of these hydrogen atoms gives 3 different alkyl groups.

These are as follows:

Question 4. Write the IUPAC names of the following compounds

Answer:

1. 3-ethyl-4-methylhept-5-en-2-one
2. 3,3,5-trlmethylhex-1 -en-2-ol.
3. l-bromo-4-metlvylheptan-3-on«.
4. l-etliyl-4-methylcyclohexane.
5. Cyclohexylcyclohexnne.
6. 1,3-dlcyclopropylpropanc.
7. 2-metliyl-2-cyclopropylpropnne.
8. N -ethyl- N -methylpropan-2-nmine.
9. 4-hydroxy-4-methylpontan-2-one.
10. 3-methylpent-l-ene.

Question 5. Arrange the given carbocations To increase stability and explain two-order

Answer:

The order of increasing stability of these carbocation Is :

Being an aromatic one [(4 n + 2)n -electron system, where n = 1], the carbocation (I) is the most stable. The carbocation (II) is effectively resonance stabilized. So, its stability is greater than that of (ill) and (IV) (which are not resonance-stabilized) but less than that of (I).

The carbocation (III) is stabilized by +1 and the hyperconjugation effect of the methyl group and its stability is less than (II). The carbocation (IV) is destabilized by the stronger -I effect of the — CF₃ group, so it is the least stable one.

Question 6. Terf-Butyl chloride (Me₃CCl) does not participate in D+ S_N2 reaction—explain with reasons
Answer:.

Due to severe steric hindrance caused by three methyl groups, the backside attack on the central carbon by the nucleophile becomes completely inhibited and it is for this reason, that terf-butyl chloride does not participate in the S_N2 reaction..

Question 7. Write the resonance structures of CH₂=CH —CHO and compare their stabilities.
Answer:

The compound is a resonance hybrid of three structures:

The stability order of these structures is I > II > III.

The uncharged structure (I) is the most stable one. The charged structure (II) is moderately stable because the more electronegative oxygen atom bears the negative charge and the less electronegative carbon atom bears the positive charge.

Also, the octet of carbon is not filled up. The charged structure (III) is the least stable because the more electronegative O-atom bears the positive charge and the less electronegative C-atom bears the negative charge. Also, the octet of the O-atom is not filled up

Question 8. Which is more stable and why: (CH₃)₃C, (CD₃)₃C
Answer:

Since D is more electron-releasing than H, — CD₃ is more electron-releasing than — CH₃. So, (CD₃)₃C⁺ is expected to be more stable than (CH₃)₃C⁺. But actually, this is true and this can be explained in terms of hyperconjugation. Since the C—H bond is weaker than the C— D bond the hyperconjugative stability of (CH₃)₃C⁺ is greater than that of (CD₃)₃C⁺

Question 9.

1. How many stereoisomers of formula, CH₃ would be possible if methane was a pyramid with a rectangular base? Draw them.
2. How many stereoisomers of formula, CH₂YZ would be possible if methane was a pyramid with a square base? Draw them.
3. What is the relationship (diastereoisomers, enantiomers, conformational isomers, homomers i.e., identical structures or constitutional isomers) between the members of given pairs of structures?

Answer:

1. Two stereoisomers Mirror (enantiomers) are Q possible if methane H / was a pyramid with a rectangular base.

2. Three stereoisomers (1, 2, and 3) are possible if methane is a pyramid with a square base. (1 and 2) are enantiomers. (1 and 3) and (2 and 3) are two pairs of diastereoisomers.

3.

1. . Diastereoisomers
2. Homomers
3. Geo¬ metrical isomers
4. Constitutional isomers
5. Conformational isomers
6. Enantiomers.

Question 10. Although Quorine is more electronegative than iQfi chlorine, fluorobenzene has a lower dipole moment (p = 1.63D ) than chlorobenzene (μ = 1.75D ).
Ans.

Both fluorine in fluorobenzene and chlorine in structure) chlorobenzene withdraws electrons from the ring by the -I effect and donates electrons to the ring by the + R effect. Because of the smaller size of fluorine, the +R effect involving orbitals of similar sizes (2p of both F and C) is much stronger. So, the moment due to the stronger -I effect of fluorine is considerably neutralized by the moment due to the +R effect.

Hence, fluorobenzene possesses a net dipole moment which is relatively low (1.63D). On the other hand, because of the larger size of chlorine, the + R effect involving orbitals of dissimilar sizes (3p of Cl and 2p of C) is much weaker than the -I effect which is somewhat lower due to lower electronegativity of chlorine. So, the moment due to the +R effect is much smaller than the moment due to the -I effect. Hence, chlorobenzene possesses a relatively high net dipole moment (1.75D

Question 11. The negatively charged carbon atom in the structure

1. Is sp² -hybridized while the negatively charged carbon atom in
2. Is sp³ -hybridised —Explain.

Answer:

The negatively charged carbon atom of a resonance-stabilized carbanion is sp² -hybridized. The carbanion (I) is resonance stabilized. So the negative carbon atom Is sp².
hybridized

On the other hand, the carbanion (2) Is not resonance stabilized because a double bond cannot be formed at the bridgehead position of small bicycle systems (Hrodt’s rule). Hence, the negatively charged carbon of the carbanion (II) is sp³ -hybridized.

Question 12. Mention the state of hybridization of the starred (*) carbon atoms in each of the following compounds.

Answer:

1. sp²
2. sp
3. sp²
4. sp
5. sp
6. sp³

Question 13. How many σ and π -bonds are present in each of the following molecules?

1. CH₃-C≡-CH= CH₂

2. CH₂=CH-CH=C=CHCH₃ 

3. 

Answer:

1. σ -bonds =10, π -bonds =3
2. σ -bonds = 13 , π -bonds = 3
3. σ -bonds =10, π -bonds = 3

Question 14.  Which atoms in each of the following molecules remain in the same plane and why?

1.  CH₃CH= CH₃
2.  C₆H₅C≡ CCH₃
3. CH₃CH=C=C=CHCH₃
4. CH₃COCH₂CH₃

Answer:

sp²-carbon atoms and the atoms attached to them lie in one plane.

sp² -carbon atoms and the atoms attached to them lie in one plane. Also, sp carbon atoms & the atoms attached to them lie not only in one plane but also in one line.

Lie in one plane

Lie in one plane

Question 15. Mention the number of primary (1°), secondary (2°), tertiary (3°), and quaternary (4°) C -atoms present in the given molecules: °
Answer:

Answer:

Question 16. Write down the IUPAC name of a hydrocarbon having a 4° C-atom with molecular formula, C₆H₁₄. How many monochrome derivatives of this hydrocarbon are possible? Write their structures
Answer:

The hydrocarbon corresponding to the molecular formula C6H14 and containing one tertiary carbon atom is CH₃C(CH₃)₂CH₂CH₃ (C-2 is a quaternary carbon atom). Its IUPAC name is 2,2-dimethylbutane.

Since the alkane contains three types of non-equivalent hydrogen atoms, three monobromo derivatives of the alkane are possible. These are: (CH₃)₃CCHBrCH₃ and (CH₃)₃CCH₂CH₂Br

Question 17. Racemic tartaric acid and meso-tartaric add are both optically inactive—why?
Answer:

Racemic tartaric add is an equimolar mixture of (+) and (-)-tartaric adds. In racemic tartaric add, therefore, the rotatory power of the (+) enantiomer is neutralized by the rotatory power of the (-) enantiomer (external compensation) and for this reason, the racemic tartaric add is optically inactive.

On the other hand, the meso-tartaric ad is optically inactive because it has a plane of symmetry and is superimposable on its mirror image. In fact, in this case, the optical rotation of one half of the molecule is exactly canceled by the optical rotation of the other half (internal compensation).

Organic Chemistry Basic Principles and Techniques Long Q&A

Question 18. How many isomers of butene are possible? What type 1 of isomerism do they exhibit?
Ans.

Three structural isomers of butene are possible: CH₃CH₂CH=CH₂ (But-l-ene), CH₃CH=CHCH₃ (But-2- ene), and (CH₃)₂C=CH₂ (2-methylpropene). Again, but-2- ene exists as two geometrical isomers (diastereoisomers):

Hence, there are in total 4 isomers of butene: but-1-ene,  cis but-2-ene, irans-but-2-ene, 2-methylpropene

Question 19. Give examples of

1. An optically inactive compound containing an asymmetric carbon atom
2. An optically active compound containing no asymmetric carbon.

Answer:

1. Meso-tartaric acid containing two asymmetric carbon atoms is optically inactive because it has a plane of symmetry, l.e., tire molecule is superimposable on its mirror Image

2. Penta-2,3-diene (an abC=C=Cab type of allene) is optically active because it is not superimposable on its mirror image.

Question 20. Name a compound having two similar asymmetric carbon atoms and give its structure. What type of isomerism does it exhibit? Draw Fischer projection formulas of these isomers and comment on their optical activity. How are they related to each other?
Answer:

Tartaric acid has two similar asymmetric carbon atoms (HOOC — *CHOH —*CHOH —COOH). The compound exhibits optical isomerism.

Fisher projection formulas of its isomers are as follows:

The relations among the isomers are as follows: I and II are enantiomers; I and III are diastereoisomers and II and III are diastereoisomers

Question 21.

1. Give the structure and IUPAC name of an optically active alkane having the lowest molecular mass. Is there another alkane of the same molecular mass also optically active?
2. Give an example of a compound that exhibits both optical & geometrical isomerism.

Answer:

Such a compound must contain an asymmetric carbon atom which will remain attached to an H-atom and three different alkyl groups (smaller size).

So, the optically active alkane having the lowest molecular mass’ is, 3-methylhexane [CH₃CH₂  *CH(CH₃)CH₂CH₂CH₃ ]. Another optically active alkane with the same molecular mass is 2,3-dimethyl pentane [CH₃*CH₂CH(CH₃)CH(CH₃)₂] which is a chain isomer of the first one.

Pent-3-en-2-oI [CH₃*CH(OH)CH=CHCH₃] exhibits both geometrical and optical isomerism because the compound contains an asymmetric carbon atom and each of the doubly bonded carbon atoms is attached to two different groups.

Question 22. The following two isomers may be called diastereoisomers but not enantiomers —why? Explain why these are optically inactive.

Answer:

The given pair of isomers have the same structure but different configurations. They are neither superimposable nor bear mirror-image relationships with each other. So they are related to each other as a pair of diastereoisomers and not as enantiomers. Each of these two isomers has a plane of symmetry, i.e., each is superimposable on its mirror image, so these are optically inactive.

Question 23. p nitrophenol is more acidic than phenol.
Answer:

In p-nitrophenol, the electron-attracting — NO₂ group by its stronger -R effect and relatively weaker -I effect makes the oxygen atom relatively more positively polarised compared to the oxygen atom of phenol. As a result, the O — H bond in nitrophenol dissociates more easily to give H⁺ ions. For this reason, p-nitrophenol is more acidic than phenol.

Answer:

In p-nitrosamine, the electron-attracting — NO₂ group by its stronger -R effect and relatively weaker -I effect makes the nitrogen atom of the — NH₂ group relatively more positive compared to the nitrogen atom of aniline. As a result, the availability of the unshared pair of electrons on nitrogen atoms in p-nitroaniline is highly reduced as compared to the unshared electron pair on nitrogen in aniline. For this reason, p-nitroaniline behaves as a weaker base compared to aniline.

Question 24. The dipole moment of vinyl chloride CH₂=CHCI) is less than the dipole moment of ethyl chloride (CH₃CH₂Cl) —explain.
Answer:

In vinyl chloride, the moment caused by the -I effect of Cl-atom (μσ) is partially neutralized by the moment caused by its +R effect (pn). As a result, the value of the net moment of vinyl chloride decreases and it is lower than that of ethyl chloride in which only the stronger -I effect of chlorine operates.

Question 25. Arrange the following ions in order of increasing basicity and explain the order

1. CH₃ ^–CH
2. CH ≡^–C
3. CH₂ =^–CH

Answer:

The order  of increasing basicity of the given ions is:

⇒ \(\mathrm{CH} \equiv \stackrel{\ominus}{\mathrm{C}}(\mathrm{II})<\mathrm{CH}_2=\stackrel{\ominus}{\mathrm{C}} \mathrm{H}(\mathrm{III})<\mathrm{CH}_3 \stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2(\mathrm{I})\)

C -atoms bearing the negative charge in carbanions (1), (2) & (3) are sp³, sp, and sp² -hybridized respectively. Percentages of s -the character of these three hybrid orbitals are 25%, 50%, and 33% respectively.

As the s -the character of hybrid orbital increases, C -atoms bearing the negative charge in carbanions (1), (2) & (3) are sp³, sp, and sp² -hybridized respectively. Percentages of s -the character of these three hybrid orbitals are 25%, 50%, and 33% respectively. As the s -s-character of hybrid orbital increases,

Question 26. Give example:

1. A non-nucleophilic anion
2. A planar carbocation
3. An aromatic carbocation
4. An aromatic carbanion
5. A reagent which acts as a source of carbanion
6. A reaction that does not proceed through intermediate
7. An aprotic polar solvent
8. An ambident nucleophile
9. A neutral electrophile
10. A group that stabilizes a carbocation
11. A group that stabilizes a carbanion
12. An alkyl group which does not supply electrons to a double bond by hyperconjugation
13. A carbocation that can be stored for years.

Answer:

1. BF^–₄

2. Benzyl cation

3. Cyclopropenvl cation

4. Cyclopentadienyl anion

5. Grignard reagents \(\left(\mathrm{R}^{\delta-}-\mathrm{M}^{\delta+} \mathrm{gX}\right)\)

6. S_N2 reaction

7. Dimethyl formamide [DMP, Me₂NCHO]

8. ^–CN \((: \stackrel{\ominus}{\mathrm{C}} \equiv \mathrm{N}: \longleftrightarrow: C=\stackrel{\ominus}{\mathrm{N}}:)\)

[Nucleophiles having two or more available sites for nucleophilic attack are called ambident nucleophiles)

9. Dichlorocarbene (: CCl₂)

10. \(-\ddot{O}:\mathrm{CH}_3\)

11. —NO₂

12. —C(CH₃)₃

13. Triphenylmethylfluoroborate Ph₃⁺CBF₄^–

Question 27. Explain the given basicity order in aqueous medium: )₂NH(2°) > CH₃NH₂(1°) > (CH₃)₃N(3°)
Answer:

The basic strength of amines in the aqueous medium depends on two factors:

Increased electron density on the N-atom makes an amine more basic.

So considering the +1 effect of different numbers of alkyl groups on the N -atom, the basic strength of amines should follow the order:

Again, the basicity of an amine increases as stabilization of the conjugate acid, through solvation, increases. The conjugate acid of primary amine attains maximum stability through intermolecular H -bond formation with three molecules of water, while the conjugate base of tertiary amine attains minimum stabilization through such H-bond formation with only one molecule of water.

Thus based on the stability of the conjugate acids, the basic strength of amines should follow the order: CH₃—NH₂ > (CH₃)₂NH > (CH₃)₃N As a consequence of these two opposite orders of basicity practically we find the given sequence of basicity in aqueous medium: (CH₂)₂NH(2°) > CH₃NH₂(1°) > (CH₃)₃N(3°)

Question 28. Which of the two: 2NCH₂CH₂O^– or CH₃CH₂O^– is expected to be more stable and why?
Answer:

O₃ NCH₂CH₃O^– is expected to be more stable than CH₃CH₂O^–. —NO₂ group by strong-I effect disperses the negative charge on O-atom in ion and stabilizes it. On the other hand, the CH₃CH₂— group by its +1 effect tends to intensify the negative charge on the oxygen atom and hence destabilizes it.

Question 29. CH₃Cl undergoes hydrolysis more easily than C₆H₅Cl. Explain.
Answer:

Unshared electron-pair on chlorine atom in chloro-benzene becomes involved in resonance interaction with the n -electrons of a benzene ring. As a result, the C — Cl bond assumes some double bond character. Thus, the C — Cl bond becomes much stronger and so the displacement of chlorine atom from the ring becomes difficult, i.e., the compound does not undergo hydrolysis easily.

On the other hand, the C —Cl bend in CH₃— Cl gets no opportunity to assume a double bond character. So r undergoes hydrolysis readily under ordinary conditions.

Question 30. Benzyl chloride participates in S_N1 action even though it is a primary (1°) substrate. Explain.
Answer:

Carbocation produced from benzyl chloride in the first step (rate-determining step) of the S_N1 reaction is stabilized by resonance. Thus, benzyl chloride participates in the S_N1 reaction even though it is a primary (1 °) alkyl halide.

Class 11 Chemistry Chapter 12 Organic Chemistry Long Questions

Question 31. The Bond Dissociation enthalpy of the C₆H₅CH₂ – H bond is much less than the CH₂-H bond Explain
Answer:

Benzyl radical (C₆H5₅C^•H₂) produced by homolytic cleavage of the C_sp³ —H bond of toluene is considerably stabilized by resonance. But, the stability of methyl radical (^•CH₃, obtained by homolytic fission of the C—H bond of methane Is not stabilized by any factor and in fact, it is very much unstable. Thus, the C—H bond dissociation enthalpy of toluene is much less than the C —H bond dissociation enthalpy of methane. ,

Question 32. N, N, 2,6-Tetramethylaniline is more basic than N- dimethylaniline. Explain.
Answer:

Because of steric interaction involving two ortho-methyl groups and two methyl groups attached to nitrogen, the unshared electron-pair on N is not involved in resonance interaction with n -electrons (steric inhibition of resonance) o: ring. To avoid steric strain, the — NMe₂ group rotates about the C — N bond axis and thereby loses coplanarity with the ring. As a result, nitrogen can easily donate its unshared electron pair to a proton.

On the other hand, no such steric inhibition occurs in N- dimethylaniline because the two ortho H-atoms are relatively much smaller in size. The unshared electron-pair on N-atom becomes involved in resonance interaction with the ring and therefore, is not fully available for taking up a proton. This explains why N, N,2, O-tetramethylaniline is more basic than N, N -dimethylaniline.

Question 33. Chloroform is more acidic than fluoroform. Explain.
Answer:

CF^–₃  the conjugate base of fluoroform (CHF₃ ), is stabilized by the -I effect of 3 F-atoms. But CCl^–₃, the conjugate base of chloroform (CHCl₃ ), is relatively more stabilized by the somewhat weaker -I effect of 3 Cl-atoms along with rf-orbital resonance (Cl has vacant rf-orbital). So chloroform is more acidic than fluoroform.

Question 34. How can you separate benzoic acid and nitrobenzene from their mixture by the technique of extraction using an appropriate chemical reagent?
Answer:

The mixture is shaken with a dilute sodium bicarbonate solution when benzoic acid gets converted to sodium benzoate and dissolves in water leaving nitrobenzene behind. The mixture is extracted with ether or chloroform when nitrobenzene goes into the organic layer. After separating the organic layer, it is distilled to get nitrobenzene. The aqueous layer is acidified with dilute HC1 when benzoic acid gets precipitated. It is obtained by filtration.

C₆H₅COOH + NaHCO₃→C₆H₅COONa(Sodium benzoate (Soluble)) + CO₂ + H₂O

Question 35.

1. Which atoms in a toluene molecule always remain in the same plane and why?
2. Which atoms in a propyne molecule remain in a straight line and why?

Answer:

1. An sp² is a hybridized carbon atom and the atoms directly attached to it always remain in the same plane. Therefore, in toluene ), all the atoms except 3 H-atoms of the methyl group {i.e., seven C – and five H -atoms) remain in the same plane.

2. An sp -sp-hybridized carbon atom and the atoms directly attached to it remain in a straight line. Therefore, in the propyne molecule (CH₃—C = CH), all the atoms except the 3 hydrogen atoms of the methyl group remain in the same straight line.

Question 36. Write the state of hybridization of C -atoms in the following compounds and predict the shape of each of the molecules :

1. H₂C=O
2. CH₂Cl
3. HC = N
4. CH₂=C=CH₂
5. CH₂=C=C=CH₂

Answer:

1. sp² -hybridised C-atom, trigonal planar;
2. sp² hybridized C-atom, tetrahedral;
3. sp -hybridized C-atom, linear;
4. sp², sp and sp² -hybridized C -atoms respectively, elongated tetrahedron;
5. sp², sp , sp and sp² -hybridised C atom respectively, planar

Question 37.

1. Expand each of the following condensed formulas into their complete structural formulas:

1. HOCH₂CH₂NH₂
2. CH₃CH=CHCOCH₃
3. CH₃C ≡ CCH₂COOH

2. Write bond-line formulas of the following two compounds

1. CH₃CH₂CH₂CH₂CHBrCH₂CHO
2. (C₂H₅)₂CHCH₂OH

Answer:

1. Condensed Formulas:

2. Bond Line Formulas:

Question 38.

1. Expand each of the following bond-line formulas to show all the atoms including C and H.

2. How many  σ and π -bonds are present in

1. CH₂=CH—CN and
2. CH₂=C=CHCH₃

Answer:

1. Bond-line formulas:

2. σ and π -bonds:

1. σ _{C -H} = 3: σ _{C -C} = 2: π _{C= C} = 1: σ _{C – N} = 1

π _{C= N}= 2 i.e., total σ -bond = 6 and total  π -bond = 3

2.  1. σ _{C -H} = 6: σ _{C -C} = 3: π _{C= C} = 2:

i.e total σ – bond = 9 and total π – bond = 2

Question 39. Which of the given compounds may exist as two or more isomeric forms? Give the structures and names of the possible isomers.

1. CHBr₃
2. C₂H₂Cl₄
3. C₃H₈
4. C₂H₅F
5. C₂H₄Br₂
6. C₆H₄Cl₂

Answer:

1. No isomer is possible

2. Two isomers are possible : ClCH₂CCl₃ (1,1,1,2-tetrachloroethane), Cl₂CHCHCl₂ (1, 2,2-tetrachloroethane)

3. No isomer is possible

4. No isomer is possible;

5. Two isomers are possible : BrCH₂CH₂Br (1, 2- dibromoethane), CH₃CHBr₂ (1,1-dibromoethane)

6. Three isomers are possible:

Question 40. Write the structures and IUPAC names of the H—C—C=C-C—C-OH compounds with the molecular formula, C₄H₈O₂. H (in) H
Answer:

The molecular formula, C₄H₈O₂conforms to the general formula of monocarboxylic acids and esters. Therefore the following structures of monocarboxylic acids and esters can be written with the given formula.

Question 41. Write the structures and IUPAC names of the compounds with molecular formula, C₄H₁₀.
Answer:

The molecular formula, C₄H₁₀ conforms to the general formula of monohydric alcohols and ethers. Therefore, the structures of the following monohydric alcohols and ethers can be written with the given molecular formula.

Question 42. Which of the following compounds will exhibit tautomerism and which do not? Give reasons.

1. CH₃COCH₃
2. C₆H₅COC₆H₅
3.  C₆H₅COCH₃
4. C₆H₅CHO
5. Me₃CCOCMe₃

Answer:

Tautomerism is possible for those carbonyl compounds which contain at least one α-H atom (the H-atom attached to a carbon atom adjacent to the C= O group) Therefore, compounds 1  and 3 containing a-H atom exhibit tautomerism while compounds 2, 4, and 5 containing no α -H atom do not exhibit tautomerism

Question 43. Designate the following pairs as metamers, chain isomers, position isomers, functional isomers, and stereoisomers. Also, indicate which are not isomers at all
Answer:

1. (CH₃)₂CHC(CH₃)₃, (CH₃)₄C

2.  CH₃CH₂CH₂OH, CH₃OCH₂CH₃ CH₃

3.

4. (CH₃)CHCOCH₃, (CH₃)₂CHCH₂CHO COOH

5. CH₃OCH₂CH₂CH₃, CH₃CH₂OCH₂CH₃

6.

Answer:

1. The molecular formulas of these two compounds are not identical. Thus, these two are not isomers.
2. Functional isomers
3. Position isomers
4. Functional isomers
5. Metamers
6. Stereoisomers (geometrical isomers)

Question 44. Which of the following compounds will exhibit geometrical isomerism and why?

Answer:

1. One of the doubly bonded carbon atoms is attached to two identical atoms (Cl). Therefore, the compound will not exhibit geometrical isomerism.
2. Each doubly bonded C-atom is attached to two different groups (C-2 is attached to CH3 and H, while C-3 is attached to Cl and C₂H₅ ). So, it will exhibit geometrical isomerism.
3. The compound will not exhibit geometrical isomerism because each of the two terminal doubly bonded carbon is attached to two identical atoms (H).
4. The compound will exhibit geometrical isomerism because each of the two ring carbons is attached to two different groups (H and CH₃).
5. The compound will exhibit geometrical isomerism because each of the two ring carbons is attached to two different groups (H and CH₃ ).
6. The compound will not exhibit geometrical isomerism because one of the two doubly bonded carbon atoms is attached to two identical groups (ring moiety.

Question 45. Which of the following compounds are optically active and why?

Answer:

1. The compound contains one asymmetric C-atom (CH₃CHBrCH₂CH₃). So, it is not superimposable on its mirror image and hence, it is optically active.
2. The molecule has a plane of symmetry and it is superimposable on its mirror image. Therefore, it is optically inactive.
3. The molecule is not superimposable on its mirror image. So, it is optically active.
4. The molecule is not superimposable on its mirror image. So, it is optically active.
5. This planar compound is superimposable on Its mirror image. So, it is optically inactive.

Question 46. Which type of stereoisomerism is exhibited by the compound, CH₃CH=CH —CH=CHC₂H₅? How many stereoisomers are possible? Draw the structures and designate them as E/Z.
Answer:

The compound exhibits geometrical isomerism because the groups attached to each of the terminal doubly bonded carbon atoms are different. The number of geometrical isomers = 2ⁿ (n = number of double bonds) =2² = 4. These are as follows:

NCERT Solutions Class 11 Chemistry Chapter 12 Long Questions

Question 47. Name a compound having; two dissimilar asymmetric carbon atoms and write Its structure. What type of isomerism does It exhibit? Draw Fischer projection formulas of the Isomers and comment on their optical activity. How are (lie Isomers related to each other? m Explain the orders of acidity of carboxylic acids:
Answer:

Compound containing two dissimilar asymmetric carbon atoms  \(\left(\mathrm{CH}_3 \stackrel{*}{\mathrm{C}} \mathrm{HOH} \stackrel{*}{\mathrm{C}} \mathrm{HBr} \mathrm{CH}_3\right)\).

The compound may have 2ⁿ(n = no. of dissimilar asymmetric carbon atom)2² = 4 possible stereoisomers. Projection formulas of these Isomers are as follows

The relations among the isomers are as follows: (1, 2) and (3, 4) are two pair of enantiomers whereas (1, 3), (1, 4), (2, 3), and (2, 4) are four pairs of diastereomer.

Question 48. Arrange cis-but-2-ene, trans-but-2-ene, and but-I-ene H3C in increasing order of their stability and give reason.
Answer:

But-2 -ene (CH₃CH=CHCH₃) contains sixa-H atoms while but-l-ene (CH₃CH₂CH=CH₂) contains only two o-H -atoms capable of participating in hyperconjugation. Therefore, because of more effective hyperconjugation, but-2- ene is thermodynamically more stable than but-l-ene.

Again, due to steric interaction between two methyl groups on the same side of the double bond in cis-but-2-ene, it is relatively less stable than the trans-isomer in which no such steric interaction operates between the methyl groups situated on the opposite sides of the double bond.

Question 49. CH₃Cl is unreactive towards S_N1 reaction—why?
Answer:

The stability of the carbocation obtained in the first step (rate¬determining step) of an S_N1 reaction determines whether the reaction will take place or not. Since methyl cation [⁺ CH₃ ] is a very unstable one, methyl chloride is unreactive toward the S_N1 reaction

Question 50. Explain the orders of acidity of carboxylic acids

1. Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH
2. CH₃CI₂COOH>(CH₃)₂CHCOOH >(CH₃)₃CCOOH

Answer:

1. -I effect explains this order of acid strength. As the number of halogen atoms on the a -carbon decreases, the overall -I effect decreases and as a consequence, the acid strength decreases

2. + 1 effect explains the given order of acid strength. As the number of methyl groups attached to the a -carbon atom increases, the overall +1 effect increases, and consequently, the acid strength decreases

Question 51. Explain why an organic liquid vaporizes below its boiling point when it undergoes steam distillation.
Answer:

In steam distillation, the sum of the vapor pressures of water and organic liquid becomes equal to the atmospheric pressure. This means that both of them distill at a pressure much lower than the atmospheric pressure, i.e., both of them will vapourize at a temperature that is less than their normal boiling points.

Question 52. 

1. Write the state of hybridization of C -atoms mentioned In each of the following compounds:

1. C-4 of Pcnt-l-cn-4-one
2. C-l of I’ropanoic acid
3. C-3 of Penta-2,3-diet,
4. C-3 of Pcntan-3-one and
5. C. -3 of 3,3-dietliylpcntane

2. Which atoms of each of the following molecules/ions always remain in the same plane?

1. CH₃CH = CHCH₃

2. C₆H₅C ≡ C—CN

3. C₆H₅CH₃

4. CH₂=C=C=CH₂

5. CH₃COCH₃

6. CH₃CONH₂

7. Cl₃ C —CH=CH—^–CH₂

8. (CD₃)₃C⁺

^{11. –}CH₂COCH₂CH₃

13. (CH₃)₂⁺CH — NH₂

Answer:

1. If four valencies of carbon atoms are satisfied by four single bonds, then it is sp³ -hybridized. Iffour valencies are satisfied. By one double bond and two single bonds, then it is sp² -hybridized. Iffour valencies are satisfied by one triple bond and one single bond or by two double bonds, then it is sp -hybridized.

2. An sp² -hybridized C -atom and the atoms directly attached to it remain in the same plane. Again, an sp -sp-hybridized C atom along with die atoms with which it is directly attached remain in a straight line. Therefore, in a molecule containing bond sp² -and sp -hybridized carbons, all the atoms remain in the same plane (except 1, 2-dienes).

A negatively charged C -atom (or a heteroatom containing lone pairs of electrons such as N, O, etc.) adjacent to a double bond is sp² – hybridized, and tire atoms attached to that carbon or heteroatom remain in the plane of the system containing the double bond.

Therefore, the atoms in the given molecules/ions that remain in the same plane are as follows:

The boiling point of its pure organic liquid is 70°C. There are two samples of tills liquid having boiling ranges:

1. 76-78°C and
2. 69-78°C respectively.

Question 53. 

1. The electronic configuration of C-atom is: ls²2s²2p², yet its valency is four —why?
2. The four C —H bonds of methane molecule are equivalent — explain with reasons.

Answer:

During a chemical reaction, the carbon atoms gain energy and promote one of the two electrons of the 2s -orbital to the higher 2p_z -orbital. Thus in the excited state, the electronic configuration of carbon becomes  ls²2s¹2p¹x2p¹y2p¹z.

At this condition, the valence shell of the C-atom contains four unpaired electrons. Thus, the C -atom can form four covalent bonds using four unpaired electrons. This explains why carbon having electronic configuration, ls²2s²2p² is tetravalent.

The equivalency of four C—H bonds in methane (CH4) be explained by the concept of hybridization of orbitals. In the excited state, the four valence orbitals of carbon, i.e., one 2s and three 2p orbitals possessing slightly different energies mix up and result in the formation of four equivalent sp3 -hybrid orbitals. These hybrid orbitals overlap with the four Is -orbitals of four H -atoms to form four C—H bonds which are also equivalent (same bond length and bond strength)

Question 54. 

1. Arrange sp, sp² & sp³ -orbitals in increasing order of:

1. Bond length
2. Bond angle
3. Bond energy
4. Size of orbitals and
5. S -character.

2. Organic compounds are usually water-insoluble. Why?

3. Write the structure of the smallest hydrocarbon having the empirical formula C₂H. What is the shape of the molecule?

4. Draw the p -p-orbitals involved in forming n -n-bonds in the molecule, CH₂=C =CH_2, and predict whether the molecule is planar or not.
Answer:

1.

1. sp—sp<sp²—sp²<sp³—sp³
2. sp³ < sp² < sp
3. sp³—sp³ < sp²—sp² < sp—sp
4. sp < sp² < sp³
5. sp³ < sp² < sp

2. Organic compounds are covalent. Thus they do not get ionised. Moreover, they are usually less polar or non-polar compounds and hence do not dissolve in highly polar solvents, or water.

3. The compound is (C₂H)₂ or C₄H₂ and its structure is HC = C —C = CH. The shape of the molecule is linear because all the C -atoms are sp -hybridized.

The molecule is non-planar because the two planes containing one C-atom and two H-atoms are perpendicular to each other

Question 55. Give the IUPAC names of the following compounds:

1. CH₃CHClCHBrCH₃

2. CH₃CHFCOCH₂CH₃

3.  (CH₃)₂CHCH₂OH

4. CH₃COOCH(CH₃)₂

5. CH₃CHBrCH(CH₃)COOII

6. CH₃CHOHCH₂CHO

7.

8.

9. HC ≡ CCH(CH₃)CH=CH₂

10. CH₃OCH(CH₃)CH₂CH₃

11. CH₃CHClCH₂CONH₂

12. BrCH₂CBr₂(CH₂)₃CHCl₂

13.

14. CH₃COCH₂COCH₃

Answer:

1. 2-bromo-3-chlorobutane
2. 2-fluoropentan-3-one
3. 2-methyldopa-l-ol
4. Isopropyl ethanoate
5. 3-bromo-2-methyl butanoic acid
6. 3-hydroxy butanal
7. 2,3,5-trimethyl-4-propylheptane (the chain containing a maximum number of substituents is considered as the principal chain).
8. 3-ethyl-2, 4, 5-trimethyl heptane
9. 3-methyl pent-l-en-4-one
10. 2-methoxyflurane
11. 3-chloro-butanamide
12. 5,5,6-tribromo-l,l-dichlorohexane
13. 5-sec-butyl-4-isopropyl decane or 4-(l-methyl ethyl)-5-  (1-methyl propyl) decane
14. Pentane-2,4-dione

Question 56. Write structures of the following: 

1. Hept-5-en-1-one
2. 1-bromo-2-ethoxyethane
3. 3-chloropropanoyl bromide
4. 1-chloroprocaine-2-amine
5. 4-iodo-3- nitro butanal
6. 3-phenyl prop-2-enoic acid
7. Ethanoic methanoic anhydride
8. 2-carbomoylpropanoic acid
9. Pentane-2,4-dione
10. 5-formyl-3-oxo pentanoic acid
11. Ferf-butyl alcohol
12. But-2-ene-1,4-dioic acid
13. Trimethylacetic acid
14. Diethylbutane-1,4-dioate
15. 3-(carboxymethyl) pentanoic acid
16. 1,3-dimethyl cyclo hex-l-ene

Answer:

Organic Chemistry Chapter 12 NCERT Long Question and Answers

 Question 57. Draw resonance structures of the following compounds.

1. C₆H₅NO₂
2. CH₃CH=CHCHO
3. C₆H₅CHO
4. C₆H₅CH₂
5. CH₃CH=CHCH₂

Question 58. 

1.  A mixture of ether and water can be separated by simple distillation.
2. Water present in rectified spirit can be removed by azeotropic distillation.
3. Benzoic add can be extracted from its aqueous solution using benzene.
4. Sugar containing NaCl as an impurity can be purified by crystallization using ethanol but not water.

Answer:

1. There is a considerable difference between the boiling points of ether and water Hence, at the boiling point of more volatile ether, the vapors almost exclusively consist of ether and at the boiling point of less volatile water, the vapors almost entirely consist of water Thus, these can be separated by simple distillation.
2. A mixture of water and rectified spirit forms an azeotropic mixture, i.e., the constituents of this mixture cannot be separated into their components by fractional distillation. So, azeotropic distillation is required to remove the water-rectified spirit.
3. Benzene is immiscible ’with water but benzoic add is highly soluble in benzene, Hence, benzoic add can be extracted from its aqueous solution using benzene.
4. Sugar is soluble in hot ethanol whereas common salt remains insoluble. Thus, impure sugar can be purified by crystallization. However, purification is not possible using water as a solvent because both components become readily soluble in water

Question 59. In the Lassa goe’s test, NH₂OH.HCl responds to the test for the element chlorine but not for the element nitrogen, explain
Answer:

As there is no carbon (C) atom in NH₂OH.HCl, NaCN is not produced in the first step by the reaction between sodium (Na) and C.

Hence, the formation of sodium ferrocyanide and ferric ferrocyanide (Prussian blue) in the subsequent step does not take place. Therefore NH₂OH HCl does not respond to Lassaigne’s test for nitrogen.

But chlorine (Cl) present in NH₂OH-HCl combines with Na metal to form soluble NaCl which reacts with AgNO₃ in the subsequent step to produce a white precipitate of AgCl which is soluble in ammonium hydroxide

Na + Cl → NaCl ; NaCl + AgNO₃→AgCl ↓(white) + NaNO₃

AgCl + 2NH₄OH → [Ag(NH₃)₂]Cl (water soluble) + 2H₂O

Question 60. What are the hybridization states of each C- atom in the compounds:

1. CH₂=C= O
2. CH₃CH=CH₂
3. (CH₃)₂CO
4. CH₂=CHCN
5. C₆H₆

Answer:

Question 61. Indicate the σ and π bonds in the following Heptan-4-one molecules:

1. C₆H₆
2. C₆H₁₂
3. CH₂CI₂,
4. CH₂=C=CH₂
5. CH₃NO₂
6. HCONHCH₃

Answer:

Question 62. Write bond-line formulas for:

1. Isopropyl alcohol,
2. 2,3-dimethyl butanal
3. Heptan-4-one

Answer:

Question 63.  Give the IUPAC names of the following compounds:

Answer:

1. Propylbenzene
2. 3-methylpentanenitrile,
3. 2,5-dimethyl heptane
4. 3-bromo:3-chloroheptane,
5. 3-chloro-propanal
6. 2,2-dichloroethanol.

Question 64. Which of the following represents the correct IUPAC CHO name for the compounds concerned?

1. 2,2-dimetliyIpentane or 2-dimethyl pentane
2. 2,4,7-trimethylolethane or 2,5,7-trimethylolethane
3. 2-chloro-4-methyl pentane or 4-chloro-2-methyl pentane
4. But-3-yne-l-ol or But-4-ol-l-yne.

Answer:

1. 2,2-dimethyl pentane (two alkyl groups are on the same carbon and hence the locant is repeated twice).
2. 2,4,7-trimethyloctane (since 2,4,7 locant set is lower than the set 2,5,7).
3. 2-chloro-4-methylpentane (alphabetical order of substituents is maintained).
4. But-3-one-l-ol (using lower locant for the principal functional group)

Question 65. Draw formulas for the first 5 members of each homologous series beginning with the given compounds:

1. HCOOH
2. CH₃COCH₃
3. H-CH=CH₂

Answer:

1. HCOOH , CH₃COOH, CH₃CH₂COOH,CH₃CH₂CH₂COOH,CH₃CH₂CH₂CH₂COOH

2. CH₃COCH₃ , CH₃COCH₂CH₃, CH₃COCH₂CH₂CH₃ ,CH₃CH₂COCH₂CH₃, CH₃COCH₂CH₂CH₂CH₃

3. CH=CH₂ , CH₃CH=CH₂ , CH₃CH₂CH=CH₂ , CH₃CH₂CH₂CH=CH₂, CH₃CH₂CH₂CH₂CH=CH₂

Question 66. Give condensed and bond line structural formulas i and identify the functional group(s) present, if any, I for:

1. 2,2,4-trimethylpentane
2. 2-hydroxyl, 2,3-propane tricarboxylic acid
3. Hexanediol

Answer:

Question 67. Identify the functional groups in given compounds:

Answer:

1.  —OH (phenolic hydroxyl), —CHO (aldehyde), — OMe (methoxy).

2. — NH₂ [1° amino (aromatic)], —CO₂—CH₂— (ester), -N(C₂H₅)₂ (3° amino)

3.  —CH=CH— (ethylenic double bond), — NO₂ (nitro)

Long Answer Questions Organic Chemistry Class 11 Chapter 12

Question 68. 0.4 g of an organic compound containing N was Kjeldahlised and NH₃ obtained was passed into 50 mL (N/2) H₂SO₄ solution. The volume of the acid solution was increased to 150 mL by adding distilled water. 20 mL of this acid solution required 31 mL (N₂O) NaOH for complete neutralization. Calculate the percentage of N.
Answer:

20 mL of (partially neutralized) diluted acid solution

= 31mL \(\frac{1}{20}\) NaOH solution.

= 15.5

Strength of(partially neutralized) diluted acid solution

= \(31 \times \frac{1}{20} \times \frac{1}{20}(\mathrm{~N})=\frac{31}{400}(\mathrm{~N})\)

Amount of H₂SO₄ present in 150 mL (partially Amount of H2S04 present in 150 mL (partially

= \(\frac{31 \times 150}{400 \times 1000}\)

Now, 50mLof \(\frac{1}{2}(\mathrm{~N}) \mathrm{H}_2 \mathrm{SO}_4\) solution contains = \(\frac{1 \times 50}{2 \times 1000}\) g-equiv. H₂SO₄

NH₃ produced by decomposition 0.4 g of the organic compound = \(\left(\frac{50}{2000}-\frac{31 \times 150}{400000}\right)\)

= 0.013375 g-equivalent

= 0.013375 X×17 g

Now, 0.013375 x 17 g NH3 = \(\frac{14}{17} \times 0.013375\)

% of the nitrogen in the organic compound

= \(\frac{14 \times 0.013375}{0.4} \times 100\)

= 46.81

Question 69. Expand each of the following condensed formulas into their complete structural formulas:

1. CH₃CH₂COCH₂CH₂CI
2. CH₃CH= CH(CH₂)₄CH₃
3.  BrCH₂CH₂C=CCH₂CH₃

Answer:

Question 70. Write down the condensed structural formula and bond¬ line structural formula for each of the following molecules:

1. ICH₂CH₂CH₂CH₂CH(CH₃)CH₂

2.

Answer:

Question 71. Expand each of the following bond-line formulas to show all the atoms including carbon and hydrogen:

Answer:

.

Question 72. Explain why alkyl groups act as electron donors when attached to a n system.
Answer:

Due to hyperconjugation (cr, n conjugation), alkyl groups act as electron donors when attached to a n -system. This is shown in the case of propane—

Question 73. Draw the resonance structures for the following compounds. Show the electron shift using curved arrow notation:

1. C₆H₅OH
2. C₆H₅NO₂
3. CH₃CH=CHCHO
4. C₆H₅— CHO
5. C₆H₅—CH₂
6. CH₃CH=CHCH₂

Answer:

Question 74. Identify the reagents underlined in the following ey=o + H2O equations as nucleophiles or electrophiles

Answer:

1. Nucleophile OH^–
2. Nucleophile (^–CN)
3. Electrophile (CH₃⁺CO)

Question 75. Classify the following reactions in one of the reaction types studied in this unit.

Answer:

1. Nucleophilic substitution
2. Electrophilic addition
3. P -Elimination
4. Nucleophilic substitution & rearrangement

Question 76. What is the relationship between the members of the following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

Answer:

1. Structural isomers (position isomers as well as metamers)
2. Geometrical isomers
3. Resonance contributors

Question 77. For the given bond cleavages, use curved arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbonation, and carbanion.

Answer:

Question 78. Write down the IUPAC names of the alkyl groups having the molecular formula, C₄H₆.
Answer:

Four alkyl groups are possible. These are

Question 79. Give the structural difference O aldehyde C & ketonic groups.
Answer:

An aldehyde group (—CHO) is a carbonyl group in which one valency of the carbonyl carbon is satisfied by a H-atom, and the other valency is satisfied by another atom or an alkyl group. On the other hand, the keto group is also a carbonyl group in which two valencies of the carbonyl carbon are satisfied by two alkyl groups.

Question 80. Both formic acid (HCOOH) and acetic acid (CH₃COOH) contain the same functional group, yet there are some differences in their chemical properties—explain.
Answer:

The structural formula of formic acid is such that it can be said to contain  said to contain a  as well as  

So it exhibits the properties of both —CHO and

—COOH groups. But acetic acid contains only

—COOH group and hence it exhibits the properties of compounds containing only carboxyl group

Question 81. Label the primary (1°), secondary (2°), tertiary (3°), and quaternary (4°) carbon atoms in the following compound:

Answer:

NCERT Class 11 Organic Chemistry Chapter 12 Long Answer Solutions

Question 82. Write down the IUPAC and common names of each of the given compounds:

1. CH₃CH= CH₂
2. CH₃C=CCH₃
3. CH₃CHOHCH₃
4. CH₃OCH₂CH₂CH₃
5. CH₃CH₂CHO
6. CH₃COC₂H₅
7. C₂H₅COOH
8. C₂H₂COCl
9. CH₃CONH₂
10. CH₃CO₂C₂H₅
11. CH₃CH₂NH₂
12. CH₃NHCH₂CH₃
13. (CH₃)₂NCH₂CH₃
14. CH₃CH₂CN

Answer:

1. Propene; Propylene
2. But-2-yen; Dimethylacetylene
3. Propan-2-ol; Isopropyl alcohol
4. 1-methoxy propane; Methyl /t-propyl ether
5. Propanal; Propionaldehyde
6. Butan-2-one; Ethyl methyl ketone
7. Propanoic acid; Propionic acid
8. Propanoyl chloride; Propionyl chloride
9. Propanamide; Propionamide
10. Ethyl ethanoate; Ethyl acetate
11. Ethanamine; Ethylamine
12. Methylethanamine; Ethylmethylamine
13. N, N-dimethylethanolamine; Ethyldimethylamine
14. Propanenitrile; Ethyl cyanide

Question 83. Write down the structures of the following compounds:

1. 2-Iodopropane
2. Hex-3-one
3. Pent-l-ene
4. 2,2-Dichloropropane
5. 1, l, 1, 2-Tetrachloroethane
6. Propan-2-ol
7. Propane-1,3-diol
8. Butane-1,2,3-triol.
9. 2-Methoxypropane
10. 2-Methylpentanoic acid
11. 2,2-Dimethylbutanal
12. Pentan-3-one 
13. Butanoyl chloride
14. Aceticformic anhydride
15. Ethylmethanoate
16. N-Methylmethanamine
17. N-Ethyl-N-methylhexanamine
18. Butanenitrile.

Answer:

1. CH₃CHICH₃
2. CH₃CH₂C=CCH₂CH₃
3. CH₃CH₂CH₂CH=CH₂
4. CH₃CCl₂CH₃
5. CH₂Cl CCl₃,
6. CH₃CH(OH)CH₃
7. HOCH₂CH₂CH₂OH
8. CH₃CH(OH)CH(OH)CH₂OH
9. CH₃CH(OCH₃)CH₃
10. CH₃CH₂CH₂CH(CH₃)COOH
11. CH₃CH₂C(CH₃)₂CHO,
12. CH₃CH₂COCH₂CH₃
13. CH₃CH₃CH₂COCl
14. CH₃COOCHO
15. HCOOCH₂CH₃
16. CH₃NHCH₃
17.  CH₃CH₂N(CH₃)CH₂CH₃
18. CH₃CH₂CH₂CN

Question 84. Write down the IUPAC names of the following compounds:

Answer:

1. 3-ethyl-5-methyl heptane,
2. 2,2-dimethylbutane
3. 2,2,4-trimethylpentane,
4. 3-ethyl-2,2,4-trimethylpentane
5. 4-(1,1-dimethyl ethyl)heptane
6. 3,4-diethyl hexane
7. 6-ethyl-2-methyl-5-(1,1-dimethyl ethyl)octane

Question 85. What is wrong with the following names? Draw the structures they represent and write their correct names.

1. 1,1-dimethyl hexane
2. 3-methyl-5-methyl heptane
3. 4, A-dimethyl-3-ethyl pentane
4. 3, 4,7-trimethylolethane
5. 3,3-diethyl-2, A, Atrimethylpentane

Answer:

1. (CH₃)₂CHCH₂CH₂CH₂CH₂CH₃ – 2-methylheptane
2. CH₃CH₂CH₂C(CH₃)₂CH₂CH₂CH₂CH₃ – 4,4-dimethyl octane
3. CH₃CH₂CH(CH₃)CH₂CH(CH₂CH₃)₂–  3-ethyl-5-methyl heptane
4. CH₃C(CH₃)₂CH(CH₂CH₃)₂ –  3-ethyl-2, 2-dimethyl pentane
5. CH₃CH₂CH(CH₃)CH(CH₃)CH₂CH₂CH(CH₃)₂ –  2,5,6-trimethylolethane
6. (CH₃)₂CHC(CH₂CH₃)₂C(CH₃)₃ –  3,3-diethyl-2,2,4-trimethylpentane

Question 86. Give the IUPAC name of the following alkane containing complex substituents:

Answer:  3-ethyl-7,7-fels(2,4-dimethylhexyl)-5,9,11-trimethyltridecane

Question 87. Write the IUPAC names of the following compounds:

Answer:

1. 3-ethyl pent-1-ene,
2. 3-ethyl hex-1-en-5-one
3. 2-ethyl-3,3-dimethyl but-1-ene
4. Pent-3-en-1-one
5. 3-methyihexa-1,5-diene
6.  3-isobutylhept-1-en-4-yne
7. 3-propylhept-l-ene,
8. 3-methyl-4-methylidenehept-1-en-6-yne
9. Hexa- 1,3-dien-5-one,
10. 5-methylhepta-1,2,6-triene

Question 88. Write down the structures of the following compounds

1. Pent-3-en-1-one
2. 3-methylpenta-1, 4-diyne
3. 3-(2-methylpropyl)hept- 1-en-4-yne
4. 3-ethylpenta-1,3-diene
5. 5-ethynylhepta-1,3,6-triene
6. 4-ethyl-4-methylhex-1-one

Answer:

Question 89. Write down the IUPAC names of the following compounds:

Answer:

Question 90. Write down the structures of the following:

1. 2-methyl butanol chloride
2. 5-chloro-3-ethylpentan-2-one
3. Diethyl butane-1, A-date
4. Methyl-2-methyl prop- 2-en-1-rate 
5. 3-phenyl prop-2-enoic acid
6. Propane- 1,2,3-tricarboxamide.

Answer:

Question 91. Give the IUPAC names of the following compounds:

1. CH₃COCH₂COOC₂H₅
2. H₂NCH₂CH₂CH₂COOH
3. CH₃CH(CN)CH₂COCH₃

Answer:

Question 92. Write down the structures of the following compounds:

1. 3-formylpentanoic acid
2. 3-hydroxyl-oxopentanal
3. 2, 3-dihydroxy butane dioic acid
4. 3-hydroxy cyclohexanone
5. 3-hydroxy-3-methyl butane-2-one

Answer: 

Organic Chemistry Techniques Long Answer Class 11 NCERT

Question 93. Write the structures of the following compounds:

1. 2-chloro-2-methyl butane-1-ol
2. 4-amino-2-ethyl pent-2-enal
3. Hex-A-in-2-one
4. 1-bromo-3-chloracyclohex-1-ene
5. But-2-ene-l, 4-dioic acid
6. 4-nitrogen-l-one Ethyl 3-methoxy – 4-nitro butanoate

Answer:

Question 94. What type of structural isomerism is exhibited by the following pairs of isomers?

1. CH₃CHCOOH and CH₃COOCH₃

2. CH₃ —C≡C — CH₃ and CH₃CH₂C≡CH

3. CH₂= CHOH and CH₃CHO

4. CH₂ = CH(CH₂)₃CH₃ and C₆H₆

6. CH₃CH₂CH₂OH and (CH₃)₂CHOH

Answer:

1. Functional group isomerism
2. Position isomerism
3. Tautomerism (special case of functional group isomerism)
4. Ring-chain isomerism
5. Position isomerism
6. Position isomerism

Question 95. Which two of the following compounds are

1. Position isomers
2. Tautomers
3. Ring-chain isomers
4. Metamers
5. Chain isomers and
6. Functional isomers

Answer:

1. Position isomers: (h) and (k),
2. Tautomers: (a) and (f)
3. Ring-chain isomers: (e) and (j)
4. Metamers : (c) and (g)
5. Chain isomers: (b) and (i)
6. Functional isomers: (d) and (l)

Question 96. Identify the optically active and optically inactive compounds:

1. CH₃CHOHC₂H₅

2. CH₃CH₂OH

3. C₂HgCHBrCH(CH₃)₂

4.

5. CH₃CH=CHC₂H₅

Answer:

(1), (3), and (4) will be optically active as each of these molecules contains one asymmetric center.

But (2) & (5) are optically inactive as they do not have a symmetric center.

Question 97. Which of the following will exhibit geometrical or cis-trans isomerism and which of them will not? Give reasons.

1. CH₃CH=CBr₂

2. BrCH=CHCH₂CH₃

3. CH₂=CH —CH=CH₂

5. CH₂=CHCH=CHCH=CH₂

Answer:

(2), (5), (6), (7), and (8) exhibit geometrical isomerism.

But (1), (3), and (4) do not exhibit geometrical isomerism.

Question 98. Draw the Fischer projectionformulas of all stereoisomers CH₃CHBrCHClCOOH. Mention how they are related to each other
Answer:

Enantiomers: 1 and 2, 3 and 4

Diastereomers: 1 and 3; 1 and 4; 2 and 3; 2 and 4

Question 99. Write down the structure and IUPAC names of two isomeric optically active alkanes having the lowest molecular mass.
Answer:

Question 100. Which of the following compounds are meso-compounds and which are not? Give reasons

Answer:

(2) and (3) are meso compounds (they are optically inactive due to the presence of a center of symmetry). (1) is optically active. It contains two asymmetric centers and it is not superimposable on its mirror image.

Question 101. Arrange in order of decreasing basic strength and show(XI) Pari-n reasons: CH₃—CH=NH,  CH₃—C=N, CH₃ — NH₂

Ongoing from 

in CH₃NH₂, CH₃CH=NH and CH₃C=N, the unshared electron pairs are in sp³ , sp² and sp -orbitals respectively. As the s -character of the hybrid orbital (containing lone pair) of N -atom increases, the electrons are drawn closer to the nitrogen nucleus, and hence electron donating ability of the amino nitrogen decreases causing a decrease in basicity.

Thus basic strength decreases in the sequence:

Question 102. Arrange in order of increasing acidity and give reasons: CH₃CH₂OH, (CH₃)₃ COH, CH₃OH, (CH₃)₂CHOH
Answer:

Since alkyl groups have a +1 effect, there will be an increased electron displacement towards the oxygen atom on going from primary to secondary to tertiary alcohol. This may be represented (qualitatively) as follows:

Question 103. Arrange the following anions in order of increasing stability and give reasons: CH₂=^–CH, CH₃^–CH₂, CH = C^–
Answer:

The greater the -ve charge on the oxygen atom, the closer the displacement of the covalent pair in the O —H bond to the hydrogen atom, hence separation of a proton becomes increasingly difficult. Thus the acid strengths of alcohols will be in the following order:

(CH₃)₃COH < (CH₃)₂CHOH < CH₃CH₂OH < CH₃OH

Ongoing from \(\mathrm{CH}_3 \stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2→\mathrm{CH}_2=\stackrel{\ominus}{\mathrm{C}} \mathrm{H}→\mathrm{HC} \equiv \stackrel{\ominus}{\mathrm{C}}\) it is seen that the unshared electron pairs of the carbanion carbons are in sp³, sp² and sp -hybrid orbitals respectively. As the s -s-character of the hybrid orbitals increases, the electrons are drawn closer to the nucleus of the carbanion carbon, and hence the ability to hold the electron pair increases, causing successive increases in stability. Stability order

⇒ \(\mathrm{CH}_3 \stackrel{\ominus}{\mathrm{C}} \mathrm{H}_2<\mathrm{CH}_2=\stackrel{\ominus}{\mathrm{C}} \mathrm{H}<\mathrm{CH} \equiv \stackrel{\ominus}{\mathrm{C}}\)

Question 104. Which of the following pairs do not represent two resonance structures and why?

Answer:

The two structures which differ in the positions of atoms are not resonance structures. Thus the pair of structures given in (1), (2), and (5) do not represent resonance structures.

Question 105. In between CH₃COOH and CH₃COO^–, which one is more resonance stabilized and why?
Answer:

Both CH₃COOH and CH₃COO^– are resonance hybrids of two canonical forms. But one of the resonance structures of CH₃COOH involves separation of charge, while none of the resonance structures of CH₃COO^– involves any separation of charge (also these structures are equivalent). Hence CH₃COO^– is more resonance stabilized compared to CH₃COOH.

Question 106. Which N -atom of guanidine is more basic and why
Answer:

Protonation on tire doubly bonded N -atoms produces a cation (conjugate acid) which is stabilized by resonance involving three equivalent canonical forms. On the other hand protonation on either of the singly bonded N-atoms produces a cation (conjugate acid) which is not stabilised by resonance. Thus the doubly bonded N-atom of guanidine is more basic.

Question 107. Which of the two N atoms of the following compound undergoes protonation and why?
Answer:

N-atom ofthe ring ‘A’ undergoes protonation because the resultant cation (conjugate acid) is stabilized by resonance. The n-atom of the ring ‘B’ does not undergo protonation because in that case the resulting cation will not be stabilized by resonance.

Question 108. Which resonance structure in each of the following compounds contributes more towards the hybrid and why?
Answer:

• The 1st structure contributes more because both C and N have octets of electrons in their valence shells.
• The 1st structure contributes more because it involves no separation of charge.
• The 2nd structure contributes more as the -ve charge is on the more electronegative O-atom.
•  The 1st structure contributes more because there is separation of charge. The 2nd structure involves the separation of charge; also the +ve charges, on the adjacent atoms, repel each other.
• The 1st structure contributes more compared to the 2nd structure. The 2nd structure is highly unstable as it contains a negative N-atom.

Question 109. Which of the following compounds can be represented as a resonance hybrid and which of them can the? Give reasons.

1.  CH₃CH₂OH,

2.CH₃CONH₂

3. CH₃CH=CHCH₂NH₂

4. H₂N—CH=CH— NO₂

5. 

Answer:

Structures and (3) can not be represented as resonance hybrids because lone pairs on the O-atom or N-atom can not undergo delocalization. However, structures (2), (4) and (5) can be represented as resonance hybrids.

Question 110.  Why are the three carbon-oxygen bonds In carbonate (CO₃^-2) ion equal in length?

This is so because the CO₃^-2 ion is a resonance hybrid of three equivalent canonical forms.

Question 111. Which one between phenol and cyclohexanol is more acidic and why?
Answer:

Phenol is a stronger acid than cyclohexanol.

It can be explained as follows:

1. Due to resonance, the O-atom of the OH group acquires a +ve charge and so the release of the proton is facilitated

2. When phenol ionizes, the formed phenoxide ion is also a resonance hybrid, but it is more stabilized by resonance than a unionized phenol molecule because of the spreading of a negative charge only. In the unionized molecule, resonance involves the separation of charge

Such effects are not possible in the case of cyclohexanol and hence proton release is not facilitated.

Long Questions for Organic Chemistry Class 11 Chapter 12 NCERT

Question 112. Arrange the following ions in order of increasing stability and give your reasons
Answer:

Question 113. Which one between 2-methylbut-2-ene and 2-methylbut-1- ene has higher heat of hydrogenation and why?
Answer:

In structure 3+ve charge on the carbon is involved in I delocalization with the benzene ring.

In structure (1), +ve charge on the carbon is involved in delocalization not only with the benzene ring but also with N-atom of the \(\ddot{\mathrm{N}}\)Me₂ group, thereby making this structure more stable than (III).

Structure (2) is the least stable because +ve charge on the carbon can not be involved in delocalization with the aromatic ring (steric inhibition of resonance)

Question 114. Arrange the following ions in order of increasing stability and give your reasons
Answer:

2-Methylbut-2-ene contains nine hyperconjugable α-H -atoms, so this molecule is involved in effective hyperconjugation. As a result, this molecule gains extra stability and it has relatively lower heat of hydrogenation. On the other hand, 2-methylbut-1-ene contains only five hypercoagulable α-H -atoms so the effect of hyperconjugation stabilizing this molecule is relatively less. Thus it has a relatively higher heat of hydrogenation.

Question 115. The C—C bond in acetaldehyde (CH₃CHO) is shorter than that in ethane while the C— C bond in trifluoro acetaldehyde (CF₃CHO) is essentially the same as that in ethane. Explain
Answer:

Acetaldehyde molecule contains three α-H -atoms. These H-atoms are involved in hyperconjugation with the double bond of the carbonyl group. So C—C bonds in acetaldehyde have a partial double bond character. In ethane the C—C bond has pure single bond character.

Thus C — C bond in acetaldehyde is shorter than that in ethane. Trifluoroacetaldehyde does not have a-H -atoms. So hyper-conjugation is not possible in CF₃CHO.

Thus, the C — C bond in CF₃CHO is essentially the same as that in ethane.

Question 116. Arrange the following isomeric alkenes in order of increasing stability and give your reasons:

1. (CH₃)₂C=C(CH₃)₂ (1)
2. CH₂=CHCH₂CH₂CH₃ (2)
3. CH₃CH=CHCH(CH₃)₂ (3)
4. CH₃CH =C(CH₃)CH₂CH₃ (4)

Answer:

The stability of an alkene is determined by the number of hyperconjugative structures, which in turn is dictated by the number of α-H -atoms (concerning the olefinic carbons) present in the molecule. The greater the number of hyperconjugative structures, the higher the stability of the alkene Now the number of ar-H -atoms in the alkene 1, 2, 3, and 4 are 12, 2, 4, and 8 respectively. Thus stability of the alkanes follows the sequence: 2 < 3< 4 < 1.

Question 117. Which one of the following two conformations of butane is more stable and why?

Answer:

Eclipsed conformation I, in which methyl groups on two adjacent carbons are just opposite to each other. In this conformation steric strain and bond opposition strain are maximum, hence this conformation is most unstable.

Anti-conformation II, in which methyl groups are as far apart as possible, is most stable due to minimum repulsion between methyl groups. Note that, there is no bond opposition strain in this conformation.

Question 118. Which of the 2 geometric isomers I of Me₃CCH=CHCMe₃ has a higher heat of combustion and why?
Answer:

Cis-isomer is less stable because of a very large steric hindrance between two bulky t-butyl groups lying on the same side of the double bond. On the other hand, transisomer is more stable because the bulky f-butyl groups are on the opposite sides of the double bond. Thus cis isomer has a higher heat of hydrogenation.

Question 119. Which one between C₆H₅CH₃ and CH₄ has lower Csp³—H bond dissociation enthalpy and why?
Answer:

The bond dissociation enthalpy of C₆H₅CH₂—H is less than (II) that of H₃C — H as  C₆H₅CH₂ is more stable (stabilized by resonance) than that of ^•CH₃ (which has no resonance stabilization).

Question 120. Arrange the following carbocations in order of increasing stability and explain the order:

Answer:

In cation in,(3) +ve charge is not delocalized due to steric inhibition of resonance. However, +ve charge is delocalized in both the cations I and But the extent of delocalization of +ve charge is higher in H due to the additional effect involving the methoxy group.

Question 121. Arrange the following carbanions in order of increasing stability and explain the order

Answer:

The stability of carbanions increases as the extent of delocalization of the -ve charge increases. Carbanion I is the most stable as the -ve charge is delocalized not only by the benzene ring but also by the -NO₂ group. Carbanion HI is moderately stable as the -ve charge is delocalized only by the benzene ring. Carbanion U is the least stable because of the +R effect of the -OMe group (although the -ve charge is delocalized by the benzene ring). Thus the sequence of stability is 2 < 3< 1

Question 122. Classify the following species as electrophile or nucleophile and explain your choice:

Answer:

Nucleophile : CH₃C^– , CHgCOO^– , CH₂=CH₂

Electrophile : Cl+ , BF₃ , (CH₃)₃C⁺ , R— X

CH₃O^–– and CH₃COO^– are negatively charged species having available unshared electron pairs on the O-atom. So these are nucleophiles. CH₂=CH₂ can also act as a nucleophile as it contains loosely bound or -electrons.

In the species Cl⁺, BF_3, and Me₃C⁺, there is electron deficiency (having sex tet of electrons) on the valence shells of Cl, B, and C-atom respectively. So these are electrophiles. In the alkyl halides (R—X) there is electron deficiency on the a -carbon due to the strong -I effect of the halogen atom. So RX can act as electrophile.

Question 123. Formulate the following as a two-step reaction and designate the nucleophile and electrophile in each step:  CH₂= CH₂ + Br₂ → BrCH₂CH₂Br
Answer:

Question 124. CN^– and NO₂^– are called ambient nucleophiles. Explain
Answer:

Nucleophiles that have more than one (generally two) suitable atoms through which they can attack the substrate are called ambident nucleophiles. Each of the ^–CN ions and NO₂^– contain two atoms through which they can be involved in nucleophilic attack (these atoms are indicated by arrows). So these are ambident nucleophiles.

Question 125. Mention the type of each of the following reactions
Answer:

Answer:

1. Nucleophilic substitution (S_N2)
2. Electrophilic addition
3. Free-radical substitution
4. Elimination reaction (E2)
5. Rearrangement reaction

Question 126. Calculate the double bond equivalent (DBE) of each of the given compounds:

1. C₁₃H₉BrS
2. C₁₂H₁₆N₂O₄

Answer:

Double bond equivalent (DBE) of C₁₃H₉BrS

= \(\frac{13(4-2)+9(1-2)+1(1-2)+1(2-2)}{2}+1\)

= 9

DBE of  C₁₂H₁₆N₂O₄

= \(\frac{12(4-2)+16(1-2)+2(3-2)+4(2-2)}{2}+1\)

= 6

Question 127. Calculate the double bond equivalent (DBE) of a compound having molecular formula, C₅H₈. On catalytic hydrogenation, the compound consumes 1 mol of hydrogen. Write the structures of all the possible isomers of the compound
Answer:

DBE of C₅H₈ \(=\frac{5(4-2)+8(1-2)}{2}+1\) = 2

On hydrogenation, it consumes 1 mol of H₂. So it contains one double bond and one ring.

Thus possible structures of the compounds are:

Question 128. Arrange the following carbocations in order of increasing stability and explain the order:
Answer:

The sequence of stability: (4) > (2) > (3) > (1) Carbocation (4) is a primary carbocation, but it is most stable due to resonance.

Question 129. Which of the carbocations is the most stable?

1. CH₃CH₂⁺CH₂,
2. CH₂=CH⁺CH₂,
3. C₆H₅ ⁺CH₂
4. All are equally stable.

Answer:

(1), (2), and (3) are all primary carbocations. Cations (1), (2), and (3) have 0, 2, and 4 resonance structures respectively. So carbocation (3) is the most stable

Class 11 Chemistry Organic Chemistry Chapter 12 Long Questions & Answers Solutions

Question 130.  Which one between the two CH₃CO^– and CH₃ CH₃COC^– HCOCH₃ is more stable and why?

Negative charge CH₃COC^– H₂ is involved in delocalization with the n electrons of only one carbonyl group. On the other hand, the carbonation-carbon of on the carbanion-carbon of charge e -ve on n CH₃COC^–HCOCH₃ is involved in delocalization with the π -electrons of two carbonyl groups. Thus the second carbanion is more stable.

NCERT Solutions For Class 11 Chemistry Chapter 12 Organic Chemistry Basic Principles And Techniques Multiple Choice Questions

Question 1. The ease of dehydrohalogenation of an alkyl halide with alcoholic KOH is—

1. 3° <2° < 1°
2. 3° < 2° > 1°
3. 3° > 2° > 1°
4. 3° > 2° < 1°

Answer: 2. 3° < 2° > 1°

Question 2. Which will exhibit optical isomerism

Answer: 2

Question 3. Which of the following is sec-butyl Phenyl Vinyl (BPV)methane

Answer: 3

Question 4. The correct states of hybridisation of C₂ and C₃ in compound H₃C —CH=C=CH —CH₃

1. sp², sp³
2. sp², sp
3. sp², sp³
4. sp, sp

Answer: 2. sp², sp

Question 5. Under identical conditions, the S_N1 reaction will most efficiently with

1. Tert-butyl chloride
2. 2-methyl-l-chloropropane
3. 2-chloroquine
4. Vinyl Chloride

Answer: 1. Tert-butyl chloride

Question 6. Which one of the following characteristics belongs to an electrophile

1. It is any species having electron deficiency that reacts at an electron-rich C-centre
2. It is any species having electron enrichment, that reacts at an electron deficient-centre
3. It is cationic
4. It is anionic

Answer:  1. It is any species having electron deficiency that reacts at an electron-rich C-centre

Question 7. The most stable enol tautomer of MeCOCH₂CO₂Et 

1. CH₂=C(OH)CH₂CO₂Et
2. MeC(OH)—CHCO₂Et
3. MeCOCH=C(OH)OEt
4. CH₂=C(OH)CH=C(OH)OEt

Answer: 2. MeC(OH)—CHCO₂Et

Organic Chemistry Basic Principles and Techniques MCQs Class 11

Question 8. Order of stability of the carbocations

1. Ph₂C⁺CH₂Me

2. PhCH₂CH₂C⁺HPh

3. PhCH⁺CHMe

4. Ph₂C(Me)CH₂  is

1. 4 > 3 > 1 > 3
2. 1 > 2 > 3 > 4
3. 2 > 1 > 4 > 3
4. 1>4>3>2

Answer: 2. 1 > 2 > 3 > 4

Question 9. MeCH₂CH=CH₂ is stable than Me₂C=CH₂ because—

1. Inductive effect of group.
2. Resonance effect of Me-group.
3. Hyperconjugative effect of Me-group.
4. Resonance and inductive effects of Me-group.

Answer: 3. Hyperconjugative effect of Me-group.

Question 10. (+) and (-)-Lactic acid has the same molecular formula, C₃H₆O₃. They are related as

1. Structure isomers
2. Geometric isomers
3. Optical isomers
4. Homomers

Answer: 3. Optical isomers

Question 11. Which of the following statements is correct for 2-butene—

1. The C₁ —C₂ bond is an sp³ —sp³ σ -bond
2. The C₂— C₃ bond is an sp³—sp² σ -bond
3. The C₁—C₂ bond is an sp³—sp³ σ-bond
4. The C₁— C₂ bond is an sp²—sp² σ -bond

Answer: 3. The C₁—C₂ bond is an sp³—sp³ σ-bond

Question 12. Basicity of aniline is less than methyl amine, because—

1. Hyperconjugation effect of Me-group in MeNH₂
2. Resonance effect of the phenyl group in aniline
3. The molar mass of methylamine is less than that of aniline
4. Resonance effect of Me-group in MeNH₂

Answer: 2. Resonance effect of the phenyl group in aniline

Question 13. Tautomerism is exhibited by—

Answer: 1, 2 and 3

Question 14. Amongst the following, the one which can exist in a free state as a stable compound is—

1. C₇H₉O
2. C₈H₁₂O
3. C₆H₁₁O
4. C₁₀H₁₇O₂

Answer:  2. C₈H₁₂O

Question 15. The correct pair of compounds that gives blue coloration/ precipitate and white precipitate, respectively, when their Lassaigne’s test is separately done is—

1. NH₂NH₂ HCl and C1CH₂COOH

2. NH₂CSNH₂ and PhCH₂Cl

3. NH₂CH₂COOH and NH₂CONH₂

Answer: 4.

Question 16. The IUPAC name of the compound X is—

1. 4-cyano-4-methyl-2-isopentane
2. 2-cyano-2-methyl-4-isopentane
3. 2,2,-dimethyl-4-oxo pentane nitrile
4. 4-cyano-4-methyl-2-pentanone

Answer: 3. 2,2,-dimethyl-4-oxo pentane nitrile

Question 17. The optically active molecule is

Answer: 3

Class 11 Chemistry Chapter 12 Organic Chemistry Multiple Choice Questions

Question 18. (+) -2-chloro-2-phenylethane in toluene racemizes slowly in the presence of the small amount of SbCl5, due to the formation of—

1. Carbanion
2. Free-radical
3. Carbene
4. Carbocation

Answer: 4. Carbocation

Question 19. The order of decreasing ease abstraction of hydrogen atoms in the following Hb molecule is—

1. H_a>H_b>H_c
2. H_a>H_c>H_b
3. H_b>H_a>H_c
4. H_c > H_b>H_a

Answer: 2. H_a>H_c>H_b

Question 20. The most likely protonation site in the given molecule is

1. C-1
2. C-2
3. C-3
4. C-6

Answer: 1. C-1

Question 21. The 4-th higher homolog of ethane is—

1. Butane
2. Pentane
3. Hexane
4. Heptane

Answer: 3. Hexane

Question 22. Among the following structures, the one which is not a resonating structure of others is—

Answer: 4.

Question 23. The correct order of decreasing length of the bond as indicated by the arrow in the following structures is—

Answer: 3

Question Question 24. IUPAC name ofthe molecule, is

1. 5,6-dimethylhept-2-ene
2. 2,3-dimethylhept-5-ene
3. 5,6-dimethylhept-3-ene
4. 5-isopropyl hex-2-ene

Answer:  1. 5,6-dimethylhept-2-ene

Question 25. The correct statement regarding the given compound is—

1. All three compounds are chiral
2. Only 1 and 2 are chiral
3. 1 and 3 are diastereomers
4. Only 1 and 3 are chiral

Answer: 4. Only 1 and 3 are chiral

Question 26. In Lassaigne’s test for the detection of nitrogen in an organic compound, the appearance of blue colored

1. Ferric ferricyanide
2. Ferrous ferricyanide
3. Ferric ferrocyanide
4. Ferrous ferrocyanide

Answer: 3. Ferric ferrocyanide

Question 27. The reaction of methyl trichloroacetate (Cl₃CCO₂Me) with sodium methoxide (NaOMe) generates

1. Carbocation
2. Carbene
3. Carbanion
4. Carbon radical

Answer: 2. Carbene

Question 28. In a mixture, two ommtlomors are found to be present in the amount of 0f*% mu) 15% respectively. The enantiomeric excess (e,o) Is—

1. 85%
2. 15%
3. 70%
4. 60%

Answer: 3. 70%

Question 29. In the following compound, the number it ‘sp’ hybridized carbon is CH₂=C=CH-CH-C≡CH –

1. 2
2. 3
3. 4
4. 5

Answer: 3. 4

Question 30. Which of the following statements Is/are correct

Answer: 2 and 4

NCERT Solutions Class 11 Chemistry Chapter 12 MCQs

Question 31. The correct order of adding strengths of benzoic acid (X), hydroxybenzoic acid (Y), and p-nitrobenzoic acid (Z) is—

1. Y> Z > X
2. Z > Y > X
3. Z > X > Y
4. Y > X > Z

Answer:  3. Z > X > Y

Question 32. In the IUPAC system, PhCH₂CH₂CO₂H is named as—

1. 3-phenylpropanoid acid
2. Benzoyl acetic acid
3. Carboxyethyl benzene
4. 2-phenylpropanoid acid

Answer: 1. 3-phenylpropanoid acid

Question 33. The major product(s) obtained in the reaction Is/are

Answer: 1 and 4

Question 34. The possible product (s) to be obtained from the reaction of cyclobutyl amine with HNOz is/are—

Answer:  1 and 3

Question 35.  Identify the compound that exhibits tautomerism—

1. 2-pentanone
2. 2-butene
3. Lactic acid
4. Phenol

Answer:  2. 2-butene

Question 36. How many chiral compounds are possible on monochlorination of 2-methylbutane—

1. 2
2. 4
3. 6
4. 8

Answer: 1. 2

Question 37. Identify the compound that exhibits tautomerism—

1. 2-pentanone
2. 2-butene
3. Lactic acid
4. Phenol

Answer: 1. 2-pentanone

Question 38. The order of stability of the following carbocations is—

1. 3>1>2
2. 3>2>1
3. 2>3>1
4. 1>2>3

Answer: 1. 3>1>2

Question 39. Arrange the compounds in order of decreasing acidity

Answer: 4.

Question 40. A solution of (-)-l-chloro-l-phenylethane in toluene racemizes slowly in the presence of a small amount of SbCl₅, due to the formation of—

1. Free radical
2. Carbanion
3. Carbene
4. Carbocation

Answer:  4. Carbocation

Question 41. In S_N2 reactions, the correct order of reactivity for the compounds: CH₃Cl, CH₃CH₂Cl, (CH₃)₂CHCl and (CH₃)₃CCl is _____

1. (CH₃)₂CHCl > CH₃CH₂Cl > CH₃Cl > (CH₃)₃CC;
2. CHgCl > (CH₃)₂CHCl > CH₃CH₂CI > (CH₃)₃CCI
3. CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl
4. CH₃CH₂Cl > CH₃CI > (CH₃)₂CHCl > (CH₃)₃CCl

Answer: 3. CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl

Question 42. For the estimation of nitrogen, 1.4 g of an organic compounejÿwas digested by the Kjeldahl method, and the evolved ammonia was absorbed in 60 mL of M/10

1. 5%
2. 6%
3. 10%
4. 3%

Answer: 3. 10%

Multiple Choice Questions Organic Chemistry Class 11 Chapter 12

Question 43. In the Carius method of estimation of halogens, 250 g of an organic compound gave 141 g AgBr. Percentage of Br in the compound(Ag = 108, Br = 80)

1. 48
2. 60
3. 24
4. 36

Answer: 3. 24

Question 44. Which of the following compounds will exhibit

1. 2-phenyl-l-butene
2. 1,1 – diphenyl 1 – propane
3. 1-phenyl-2-button
4. 3 – phenyl – 1 – butene

Answer: 3. 1-phenyl-2-button

Question 45. The increasing order of S_N1 reactivity of the following compounds is—

1. 1<3<2
2. 2<3<1
3. 3<2<1
4. 2<1<3

Answer:  4. 2<1<3

Question 46. The resonance stability is minimal for the compound—

Answer: 2.

Question 47. Which of the following compounds will be suitable for Kjeldahl’s method of nitrogen estimation—

Answer: 4

Question 48. The increasing order of basicity of the following compounds is—

Answer: 1

Question 49. Consider the reactions:

Answer: 3

Question 50. Which undergoes nucleophilic substitution most easily—

Answer: 1

Question 51. IUPAC name ofthe compound,

1. Trans-2-chloro-3-iodo-2-pentene
2. Cis-3-iodo-4-chloro-3-pentane
3. Trans-3-iodo-4-chloro-3-pentene
4. Cis-2-chloro-3-iodo-2-pentene

Answer: 1. Trans-2-chloro-3-iodo-2-pentene

Question 52. Considering the state of hybridization of C-atoms, which one among the following is linear

1. CH₃—C₂—CH₂—CH₃
2. CH₃—CH=CH—CH₃
3. CH₃-C = C-CH₃
4. CH₂=CHCH₂C= C

Answer: 3. CH₃-C = C-CH₃

Question 53. Which is a nucleophilic substitution reaction—

Answer: 4

Question 54. Which is most reactive towards an electrophilic reagent—

Answer: 3.

NCERT Class 11 Organic Chemistry Basic Principles MCQs

Question 55. The correct order of increasing bond length of C—H, C — O, C—C and C=C is _

1. C—H < C—O < C—C < C=C
2. C—H<C=C<C—O<C—C
3. C—C < C=C < C—O < C—H
4. C—O<C—H<C—C<C=C

Answer: 2. C—H<C=C<C—O<C—C

Question 56. RCHO + NH₂NH₂→RCH=N—NH₂ What sort of reactions it—

1. Electrophilic addition-elimination reaction
2. Free radical addition-elimination reaction
3. Electrophilic substitution-elimination reaction
4. Nucleophilic addition-elimination reaction

Answer: 4. Nucleophilic addition-elimination reaction

Question 57. Which of the following acids do not exhibit optical

1. Maleic acid
2. Amino acids
3. Lactic acid
4. Tartaric acid

Answer: 1. Maleic acid

Question 58. The correct order of decreasing acid strength of trichloroacetic acid (I), trifluoroacetic acid (n), acetic acid (in), and formic acid (IV) is—

1. 2>1> 4>3
2. 2>4>3 >1
3. 1> 2>3>4
4. 1>3>2>4

Answer: 1. 2>1> 4>3

Question 59. Which nomenclature is not according to the IUPAC system

Answer: 3

Question 60. Structure of the company whose IlIPAC name is 3-ethyl- 2-hydroxy-4-methylhex-3-en-5-yonic acid is

Answer: 4

Question 61. The structure of the isobutyl group an organic compound is —

Answer: 2

Question 62. The order of stability of the following tautomeric forms is—

1. 2>3>1
2. 1>2>3
3. 3>2>1
4. 2>1>3

Answer: 3. 3>2>1

Question 63. Which of the following compounds will undergo racemization when a solution of KOH hydrolyses—

1. 1 and 2
2. 2 and 4
3. 3 and 4
4. 1 and 4

Answer: None of  these

Organic Chemistry Techniques MCQs Class 11 NCERT

Question 64. Most reactive towards nucleophilic addition reaction is—

Answer:  4

Question 65. In Kjeldahl’s method for estimation of nitrogen present in a soil sample, ammonia evolved from 0.75 g of sample neutralized 10 mL of 1M H₂SO₄ The percentage of nitrogen in the soils—

1. 37.33
2. 45.33
3. 35.33
4. 43.33

Answer: 1. 37.33

Question 66. The number of structural isomers possible from the molecular formula C₃H₉N is

1. 4
2. 5
3. 2
4. 3

Answer:  1. 4

Question 67.  In an S_N1 reaction on chiral centres there is__________

1. 100% racemisation
2. Inversion more than retention leading to partial racemization
3. 100% retention
4. 100%Inversion

Answer:  2. Inversion more than retention leading to partial racemization

Question 68. Which of the following statements is not correct for a nucleophile—

1. A nucleophile is a Lewis acid
2. Ammonia is a nucleophile
3. Nucleophiles attack electron density sites
4. Nucleophiles are not electron-seeking

Answer: 1, 2,3 and 4

Question 69. Two possible stereo-structures of CH3CHOH-COOH, which are optically active, are called—

1. Diastereomers
2. Atropisomers
3. Enantiomers
4. Mesotners

Answer: 3. Enantiomers

Question 70. In which of the following molecules, all the atoms are coplanar—

Answer: 4.

Question 71. The correct order of add strength of the given carboxylic acid is—

Answer: 3

Question 72. Which among the given molecules can exhibit tautomerism

Answer: 4

Question 73.  Which of the following biphenyls is optically active

Answer: 2

Question 74. The pair of electrons in the given carbanion is present in which ofthe following orbitals 

1. sp
2. 2p
3. sp
4. sp

Answer: 1. sp

Question 75. The correct statement about the basicity of aryl amines is—

1. Aryi amines are in general more basic than alkyl amines because the N-atom in aryl amines is sp hybridized
2. Aryi amines are in general less basic than alkyl amines because the unshared pair of electrons on nitrogen in aryi amines undergoes effective delocalization with the ring π-electrons
3. Aryi amines are in general more basic than alkyl amines because the unshared pair of electrons on nitrogen in aryi amines does not undergo delocalization with the ring electrons
4. Aryi amines are more basic than alkyl amines due to

Answer: 2. Aryi amines are in general less basic than alkyl amines because the unshared pair of electrons on nitrogen in aryi amines undergoes effective delocalization with the ring π-electrons

Question 76.  Which one of the following statements for the given
reactions are correct—

1. Is a substitution reaction but (2) and (3) are addition reactions.
2. (1) and (2) are elimination reactions, but (3) is an addition reaction.
3. (1) is an elimination reaction, (2) is a substitution reaction, and (3) is an addition reaction.
4. (1) is an elimination reaction, but (2) and (3) are substitution reactions.

Answer:  3. (1) is an elimination reaction, (2) is a substitution reaction, and (3) is an addition reaction.

Class 11 Chemistry Organic Chemistry Chapter 12 MCQ Solutions

Question 77. The IUPAC name ofthe compound 

1. 5-formylhex-2-en-3-one
2. 5-methyl-oxohex-2-en-5-al
3. 3-keto-2-methylhex-5-enal
4. 3-keto-2-methylhex-4-enal

Answer:  4. 3-keto-2-methylhex-4-enal

Question 78. Which one is the most acidic compound—

Answer: 3

Question 79. The most suitable method of separation of 1: a 1 mixture of ortho and para- nitrophenols is—

1. Chromatography
2. Crystallization
3. Steam distillation
4. Sublimation

Answer:  3. Steam distillation

Question 80. The correct statement regarding electrophile is—

1. Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from another electrophile
2. Electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile
3. Electrophiles can be either neutral or positively charged species and can form a bond by accepting a pair ofelectrons from a nucleophile
4. Electrophile is a negatively charged species and can form a bond accepting a pair of electrons from a nucleophile

Answer: 3.  Electrophiles can be either neutral or positively charged species and can form a bond by accepting a pair ofelectrons from a nucleophile

Question 81. Which of the following is correct concerning -I effect of the substituents (R = alkyl) —

1. —NR₂ > —OR > —F
2. —NH₂ < —OR < —F
3. —NH₂ > —OR > —F
4. —NR₂ < —OR < —F

Answer: 2. —NH₂ < —OR < —F

Question 82. Which ofthe following carbocations is expected to be most stable—

Answer: 3.

Question 83. 3. Which of the following molecules represents the order of hybridization sp², sp², sp, sp from left to right atoms—

1. CH₃ —CH=CH —CH₃
2. HC=C —C=CH
3. CH₃=CH—CH=CH₂
4. CH₂=CH-C=CH

Answer: 4. CH₂=CH-C=CH

Question 84. S_N2 reaction readily occurs in—

Answer: 1.

Question 85. The correct decreasing order of pK_a is

1. 2>4>1>3
2. 4>2>3>1
3. 3>2>4>1
4. 4>1>2>3

Answer:  1. 2>4>1>3

Question 86.  The correct decreasing order of pK_b is—

1. 1>2>3>4
2. 3>4>2>1
3. 2>3>4>1
4. 4>2>1>3

Answer: 4. 4>2>1>3

Question 87. Find the dihydroxy cyclopentane—

1. 1
2. 2
3. 3
4. 4

Answer: 3. 3

Question 88. Decreasing order of nucleophilicity is—

1. OH^–>NH^–>CH₃O^–>RNH₂
2. NH^–>OH^–>CH₃O^–>RNH₂
3. NH^–>CH₃O^–>OH^–>RNH₂
4. CH₃O^–>NH^–>OH^–>RNH₂

Answer:  3. NH^–>CH₃O^–>OH^–>RNH₂

Question 89. pK_a increases in benzoic acid when substituent “x” is bonded at para-position, then “x” is—

1. —COOH
2. —NO₂
3. —CN
4. —OCH₃

Answer: 4. —OCH₃

Question 90. The IUPAC name of the given compound is (CH₃)₃CCH₂C(CH₃)₃

1. 2,3,4,4-tetramethylpentane
2. 1,2,2,4-tetramethylpentene
3. 2,2,4,4-tetramethylpentane
4. 3,3-dimethyl pentane

Answer: 3. 2,2,4,4-tetramethyl pentane

Question 91. The purity of an organic compound is determined by—

1. Chromatography
2. Crystallization
3. Melting or boiling point
4.  Both (1) and (3)

Answer: 4.  Both (1) and (3)

Question 92. Lassaigne’s test for the detection of nitrogen fails in—

1. H₂N—CO—NHNH₂.HCI
2. NH₂—NH₂HCI
3. C₆H₅—NH —NH₂. HCl
4.  C₆H₅CONH₂

Answer: 2. NH₂—NH₂HCI

Question 93. Among the following, the achiral amino acid is—

1. 2-ethylamine
2. 2-dimethylglycine
3. 2-hydroxymethyl serine
4. Tryptophan

Answer: 3. 2-hydroxymethyl serine

Question 94. Arrange the following nucleophiles in the decreasing order of nucleophilicity—

Answer:  4

Organic Chemistry MCQs NCERT Class 11 Chapter 12

Question 95. Which ofthe following is an electrophile—

1. CCl₂
2. CH₃^–
3. H₂O
4. NH₃

Answer: 1. CCl₂

Question 96. Give IUPAC the name ofthe following compound—

1. 5-hydroxy cyclo hex-3-en-1-one
2. 3-hydroxy cyclo hex-5-en-1-one
3. 8-hydroxy cyclo hex-3-en-1-one
4. 7-hydroxy cyclo hex-5-en-1-one

Answer:  1. 5-hydroxy cyclo hex-3-en-1-one

Question 97. Which of the following is the correct order of acidic strength of the following compounds

1. 1 > 2> 3
2. 2 > 3> 1
3. 1 > 3> 2
4. 3 > 2> 1

Answer:  2. 2 > 3> 1

Question 98. IUPAC name ofthe given compound is

1. 2-methoxy-4-bromonitrobenzene
2. 3-bromo-6-nitro-l-methoxybenzene
3. 3 -methoxy-4-nitrobromobenzene
4. 5-bromo-2-nitro-l-methoxybenzene

Answer: 4. 5-bromo-2-nitro-l-methoxybenzene

Question 99. The number of <r- and n -bonds in the pent-1-en-4-one molecule is respectively—

1. 8 and 2
2. 10 and 3
3. 6 and 4
4. 7 and 2

Answer: 2. 10 and 3

Question 100. The hybrid orbitals involved in the formation of the C₂— C₃ bond in the following compound, CH₂=CH — CH₂—CH₂—C≡CH are—

1. sp-sp²
2. sp-sp³
3. sp²-sp³
4. sp³-sp³

Answer: 3.  sp²-sp³

Question 101. The increasing order of electronegativity of the carbon atoms C-2, C-3 and C-4 in the compound CH₃ — C=C — CH₂ — CH=CH₂ is—

1. C-3 < C-2 < C-4
2. C-4 < C-3 < C-2
3. C-2 < C-4 < C-3
4. C-3 < C-4 < C-2

Answer: 1.  C-3 < C-2 < C-4

Question 102. CH₃CONH₂→CH₃CN; In this conversion, the change in the hybridization state of the carbon atom of the functional group is—

1. sp³-sp
2. sp³-sp
3. sp-sp³
4. sp³-sp³

Answer: 2. sp³-sp

Question 103. The correct shapes of CCl₄ and CCl₂=C=C=CCl₂ molecules are respectively—

1. Linear and tetrahedral
2. Planar and pyramidal
3. Tetrahedral and planar
4. Tetrahedral and linear

Answer: 3. Tetrahedral and planar

Question 104. The number of C and H-atoms that lie in the same plane in a toluene (C₆H₅CH₃) molecule is respectively—

1. 7 and 5
2. 6 and 5
3. 7 and 3
4. 6 and 3

Answer: 1.  7 and 5

Question 105. The number of primary, secondary, tertiary, and quaternary carbon atoms in 2,2,4-trimethylpentane is respectively—

1. 5,1,1 and 1
2. 1,1,1 and 5
3. 4,1,1 and 2
4. 1,5,1 and 1

Answer: 1. 5,1,1 and 1

Question 106. The number of primary, secondary, and tertiary H-atoms in the molecule is respectively—

1. 14,9 and 2
2. 15,9 and 1
3. 15,8 and 2
4. 15,8 and 1

Answer: 4.  15,8 and 1

Question 107. In which of the following molecules does all the atoms lie on the same straight line—

1. HC = C—C = CH
2. HC = C—CH₃
3. HC≡CN
4. C₃O₂

Answer: 2. HC = C—CH₃

Question 108. Hybridisation states of C-2, C-3, C-5 and C-6 in the com¬pound, (CH₃)₃CCH =CHCHC=CH are respectively—

1. sp, sp³, sp² and sp³
2. sp³, sp², sp² and sp
3. sp, sp², sp² and sp³
4. sp, sp², sp³ and sp²

Answer: 1. sp, sp³, sp² and sp³

Question 109. IUPAC name of the compound,

1. 1,1-diethyl-2,2-dimethyl pentane
2. 4,4-dimethyl-5,5-diethyl pentane
3. 5,5-diethyl-4,4-dimethyl heptane
4. 3-ethyl-4,4-dimethyl heptane

Answer: 4.  3-ethyl-4,4-dimethyl heptane

Question 110. IUPAC name of the compound,

1. 4-propyl-3-methylhex-5-en-2-one
2. 3-propyl-5-methylhex-l-en-5-one
3. 3-methyl-4-propyl hex-5-en-2-one
4. 3-methyl-4-vinylheptan-2-one

Answer:  3.  3-methyl-4-propyl hex-5-en-2-one

Question 111. Bond lengths of C—H, C—O, C—C and C follow the sequence—

1. C—H < C — O < C—C < C=C
2. C—H < C=C < C—O< C — C
3. C—C<C=C<C—O<C—H
4. C—O<C—H<C—C<C=C

Answer: 2. C—H < C=C < C—O < C — C

Question 112. IUPAC name of the compound

1. 1-chloro-2-nitro-4-methylbenzene
2. 1-chloro-4-methyl-2-nitrobenzene
3. 2-chloro-1-nitro-5-methylbenzene
4. m-nitro-p-chlorotoluene

Answer:  2. 1-chloro-4-methyl-2-nitrobenzene

Question 113. The correct IUPAC name of the alkyl group is—

1. 2-ethyl-3-sec-butyl propyl
2.  2,4-diethyl pentyl
3. 2-ethyl-4-methyl hexyl
4. 2-methyl-4-Ethylhexyl

Answer: 3.  2-ethyl-4-methyl hexyl

Question 114. The hybridization states of the carbon atom of the amido and cyano groups are respectively—

1. sp³ and sp²
2. sp² and sp
3. sp and sp²
4. sp³ and sp

Answer: 2.  sp² and sp

Question 115. Which of the following pair of compounds are isomers—

1. CH₃CH₂OH, CH₃OCH₃

2. CH₃OC₃H₇, C₂H₅OC₂H₅ .

3. CH₃CH₂CHO, CH₃COCH₃

4.

Answer: 4.

Question 116. Two aliphatic compounds will not be considered isomers if they are—

1. Aldehyde and ketone
2. Ether and alcohol
3. Ether and aldehyde
4. Carboxylic acid and ester

Answer: 3.  Ether and aldehyde

Question 117. The number of organic compounds with molecular formula C₄H₁₀ is—

1. 7
2. 5
3. 6
4. 8

Answer: 1. 7

Question 118. Only a monosubstituted compound is obtained when an alkane reacts with chlorine in the presence of UV light. The alkane is—

1. Propane
2. Pentane
3. Butane
4. Cyclohexane

Answer: 4.  Cyclohexane

Question 119. Two enantiomers rotate the plane of polarisation of plane polarised light

1. In different directions but keeping the angle the same
2. In the same direction but with different angles
3. In the same direction and the same angle
4. In different directions with different angles

Answer: 1.  In different directions but keeping the angle the same

Question 120. Which of the following is an optically active compound—

1. CH₃CHClCH=:CH₂
2. CH₃CHCl₂
3. Meso-tartaric acid
4. CH₃CH=C=CH₂

Answer: 1.  CH₃CHClCH=:CH₂

Question 121. Which of the following compounds exhibit both geometrical and optical isomerism—

1. CH₃CHClCH=C(CH₃)₂
2. CH₃CH=CH — CHBrCH₂CH₃
3. CH₂=C=CH—CH=CHCH₃
4. CH₃CH₂CH=CH₂

Answer: 2.  CH₃CH=CH — CHBrCH₂CH₃

NCERT Class 11 Chapter 12 Organic Chemistry Multiple Choice Questions & Answers

Question 122. Which of the following compounds does not exhibit tautomerism—

1. CH₃CH₂N=O
2. CH₃NO₂
3. CH₃COCH₃
4. (CH₃)₃CCOC₆H₅

Answer: 4.  (CH₃)₃CCOC₆H₅

Question 123. The enol content in which of the following compounds is maximum—

1. CH₃COCH₂COCH₃

3. CH₃COCH₃

4. CH₃CHO

Answer: 1. CH₃COCH₂COCH₃

Question 124. The optically active alkane of the lowest molecular mass which is also chiral is—

1. 3-methyl hexane
2. 2, 3-dimethyl pentane
3. 2-methyl hexane
4. 2,5-dimethyl hexane

Answer: 1.  3-methylhexanec

Question 125. Which of the following compounds possesses a center of symmetry—

1. Trans-1,3-dimethylcyclobutane
2. Cis-1,3-dimethylcyclobutane
3. Trans-1-ethyl-3-methyl cyclobutane
4. Cis-l-ethyl-3-methyl cyclobutane

Answer: 1.  Trans-1,3-dimethylcyclobutane

Question 126. The compounds ds-2-butene and Frans-2-butene can be differentiated by

1. The number of products obtained due to their chlorination
2. The number of products obtained due to their bromination
3. Their reaction with H₂ in the presence of a nickel catalyst
4. Their respective boiling points

Answer:  4. Their respective boiling points

Question 127. Which of the following is optically active—

Answer: 4.

Question 128. Which of the following is non-superimposable on its mirror image—

Answer: 3

Question 129. The number of structural isomers that are formed when one H-atom of diphenylmethane   is substituted by chlorine is—

1. 4
2. 6
3. 8
4. 7

Answer: 1. 4

Question 103. The number of isomers formed by a compound whose molecular formula is C₂BrClFI is—

1. 3
2. 4
3. 5
4. 6 
   

Answer: 4. 6

Question 131.   In this reaction, the hydroxy acid obtained is—

1. (+)-enantiomer
2. (-)-enantiomer
3. 50% (+) and 50% (-)- enantiomer
4. 20% (+) and 80% (-)-enantiomer

Answer: 3.  50% (+) and 50% (-)- enantiomer

Question 132. Which of the following compounds will produce the most stable carbocation in the presence of an acid—

1. (CH₃)₂CHCH₂OH
2. CH₂=CH—CH₂OH
3. (CH₃)₂CHOH
4. (CH₃)₃COH

Answer: 2. CH₂=CH—CH₂OH

Question 133. The correct order of stability of the given carbanions:

1. 1>2>3
2. 2>1>3
3. 3>2>1
4. 2>3>1

Answer: 1. 1>2>3

Question 134. The most stable carbocation is—

Answer: 4.

Question 135. The most stable carbanion is—

Answer:  2

Question 136. Carbocation which does not undergo rearrangement is-

1. (CH₃)₂CH⁺CH₂
2. (CH₃)₂CH⁺CHCH₃
3.  (CH₃)₃⁺C
4. (CH₃)₃C⁺CH₂

Answer: 3.  (CH₃)₃⁺C

Question 137. Which of the following carbocations is quite stable and can even be stored in the laboratory as a salt— in which hyperconjugation does not occur—

1. (CH₃)₂ ⁺CH
2. (C₆H₅)₃⁺C
3.  CH₂=CH-⁺CH₂
4. ⁺CH₂CH₂C₆H₅

Answer: 2. (C₆H₅)₃⁺C

Question 138. Which of the given alicyclic compounds is most active—

1. C₆H₅C(CH₃)₃
2. C₆H₅CH₃
3. (CH₃)₂C=CH₂
4. CD₃CH=CH₂

Answer: 1. C₆H₅C(CH₃)₃

Question 139.  Which of the given resonance structures is most stable—

Answer: 4

Question 140. Which ofthe given alicyclic compounds is most active—

Answer: 2.

Question 141.

1. ^•CH₃
2. CH₃^•CH₂
3. CH₂=CH^•CH₂
4. C₆H₅^•CH₂

The correct order of stability of these free radicals is—

1. 1>2>3>4
2. 3>2>1>4
3. 4>3>2>1
4. 4>7>2>3

Answer: 3. 4>3>2>1

Question 142. The structures of carbocation and carbanion are respectively—

1. Linear and planar
2. Trigonal planar and trigonal pyramidal
3. Tetrahedral and trigonal planar
4. Trigonal pyramidal and tetrahedral

Answer: 2.  Trigonal planar and trigonal pyramidal

Question 143. The correct order of stability of the given three carbanions is—

Answer: 3

NCERT Class 11 Chapter 12 Organic Chemistry Multiple Choice Questions & Answers

Question 144. Which of the following alkenes is most stable—

1. (CH₃)₂C=C(CH₃)₂
2. (CH₃)₂C=CHCH₂CH₃
3. CH₃CH₂CH=CHCH₂CH₃
4. CH₃CH₂CH₂CH₂CH=CH₂

Answer: 1.  (CH₃)₂C=C(CH₃)₂

Question 145. In which of the following compounds, the extent of resonance between the benzene ring and halogen atom is maximum—

Answer: 1

Question 146. The compound whose basicity is maximum in gaseous and aqueous medium is—

1. NH₃
2. CH₃NH₂
3. (CH₃)₂NH
4. (CH₃)₃N

Answer: 3. (CH₃)₂NH

Question 147. (+)-1-Chloro-1-phenylethane undergoes racemization in the presence of a small amount of SbCl₅.The intermediate formed in this process is—

1. A carbene
2. A carbocation
3. A carbanion
4. A free radical

Answer: 2.  A carbocation

Question 148. In which of the following compounds, the presence of nitrogen cannot be detected by Lassaigne’s test

1. NH₂NH₂.HCI
2. C₆H₅NHNH₂.HCI
3. PhN≡NPh
4. NH₂CONH₂

Answer: 1.  NH₂NH₂.HCI

Question 149. Which of the following compounds is responsible for the formation of Prussian blue during the detection of nitrogen by Lassaigne’s test

1. Na₄[Fe(CN)₆]
2. Fe₄[Fe(CN)₆]
3. Fe₂[Fe(CN)₆]
4. Fe₃[Fe(CN)6]₂

Answer: 2.  Fe₄[Fe(CN)₆]

Question 150. The process by which essential oils can be extracted from flowers is

1. Distillation
2. Crystallization
3. Vacuum distillation
4. Steam distillation

Answer: 4.  Steam distillation

Question 151. The process that is suitable for detecting two different types of ink in any handwritten ancient document is

1. Column chromatography
2. Solvent extraction
3. Distillation
4. Thin layer chromatography

Answer: 4.  Thin layer chromatography

Question1 52. Detection of which of the following functional groups is required to confirm the presence of nitrogen in the corresponding compound—

1. Amido
2. Carboxyl
3. Carbonyl
4. Alkoxycarbonyl

Answer: 1.  Amido

Question 153. Which of the following compounds does not exhibit geometrical isomerism—

Answer: 3.

Question 154. The number of geometrical isomers of the compound, CH₃CH=CH — CHCH₃ is_____

1. 3
2. 2
3. 4
4. 5

Answer: 1.  3

Question 155. Types of non-equivalent H-atoms in CH₃CH(OH)CH(OH)CH₃

1. 2
2. 4
3. 3
4. 6

Answer: 3. 3

Question 156. The number of compounds formed on monobromination

  is_________ 

1. 3
2. 2
3. 5
4. 4

Answer: 4.  4

Question 157.  Types of non-equivalent H-atoms in 

1. 2
2. 4
3. 3
4. 6

Answer: 2. 4

Question 158. The double bond equivalent of C₈H₉CIO is—

1. 4
2. 6
3. 3
4. 2

Answer: 1.  4

Question 159. The correct order of basicity of

1. CH₃NH₂
2. (CH₃)₂NH and
3. (CH₃)₃N

in chlorobenzene is—

1. 1<3< 2
2. 2 <3< 1
3. 1< 2< 3
4. 2< 1< 3

Answer: 3. 1< 2< 3

Question 160. The correct order of stability of 

1. (CH₃)₂⁺CH

2. CH₃⁺CHOCH₃

3. ClCH₂⁺CHCH₃ is—

1. 1> 2 > 3
2. 2 > 3 > 1
3. 2 > 1> 3
4. 3 > 2 > 1

Answer: 3.  2 > 1> 3

Question 161. The correct order of stability of these carbanions is—

1. CH₃^–CH —CO —CH₃

2. CH₃CH₂-CO- ^–CH₂ and

3.  ^–CH₂CH₂— CO — CH₃ 

1. 3<1<2
2. 1<3<2
3. 2<3<1
4. 3<2<1

Answer: 1. 3<1<2

Question 162. In Kjeldahl’s method, CuSO₄ is used to—

1. Catalyze the reaction
2. Oxidise the reaction
3. Reduce the reaction
4. Increase boiling point

Answer: 1. Catalyse the reaction

Question 163. The number of optically active isomers among five probable alcohols of molecular formula C₄H₁₀O is—

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Question 164. Which compound gives the most unstable enol—

Answer: 1.

Question 165. If 3.4% sulfur is present in insulin, then the minimum molecular mass of insulin will be—

1. 350
2. 470
3. 560
4. 940

Answer: 4. 940

Question 166. The number of enantiomer pairs formed on monochlorination of 2,3-dimethylbutane is—

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Question 167. There are three different asymmetric carbon atoms in a compound. The number of possible stereoisomers of this compound is—

1. 8
2. 3
3. 9
4. 6

Answer: 1. 8

Question 168. In which of the following compounds, the nucleophilic character of N-atom is maximum—

Answer: 1

Question 169. Which of the following resonance structures is incorrect 

1. x>z>y
2. z>x>y
3. x>y>z
4. z>x>y

Answer: 3. x>y>z

NCERT Class 11 Chapter 12 Organic Chemistry Multiple Choice Questions & Answers

Question 170. The number of electrons in the p-orbital of methyl cation is—

1. 2
2. 3
3. 4
4. None of these

Answer: 4. None of these

Question 171. Which compound can exhibit geometrical isomerism—

1. Acetone-oxime
2. Isobutene
3. Acetophenone-oxime
4. Benzophenone-oxime

Answer: 3. Acetophenone-oxime

Question 172. In the given compound, the hydrogen atom is linked to which carbon atom is most acidic—

1. C_α
2. C_β
3. C_γ
4. C_δ

Answer:   4. C_γ

Question 173. In which of the following molecules, all the constituent carbon atoms have the same state of hybridization—

1. HC≡C —C≡N
2. CH₃—C≡C —CH₂CH₃
3. CH₂=C=C=CH₂
4. CH₂=CH —CHO

Answer: 4.  CH₂=CH —CHO

Question 174. In which of the following molecules, all the atoms lie in the same plane—

1. CH₂=C=CH₂
2. CCI₂=C=C=CH₂
3. C₆H₅C=CH
4. CH₂=CH—C=CH

Answer: 2 and 3

Question 175. Which of the following express a homologous series—

1. Methanol, ethanol, 1-propanol
2. 1-hexene, 2-hexene, 3-hexene
3. Formic acid, acetic acid, propionic acid
4. Methane, methanol, methanal

Answer: 1 and 3

Question 176. Which of the following statements is incorrect—

1. Heat of hydrogenation of CH₃CH₂CH=CH₂ is less than that of (CH₃)₂C =CH
2. CCl₃ is more stable than CF₃
3. The bond lengths of three carbon-oxygen bonds in carbonate (CO^2-₃) ions are not equal.
4. Free radicals are paramagnetic

Answer: 1 and 3

Question 177.  Which of the following orders of stability are correct—

1. CH₃⁺CHOCH₃ > CH₃⁺CHCH2OCH₃
2. F⁺CH₂>FCH₂⁺CH₂
3. PhCH₂⁺CH₂>Ph⁺CH₂
4. FCH₂COO^–< CH₃COO^–

Answer:   a and 2

Question 178. Which process is not represented correctly—

Answer: 1 and 3

Question 179. Which ofthe following sets represent only electrophiles ______

1. BF₃, NH₃, H₂O
2. AICI₃, SO₃, ⁺NO₂
3. ⁺NO₂, ⁺CH₃, CH₃⁺CO
4. C₂H₅^–, ^•C₂H₅, C₂H⁺₅

Answer:  2 and 3

Question 180. Delocalisation in hyperconjugation occurs—

1. In the case of rr -bond electrons of the C — H bond of any alkyl group directly linked to a double bond (f)
2. In the case of cr -bond electrons of the C — H bond of any alkyl group directly linked to a positive carbon atom
3. In the case of a-electrons of C=C
4. In the case of a lone pair of electrons

Answer:  1 and 2

Question 181. Which of the following statements is incorrect—

1. Sodium extract is first boiled with dilute HC1 during the detection of halogens by Lassaigne’s test
2. If in an organic compound, both nitrogen and sulfur are present, then blood-red coloration is observed during the detection of nitrogen by Lassaigne’s test
3. Organic compounds that dissociate at their melting points are purified by vacuum distillation
4. In paper chromatography, the stationary phase is solid while the mobile phase is liquid

Answer: 1 and 4

Question 182. Which of the following compounds give a red coloration in Lassaigne’s test during the detection of nitrogen—

Answer:  3 and 4

Question 183. Which of the following mixtures is responsible for blue coloration in Lassaigne’s test during the detection of N—

1. NH₂NH₂ + charcoal
2. NH₄Cl + NaNO₃
3. C₆H₅COOH + NaNO₃
4. NH₂NH₂ + NH₄Cl

Answer: 1 and 3

Question 184. In which of the following structures, the arrangement of the four groups/atoms is different from that of  _________ 

Answer:   1 and 3

Question 185. Which ofthe following are optically active compounds ___________

Answer:  2 and 3

Question 186. In which of the following compounds, is the number of hypercoagulable hydrogen atoms is same—

Answer:  2 and 4

Question 187. For which of the following compounds, is the number of compounds formed on monobromination are same—

Answer:   1 and 2

Question 188. The carbocations which attain stability by resonance are___

Answer: 1,3, and 4

Question 189. Mixtures that can be separated by simple distillation are

1. A mixture of ether and toluene
2. A mixture of hexane and toluene
3. A mixture of benzene and chloroform
4. A mixture of 95% ethanol and 5% water

Answer:  1 and 3

NCERT Class 11 Chapter 12 Organic Chemistry Multiple Choice Questions & Answers

Question 190. Compounds that can be purified by steam distillation are—

1. Acetic acid
2. o-nitrophenol
3. Ethanol
4. Nitrobenzene

Answer:  2 and 4

Question 191. Which of the following statements is incorrect—

1. Quantitative estimation of nitrogen in any compound can be done by Kjeldahl’s method
2. Quantitative estimation of sulfur in organic compounds can be done by the Dumas method
3. Quantitative estimation of halogens in organic compounds can be done by the Carius method
4. In Liebig’s method of detecting carbon and hydrogen in organic compounds, carbon is converted into carbon dioxide while hydrogen is converted into water

Answer: 1 and 3

Question 192. In which of the following processes, any organic liquid vapourises below its boiling point—

1. Vacuum distillation
2. Distillation
3. Steam distillation
4. Sublimation

Answer: 1 and 3

Question 193. Which exhibit optical and geometrical isomerism—

1. CH₃CHCICH=CH₂
2. CH₃-CHCI —CH=CH₂
3. CH₃CH=CH— CH(CH₃)₂
4. CH₃CH=CH—CH=C=CHCH₃

Answer: 2 and 4

Question 194. Which of the following are correct statements—

1. (CF₃)₃C⁺ is more stable than (CH₃)₃C⁺ ,
2. Na⁺does did not act as an electrophile
3. Ph₃C⁺ can be stored in the form of Ph₃C⁺BF^–₄ salt
4. CH₃CH₂O^– is less stable than O₂NCH₃CH₂O^–

Answer:  2 and 3

Question 195.  The nucleophilic reagents are—

1. OH^–
2. : NH₃
3. CCl₂
4. CN^–

Answer: 1,2 and 4

Question 196. The electrophilic reagents are—

1. ⁺NO₂
2. Cl⁺
3. H₂O
4. SO₃

Answer: 1, 2, 4

Question 197. Which of the following statements regarding Lassaigne’s test are correct—

1. N, S, and halogens are detected by converting them into their corresponding inorganic salts
2. This test is done to detect N, S, and hydrogen
3. Organic compounds are fused with sodium metal
4. Different halogens can be distinguished

Answer:  1,3 and 4

Question 198. Which of the following exhibits keto-enol tautomerism—

1. C₆H5COC₆H₅
2. C₆H₅COCH=CH₂
3. C₆H₅COCH₂COCH₃
4. CH₃COCH₂COCH₃

Answer:  2. C₆H₅COCH=CH₂

Question 199. Which of the following do not exhibit optical activity—

1. 3-methyl-l-pentene
2. 2-methyl-2-pentene
3. 4-methyl-l-pentene
4. 3-methyl-2-pentene

Answer: 2,3, and 4

Question 200. The correct statements are—

1. A racemic mixture is an equimolecular mixture of a pair of enantiomers
2. Configuration of a molecule means a stable three-dimensional arrangement of the groups attached to a specific atom of the molecule
3. The melting & boiling points of 2 enantiomers are different
4. A molecule may be optically active or inactive if more than one asymmetric carbon is present in the molecule

Answer: 1,2 and 4

Question 201. Which of the following are planar—

1. Ferf-butyl radical
2. Ferf-butyl carbocation
3. Ferf-butyl carbanion
4. Allyl carbanion

Answer: 1,2 and 4

Question 202. Which can act as both electrophile and nucleophile—

1. CH₃OH
2. CH₃Cl
3. CH₃CN
4. HCHO

Answer: 3 and 4

Question 203. Which of the following can act neither as an electrophile nor as a nucleophile

1. H₃O⁺
2. R₄N⁺
3. CN^–
4. SO₃

Answer: 1 and 2

Question 204. Which of the following conditions favor E2 reaction—

1. A strong base of high-concentration
2. A solvent of low polarity
3. 3° alkyl halide as the substrate
4. Alkyl iodide

Answer: 1,2,3,4

Question 205. Compounds that will not exhibit geometrical isomerism—

Answer:  1,2,3,

Question 206. In which ofthe following, a plane of symmetry is present—

Answer: 1,2,4

Question 207. Three stereoisomers of CH₂YZ are possible if the tire structure of methane is—

1. Rectangular planar
2. Square planar
3. square pyramidal
4. Octahedral

Answer: 1 and 2

NCERT Solutions For Class 11 Chemistry Chapter 12 Organic Chemistry Basic Principles And Techniques Short Question And Answers

Question 1. Why does distillation purify impure glycerol-reduced pressure?
Answer:

The boiling point of glycerol is 563K under normal pressure and it undergoes decomposition below this temperature. Thus, simple distillation cannot be used for its purification. Under a reduced pressure of 12 mm Hg, the boiling point of glycerol is reduced and then at 453 K, it can be distilled without getting decomposed.

Question 2. Why is it necessary to use acetic acid and not sulphuric acid for the acidification of sodium extract for testing sulfur by lead acetate test?
Answer:

To detect the presence of sulfur, sodium extract is acidified with acetic acid because lead acetate being soluble does not interfere with the test. If sulphuric acid is used, lead acetate will react with sulphuric acid to form a white precipitate of lead sulfate which will interfere with the test.

Pb(CH₃COO)₂  ( Lead acetate )+ H₂SO₄ →PbSO₄  (Lead sulphate)↓(White) + 2CH₃COOH

Read and Learn More NCERT Class 11 Chemistry Short Answer Questions

Question 3. The presence of N in hydroxylamine hydrochloride cannot be detected by Lassaigne’s test—why?
Answer:

When hydroxylamine hydrochloride (NH₂OH.HCl) is fused with metallic sodium, NaCN is not obtained as the compound contains no carbon. Thus, the presence of nitrogen in this compound cannot be detected by Lassaigne’s test.

Question 4. How can it be possible to detect the presence of nitrogen in hydrazine hydrochloride?
Answer:

During fusion of the compound with metallic sodium if some starch or charcoal is added, carbon of starch or charcoal combines with nitrogen of the compound to form NaCN which will further indicate the presence of nitrogen in the compound a proton. On the other hand, no such steric inhibition occurs in N, N-dimethylaniline because the two orlp H-atoms are relatively much smaller in size. The unshared electron-pair on N-atom becomes involved In resonance interaction with the ring and therefore, is not fully available for taking up a proton. This explains why N, N,2,6-tetramethylsilane is more basic than N, N -dimethylaniline.

Short Question and Answers Class 11 Organic Chemistry Chapter 12

Question 5. Chloroform is more acidic than fluoroform. Explain.
Answer:

CF₃^–, tire conjugate base of fluoroform (CHF₃^–), is stabilized by the -I effect of 3 F-atoms. But CCl^–₃ the conjugate base of chloroform (CHCl₃), is relatively more stabilized by the somewhat weaker -I effect of 3 Cl-atoms along with d-orbital resonance (Cl has vacant d-orbital). So chloroform is more acidic than fluoroform.

Question 6. How can you separate benzoic acid and nitrobenzene from their mixture by the technique of extraction using an appropriate chemical reagent?
Answer:

The mixture is shaken with a dilute sodium bicarbonate solution when benzoic acid gets converted to sodium benzoate and dissolves in water leaving nitrobenzene behind. The mixture is extracted with ether or chloroform when nitrobenzene goes into the organic layer. After separating the organic layer, it is distilled to get nitrobenzene. The aqueous layer is acidified with dilute HCl when benzoic acid gets precipitated. It is obtained by filtration.

Question 7. Why is impure glycerol purified by distillation under reduced pressure?
Answer:

The boiling point of glycerol is 563K under normal pressure and it undergoes decomposition below this temperature. Thus, simple distillation cannot be used for its purification. Under a reduced pressure of 12 mm Hg, the boiling point of glycerol is reduced and then at 453 Kit can be distilled without getting decomposed.

Question 8. Why is it necessary to use acetic acid and not sulphuric acid for the acidification of sodium extract for testing sulfur by lead acetate test?
Answer:

To detect the presence of sulfur, sodium extract is acidified with acetic acid because lead acetate being soluble does not interfere with the test. If sulphuric acid is used, lead acetate will react with sulphuric acid to form a white precipitate of lead sulfate which will interfere with the test.

Question 9. The presence of N in hydroxylamine hydrochloride cannot be detected by Lassaigne’s test—why? p
Answer:

When hydroxylamine hydrochloride (NH₂OH HCl) is fused with metallic sodium, NaCN is not obtained as the compound contains no carbon. Thus, the presence of nitrogen in this compound cannot be detected by Lassaigne’s test

Question 10. How can it be possible to detect the presence of nitrogen in hydrazine hydrochloride?
Answer:

During the fusion of the compound with metafile sodium if some starch or charcoal is added, carbon of starch or charcoal combines with nitrogen of the compound to form NaCN which will further indicate the presence of nitrogen in the compound.

Question 11. Can you separate two liquids A (b.p. 413 K) and B (b.p. 403 K) present in a mixture by simple distillation?
Answer:

The two components cannot be separated by simple distillation. This is because the vapors of both liquids will be formed simultaneously and will condense together in the receiver. The separation of these two liquids can be done by fractional distillation.

Question 12. Will CCl₄ give a white precipitate of AgCl on heating it with silver nitrate solution? Give a reason for your
Answer.

When CCl₄ is heated with AgNO₃ solution, a white precipitate of AgCl will not be formed. This is because CCl₄ being a covalent compound with a strong C— Cl bond does not ionize to give the Cl” ions required for the formation of the precipitate of AgCl.

Question 13. Is it possible to distinguish between phenylhydrazine hydrochloride and hydrazine hydrochloride by Lassaigne’s test? Give reason.
Answer:

Lassaigne’s test can be used to distinguish between phenylhydrazine hydrochloride (C₆H₅NHNH₂ HCl) and hydrazine hydrochloride (NH₂NH₂ HCI). This is because the former containing carbon and nitrogen produces NaCN when fused with metallic sodium while the latter containing no carbon does not produce NaCN when fused with sodium.

Question 14. Define R_f value. What is called descending paper chromatography
Answer:

If the solvent is placed at the top and the upper end of the chromatography paper is dipped in it, then the solvent moves downwards. This is called descending paper chromatography.

Organic Chemistry Basic Principles and Techniques Short Q&A

Question 15. The tendency of carbon to exhibit catenation is much higher than that of Si and S—why
Ans.

The C— C bond dissociation enthalpy (348.6 kj. mol^-1) is much higher than that of Si—Si (228.4 kj -mol-1) and S—S (224.2 kj- mol^-1) bond dissociation enthalpies and since the formation of C—C bond is thermodynamically much favorable, the tendency of carbon to exhibit catenation is much higher than that of silicon and sulfur

Question 16. Melting and boiling points of organic compounds are usually very low— Why?
Answer:

Covalent organic compounds usually exist as single molecules. The attractive forces operating among these less polar or non-polar molecules are very low. As a result of this, the melting and boiling points of these compounds are usually very low

Question 17. Write the IUPAC name of the compound,   mentioning the secondary prefix, primary prefix, a word root, primary suffix & secondary suffix respectively
Answer:

In the compound, the secondary prefix: is bromo (at C-3); the primary prefix: is cyclo; the word root: is pent; the primary suffix: is ane (e is to be omitted), and the secondary suffix: is ol (at C-l). Therefore, the IUPAC name of the compound is

Question 18. Which one of them is more pure and why?
Answer:

The first sample (boiling range: 76-78°C) Is more pure because its boiling range is shorter.

Question 19. The wind is on an azeotropic mixture? Give example.
Answer:

An azeotropic mixture is a mixture of two or more liquids having a constant boiling point. The most familiar example of an azeotropic mixture is a mixture of ethanol and water in the ratio of 95.6: 4.4. It boils at a temperature of 78.5°C.

Question 20. A mixture contains two organic solids, A and B. The solubilities of A and B in water are 12 g per 100 mL and 3 g per 100 mL respectively. How will you separate A and B from this mixture?
Answer:

The two components can be separated by fractional crystallization. When the saturated hot solution of this mixture is allowed to cool, the less soluble compound, B crystallizes out first leaving the more soluble compound, A in the mother liquor. The mother liquor is then concentrated and allowed to cool when the compound A crystallizes out.

Question 21. What is seeding?
Answer:

Seeding is a process of inducing crystallization by adding a crystal of the pure substance into its saturated solution.

Question 22. Suggest methods for the separation of the components in each of the following mixtures:

1. A mixture of liquid A (b.p. 366 K) and liquid B (b.p. 355.5 K).
2. A mixture of liquid C (b.p. 360 K) and liquid D (b.p. 420 K).

Answer:

1. The two liquids, A and B can be separated by fractional distillation because the boiling points of them differ by just 10.5 K.
2. The two liquids, C and D can be separated by simple distillation because the boiling points of them differ widely by 60 K.

Question 23. A mixture contains three amino acids. How can they be Identified?
Answer:

When the mixture is subjected to separation by paper chromatography, three spots on the paper become visible at different heights from the starting line by placing the paper under UV light. Then they can be identified by determining their R_f values and comparing these values with the R_f value of the pure compounds.

Question 24. The R_f values of X and Fin a mixture determined by TLC method in a solvent mixture are 0.75 and 0.52 respectively. If the mixture is separated by column chromatography using the same solvent mixture as the mobile phase, which of the two components, will elute first and why?
Answer:

The higher R_f value of X (0.75) indicates that it is less strongly adsorbed as compared to compound Y having a lower R_f value (0.52). Therefore, if the mixture is separated by column chromatography using the same solvent mixture as the mobile phase, X will be eluted first.

Question 25. Why is an organic compound fused with sodium for
Answer:

When an organic compound is fused with metallic sodium, these elements present in the compound are converted into water-soluble sodium salts (NaCN, NaX, and Na₂S). The presence of cyanide ion (CN^–), halide ion (X^–), and sulfide ion (S^2-) in the solution can then be 40 What is seeding? confirmed by using suitable reagents

Question 26.  What type of fission of a covalent bond produces free radicals? Give an example with a proper sign.
Answer;

Homolytic fission of covalent bonds produces free radicals.

Question 27.

1. Write down the IUPAC name of the following compound

Question 28. Draw the structure of the following compound: 3,4-dimethyl pentanoic acid
Answer:

1. 2-bromo-2-chloroethanol.

2.

Question 29. Draw the canonicals of CH₃COOH and CH₃COO^–. In which case resonance is more important?
Answer:

Equivalent structures (more stable) Resonance is more important for CH₃COO^– as it involves two equivalent resonating structures and the negative charge is always on the electronegative O-atom.

Question 30. Write the principle of estimation of carbon and hydrogen in an organic compound.
Answer:

A known amount of dry and pure organic compound is heated with cupric oxide(CuO) in a hard glass test tube. As a result, the carbon (C) and hydrogen (H) present in the compound are oxidized to carbon dioxide (CO₂) and water (H₂O ) respectively.

Knowing the amounts of CO₂ and H₂O formed in the reaction it is possible to calculate the percentages of C and H present in the compound.

NCERT Solutions Class 11 Chemistry Chapter 12 Short Questions

Question 31.

1. Write down the IUPAC name of the following compound: CH₃COCH₂CH – Cl – COCH₃ sodium nit
2. Write down the structural formula of the following compound: Hex-1 -en-4-one

Answer:

Question 32.

1. Arrange the following radicals in increasing order of-I effect: I, Br, Cl, F’
2. Write the structural formula of the following compound: 5-amino pent-3-enoic acid

Answer:

1. I <Br<Cl<P

2.   \(\mathrm{HOO} \stackrel{1}{\mathrm{C}}-\stackrel{2}{\mathrm{C}} \mathrm{H}_2-\stackrel{3}{\mathrm{C}} \mathrm{H}=\stackrel{4}{\mathrm{C}} \mathrm{H}-\stackrel{5}{\mathrm{C}} \mathrm{H}_2-\mathrm{NH}_2\)

Question 33.

1. Whyis(CH₃)₃C⁺ more stable than CH₃CH⁺₂?
2. Indicate the electrophilic center of the following compounds: CH₃CHO, CH₃CN.

Answer:

1. Due to the +1 effect of three electron-donating CH₃ groups, (CH₃)₃C⁺ is more stable than CH₃⁺CH₂ (which contains only one CH₃ group attached to C⁺).  Furthermore (CH₃)₃ C⁺ is stabilized by 9 hyper conjugative structures, while CH₃CH₂ is stabilized by only three hyper conjugative structures

2.

Question 34. Name IUPAC name of the following:

Answer:

1. Propan-1,2,3-trio
2. 3,3-dichlorobutanoic acid

Question 35. Explain the order of basicity of the following compounds:

1. CH₃—CH₂— NH₂
2.  CH₃ —CH=N H
3. CH₃—CH₂—CN

Answer: CH₃— CH₂—CN < CH₃—CH=N-H < CH₃—CH₂— NH₂

Question 36. A compound having molecular formula CgH18 can form only one monobromo derivative. Draw the structure of the compound.
Answer:

Since, the compound forms only one monobromo derivative, all hydrogen atoms are equivalent. Electrophilic center CH₃—C— H.

Thus the compound will be

Question 37. Is 2-hydroxypropanoic acid optically active? Explain.
Answer:

2-hydroxy propanoic acid \(\stackrel{3}{\mathrm{C}} \mathrm{H}_3-\stackrel{2 *}{\mathrm{C}} \mathrm{H}(\mathrm{OH})-\stackrel{1}{\mathrm{C}} \mathrm{OOH}\)1 is optically active as there is one chiral center situated at C-2

Question 38. Write the IUPAC names of the compound CH₂=CHCH₂CH₂C=CH & CH₃CH=CHCH₂C=CH

NCERT Solutions Class 11 Chemistry Chapter 12 Short Questions

Question 39. Which of the two : O₂NCH₂CH₂O^– or CH₃CH₂O^– is expected to be more stable and why?_
Answer:

  Ion is more stable than the effect of the — NOz group causing the dispersal of ~ve charge on the O-atom. On the other hand, the —CH₂CH₃ group has a +1 effect which tends to intensify the -ve charge on the O -atom leading to the destabilization of the ion.

Question 40. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulfur, and halogens.
Answer:.

Nitrogen, sulfur, and halogen atoms, present in organic Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulfur, and halogens. Ans. Nitrogen, sulfur, and halogen atoms, are present in organic.

Question 41. Name a suitable technique to separate the components from a mixture of calcium sulfate & camphor.
Answer:

Camphor is sublimable but CaSO₄ is not Therefore sublimation of the mixture gives camphor on the inner surface of the funnel while CaSO₄ is left in the china dish.

Question 42. Will CCl₄ give a white precipitate of AgCl on heating it with silver nitrate? Give a reason for your answer.
Answer:

CCl₄ is a covalent compound. Thus, it does not ionize to give Cl- ions. Hence AgNO₃ does not react with CCl₄ even under hot conditions to form a white precipitate of AgCl.

Question 43. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
Answer:

CO₂ is an acidic oxide. Thus, it reacts with the strong base KOH to form K₂CO₃ (salt). 2KOH + CO₂ →K₂ CO₃ + H₂ O

So, during the estimation of carbon, an increase in the mass of the U-tube containing KOH solution is produced from the organic compound. Thus % of carbon in the organic compound can be estimated by using the equation:

⇒ \(\% \text { of carbon }=\frac{12}{44} \times \frac{\text { Mass of } \mathrm{CO}_2 \text { formed }}{\text { Mass of organic compound }} \times 100\)

Question 44. Why is it necessary to use acetic acid and not sulphuric acid for the acidification of sodium extract for testing sulfur by lead acetate test?
Answer:

For testing sulfur, the sodium extract is acidified with acetic acid because lead acetate remains soluble in an acetic acid medium and hence does not interfere with the test. If H₂SO₄ is used, lead acetate itself will react with H₂SO₄ to form a white precipitate of lead sulfate.

Pb(OCOCH₃) + H₂SO₄→PbSO₄ ↓ (white ppt)+ 2CH₃COOH

Class 11 Organic Chemistry Basic Principles Short Questions and Answers

Question 45. Explain why chlorine but not nitrogen in hydroxylamine hydrochloride (NH₂OH-HCl) can be detected by Lassaigne’s test
Answer:

Hydroxylamine hydrochloride (NH₂OH-HCl) contains nitrogen but does not contain carbon as an element. So, on fusion with metallic sodium, it cannot form sodium cyanide. Cyanide ion is essential to produce Prussian blue. Thus nitrogen cannot be detected by Lassaigne’s test. However hydroxylamine hydrochloride contains chlorine as an element and so on fusion with Na, it forms NaCl. Thus, chlorine can be detected by Lassaigne’s test.

Question 46. Differentiate between the principles of Dumas’s method & Kjeldahl’s method.
Answer:

In Dumas’s method, the nitrogenous organic compound is decomposed to produce gaseous nitrogen. The measured volume of N₂ is used to calculate the % of nitrogen in the given compound. In Kjeldahl’s method, the nitrogenous organic compound is decomposed to produce

(NH₄)₂ SO₄ which is further decomposed to give NH₃. The amount of NH3 so formed is used to calculate the % of nitrogen in the compound.

Question 47. 0.495 g of organic compound on combustion gave 0.99 g of CO₂ and 0.405 g of water. Calculate the percentages of carbon and hydrogen in the compound
Answer:

Amount of C in die compound = \(\frac{12}{44} \times 0.99\) 0.99 = 0.27

Amount of H in die compound = \(\frac{2}{18} \times 0.405\) x 0.405 = 0.045

v % of C in the compound = \(\frac{12}{44} \times 0.99 \times \frac{100}{0.495}\) = 54.54

% of H in the compound = \(\frac{2}{18} \times 0.405 \times \frac{100}{0.495}\) = 9.09

Question 48. 0.50 g of an organic compound when analysed by Dumas method produced 62.0 mL of nitrogen at STP. Determine the percentage of nitrogen in the compound.
Answer:

Mass of 62.0 mL (STP) of N₂ = \(\frac{28 \times 62.0}{22400}\)

-. % of N in the compound = \(=\frac{28 \times 62.0}{22400} \times \frac{100}{0.50}\)

= 15.5

Question 49. Is it possible to distinguish between hydrazine and phenylhydrazine by Lassaigne’s test? Give your reason.
Answer:

For the detection of nitrogen in an organic compound by Lassaigne’s test, the compound must contain both C and N, to permit the formation of NaCN.

Phenylhydrazine contains both C and N. So it gives a positive test for nitrogen in Lassaigne’s test. On the other hand, hydrazine contains nitrogen but does not contain carbon so it gives a negative test for nitrogen. Thus the two compounds can be distinguished by Lassaigne’s test.

Question 50. Give an example of a ketone that does not exhibit tautomerism.
Answer:

2,2,4,4-tetramethylpentan-3-one

Organic Chemistry Chapter 12 NCERT Short Questions and Answers

Question 51. Arrange in the order of increasing enol content and give
CH₃COCH₃ ,   , CH₃ COCH₂COCH₃
Answer:

Question 52. Write the structure and the IUPAC name of the alkane having the lowest molecular mass and which on bromination produces three monobromo derivatives.
Answer:

Pentane (CH₃CH₂CH₂CH₂CH₃) is the desired hydrocarbon with the lowest molecular mass which contains three types of non-equivalent H-atoms.

Question 53. How many types of non-equivalent H -atoms are there in each of the following compounds

Answer:

(1) 2 types

(2) 3 types

(3) 4 types

(4) 6 types

(5) 5 types

(6) 7 types

(7) 2 types.

Question 54.  Write the structure and the IUPAC name of an alkane (C₁₈H₃₆) which on bromination produces only 1 monobromo derivative.
Answer:

Question 55. Mention the type of substitution reactions in which the attacking reagents are NO₂⁺, OH^-, or ^•Cl
Answer:

Attacking reagent NO₂⁺ : Electrophilic substitution

Attacking reagent OH^–: Nucleophilic substitution.

Attacking reagent Cl^• : Free-radical substitution

Question 56. Suggest a method to purify:

1. Iodine containing traces of common salt,
2. Kerosene containing a little of water and
3. Camphor contains little benzoic acid.

Answer:

1. On sublimation, I₂ sublimes leaving behind NaCl. Alternatively, I₂ can be extracted with CCl₄ and the extract on evaporation gives I₂.
2. Kerosene and water are immiscible liquids having different densities. So they can be separated by using a separating funnel.
3. On boiling with water, benzoic acid dissolves but camphor remains insoluble, which can be separated by filtration (under hot conditions).

Question 57.  Suggested method for the separation of each of the following mixtures:

1. A mixture of liquid (b.p. 365 K) and liquid B (b.p. 356K)
2. A mixture of liquid C (b.p. 395 K) and liquid D (b.p. 360 K)

Answer:

1. By fractional distillation
2. By ordinary distillation.

NCERT Class 11 Organic Chemistry Chapter 12 Question & Answers

Question 58. The Rj values of two compounds, X and Y in a mixture determined by TLC are 0.66 and 0.41 respectively. If the mixture is separated by column chromatography using the same solvent mixture as the mobile phase, which one of the two compounds will be eluted first and why?
Answer:

Compound X (Rj- = 0.66) will be eluted first. The component having a higher R_f value is adsorbed less strongly by the stationary phase (adsorbent) and hence it is eluted first.

Question 59. Give an example of each of

1. Adsorption chromatography and
2. Partition chromatography

Answer:

1. Thin layer chromatography
2. Paper chromatography

Question 60. Is it possible to get pure benzoic acid from a sample containing impurities of naphthalene through the process of crystallization using benzene as a solvent? Give reason.
Answer:

It is not possible to purify impure benzoic acid by recrystallization using benzene as a solvent because both naphthalene and benzoic acid are quite soluble in benzene. Purification is however possible if hot water is used as the solvent because benzoic acid is soluble in hot water but naphthalene is not

Question 61. Write down the bond-line structural formulas

1. 2-methylbutane
2. 3,3 -dimethyl hexane
3. 2 -bromooctane and
4. Chlorocyclopentane.

Answer:

Question 62. Arrange in increasing order of strength and give reasons: CH₂=CHCOOH, HC=CCOOH, CH₃CH₂COOH
Answer:

Electronegativity of the hybridized carbon atoms increases in the sequence C_sp²<C_sp²<C_sp

The strength of carboxylic acid increases as the electronegativity of a carbon increases. This sequence of acid strength

⇒ \(\stackrel{\beta}{\mathrm{C}} \mathrm{H}_3 \stackrel{\alpha}{\mathrm{C}} \mathrm{H}_2 \mathrm{COOH}<\stackrel{\beta}{\mathrm{C}} \mathrm{H}_2=\stackrel{\alpha}{\mathrm{C}} \mathrm{HCOOH}<\stackrel{\beta}{\mathrm{C}} \equiv \stackrel{\alpha}{\mathrm{C}} \mathrm{COOH}\)

Question 63.  Arrange the following free radicals in order of increasing stability and explain the order

The stability of alkyl free radicals increases as the number of a -hydrogens increases. This is so because the extent of delocalization of the unpaired electron of any free radical increases with an increase in the number of or-H atoms. Since the number of or-H atoms in (1), (2), and (3) are 1, ,5 and 0 respectively, the sequence of stability is given by: (3)< (1)< (2).

Question 64. Designate the species as electrophile or nucleophile obtained on heterolytic cleavage of the C— C bond in ethane.
Answer: 

Ethane undergoes heterolytic bond fission to give a carbocation (methyl cation, ⁺CH₃) and a carbanion (methyl anion, ^–CH₃ ). Methyl cation is an electrophile, while methyl anion is a nucleophile.

Question 65. Although BF₄^– is an anion, it is not a nucleophile—why?
Answer:

In BF₄^– ion the central boron atom is negatively charged but it does not have any unshared electron-pair to act as a nucleophile.

NCERT Solutions For Class 11 Chemistry Chapter 12 Organic Chemistry Basic Principles And Techniques Warm-Up Exercise Question And Answers

Question 1. Why are the four C —Cl bonds in CCI, equivalent?
Answer:

In forming a CCI₄ molecule, a carbon atom with sp³ -hybridization (containing four equivalent sp³ – hybrid orbitals) uses its hybrid orbitals to form four C — Cl bonds with four Cl -atoms. So these C— Cl bonds are all equivalent.

Question 2. Which atoms in each of the following molecules lie in the same line and why?
Answer:

1. SP -carbon atoms and the atoms attached to them lie in the same line 

2.

Short Questions and Answers Organic Chemistry Class 11 Chapter 12

Question 3. A π-bond is weaker and more reactive than a σ-bond. Sp -carbon atoms and the atoms attached to them lie in
Answer: 

End-on overlap gives rise to σ -bonds, and lateral overlap gives rise to n -bonds. Die lateral overlap in a π-bond cannot be as effective as the overlap in a σ bond. Hence, a σ -bond is always stronger than a π -bond.

Question 4. What is the shape of each of the given compounds?

1. H₂C=O
2. CH₃CI
3. HCN

Answer:

1. Planar trigonal
2. Tetrahedral
3. Linear

Question 5. Arrange C_sp—C_sp, C_sp² —C_sp², and C_sp³ —C_sp³ σ -bonds in order of increasing bond length and explain the order.
Answer: C_sp—H < C_sp²— H < C_sp³ —H. For an explanation, see bond lengths

Question 6. Arrange C_sp—C_sp, C_sp² —C_sp², and C_sp³ —C_sp³ σtrbonds in order of increasing bond dissociation enthalpy and explain the order
Answer:

C_sp³—C_sp³, C_sp² —C_sp², and C_sp —C_sp  – For explanation see bond strength

Question 7. Which is the correct bond-line structural formula of CH₃CH₂C ≡ CCH₂CH₃: 
Answer:

Question 8. Identify the saturated compounds:

1. CH₃CH₂CH=O
2. C₂H₅OH

Answer:

1. CH₃CH₂CHO and
2. CH₃CH₂OH

Question 9. Write down the structure of an alkane that contains only primary (1°) carbon atoms and primary (1°) hydrogen atoms.

⇒ \(\mathrm{H}_3 \stackrel{\mathrm{l}^{\circ}}{\mathrm{C}}-\stackrel{1^{\circ}}{\mathrm{C}} \mathrm{H}_3\) – Ethane

Question 10. Give examples of the following:

1. A mixed ether
2. A tertiary alcohol,
3. An aromatic aldehyde
4. A mixed anhydride and (v)a secondary amine

Answer:

1. Ethyl methyl ether (CH₃CH₂OCH₃)
2. Tert-butyl alcohol [(CH₃)₃COH]
3. Benzaldehyde (C₆H₅CHO)
4. Dimethylamine [(CH₃)₂NH]

Question 11. Arrange the following functional groups in order of preference as the principal functional groups:
Answer:  —CONH₂,—NH₂,—CHO, —CN, —COOH, —O

Question 12. Write the structures and IUPAC names of two metamers having molecular formula, C₅H₁₀O.
Answer: CH₃COCH₂CH₂CH₃ (but-2-one) and CH₃CH₂COCH₂CH₃ (pentane-3-one)

Question 13. Arrange the following atoms or groups in increasing order of -l effect: — L, —Br, —Cl, —F
Answer: —I < —Br < —Cl < — F.

Question 14. Arrange in decreasing order of their strength and give Newtons; CH₃CH₂COOH > Me₂CHCOOH > Me₃CCOOH
Answer:

The strength of carboxylic acid decreases as the electron-releasing effect of the alkyl group attached to the —COOH group increases. Thus, acid strength decreases in the order: CH₃CH₂COOH > Me₂CHCOOH > Me₃CCOOH.

Organic Chemistry Techniques Short Q&A Class 11 NCERT

Question 15. Why is BU₃N more basic than BuNH₂, in the C₆H₅Cl medium?
Answer:

Chlorobenzene is an aphotic solvent. In such solvents, the basic strength of the amine increases as the number of electron-donating alkyl groups on the amino nitrogen increases.

Question 16. Explain the following observation
Answer:

Resonance is inhibited due to steric hindrance. So, electrophilic substitution at the p -position does not occur.

Question 17. Label the following carbonations as 1°, 2° or 3°:

Answer:

(1)2° (2) 3° (3) 2° (4) 1°

Question 18. Which one of the two carb anions is less stable and why?
Answer: The second one is less stable as it is an antiaromatic 4π -electron system).

Question 19.  What are the shapes of the free radicals ^•CH₃, ^•CF₃ and why?
Answer:

CF₃: Pyramidal (sp³ -hybridised C-atom),

CH₃ : Planar (sp² -hybridised C-atom)

Question 20. Why does: CCl₂ act as an electrophile?
Answer: Dichlorocarbene is an electron-deficient molecule. There is only a sextet of elections in the valence shell of the carbon atom of this molecule (: CCl₂). So this molecule acts as an electrophile (to fulfill the octet of carbon)

Question 21. Which one out of the S_N1 and S_N2 reactions is more susceptible to steric effect and why?
Answer: The S_N2 reaction is susceptible to steric effect because, in the transition state, the carbon atom undergoing nucleophilic attack is attached to five atoms or groups.

Question 22. Which of the following reactions do not involve an intermediate and why?— S_N1, S_N2, E1, E2
Answer: S_N2 and E2 reactions are one-step processes and hence intermediate is not involved in such reactions

Question 23. Explain the reason for the fusion of an organic compound with metallic sodium in Lasagne’s test
Answer: The purpose of the fusion of an organic compound with metallic sodium is to convert nitrogen, sulfur, and halogen present in the organic compound to water-soluble sodium cyanide, sodium sulfide, and sodium halide respectively

Question 24. How will you purify a sample of benzoic acid that contains traces of common salt? 
Answer: By sublimation. Benzoic acid sublimes by, leaving behind NaCl.

Question 25. Explain why glycerol cannot be purified by simple distillation. Mention a method that can be useful.
Answer: Glycerol cannot be purified by simple distillation because it decomposes at its boiling point. It can however be purified by distillation under reduced pressure.

Question 26. How do you separate a mixture of o-nitro phenol and p-nitro phenol?
Answer: o -and p -nitro phenol can be separated from a mixture by steam distillation.

Note: o-nitro phenol is steam volatile

Question 27. In the fusion test of organic compounds, the nitrogen of an organic compound is converted to—sodium nitrate, sodium nitrify,  sodium amide, and sodium cyanide.
Answer: Sodium cyanide

Question 28. Why is Lasagne’s extract not prepared with tap water?
Answer: Tap water often contains chloride ions, which will interfere with the test for halogen.

Question 29. Write down the formula of Prussian blue.
Answer: Fe₄[Fe(CN)₆]₃

Question 30. Why do diazonium salts not respond to Lasagne’s test?
Answer: Under hot conditions, diazonium salts decompose to liberate N₂ gas, and hence they do not respond to Lasagne’s test for nitrogen.

Question 31. Beilstein test cannot be considered as a confirmatory test for the presence of halogen in a compound—why?
Answer: Many halogen-free compounds,

For example – Certain derivatives of  Pyridine and quinoline, purines, acid amides, urea, thiourea, cyano compounds, etc. Give this test, presumably owing to the formation of volatile copper cyanides.

Question 32. What is the role of CuSO₄ and K₂ SO₄ used in Kjeldahl’s method for the estimation of nitrogen?
Answer: Potassium sulfate increases the boiling point of H₂ SO₄ and thus ensures a complete reaction, while copper sulfate catalyzes the reaction.

NCERT Chapter 12 Organic Chemistry Basic Principles Short Questions & Answers

Question 33. Which method is used to estimate N in foodstuffs
Answer: Kjeldahl’s method is largely used for the estimation of nitrogen in foodstuffs, drugs, and fertilizers

Question 34. For which compounds, Kjeldahl’s method is not applicable for the estimation of nitrogen?
Answer: Kjeldahl’sand Techniques method does not apply to compounds] containing nitrogen in the ring

For example –  Pyridine, quinoline, etc.) and compounds containing nitrogen directly linked to an oxygen atom, NO_2, or another nitrogen atom i.e., azo (— N=N— ) compounds.

Question 35. The weight of the compound is finally taken in the Carius method for the estimation of phosphorus
Answer: In this method, the amount of ammonium phosphomolybdic formed is weighed in the final step

NCERT Solutions For Class 11 Chemistry Chapter 12 Organic Chemistry Basic Principles And Techniques Very Short Question And Answers

Question 1. Which property of carbon is responsible for forming straight chains, branched chains, or rings?
Answer: Catenation property of carbon.

Question 2. Wind are the reason fort the existence of a large number of organic compounds.
Answer:

Reasons are: 

1. Catenation property of carbon
2. Its tendency to combine with other non-metals and
3. Phenomenon of isomerism exhibited by carbon compounds.

Read and Learn More NCERT Class 11 Chemistry Very Short Answer Questions

Question 3. Find the number of <r -and ;r -boud(s) in the molecule: CH₃CH₂CH==CH—C = 
Answer: Number of σ -bonds = 13 and number of π-bonds = 3

Question 4. Predict the state of hybridization of the starred carbon atoms: 

1. H*C ≡ CCH₃

2. 

Answer:

1. sp
2. sp².

Question 5. What is the shape of the molecule: C₆H₆CN?
Answer: Planar.

Question 6. Give the shape of the molecule: HC ≡ C — C ≡CCl
Answer: Linear

Question 7. What is the state of hybridization of a carbon atom linked to two other atoms by two double bonds?
Answer: Sp -hybridization.

Question 8.  Arrange the following in order of increasing carbon-carbon bond length: ethane, ethylene, and acetylene.
Answer:

Acetylene (HC ≡ CH) < Ethylene (CH₂=CH₂) < Ethane (CH₃—CH₃).

Organic Chemistry Basic Principles and Techniques Class 11 Short Questions & Answers

Question 9. What will be the shape of a hydrocarbon molecule containing two sp² – & one sp³ -hybridized C-atom?
Answer: Three-dimensional.

Question 10.  Arrange in order of increasing bond dissociation enthalpy:  C_sp— C_sp  —C_sp³ – C_sp³ —, C_sp² —C_sp² 
Answer:

C_sp³ — C_sp³ < C_sp² — C_sp²  < C_sp— C_sp

Question  11. Arrange the starred C-atoms in the following compound in order of increasing s -character of their hybridization states:
¹C*H₃—²CH=³C*H—⁴CH=⁵*C=⁶CH—⁷CH₂—⁸CH₃
Answer:

C-1 <C-3<C-5.

Question  12. Which is the correct bond-line structural formula of CH₂=CH—C = CCH₂CH₃ ?

Answer:

No. 2 s the correct bond-line structural formula.

Question 13. Write the names of an alicyclic compound and a heterocyclic compound.
Answer:

Cyclohexane and pyridine, respectively.

Question 14. Give one example of each benzenoid and non-benzenoid aromatic compound.
Answer: Toluene  and Azulene

Question 15. Write down the IUPAC name of the compound represented by the swastika sign.
Answer: 3,3-diethyl pentane

Question 16. Which one is the correct name of an alkyne containing five carbon atoms? Pent-2-yne or Pent-3-yne
Answer: Pent-2-yne.

Question 17. Mention the name of the alkyl group that may be obtained by removal of one 2° H-atom from propane.
Answer:  Isopropyl.

Question 18. How many alkyl groups are expected to be obtained from CH₃CH₂CH₂CH(CH₃)CH₂CH₃ by the removal of different non-equivalent H -atoms?
Answer: Seven alkyl groups because there are seven types of non-equivalent H -atoms.

Question 19. Which of the following has no existence? 

1. 1° H – atom
2. 3° C -atom
3. 2° H -atom
4. 4°H -atom

Answer: 4°H-atom has no existence

Question 20. How many 4°C-atoms are there in 2,2.3,3- tetramethylbutane?
Answer:. The number of 4°C- atoms is 2.

Question 21. How many 3°H-atoms are there in 4-ethyl-2- methylhexan e?
Answer: The number of 3° H-atoms is 2.

Question 22. Write the bond-line structural formula of an alkane with five carbon atoms which has only primary hydrogen atoms.
Answer: Bond-line structure of the alkane:

Question 23. Give examples of two terminal functional groups.
Answer: . —COOH, — CHO

Question 24. How many univalent groups are expected to be obtained from toluene?
Answer: 4 univalent groups can be expected

Question 25. What are the primary suffixes used to write IUPAC names of CH₃CH₃, CH₂=CH_2, and HC ≡CH?
Answer: ‘anej ‘ene’ and ‘yne’ respectively.

Very Short Questions and Answers Class 11 Organic Chemistry Chapter 12

Question 26. Give an example of a saturated hydrocarbon that can be represented by the general formula, C_nH_2n.
Answer: Cyclopentane

Question 27. How many 7r -bonds are there in 3-methylidene-l, 4- pentadiene?
Answer: The number of n -bonds is 3.

Question 28. Write the names of the alkyl group(s) that may be obtained from (CH₃)₄C.
Answer: Neopentyl or 2,2-dimethylpropyl.

Question 29. Write the IUPAC name of a hydrocarbon containing one sp, two sp² & two sp³ -hybridized C-atoms
Answer: Penta-2,3-diene or 3-methyIbuta-l,2-diene.

Question 30. How many alkyl groups are possible having the molecular formula, C₄H₉?
Answer: 4 different alkyl groups are possible.

Question 31. Which type of isomerism is exhibited by n-pentane and neopentane?
Answer: Chain isomerism.

Question 32. Write down the structure and the IUPAC name of the tautomer of butanal
Answer: CH₃CH₂CH=CH— OH (But-l-en-l-ol).

Organic Chemistry Chapter 12 NCERT Solutions for Very Short Questions

Question 33. How many structural isomers will be obtained by the displacement of two H-atoms of propane by two Cl- atoms? Write their structures.
Answer: Question 65. A compFour (ClCH₂CH₂CH,Cl, ClCH₂CHClCH₃ , Cl₂CHCH₂CH₃ and CH₃CCl₂CH₃).

Question 34. Write structures and names of two compounds which are position isomers as well as metamers
Answer:

Question  35. How are the two compounds, CH₂=CHCH₂CH₃ and related to each other?
Answer: These two compounds are ring-chain isomers.

Question 36. Which two of the following are geometrical isomers?

Answer: 2 and 4 are geometrical isomers.

Question 37. Which of the following compounds do not exhibit geometrical isomerism

1. 

2. PhCH=CHPh

3. Me2C=NOH

4. CH₂=CH—CH=CH —CH=CH₂

Answer:  1 and 3 do not exhibit geometrical isomerism

Question 38. Which of the given compounds are optically active?

Answer: 3 and 4 are optically active compounds

NCERT Class 11 Organic Chemistry Short Q&A Chapter 12

Question 39. Mention the type of the following reaction: Me₃CCH₂OH + HBr Me₂CBrCH₂CH₃ + H₂O
Answer: It is a substitution and rearrangement reaction.

Question 40. How many types of non-equivalent H-atoms are present in the given compound?

Answer: Two types of non-equivalent H-atoms are present

Question 41. In which of the given compounds, all the H-atoms are; equivalent?

Answer: 1 and 3

Question 42. Calculate the double bond equivalent (DBE) of the compound having the molecular formula, C₆H₈ Is the compound aromatic?
Answer: DBE = 3 ; The compound is not aromatic

Question 43. How many monobrow derivatives are possible for each of Ortho, meta, and para-xylene?
Answer: Three, four, and two respectively

Question 44. Arrange the following groups in order of decreasing strength of -I effect groups in order of decreasing — ⁺NR₃, —NO₂, —F, —CN.
Answer:  — ⁺NR₃ >NO₂>—CN >—F

Question 45. Arrange the following free radicals in the decreasing order of their stability:

Answer: 5 >4>1>2>3

Question 46. In which C—C bond of CH₃CH₂CH₂Br, the inductive effect is expected to be the least?
Answer: C-2 —C-3 bond.

Question 47. Arrange the following carbocations in increasing order of stability

Answer: 4<2<1<3

Question 48. Arrange the following compounds in increasing order of number of hypercoagulable hydrogen atoms:

Answer: 6<2<3<5 <4<1

Question 49. Arrange the following compounds in order of increasing bond dissociation enthalpy:

1. CH₃ — H
2. (CH₃)₂CH-H
3. (CH₃)₃C—H
4. CH₃CH₂ —H

Answer: 3<2<4<1

Question 50. Arrange the following in increasing order of stability:

Answer: 4<2<3<1< 5

Question 51. How can aniline be purified?
Answer: By steam distillation as well as vacuum distillation.

Question 52. How can glycerol be purified?
Answer: By distillation under reduced pressure.

Chapter 12 Organic Chemistry Basic Principles Short Questions Class 11

Question 53. Suggest a method to separate a mixture of o-hydroxybenzaldehyde and p-hydroxybenzaldehyde.
Answer: Steam distillation.

Question 54. How will you separate a mixture of two solid pounds of different solubilities in the same solvent?
Answer: By fractional crystallization.

Question  55. An organic liquid decomposes below its boiling point. How can it be purified?
Answer: By distillation under reduced pressure.

Question 56. Which technique can be used to purify iodine-containing traces of common salt?
Answer: Sublimation.

Question 57. Suggest a method for the purification of a liquid containing non-volatile Impurities.
Answer: Simple distillation

Question 58. How can aniline (b.p. 184°C) be separated from petroleum ether (b.p. 40-60°C)?
Answer: By simple distillation

Question 59. Out of water and benzene, which can be used to purify benzoic acid containing naphthalene by fractional crystallization?
Answer: Water (because both are soluble in benzene).

Question 60. Give an example of a chromatographic technique in which both the mobile and stationary phases are liquids
Answer: Paper chromatography.

Question 61. Mention two distillation processes in which organic liquids boil at temperatures below their respective boiling points. 
Answer: Distillation under reduced pressure & steam distillation.

Question 62. Explain why the Lassaigne’s extract should not be prepared by using tap water.
Answer: Because tap water contains chloride ion (Cl )

Question 63. Give an example of a compound that does not contain halogen but gives Beilstein’s test
Answer:  Pyridine (C₆H₅N)

Question 64. In Carius’s method for estimation of phosphorus, the precipitate of which compound is finally obtained?
Answer: Ammonium phosphomolybdate [(NH₄)₃PO₄-12MoO₃] or magnesium pyrophosphate (Mg₂P₂O₇).

Organic Chemistry Basic Principles and Techniques Class 11 Very Short Questions & Answers

Question 65. Give an example of a nitrogenous organic compound to which Kjeldahl’s method for the estimation of nitrogen is not applicable.
Answer: Azobenzene (C₆H₅— N=N—C₆H₅)

Question 66. What is the C — C=C bond angle value in a benzene (C₆H₆) molecule?
Answer: 120°

Question 67.  What is the H — C = C bond angle value in an acetylene (C₂H₂) molecule?
Answer: 180°

Question 68.  Mention the state of hybridization of C and N-atoms in
Answer: sp, sp²

Question 69. Mention the state of hybridization of the carbon atoms present in the molecule, CH₃CH =C =CHCH₂CH₃.
Answer: sp, sp, sp, sp, sp, sp

Question 70.  Give the name of a simple organic molecule that has a cylindrical n -electron cloud.
Answer: Acetylene

Question 71.  Give an example of a molecule in which all atoms lie in the same plane.
Answer: CH₂=CH₂

Question 72.  Give an example of a molecule in which all the atoms lie in a straight line.
Answer: HC = CH

Question 73. Calculate the number of cr and n -bonds in the molecule, CH3CH=CH—C = C —CHO.
Answer: σ -bond:12, π -bonds

Question 74. What are the possible values of n if CH₂=(C)_n—CH₂ is a planar molecule?
Answer: n = 0,2,4, — etc

Question 75. What are the possible values of n if CH₂— (C)n_n=CH₂ is a non-planar molecule?
Answer: n = 1, 3, 5, etc.

Class 11 Chemistry Organic Chemistry Chapter 12 Short Answer Solutions

Question 76. Write the structure of a hydrocarbon molecule that contains one 4° carbon atom.
Answer:  (CH₃)₄C;

Question 77. Give an example of a compound that contains primary (1°), secondary (2°), and tertiary (3°) H -atoms.
Answer: CH₃CH₂CH(CH₃)₂

Question 78. How many alkyl groups can be derived from the alkane, CH₃(CH₂)₃CH₃?
Answer: Three

Question 79.  Write the group prefix used for the — COOH group.
Answer: ‘oic acid’;

Question 80.  Write the IUPAC name:
Answer: (3-ethyl-l-methyl)-pentyl -CH(CH₃)CH₂C(C₂H₅)CH₂CH₃

Question 81.  Write the IUPAC name: CH₂=CH—CH=CH — C = CH
Answer:  Hexa-l,3-dien-5-yne

Question 82.  Write the structure and name of an alkane having five C- atoms which on bromination gives only one monobromo derivative.
Answer:  Neopentane [(CH₃)₄C];

Question 83. Write structures of two compounds which are metamers as well as position isomers.
Answer: CH₃CO(CH₂)₂CH₃, C₂H₅COC₂H₅

Question 84. Give an example of a carbonyl compound in which tautomerism does not take place.
Answer: Benzaldehyde (C₆H₅CHO)

Question 85.  Give an example of a 3° free radical containing six hyperconjugable hydrogens.
Answer: (CH₃CH₂)₃C

Question 86.Which is the most stable carbocation having the formula,
Answer: Terf-butyl cation, Me₃C

Question 87. Which has greater resonance stabilization—PhNH₂ or PhNH₃?
Answer: PhNH₂

Question 88. Write the name of a cyclic compound that is isomeric with but-l-ene.
Answer: Cyclobutane

Question 89.  Write the names of two non-polar solvents that are commonly employed for crystallization.
Answer: Benzene and carbon tetrachloride

Question 90.  What type of furnace is used in the Carius method for the estimation of halogens?
Answer: Bomb furnace

Question 91. Mention the type of chromatography in which both the mobile and stationary phases are liquid.
Answer: Paper chromatography

Question 92. With the help of which type of distillation process glycerol can be purified?
Answer: Distillation under reduced pressure

Very Short Questions on Organic Chemistry Chapter 12 NCERT Class 11

NCERT Solutions For Class 11 Chemistry Chapter 12 Organic Chemistry Basic Principles And Techniques

Fill In The Blanks

Question 1. When four alkyl groups are attached to a carbon atom, that particular C-atom is called _______________
Answer: 4°

Question 2. The shape of the molecule containing only sp² -hybridized carbon atoms is _______________
Answer: Planar

Question 3. The C-2 atom ofpropa-1,2-dieneis ________hybridised.
Answer: sp

Question 4. The shape of the molecule containing only sp-hybridized carbon atoms is_______________
Answer: Linear

Question 5. The successive members of a homologous series differ by _______________ mass units.
Answer: 14

Question 6. The molecule, HC = C—CH —CH — CH₃ contains _______________ a -bonds.
Answer: 10

Question 7. The molecule contains_______________ 2°H -atoms.
Answer: 4

Question 8. The general formula of dihydric alcohol is _______________
Answer: C_nH_2n(OH)₂

Question 9. The compound, 5-(l,2-dimethylpropyl)-6-ethyldecane contains _______________ 3° carbon atoms.
Answer: 4

Question 10. Stereoisomers have _______________ atom-to-atom bonding sequence or connectivity.
Answer: Same

Question 11. The number of isomers of a benzenoid aromatic compound having molecular formula, C₇H₈O is _______________
Answer: 5

Question 12. Ethoxyethane and 2-methoxy propane are related as _______________
Answer: 12 metamers

Question 13. The amount of negative charge present on each O-atom of carbonate ion is _______________
Answer: =-2/3

Very Short Questions on Organic Chemistry Chapter 12 NCERT Class 11

Question 14. The homolytic fission of a covalent bond requires _______________ energy than that required by its heterolytic c fission.
Answer: Less

Question 15. _______________ involves delocalization of σ -electrons of C — H bond of an alkyl group directly attached to an unsaturated system or to an atom with a vacant or singly p-orbital.
Answer: Hyperconjugation

Question 16. In paper chromatography, both the stationary and mobile phases are _______________
Answer: Liquid

Question 17. An impure sample of benzoic acid containing a little sodium chloride can be purified by _______________
Answer: Sublimation

Question 18. In steam distillation, the organic liquid boils at a _______________ temperature than its normal boiling point.
Answer: Lower

Question 19. In Cariu’s method of estimation, chlorine present in an organic compound is converted into _______________.
Answer: AgCl

Question 20. _______________ distillation is used to remove water from rectified spirit.
Answer: Azeotropic

CBSE Class 11 Chemistry Notes For Chapter 12 Organic Chemistry Basic Principles And Techniques Classification And Nomenclature

It has been known from ancient times that minerals, plants, and animals are the three major sources of naturally occurring substances. However, very little information was known regarding their chemistry until the beginning of the eighteenth century. In 1675, Lemery classified the natural substances into three classes such as mineral substances, vegetable substances, and animal substances based on the sources from which they were obtained, and it was readily accepted.

• In 1784, it was Lavoisier who first showed that all compounds derived from vegetable and animal sources always contained carbon and hydrogen and sometimes oxygen, nitrogen, sulfur, and even phosphorus. So, there is a dose relationship between the vegetable and animal products.

This led to the classification of natural substances into two categories: 

1. Organic compounds: All those substances which were obtained from plants and animals, i.e., the substances which were obtained from living organisms and
2. Inorganic compounds: All those substances that were obtained from non-living sources, such as rocks and minerals etc. This classification, however, found justification in the fact that in several cases, the same compound could be derived from both vegetable and animal sources.

A detailed investigation of the structure of the organic compounds revealed that almost all of them essentially contain both carbon and hydrogen atoms (hydrocarbon) and some of them also contain the atoms of a few other elements such as nitrogen, oxygen, phosphorus, halogens etc.

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Since these are formed by replacing the hydrogen atoms in the hydrocarbons by these atoms, therefore, these are regarded as derivatives of hydrocarbons.  Organic compounds are, therefore, hydrocarbons and their derivatives and the branch of chemistry that deals with the study of these compounds is known as organic chemistry.

Inorganic chemistry, on the other hand, is defined as the chemistry of all elements other than carbon and their compounds

Organic Chemistry Basic Principles and Techniques Class 11 Notes

Tetrahedral Arrangement Of The Four Valencies (Bonds) Of Carbon

In 1859, Kekule proposed that carbon exhibits the normal valency of four units in simple as well as complicated organic molecules and an atom of carbon is joined to other atoms by four covalent bonds which remain in a plane. Also, the angle between any two adjacent bonds is 90°.

• This proposal, however, could not explain some special characteristics of organic compounds, and a change in Kekule’s theory was urgently required.
• In 1874, both can’t Hoff and Le Bel predicted that the four bonds of a carbon atom are directed towards the four corners of a regular tetrahedron, i.e., the angle between any two adjacent bonds is I09°28′ (tetrahedral angle).
• This representation is regarded as a tetrahedral model or space model. It was supported by electron diffraction and spectroscopic studies.
• The tetrahedral arrangement of four bonds of carbon laid the foundation for a fascinating field of ‘stereochemistry It is for this reason, that the first Nobel Prize in chemistry was awarded to van’t Hoffin 1901

Explanation for the existence of a large number of organic compounds:

An important and interesting property of a carbon atom is its unique capacity to form bonds with other carbon atoms. This property of forming bonds with atoms of the same element is called catenation. Carbon shows maximum catenation in its group (group 14) in the periodic table.

This is because of the greater strength of the C —C bond as compared to other atoms. For example, the C— C bond is very strong (335 kj- mol^-1 ) in comparison to the Si—Si bond (220 kj-mol^-1 ) or Ge—Ge bond (167 kj – mol^-1). As a result, carbon atoms can link with each other to form either linear chains of various lengths or branched chains and also rings of different sizes.

Again, two or more organic compounds having the same molecular formula may be formed. Compounds having the same formula but different properties are called isomers and the phenomenon is called isomerism.

Catenation and isomerism are the two properties of carbon that are responsible for the existence of a large number of organic compounds. Carbon is also involved in forming multiple bonds with other carbon atoms (C=C, C≡C) and also with oxygen (C—O) and nitrogen atoms (C=N, C≡N). Themultiplebondingis also responsible for the existence of a variety of carbon compounds.

Electronic explanation of the tetra covalency of carbon: Lewis’ theory:

In 1916, G.N. Lewis put forward the electronic concept regarding the formation of bonds. According to him, atoms with similar or almost similar electrochemical nature combine by forming one or more electron pair using odd electron(s) in their valence shells, and in this way, they attain stable electronic configurations (electron octets) of the nearest noble gases.

Carbon exists in group IVA of the second period of the periodic table. In the perspective of its position just between the electropositive and electronegative elements, it can be said that it is neither an electropositive nor an electronegative element. Its ionization enthalpy is sufficiently high and its electron affinity is of moderate magnitude.

Therefore, carbon (ground state electronic configuration: ls²2s²2p_x¹2p_y¹) can neither produce C⁴⁺ ion by losing 4 electrons from its valence shell nor form C^-4- ion by gaining 4 electrons.

Both processes require large amounts of energy which are not ordinarily available during a chemical reaction. Thus, carbon does not usually form electrovalent compounds

Exception:

During the reaction between carbon and highly electropositive sodium, carbon forms the C^4- ion thereby producing the electrovalent compound sodium carbide, Na₄C ).

According to the electronic configuration, carbon should be bivalent, i.e., it should exhibit a valency of two because of the presence of two odd electrons in 2p_x^– and 2p_y^–orbitals.

To account for tetracovalency, it is believed that during the process of bond formation (an energy-releasing process), the two electrons in the 2s -orbital get unpaired, and one of them is promoted to the empty 2pz -orbital. Therefore, the electronic configuration of carbon in the excited state is ls²2s²2p_x¹2p_y¹ 2p_z¹) So, carbon can complete its octet by sharing the four odd electrons in its valence shell

With the 4 electrons of the valence shells of the other 4 atoms. Hence, carbon fulfills its octet by covalency and In the formation of covalent compounds, its valency is always 4

Example:

One C -atom forms four electron pairs by combining with four H-atoms and produces one methane (CH₄) molecule with consequent completion of its octet According to Lewis’s theory, 1 electron pair stands for 1 covalent single bond, represented by dash sign(-).

Now, two carbon atoms can form 1, 2, or 3 electron pairs by contributing 1, 2, or 3 electrons respectively from each carbon atom and use them equally to form a carbon-carbon single bond (C—C), carbon-carbon double bond (C= C) or carbon-carbon triple bond (C = C) respectively. For example, ethane (CH₃ —CH₃), ethylene (CH₂=CH₂), and acetylene (HC = CH) contain one carbon-carbon single, double, and triple bond respectively.

Although Lewis’s theory gives a satisfactory explanation to the tetracovalency of carbon, it fails to offer any idea about how these valencies are oriented in three-dimensional space. Moreover, this concept of covalent bond formation by sharing of electrons is purely a qualitative approach that tells nothing about the attractive forces involved in bond formation

Class 11 Chemistry Organic Chemistry Basic Principles NCERT Notes

Hybridization Of Carbon And Shapes Of Some Simple Organic Molecules

For sp³, sp² and sp -hybridisations of carbon and the shapes ofmethane, ethane, ethylene and acetylene see article no. 4.8 of the chapter “Chemical Bonding and Molecular Structure”. From the knowledge of the state of hybridisation of atoms in an organic molecule, an idea about shapes ofthe molecules can be obtained.

These maybe summarised as follows:

1. Atoms bonded to an sp³ -hybridised carbon atom are tetrahedrally oriented. So, any molecule containing an sp³ -hybridised carbon atom must have threedimensional shape (irrespective of the presence of any sp2 or sp -hybridised carbon atom in it).

2. The two sp² -hybridized C-atoms and the atoms directly attached to them remain in the same plane. So, if any organic molecule contains only sp² -hybridized C-atoms which are arranged in a chain or cyclic pattern, then the whole molecule becomes planar.

For example: 1,3-butadiene (CH₂=CH—CH=CH₂) and benzene molecules are planar.

If a molecule contains both sp and sp² – hybridized C-atoms, then the molecule will also be planar. For example, the but-l-en-3-yne (CH₂= CH—C= CH) molecule is planar. If sp² or sp -hybridized carbon atoms remain attached to a benzene ring, then the molecule will also be planar. For example, vinylbenzene and ethylbenzene molecules are found to be planar. When an atom is present as a substituent in benzene, even then the molecule becomes planar.

For example: Fluorobenzene, chlorobenzene, bromobenzene etc…  are planar molecules.

3. The two sp -hybridized carbon atoms and the atoms directly attached to them remain on the same line. So, if a

A molecule contains only sp -hybridized carbon atoms which remain bonded one after another, and then the molecule will be linear in shape. But 1,3-diyne, for example, is a linear molecule.

Structure of some familiar organic compounds:

Prediction of the state of hybridization of a carbon atom from the nature of bonding:

If a carbon atom is bonded to four atoms by four single bonds (i.e., 4 cr -bonds), then that carbon atom is sp³ -hybridized;

• If a carbon atom is bonded to three atoms by two single bonds [i.e., 2 σ – bonds) and one double bond [i.e., 1 σ and 1 π -bond), then that carbon atom is sp² -hybridized; and
• If a carbon atom is bonded to two atoms by one single bond [i.e., one bond) and one triple bond [i.e., one σ -bond and two π bonds) or by two double bonds [i.e., two car and two n bonds), then that carbon atom is sp -sp-hybridized

Basic Principles of Organic Chemistry Class 11 NCERT Notes

Effect Of Hybridisation On Bond Lengths And Bond Strengths

The bond length as well as bond strength of any bond depends upon the size of the hybrid orbitals involved in bond formation.

1. The s -characters of sp3, sp², and sp -hybrid orbitals are 25%, 33.33%, and 50% respectively. As s -the orbital is more closer to the nucleus as compared to p -orbital, more the percentage of s -the character of the hybrid orbital, more it will be attracted by the nucleus, and as a result, its size will

The s -characters of sp³, sp² and sp -hybrid orbitals are 25%, 33.33%, and 50% respectively. As s -orbital is closer to the nucleus as compared to p -orbital, the more the percentage of s -character of the hybrid orbital, the more it will be attracted by the nucleus, and as a result, its size will decrease more.

Therefore, the sizes of sp³, sp², and sp hybrid orbitals follow the order: sp < sp² < sp³. Consequently, the bond formed by an sp³ -hybrid orbital is longer than the bond formed by an sp² hybrid formed orbital by an which-hybrid orbital turn is longer than the bond

Example:

The C_sp³   bond is longer than the C_sp²  —H bond which in turn is longer than the C_sp —H bond. Similarly, the C_sp³ —C_sp³  bond is longer than the C_sp² —C_sp²  bond which turns longer than the C_sp — C_sp bond. For the same reason, the lengths of single, double, and triple bonds follow the order: C —C>C=C>≡ C. The presence and the increase in the number of n -bonds between two carbon atoms are also responsible for the decrease in bond length.

2. Shorter the bond, the greater is its strength i.e., stronger itis. Therefore, a <r -bond formed by sp -hybrid orbitals is stronger than a σ -bond formed by sp² -hybrid orbitals which in turn forms more stronger σ -bond than formed by sp³ -hybrid orbitals. Again, the bond strength increases with an increase in bond multiplicity and so, a triple bond is stronger than a double bond whichin turn is stronger than a single bond

Bond lengths and bond dissociation enthalpies of different bonds:

Different Structural Representations Of Organic Compounds

Lewis Structure Or electron dot Structure

In Lewis structure or electron dot structure, only the electrons of the valence shell of each atom of a molecule is H shown. For example, the electron dot structure of propene is shown at the right side. Since this style of writing is time-consuming and inconvenient, it is not generally followed

Dash formula

The Lewis structure, however, can be simplified by representing each pair ofelectrons involved in forming a covalent bond by a dash(—). In this representation, a single dash represents a single bond, a double dash (=) represents a double bond, and a triple dash (≡) represents a triple bond.

The lone pair of electrons on the heteroatoms e.g., oxygen, nitrogen, halogens, sulfur, etc. May or may not be shown. Thus, ethane, ethylene, acetylene and ethanol can be represented by their structural formulas as given above. Such structural representation is also called complete structural formula or graphic formula.

Condensed structural formula

The complete structural formulas can be further abbreviated by deleting some or all of the covalent bonds and by indicating the number of identical groups attached to an atom by a subscript. Such simplified structural representations of molecules are called condensed structural formulas. Thus, ethane, ethylene, acetylene and ethanol can be represented as: CH₃CH₃ (ethane); H₂C — CH₂ (ethylene); HC=CH (acetylene); CH₃CH₂OH (ethanol)

Similarly, a long chain of carbon atoms such as CH₃CH₂CH₂CH₂CH₂COOH can be represented as CH₃(CH₂)₄COOH.

Bond-line Structural formula

It is a very simple, short and convenient method of representing the structures of organic molecules. In this representation, only the carbon-carbon bonds are shown by lines drawn in a zig-zag fashion.

The line ends and the intersection of lines represent carbon atoms carrying an appropriate number of H -atoms so that its tetravalency is fulfilled (i’.e., the terminals denote CH₃ -groups, and the unsubstituted intersections or line junctions denote CH₂ -groups). The heteroatoms and the H-atoms attached to them are, however, specifically shown. A single bond is represented by a single line ( —), a double bond is represented by two parallel lines (= ) and a triple bond is represented by three parallel lines (≡).

In bond-line notation, the cyclic compounds are represented by an appropriate ring or polygon without showing carbon and hydrogen atoms. Each comer of a polygon represents a carbon atom while each side of the polygon represents a carbon-carbon bond.

For example:

Three-dimensional representation of organic molecules

The three-dimensional structure of organic molecules is quite difficult to represent on paper (two-dimensional). So, certain graphic conventions have been proposed. For example, by using a solid wedge and dashed wedge the threedimensional structure of a molecule from a two-dimensional presentation can be perceived.

In these formulas, the solid wedge (i.e., the thick solid or heavy line) is used to indicate a bond lying above the plane ofthe paper and projecting towards the observer, and the dashed wedge is used to represent a bond lying below the plane ofthe paper and projecting away from the observer. The bonds lying in the plane of the paper are shown by normal or ordinary lines (—).

A three-dimensional representation of a methane molecule, for example, is shown below:

Organic Chemistry Techniques Class 11 Notes

Classification Of Organic Compounds Based On Carbon Skeleton

Due to the existence large number of organic compounds is rather inconvenient to study the chemical nature of these compounds individually. To simplify and systematize the study of organic chemistry, organic compounds have been broadly classified into two categories depending on the nature of their carbon skeleton or structures.

The flow sheet given here is the general classification of organic compounds:

Acyclic or open chain or aliphatic compounds:

The compounds containing open chains of carbon atoms with the appropriate number of H -atoms and functional groups are called acyclic or open-chain compounds. These compounds are also called aliphatic compounds because the earlier members of this class were obtained either from animal or vegetable fats (Greek, aliphatic = fat). Due to the variation in the structure of carbon chains, two types of open-chain compounds have been formed.

These are:

• Unbranched or straight-chain compounds and
• Branched chain compounds.

1. Straight chain compounds

The open chain or aliphatic compounds in which no carbon atom (except the two terminal C -atoms) is attached to more than two carbon atoms are called unbranched or straight chain compounds.

Examples:

2. Branched-chain compounds

The open-chain or aliphatic compounds in which at least one carbon atom is attached to three or more carbon atoms are called branched-chain compounds

Examples:

Cyclic or closed chain or ring compounds

Compounds containing one or more closed chains or rings of atoms in their molecules are called cyclic or closed chain or ring compounds. Depending upon the nature of the atoms present in the ring, the cyclic compounds may be divided into the following two classes

1. Carbocydic or homocyclic compounds:

The compounds containing rings that are made up ofonly carbon atoms are called carboxylic or homocyclic compounds.

These are further divided into two sub-classes as follows:

1. Alicyclic compounds:

The carboxylic compounds which show resemblance in properties with the aliphatic compounds are called alicyclic compounds

Examples:

2. Aromatic compounds: For aromatic compounds, see the aromatic hydrocarbon portion

2. Heterocyclic compounds:

Cyclic compounds containing more heteroatoms (atoms other than C and H i.e., O, N, S, etc.) in their rings are called heterocyclic compounds. Depending upon their chemical behaviour, they are further divided into the following two categories

1. Alicyclic heterocyclic compounds:

Aliphatic cyclic compounds containing one or more heteroatoms in their rings are known as alicyclic heterocyclic compounds

Examples:

2. Aromatic heterocyclic compounds:

Aromatic compounds containing one or more heteroatoms in their molecules are called aromatic heterocyclic compounds.

Examples:

Saturated and unsaturated compounds

1. Saturated compounds:

The organic compounds in which the carbon atoms are linked with each other only by single covalent bonds are called saturated compounds.

Examples: Methane (CH₄), ethane (CH₃—CH₃), ethyl alcohol (CH₃ —CH₂ —OH), methylamine (CH₃ —NH₂) etc.

2. Unsaturated compounds:

The organic compounds which contain at least one carbon-carbon double bond or triple bond are called unsaturated compounds

Examples:

Ethylene (H₂C= CH₂), l-butene(C₂H₅CH= CH₂)

1,3-butadiene (H₂C=CH—CH=CH₂) etc.

Organic compounds containing unsaturated groups such as are considered saturated compounds. -CHO, -COOH, -COOR etc. But no C =C or C = C For example, propionic acid (CH₃CH₂COOH) is a saturated compound but acrylic acid (CH₂=CH—COOH) is an unsaturated compound.

Functional Groups & Homologous Series

Functional group Definition

A functional group may be defined as an atom or group of atoms present in an organic compound, which is responsible for its characteristic chemical properties.

Generally, compounds having the same functional group have similar properties while compounds with different functional groups have different chemical properties. For this reason, organic compounds are classified into different classes or families based on the functional groups present In the reactions of organic compounds, the organic groups or radicals generally do not suffer any change but the functional groups actively participate:

Some common functional groups present in various organic compounds along with their classes:

1. Though the functional groups govern the chemical properties, they are also found to influence the physical properties in some cases. For example, alcohols (ROH) due to the presence of —OH group remain associated through intermolecular H -bonding and as a result, the boiling points of alcohols are much higher than that of ethers having similar molecular masses.

2. Organic compounds with 2 or more different functional groups exhibit characteristic properties of all the functional groups present in it. For example, aldol [CH₃CH(OH)CH₂CHO] exhibits the properties of ; both alcohol and aldehyde.

3. Although different compounds having the same functional group show similarities in their chemical properties, their properties are not identical. For example, formaldehyde (HCHO) and acetaldehyde (CH₃CHO) containing the same functional group ( —CHO), do not respond to the same type of reaction. The former participates in the Cannizzaro reaction but not in Aldol condensation reaction while the latter takes part in Aldol condensation reaction but the Cannizzaro reaction

Homologous series

Homologous series definition:

A homologous series is defined as a series or group of similarly constituted organic compounds which have the same functional group and thus similar chemical properties and any two successive members differ in their molecular formula by a CH₂– group. Members of such a series are called homologues.

Some homologous series, their general formulas and structures of compounds upto C₄

Characteristics of homologous series:

1. All the members of a homologous scries can be represented by the same general formula. For example, C_nH_n+1 OH is the general formula of alcohols.

2. Same functional group is present in ail members of any homologous series. So, the members of any homologous series have almost identical chemical properties (the phenomenon of such resemblance in properties among the compounds of the same homologous series is called homology). However, with an increase in molecular mass, the chemical reactivity ofthe compounds usually decreases.

3. Any two successive members of a particular series differ in molecular formulary CHHomologous series group or 14 mass units.

The physical properties such as density, melting point, and boiling point of the members of a homologous series increase gradually with the increase in molecular mass. However, solubility and volatility show a declining trend with a rise in molecular mass.

4. The members of a homologous series can be prepared by almost identical methods, known as the general methods of preparation.

Significance of homologous series:

• From the knowledge ofthe method of preparation and the properties of a particular member of a homologous series, the method of preparation and properties of the other members of the same series can easily be predicted.
• Therefore, by dividing the vast number of organic compounds into homologous series followed by the study of the method of preparation, properties, and reactions of a representative member, an overall idea about the whole family can be obtained.
• However, the first member of a series often differs from the other members in the method of preparation and properties.
• For example, the method of preparation and properties of formic acid (HCOOH), the first member ofthe carboxylic acid family, are different from that ofthe other members of the family.

NCERT Solutions Class 11 Chemistry Organic Chemistry Principles

IUPAC Nomenclature Of Aliphatic Organic Compounds

According to the IUPAC system, the name of an organic compound consists of 3 parts:

1. Word root
2. Suffix and
3. Prefix.

1. Word root:

Word root, the basic unit of the name, denotes the number of carbon atoms present in the parent chain (the longest possible continuous chain of carbon atoms including the functional group and multiple bonds) of the organic molecule.

For chains containing up to four carbon atoms, special word roots (based upon the common names of alkanes) such as ‘meth’ for C₁, ‘eth’ for C₂, ‘prop’ for C_3, and ‘but’ for C₄ are used where C₁, C₂, C₃ and C₄ represent the number of carbon atoms in the chain. For chains of five or more carbon atoms, Greek numerals or number roots are used to represent the word roots.

For example:

‘Pent’ is used for the C₅ chain, ‘hex’ is used for C₆ chain etc. The general word root for any carbon chain is ‘alk’

2. Suffix

These are of the following two types

1. Primary suffix:

A primary suffix is always added to the word root to indicate whether the carbon chain is saturated or unsaturated. There are three basic primary suffixes. If the carbon atoms are linked only by single covalent bonds (C —C) ‘-ane’ is used.

If there is at least one double bond (C= C) present in the chain, the primary suffix ‘- ene’ and if there is at least one triple bond (C = C) in the chain, the primary suffix ‘-yne’ is used. Hence, the name of

1. CH₃CH₂CH₁₃ is prop + ane = propane and
2. CH₃CH=CH₂ is prop + ene = propene.

If the parent carbon chain contains 2, 3, or more double or triple bonds, numerical suffixes such as ‘di’ (for two), ‘tri’ (for three), ‘tetra’ (for four), etc.

Are added to the primary suffix. If the primary suffix begins with a consonant then an extra ‘a’ is to be added to the word root. For example, the primary suffix used for two double bonds is diene. Now if it is added to the word root ‘but’ for C₄ chain, then the name of the compound will be butadiene.

2. Secondary suffix:

A secondary suffix is used to indicate the functional group presents an organic molecule and is to be added to the primary suffix while writing the IUPAC name. Secondary suffixes of some important functional groups are listed:

It is to be noted that while adding the secondary suffix to the primary suffix, terminal ‘e’ of the primary suffix [i.e., ‘ane,’ ‘ene’ or ‘yne’) is dropped if the secondary suffix begins with a vowel but is retained if the secondary suffix begins with a consonant.

For example: The name of CH₃CH₂OH is: eth + ane + ol = ethanol and that of CH₃CH₂CN is: prop + ane + nitrile = propanenitrile.

3. Prefix

It is a part of the IUPAC name of a compound that appears before the word root. Prefixes are of two types

1. Primaryprefix: A primary prefix is used to differentiate a cyclic compound from an acyclic compound

For Example:

In case of carbocyclic compounds, the primary prefix ‘cyclo’ is used just before the word root. For example, the cyclic hydrocarbon, maybe named as

2. Secondaryprefix:

In the IUPAC system of nomenclature, some groups are not considered functional groups or secondary suffixes. These are treated as substituents and are called secondary prefixes. These are added just before the word root (or the primary prefix in the case of alicyclic compounds) in alphabetical order.

The secondary prefixes of a few substituents are given in the following table:

Therefore, while writing the IUPAC name of an aliphatic organic compound the various parts are to be added in the following sequence: Secondary prefix + Primary prefix+ Wordroot+ Primary suffix + Secondary suffix. It is not necessary that all the parts may be present in a particular compound. However, the word root and primary suffix must be present.

Example:

In case ofthe compound, \(\stackrel{4}{\mathrm{C}} \mathrm{H}_3 \stackrel{3}{\mathrm{C}} \mathrm{HCl} \stackrel{2}{\mathrm{C}} \mathrm{H}_2 \stackrel{1}{\mathrm{C}} \mathrm{H}_2 \mathrm{OH}\), the word roots ‘but’, the primary suffix ‘ane’, the secondary suffix is ‘ol’ and the secondary prefix is ‘chloro’. As the compound is acyclic, the primary prefix is absent. Therefore, the IUPAC name of the compound is chloro (at position C-3 ) + but + an (e is omitted) + ol (at position C-l )- 3-chlorobutan-l-ol.

IUPAC Nomenclature of some organic compounds

Organic Chemistry Basic Principles Class 11 Summary

Common & IUPAC Nomenclature Of Some Important Classes Of Organic Compounds

Saturated hydrocarbons (Alkanes)

In the IUPAC system, saturated acyclic hydrocarbons are called alkanes. IUPAC names of alkanes are obtained by adding the suffix ‘ane’ to the word root indicating the number of C-atoms present in the chain. The first 4 alkanes (CH₄ to C₄H₁₀) have their special names, i.e., methane, ethane, propane, and butane.

The names of alkanes containing 5 or more C-atoms are obtained by adding prefixes such as ‘pent’ (5), ‘hex’ (6), ‘hept’ (7), ‘Oct’ (8), etc., indicating the number of C-atoms in the molecule to the suffix ‘ane! Although common and IUPAC names of alkanes are the same, the prefix (normal)is added to the common name of alkanes containing 4 or more C-atoms.

IUPAC and common names of first ten members of the alkane family:

Names of some higher members of alkane family (CH₃(CH₂)_n CH₃ ,n= 9,10,11,….etc..are as follows:

Alkyl groups

An organic group produced by the removal of one H -atom from an alkane molecule is called an alkyl group or alkyl radical.

For example: 

1. Removal of one H-atom from methane (CH₄) produces methyl group (-CH₃),
2. An ethyl group ( —C₂H₅) is formed by the removal of any one of six equivalent H -atoms from ethane (C₂H₆)

All the H -atoms of alkanes containing more than two carbon atoms are not always equivalent. So, two or more alkyl groups can be derived from these alkanes.

For example:

In the case of propane (CH₃CH₂CH₃), the removal of one hydrogen atom attached to the terminal carbon yields an unbranched propyl group. But when one H -atom linked to the middle carbon is removed, an isopropyl group is obtained.

The alkyl groups are generally represented by the letter R. So, an alkane is represented as R —H. The monovalent alkyl groups have the general formula: C_nH_2n+1 [n = 1.2,3 etc.]

Nomenclature of alkyl groups

The names of the alkyl groups are derived by replacing the suffix ‘ane’ of the corresponding alkane by the suffix ‘yl’

Classification of alkyl groups

1. Primary (1°) alkyl groups:

The removal of one 1° H -atom from an alkane gives a primary or 1° alkyl group. Ethyl (CH₃CH₂-), isobutyl (Me₂CHCH₂—), neopentyl (Me₃CCH₂—), etc. Are some examples of primary alkyl groups. The primary alkyl groups obtained from simple straight-chain alkanes are called normal alkyl groups

In the common nomenclature system, those are designated as n -alkyl group Imt In the IUPAC system, n Is omitted

For example:

In the common or trivial system of nomenclature, CH,CH₂CH₂– IS written as n -propyl group hut In the IUPAC system, it Is designated as propyl group. The primary alkyl groups In which the second last carbon in the chain is branched to one

The group are named by using the prefix Mso’

For example:

The —CH₂CH(CH₃)₃ group is called the isobutyl group. (As a group, up to isohexyl and as an alkane up to isohexane, the use ofthe ‘iso’ has been accepted by the IUPAC system).

The primary alkyl groups in which the second last carbon in the chain is branched to two —CH₃ groups are named by using the prefix ‘nco’.

For example: The —CH₂C(CH₃ )₃ group is called neopentyl group. (As a group, upto neo hexyl and as an alkane upto neohexane, the use of‘neo’ is accepted by die IUPAC system.)

2. Secondary (2°) alkyl group:

Removal of one 2° H-atom from an alkane forms a secondary (2°) alkyl group. In both trivial & IUPAC system,itis written as sec-alkyl (pronounced as secondary alkyl group),

For example: CH₃ CH₂ CHCH₃ is a 2° alkyl group named as a sec-butyl group.

3. Tertiary (3°) alkyl group:

The removal of one 3° H -atom from an alkane leads to the formation of a tertiary or 3° alkyl group. In both the trivial and IUPAC systems, it is written as ferf-alkyl group or f-alkyl group (pronounced as tertiary alkyl group),

For example:—C(CH₃ )₃ is a tertiary alkyl group whose name is a fert-butyl group or a t-butyl group

Structures and IUPAC names of some alkyl groups

Monovalent radicals derived from unsaturated acyclic hydrocarbons end with ‘-enyl ‘-ynyl,’ ‘-dienyl,’ etc., depending on the nature of the radicals or groups. Positions of double and triple bonds are indicated by numerals where necessary.

The c-atom of any radical containing free valence is always numbered as ‘1’

For example:

CH ≡ C— (ethynyl), \(\stackrel{3}{\mathrm{C}} \mathrm{H} \equiv \stackrel{2}{\mathrm{C}} \stackrel{1}{\mathrm{C}} \mathrm{H}_2\) (prop-2-any)

⇒ \(\stackrel{3}{\mathrm{C}} \mathrm{H}_3 \stackrel{2}{\mathrm{C}} \mathrm{H}=\stackrel{1}{\mathrm{C}} \mathrm{H}-\) (prop-l-enyl) etc. The following trivial names are retained in the IUPAC system: CH₂=CH— (vinyl),

CH₂=CHCH₂— (allyl), etc. The radical (CH₂= ) is called ‘methylene’ and (=CH—)is called ‘methine.

The presence of one or more free valency in radicals derived from parent alkenes is indicated by suffixes like monovalent(-yl), divalent (-diyl), trivalent (-triyl) etc.

For example: CH₃CH< is ethane-1, 1-diyl; (CH₃)₂C< is propane-2,2-diyl etc.

Aryl groups:

The organic groups derived from benzene and other benzene derivatives are termed as aryl groups. Aryl groups are generally represented by Ar. The simplest aryl group is phenyl group (—C₆H₅ ). It can be obtained by removing one hydrogen atom from a molecule of benzene (C₆H₆ )

Trivial or common system of nomenclature of other classes of compounds:

IUPAC nomenclature of different classes of compounds at a glance:

Class 11 Chemistry Organic Chemistry Techniques and Principles

Rules Branched For Iupac Nomenclature Of Chain Alkanes

Longest chain rule

The longest continuous chains of the alkane is to be identified first. It is known as the parent or root chain. The number of carbon atoms in the parent chain determines the word root

It is to be noted that the longest chain may or may not be straight but it must be continuous. All other carbon atoms that are not included in the parent chain are called branched chains side chains or substituents. The branched-chain alkane is, therefore, named as a derivative ofthe parent chain

Example:

The parent chain in the compound (I) contains 6 carbon atoms and the CH₃ -group is a side chain or substituent. Therefore, it is to be named as a derivative of hexane. The parent chain in the compound (II) contains 8 C-atoms but is not straight so, it is named as a derivative of octane. CH₃ -and C₂H₆ -groups are the two substituents here. If two chains of equal lengths are possible, then the one with a maximum number of side chains or substituents is to be considered as the parent chain.

Example:

In the compound (III), the parent chain is the horizontal six-carbon chain containing two alkyl substituents ( —CH₃ and C₂H₅) but not the other six-carbon chain containing only one alkyl substituent [(CH₃)₂CH-]

Lowest number rule

The carbon atoms of the parent chain is to be numbered as 1, 2, 3, 4, . etc. from one end in such a way that the carbon atom carrying the substituent gets the lowest possible number.

Example:

In the following compound, the numbering can be done in two different ways. The numbering of the carbon chain as given in the structure (IVA) is correct because the carbon carrying the substituent gets a lower number i.e., 3. However, the numbering of the carbon chain as given in the structure (IVB) is incorrect because the carbon carrying the substituent gets a higher number i.e., 5.

The number indicating the position of the substituent in the parent chain is called its positional number or locant. Thus, the correct locant for the methyl side chain in the above compoundis 3.

When two substituents are present in the chain, then the lowest set of locant rule is applied. It states that when two or more different sets of locants are possible, then that set of locants will be the lowest which (when compared term by term with other sets, each in order of increasing magnitude) has the lowest term at the first point of difference. This rule is used irrespective ofthe nature ofthe substituent.

Example:

When the carbon atoms of the parent chain of the following compound (V) are numbered from the sides, two sets of locants are obtained. Out of the two sets of locants (2,3) and (3,4), the first set is lower and hence preferred because the first term, i.e., 2 in the first set is lower than the first term, i.e., 3 in the second set

Similarly, for the compound (6):

Out ofthe two sets of officiants (2,2,4) and (2,4,4), the first set is lower and hence preferred as the second term in the first set i.e.,2 is lower than the second term 4in the second set.

According to the latest IUPAC recommendations of nomenclature (1993), the lowest set of locant rules Is preferred even If it violates the lowest sum rule.

For example:

In the case of the following compound, the numbering of the carbon chain from left and right gives two different set of locants with two different sum of locants.

The numbering from the left is correct because the first term CH₃ i.e., 2 in the set (2, 7, 8) is lower than the first term i.e., 3 in the set (3, 4, 9), even though the sum of locants is lower when the numbering is done from the right. Thus, the correct name ofthe alkanes 2, 7, 8-trimethyIdecane.

Name of the branched chain alkanes

When there is one alkyl group in the parent chain, its name is to be prefixed to the name of the parent alkane, and its position on the chain is to be indicated by writing before it the number of the carbon atom carrying the substituent.

The name of the substituent is separated from its positional number or locant by a hyphen (-). The final name of the alkane is to be written as one word, i.e., there will be no gap between the name of the substituent and the parent alkane.

Example:

Alphabetical order of the side chains or substituents

When two or more different alkyl groups (side chains or substituents) are present on the parent chain, such groups prefixed by their positional numbers or locants, are to be arranged in alphabetical order irrespective of their positional numbers and written before the name of the parent alkane.

Example:

In the given compound, between ethyl and methyl groups, ethyl comes first in the alphabetical order and therefore, its name is 3-ethyl-2-methylhexane. When a number appears between two substituent groups then hyphens are used on both sides ofthe number

It is to be noted that while deciding the alphabetical order of various alkyl groups, prefixes such as ‘iso’ or‘neo’ are to be considered as a part of the fundamental name of the alkyl group while the prefixes, ‘second ‘tert’ are not

Examples:

Numbering of different alkyl groups at equivalent positions

If two different alkyl groups or substituents are present at equivalent positions, i.e., at the same position from the two ends of the parent chain, then numbering of the chain is to be done in such a way that the alkyl group which comes first in the alphabetical order gets the lower number.

Example: 

Naming of same alkyl groups at different positions

When the parent chain contains two or more same alkyl groups at different positions (or at the same position), the positional number of each alkyl group is separated by commas, and suitable prefixes such as ‘di’ (for two), ‘tri’ (for three), ‘tetra’(for four) etc., and then they are to be attached to the name of the alkyl group. When two alkyl groups are attached to the same carbon atom, the positional number or locant is to be written twice. It is to be noted that the prefixes like ‘di’, ‘tri’, ‘tetra’, etc. are not considered while deciding the alphabetical order ofthe alkyl groups

Naming of complex substituents/substituted substituents

1. If the alkyl group on the parent chain is complex, i.e., if it has a branched chain, it is named as a substituted alkyl group by numbering the C-atom of this group attached to the parent chain as 1. The name of the complex substituent is generally enclosed in brackets to avoid any confusion with the numbering of the parent chain.

2. When the same complex substituent occurs more than once on the parent chain, it is indicated by multiplying the prefixes such as “bis’ (for two), ‘tris’ (for three), ‘tetrakis’ (for four), ‘pentakis'(for five), etc

Example:

IUPAC nomenclature of bicyclic compounds

The name of a bicyclic compound in the IUPAC system is based on the name ofthe alkane having the same number of carbons as there in the ring system. The name follows the prefix bicyclo and a set of brackets enclosing numbers indicating the number of carbons in each of the three bridges connecting the bridgehead carbons in order of decreasing size.

For example:

The following bicyclic compounds containing nine and eight carbon atoms are named bicyclo [4.3.0]nonane and bicyclo[3.2.1]octane respectively.

What is wrong with the following names? Draw the structures they represent and write their correct names. (i)1,1-dimethylhexane (iii)3-methyl-5-ethylheptane (iv) 4, A-dimethyl-3-ethylpentane (v) 3, 4,7-trimethyloctane (vi) 3,3-diethyl-2,A,Atrimethylpentane 36. Give the IUPAC name of the following alkane containing complex substituents:

Rules For Iupac Nomenclature Of Unsaturated Hydrocarbons

For naming the compounds containing multiple (double and triple) bonds, the following additional rules are to be applied:

1. The parent chain must contain multiple bonds (double or triple) regardless of the fact whether it denotes the longest continuous chain or not

For example:

In compound (I), the parent chain contains 5 carbon atoms and not 6 carbon atoms since the latter does not include the double bond

Example:

2. While naming a particular member ofthe alkene or alkyne family, the primary suffix ‘ane’ of the corresponding alkanes to be replaced by ‘ene’ or ‘yne’ respectively.

3. The numbering of the parent chain is to be done in such a way that the first C-atom associated with the multiple bond gets the lowest possible number

Examples:

4. If the parent chain contains a side chain, then also the multiple bonds get priority in numbering.

If the parent chain contains 2 or 3 double or triple bonds, then the primary suffix ‘diene’ (or ‘triene’) or ‘diyne’ (or ‘triyne’) are to be used to represent them. In these cases, terminal ‘a’ is also added to the wordroot For numbering, the lowest set oflocants rule is to be followed.

Examples:

5. If the parent chain contains both double and triple bonds, the following points are to be remembered while writing their names:

The unsaturated hydrocarbon is always named as a derivative of alkyne, i.e., the primary suffix ‘ene’ is always to be written before ‘yne’.

In all these cases, the terminal ‘e’ of the one is dropped. The numbering of the parent chain is to be done from that end which is nearer to the double or triple bond, i.e., the lowest set officiants rule is to be followed

Examples:

If the positions of double and triple bonds are identical, i.e., if  the set officiants from both sides ofthe chain is the same, then the double bond is always given preference over the triple bond

Example:

6. If the unsaturated hydrocarbon contains a side chain along with double and triple bonds, then the numbering of the parent chain is to be done in such a way that the multiple bonds get the lowest set of locants. However, if the numbering from both ends of the parent chain gives the same set of locants to the multiple bonds, then the locant for the side chain must be minimum.

Examples:

When there are more than two double bonds in the hydrocarbon and it is impossible to include all of the min the parent chain, then the double bond which is not included in the parent chain is treated as a substituent

NCERT Class 11 Chemistry Organic Chemistry Techniques Notes

IUPAC Nomenclature Of Compounds With Functional Groups, Multiple Bonds And Substituents

The following additional rules are to be followed while naming organic compounds containing one functional group, double and triple bonds and substituents:

1. In these compounds, the longest chain of carbon atoms containing the functional group and the maximum number ofdouble and triple bonds are to be considered as the parent chain.It may or maynot be the longest possible carbon chain.

For example:

In the following compound (I), the parent chain containing the functional group and the double bond has 6 carbon atoms while the longest possible carbon chain has 7 carbon atoms.

Example:

2. The parent chain is to be numbered in such a way that the functional group gets the lowest locant, even if it violets the lowest set oflocants rule for substituents. For example, in the following compound, the lowest locant for the functional group >C= O is 3 and not 5

Example:

3. If the organic compound contains a terminal functional group such as —CHO, —COOH, —COCl, —CONH₂, —COOR, —C =N, etc., The numbering of the parent chain must be started from the functional group, we., it is always given number but the number is usually omitted from the final name ofthe compound

Example:

4. If a compound contains two or more similar functional groups, numerical prefixes [di, tri, tetra etc.) are used to indicate their numbers, and the terminal ‘e’ of the primary suffix (ane, ene or yne) is retained while writing the name.

Examples:

5. If an organic compound contains more than two similar terminal functional groups and all of them are directly attached to the parent chain, then none of them are considered as a part of the parent chain. Special suffixes such as carboxylic acid (for —COOH), carbaldehyde (for—CHO), carbonitrile (for —C=N), carboxamide (for —CONH₂) etc. are used to name these

Examples:

IUPAC Nomenclature Of Compounds With Pentane-2,4-Dione Two Different Functional Groups

When an organic compound contains two or more different functional groups, then one of the functional groups is to be selected as the principal functional group while all other functional groups (also called secondary 7 functional groups) are to be treated as substituents. The choice of principal group is made on the basis ofthe following order of preference:

—COOH>—SO₃H >— COOCO — >— COX (X=halogen) >—CONH₂> —C≡ N>—CHO>C= O> —OH> —SH>—NH₂

All the remaining groups such as halo (fluoro, chiro, bromo, and iodo), nitroso (—NO), nitro (—NO₂), alkoxy (—OR), alkyl (—R), aryl {Example: C₆H₃), etc., are always treated as substituents or simply as prefixes.

Suffixes and prefixes of some Important functional groups [with decreasing priority

Polyfunctional compounds are named as follows:

• The chain containing the principal functional group, the maximum number of secondary functional groups, and multiple (double or triple) bonds, if any, is to be considered as the principal chain in the compound.
• The principal chain is to be numbered in such a way that the principal functional group gets the lowest possible number followed by double bond, triple bond, and substituents.
• The prefixes for the secondary functional groups and other substituents are to be placed in alphabetical order before the word root. If two or more identical secondary functional groups are present, these are to be indicated by using di, tri, tetra, etc. as prefixes
• The principal functional group is to be written after the word root and the compound is to be named as a member of that particular class of compound.

Examples of some functional compounds:

IUPAC nomenclature of other classes of compounds (According to 1993 Recommendations)

Structural Formulas Of Organic Compounds From Their Iupac Names

• To write the structure of an organic compound if its IUPAC name is given, the given steps are to be foin (parent chain) is to be selected from the word root ofthe IUPAC name ofthe compound
• The parent chain is to be numbered from either end.
• If the name of the compound contains primary suffix ‘ene’ or ‘yne\itis placed at the indicatedposition along the chain. Iv] Name and position of the functional group (secondary suffix) is to be identified from the IUPAC name andit is to be placed atits rightpositionin the carbon chain.
• The names and positions of other prefixes, if any, are to be identified from the IUPAC name and to be attached at proper positionin the carbon chain.
• Finally, the required number ofH- atoms are added, wherever
  necessary, to satisfy tetra covalency of each C-atom

Examples 1: Letus write the structural formula of 5-hydroxy- 2-methylhex-3-enoic acid.

Step 1: The word ‘hex’ indicates that the parent chain contains 6 carbon atoms: C —C —C—C —C—C

Step 2: Numbering ofthe carbon chain 6 5 is done as indicated: \(\stackrel{6}{\mathrm{C}}-\stackrel{5}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}\)

Step 3: TheIUPACname of the compound has the primary C suffix ‘ene’ at position 3. Therefore, C-3 and C-4 ofthe parent chain are linked by a double bond. \(\stackrel{6}{\mathrm{C}}-\stackrel{5}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}=\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}\)

Step 4: The secondary suffixes ‘oic acid’. Therefore, the carbon atom of the —COOH group is indicated as 1. \(\stackrel{6}{\mathrm{C}}-\stackrel{5}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}=\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}} \mathrm{OOH}\)

Step 5: The prefixes ‘hydroxy’ and ‘methyl’ are attached at the positions 5 and 2 respectively

Step 6: A required number of H- atoms are added to various carbon atoms to get the final structure ofthe compound.

2. IUPAC name: But-2-en-l-o

Step 1: C —C —C—C —C—C

Step 2: \(\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}\)

Step 3: \(\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}=\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}\)

Step 4: \(\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}=\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}-\mathrm{OH}\)

Step 5: \(\mathrm{H}_3 \stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}} \mathrm{H}=\stackrel{2}{\mathrm{C}} \mathrm{H}-\stackrel{1}{\mathrm{C}} \mathrm{H}_2 \mathrm{OH}\)

3. UPAC name: 3-amino-4-methylpentanoic acid

Step 1: C —C —C—C —C—C

Step 2: \(\stackrel{5}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}}\)

Step 3: \(\stackrel{5}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{1}{\mathrm{C}} \mathrm{OOH}\)

Step 4: 

Step 5:

4. IUPAC name: 3-ethyl-4, 5-dimethyl hex-l-yn-3-o

Step 1: C —C —C—C —C—C

Step 2: \(\stackrel{1}{\mathrm{C}}-\stackrel{2}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{5}{\mathrm{C}}-\stackrel{6}{\mathrm{C}}\)

Step 3: \(\stackrel{1}{\mathrm{C}} \equiv \stackrel{2}{\mathrm{C}}-\stackrel{3}{\mathrm{C}}-\stackrel{4}{\mathrm{C}}-\stackrel{5}{\mathrm{C}}-\stackrel{6}{\mathrm{C}}\)

Step 4: 

Step 5:

Step 6:

Class 11 Chemistry Organic Chemistry Basic Principles And Techniques Organic Chemistry Isomerism And Organic Reaction Mechanism Introduction

Organic compounds tend to exist as isomers. There are two or more compounds which have the same molecular formula (or molecular mass) but different physical and chemical properties, i.e., these are isomers of each other. This phenomenon is known as isomerism.

Again, organic compounds being covalent normally participate in molecular reactions. The mechanism of a reaction is the path followed by the substrate and reagent (the reacting species) while forming the products and in fact, it explains how the bonds in the reacting molecules break and new bonds in the product molecules are formed. To understand the mechanisms of organic reactions, some fundamental concepts are to be conceived. In this chapter, the isomerism in organic compounds and some basic concepts of organic reaction mechanisms have been discussed.

Isomerism In Organic Compounds

The property of isomerism in organic compounds is due to different sequence of bonding of their atoms or due to different arrangements of their atoms or groups in space when the sequence of bonding is same

Isomerism in organic compounds Definition:

The phenomenon of the existence of two or more compounds possessing the same molecular formula but different physical and chemical properties is known as isomerism. Such compounds are called isomers

Example:

Two compounds having the same molecular formula, C₂H₅O but with completely different physical and chemical properties are found to exist. One compound is a liquid which boils at 78°C and reacts with metallic sodium to liberate hydrogen gas and the other compound is a gas which boils at -24°C and does not react with sodium. Therefore, these two compounds are entirely different in nature.

The first compound is ethyl alcohol belonging to the class of compounds known as alcohols and the second compound is dimethyl ether belonging to the class of compounds known as ethers. Because of the difference in their structures, they are completely different in their properties.

These two compounds are, therefore, isomers and the phenomenon of the existence of these two compounds of identical molecular formulas but belonging to different families is known as isomerism

CH₃— CH₂— OH(  Ethyl alcohol)

CH₃— O—CH₃( Dimethyl Ether)

Structural Isomerism

The compounds having the same molecular formula but different structures or molecular constitutions, i.e., different atom-to-atom bonding sequences or atomic connectivity are called structural or constitutional isomers and the phenomenon Is known as structural isomerism. Structural isomerism can be further subdivided into five different categories. Besides, tautomerism is also considered structural Isomerism.

1. Chain Isomerism

Chain Isomerism Definition:

Two or more compounds (belonging to the same family) having the same molecular formula but different carbon skeletons are called chain or nuclear isomers and the phenomenon Is called chain or nuclear isomerism

Example:

1. n -butane and isobutane are two chain isomers because they have die same molecular formula (C₄H₁₀) but differ in their carbon skeletons.

2. n-pentane, isopentane and neopentane are the three yJJJj function, group isomerism Chain isomers because they (molecular formula C₅H₁₂) differ in their carbon skeleton

2. Position isomerism

Position isomerism Definition:

When two or more compounds having the same structure of the carbon chain, i.e., the same carbon skeleton, differ in the position of substituent, multiple (double or triple) bond or functional group, these are called position isomers and this phenomenon is called position isomerism

Examples:

1. n -propyl alcohol and isopropyl alcohol are two position isomers. They have the same molecular formula (C₃H₈O) and have identical carbon skeletons. But in n propyl alcohol, hydroxy (-OH) group is at the terminal C-atom of the chain while in isopropyl alcohol the hydroxy (-OH) group is attached to the middle C -atom.

3. Functional group isomerism

Functional, group isomerism Definition:

Two or more compounds having the same molecular formula but different functional groups [i.e., belonging to different families) are called functional isomers and this phenomenon is called functional group isomerism or functional isomerism

1. Alcohols & ethers (C_nH_{2n+ 2}O) exhibit functional group isomerism. Ethyl alcohol and dimethyl ether having the same molecular formula, C₃H₆O represent two functional isomers. The functional group of ether is divalent oxygen (— O —) while the functional group of alcohol is the alcoholic hydroxy (— OH) group.

2. Aldehydes, ketones, unsaturated alcohols and unsaturated ethers exhibit functional group isomerism. The molecular formula, C₃H₆O for example, represents the following two functional isomers:

3. Carboxylic acid and esters (C_nH_2nO₂) exhibit functional group isomerism. For example, the molecular formula C₂H₄O₂ represents the following two functional isomers

CH -COOH (Acetic acid) (Func. gr: —COOH)

H – COOCH(Methyl formate)( Func. gr:—COOCH₃)

4. 1°, 2° and 3° amines (C_nH_3nN) exhibit functional group isomerism. The molecular formula C₃H₉N, for example, represents

The following three functional isomers:

CH₃CH₂CH₂NH₂ Propan-l-amine (1° amine)

CH₃CH₂NHCH₃  N-methylhexanamine (2° amine)

(CH₃)₃N [N,N-dimethylethanolamine (3° amine)]

5. Aromatic alcohols, phenols and ethers exhibit functional group isomerism. For example, the molecular formula, C₇H₈O represents the following three functional isomers,

6. Dienes, alienes & alkynes (C_nH_2n-2) exhibit functional group isomerism. The molecular formula C5H8, for example, represents four functional group isomers

7. Cyanides & isocyanides (C_nH_2n-1N) exhibit functional group isomerism. The molecular formula C₃H₅N, for example, represents two functional group isomers:

CH₃CH₂CN – Propanenitrile

CH₃CH₂NC – Ethyl isocyanide

5. Metamerism

Metamerism Definition:

When two ‘or more compounds having the same molecular formula but different numbers of carbon atoms (or alkyl groups) on either side of the functional groups such as —O —S—, — NH—, —CO — etc. are called metamers and the phenomenon is called metamerism. Metamerism occurs among the members of the same homologous series

Examples: The molecular formula, C₄H₁₀O represents the following metamers of ether family:

Examples of some metamers:

6. Ring-chain isomerism

 Ring-chain isomerism Definition:

Compounds having the same molecular formula but possessing open-chain and closed-chain structures are called ringchain isomers and the phenomenon is called ring-chain isomerism.

Examples:

1. Propene and cyclopropane are ring-chain isomers with the molecular formula C₃H₆

2. Propyne and cyclopropene are ring-chain isomers with the molecular formula

7. Tautomerism

This is a special kind of functional group isomerism involving dynamic equilibrium between the isomers.

Tautomerism Definition:

The functional group isomerism which arises due to the reversible transfer of a group or atom from one polyvalent atom to the other within the same molecule witth necessary rearrangement of linkages and the resulting isomers exist in dynamic equilibrium with each other is called tautomerism. The interconvertible isomers are called tautomers or tautomerides. Tautomerism is also called desmotropism

Conditions for tautomerism:

1. There must be at least one or H -atom present concerning each functional group in the compound.
2. The compound must contain an electronegative atom bonded by a double or triple bond e.g., C=O, N=O, C=NH etc.

Keto-enol tautomerism:

In this type of tautomerism, one form is the keto-form containing a keto group >C=O ) while the other form is the enol-form containing an enolic group >C=C—OH). The term ‘enol’ comes from ‘ene’ of the double bond and ‘ol’ of the hydroxy (—OH) group. Ketoenol tautomerism is possible only for those carbonyl compounds which contain at least one a -H -atom

Example:

Ethyl acetoacetate is an important example of this type. Tautomeric equilibrium generally (if no other factors ! operate) favours the structure in which the H-atom is bonded to the C -atom rather than the more electronegative O -atom, i.e., equilibrium favours the weaker acid. However, in this case, because of conjugation and intramolecular H-bonding the percentage of enol-form is much higher (8%) as compared ! to that of acetone where no such factors operate

Keto and enol-forms are, in reality, two dynamic structural isomers. These isomers are always interconvertible and . one H-atom is shifted from one C-atom to an O-atom and vice-versa to establish the equilibrium.

Factors affecting the percentage of enol content:

1. Intramolecular H-bonding (chelation) stabilises the enol and thus increases the amount of enol-form.
2. Conjugation resulting in resonance stability of the enol-form helps to increase the enol content.
3. Polar protic solvents usually decrease the percentage of enol form because the more polar keto-form becomes relatively more stabilised in this medium

The percentage of the enol-form is greater because of its much higher stability caused by strong intramolecular H-bonding and effective resonance.

1. The following compounds do not exhibit keto-enol tautomerism due to lack of a -H-atom

2. Keto-enol actually involves interconversion of group and a tautomerism —C=C(OH) group.

The sum of bond energies of the is 347.9 kcal-mol-1 and that of the —C=C(OH) group is 336.0 kcal. mol-1 . So, the keto form is thermodynamically more stable than the enol form in the absence of other factors which can stabilise the enol form.

3. Pseudotropism: Tautomerism in which there is practically; no existence of one tautomer is called pseudotropism. For example

Nitro-acinitro tautomerism:

Nitroso-oximino tautomerism:

CH₃—CH₂—N Nitrosoetane(nitroso form) ⇌  CH₃ —CH—N — OH(Ethanal oxime (oximino form)

Organic Chemistry Techniques for Class 11 Chemistry

Stereoisomerism

Stereoisomerism Definition:

The isomers having the same structure formula, i.e., er same atom-to-atom bonding sequence or connectivity which differ in the relative arrangement of atoms or groups in space are called stereoisomers and the phenomenon is called stereoisomerism. It is the specific directional property of covalent bonds in threedimensional space which gives rise to stereoisomerism.

Stereoisomerism is broadly classified into two categories: 

1. Configurational isomerism
2. Conformational isomerism.

The isomerism that arises due to different spatial arrangements of groups or atoms (which are not interconvertible) in the same structural isomer is called configurational isomerism.

Configurational isomerism can be subdivided into two types: 

1. Optical isomerism and
2. Geometrical isomerism.

1. Optical isomerism

Optical isomerism Definition: 

If two molecules having the same atom-to-atom bonding sequence or connectivity (i.e., the same constitution) are mirror images of each other and are non-superimposable, then these are called enantiomers. Enantiomers rotate the plane polarised light to an equal degree but in opposite directions. Because of their effect on plane polarised light, separate enantiomers are said to be optically active and because of this property, they are called optical isomers and the phenomenon is called optical isomerism

Example:

As shown in the figure, lactic acid or 2-hydroxypropanoic acid (CH₃CHOHCOOH) exists as a pair of enantiomers. In (-)-lactic acid, the sequence of occurrence of the groups: OH->COOH→CH₃ appears in a clockwise direction but in (+)-lactic acid, it appears in an anticlockwise direction when viewed along the C — Ii bond axis from the side opposite to that of the H-atom. Both the enantiomers rotate the plane of plane polarised light to the same extent but in opposite directions. Enantiomer which rotates the plane of polarised light towards the right is called dextrorotatory or d- or (+)-lactic acid and which rotates the plane of polarised light towards the left is called laevorotatory or l- or (—)-lactic acid

Compounds which can rotate the plane of plane polarised light are called optically active compounds and those which cannot are called optically inactive compounds. Both the enantiomers of lactic acid are optically active.

Symmetric and asymmetric molecules:

If a molecule has at least one of the following elements of symmetry:

1. Plane of symmetry,
2. Centre of symmetry
3. Alternating axis of symmetry

Then it is a symmetric or achiral molecule. Symmetric molecules are optically inactive. However, if a molecule has none of these symmetry elements, it is called an asymmetric or chiral molecule. Asymmetric molecules are optically active.

A symmetric molecule is superimposable on its mirror image while an asymmetric molecule is not. Therefore, whether a molecule is symmetric or asymmetric may also be verified by constructing models of the molecule & its mirror image and then placing one model on the other. If they are found to be superimposable, then the molecule must be symmetric; if not, then the molecule is asymmetric.

Example:

The two mirror-image forms of lead-acid are asymmetric and so, they are optically active. In our daily lives, we come across many things which are related to mirror images and do not get superimposed,

For example: Our left hand is the mirror image of our right hand but the left hand and the right hand are non-superimposable. So, the glove of the left hand do not fit in the right hand

Plane of symmetry or Sigma plane (σ ):

The plane of symmetry (σ) is defined as an imaginary plane that bisects a molecule in such a way that one half of the molecule is the mirror image of the other half (the plane acting as a mirror). The plane is also called a mirror plane.

Example: Meso-tartaric acid has a plane of symmetry.

Centre of symmetry or Centre of inversion (i):

A centre of symmetry is a point from which lines, when drawn on one side and produced an equal distance on the other side, will meet identical points (i.e., atoms) in the molecule

Asymmetric carbon atom and optical activity:

When a carbon atom of an organic molecule is attached to four different atoms or groups (Cabde), then that carbon atom is called an asymmetric carbon atom or chiral carbon. It is also called a stereogenic centre. Molecules containing only one asymmetric carbon atom are asymmetric and optically active. If a molecule contains more than one asymmetric carbon atom, it may be asymmetric or symmetric, i.e., it may be optically active or inactive.

Example:

(+) tartaric acid having two asymmetric C-atoms is optically active but meso-tartaric acid is optically inactive.

The presence of asymmetric carbon atoms is not essential for exhibiting optical activity:

Some of the substituted allenes (abC=C=Cab) are found to be optically active, even though they contain no asymmetric carbon atom. The reason for their optical activity is the overall asymmetry of their molecules.

Example: The two mirror images of 1,3-dichloro propadiene are overall asymmetric and so, they are optically active

Therefore, the presence of asymmetric C-atom in a molecule is not essential for exhibiting optical activity. If the structure is overall asymmetric, then the molecule will be optically active.

Conditions necessary for optical activity:

1. The compound is non-superimposable on its mirror image, or
2. It contains only one asymmetric C-atom or
3. Plane, centre and alternating axis of symmetry are absent in the molecule.

E. Eliel in his book, has mentioned a molecule which does not contain any element of symmetry, yet it is not optically active. He further stated that principally possibility of the existence of such molecules cannot be discarded outright. Hence, the necessary and sufficient condition for the optical activity of any compound is the non-superimposability ofthe molecule on its mirror image

• Meso-compounds: If a compound having more than one chiral centre is found to be optically inactive, then it is called a meso-compound. Forexample, meso-tartaric acid.
• Enantiomers:  Stereoisomers that are not superimposable on each other but related to each other as mirror images are called enantiomers. Thus (+) & (-)-lactic acid form a pair of enantiomers. Similarly, stereoisomers I & II of 3- bromopentan-2- ol form a pair of enantiomers. Enantiomers are optically active molecules having equal but opposite specific rotations. All other physical and chemical properties of enantiomers are the same.
• Diastereoisomers: Stereoisomers that are not mirror images of each other are called diastereoisomers. Stereoisomers I & III of 3-bromopentan-2-ol are diastereoisomers of each other

Calculation of no. of stereoisomers:

No. of stereoisomers, both optically active and inactive (meso-form) can be obtained from the number of chiral centres present in the molecule

Examples:

1.  3-bromopentan-2-ol (CH₃CHOHCHBrCH₂CH₃) has two dissimilar chiral centres. Therefore, it has 2² = 4 optically active stereoisomers and no optically inactive meso-isomer. Fischer projections of these four stereoisomers are shown below

(I, II) and (III, IV) represent two pairs of enantiomers and they are optically active. Each of (I, III), (I, IV), (II, III), and (II, IV) represents a pair of diastereoisomers.