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

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

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.

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.

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.

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

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

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

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.

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.

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.

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.

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