NCERT Class 11 Chemistry Chapter 13 Hydrocarbons Short Question And Answers

NCERT Solutions For Class 11 Chemistry Chapter 13 Hydrocarbons Short Question And Answers

Question 1. The hydrolysis of n-propyl magnesium bromide and another alkyl magnesium bromide may obtain propane. Write the name and structure of that alkylmagnesium bromide.
Answer:

Propane can be obtained by hydrolysis of isopropyl magnesium bromide (Me₂CHMgBr)

Question 2. When a concentrated aqueous sodium formate solution is used in Kolbe’s electrolysis method, no alkane is obtained—why?
Answer:

In Koibe’s electrolysis method, two— R groups of two RCOONa molecules combine to form the alkane, and R— R and two CO₂ molecules are obtained from two COONa groups. As, there is no alkyl group in the salt, sodium formate (HCOONa), no alkane is formed on electrolysis of its concentrated aqueous solution.

Question 3. Prepare 2,2-dimethylpropane by Corey-House synthesis.
Answer:

Question 4. How will you convert: \(\left(\mathrm{CH}_3\right)_2 \mathrm{CHBr} \longrightarrow \mathrm{CH}_3 \mathrm{CHDCH}_3\)
Answer:

Question 5. Identify RI and R’l in the following reaction \(\mathrm{RI}+\mathrm{R}^{\prime} \mathrm{I} →{\mathrm{Na} / \text { ether }} \text { Butane + Propane + Ethane }\). What is the role of ether in this case?
Answer:

When the Wurtz reaction is carried out with a mixture of HI and R’l, two hydrocarbons with an even number of carbon atoms (R— R and R’—R’) and one hydrocarbon with an odd number of carbon atoms (R—R’) are formed. Among the formed alkanes, butane (CH₃CH₂CH₂CH₃) and ethane (CH₃CH₃) are respectively R— R and R’—R’ whereas, propane (CH₃CH₂CH₂) is R — R’.

So, it is evident that R is an alkyl group containing two carbon atoms, i.e., ethyl group (CH₃CH₂ — ) and Rr is a one carbon atom-containing alkyl group, i.e., methyl group (—CH₃). Therefore, RI is ethyl iodide (CH₃CH₂I) and R’l is methyl iodide (CH₃I). The role ofether in this case is that it acts as a solvent.

Question 6. How will you convert methyl bromide to (1) methane and (2) ethane in one step?
Answer:

1.

2.

Question 7. Arrange the following compounds in order of their increasing stability and explain the reason: 2-butene, propene, 2-methyl but-2-ene
Answer:

The order of increasing stability of the given alkenes is— propene < 2-butene < 2-methyl but-2-ene

The number of hypercoagulable hydrogen in propene, 2-butene, and 2-methylbut-2-ene molecules are 3, 6, and 12 respectively. With an increase in several hypercoagulable hydrogens, stability due to the hyperconjugation effect of the molecules increases.

Question 8. Between the position and chain isomer of but-1ene, which exhibits geometrical isomerism and why?
Answer:

Position isomer of but-1-ene is but-2-ene (CH₃CH=CHCH₃) and chain isomer of but-1-ene is 2-methylpropene [(CH₃)₂C=CH₂] . In a molecule of but-2-ene, different groups are attached to the C-atom which is linked to the double bond.

So, but-2-ene exhibits geometrical isomerism. On the other hand, in a molecule of 2-methylpropene, two similar atoms (H-atom) are attached to the C-atom linked to the double bond. So, 2-methylpropene does not exhibit geometrical isomerism.

Question 9. Write the structures of the two alkenes obtained when 2-butanol is heated with excess ofconcentrated H₂SO₄. Which is obtained predominantly?
Answer:

Stability due to hyperconjugation is higher in the case of but-2-ene compared to that it is obtained predominantly.

Question 10.

Answer:

Question 11. Identify the compounds obtained on heating (CH₃)₄NOHO
Answer:

Compounds formed: trimethyl amine and methyl alcohol.

Question 12. What is the major product formed in the reaction between CH₂=CH—NMe₃Ie and HI? Write its structure.
Answer:

Question 13. What happens when a mixture of ethylene and O₂ gas is passed through a solution of PdCI₂ in the presence of CuCl₂  at high pressure and 50°C?
Answer:

When a mixture of ethylene and O₂  gas is passed through a PdCl₂  solution in the presence of CuCl₂  at high pressure and 50°C, acetaldehyde is formed as the product.

Question 14. Write the IUPAC names of the following compounds:

1. HC=C-CH(CH₃)₂
2. CH₃-C=C-C(CH₃)₃

Answer:

IUPAC name: 3 – Methylpent-1-yen

IUPAC name: 4,4-dimethyl pent-2-yen

Question 15. C = C bond length is shorter than C and C —C —why?
Answer:

The σ -bond In C≡C is formed due to the overlapping of two small sp -hybridized orbitals. In C—C, the σ -bond is formed due to the overlapping of two bigger sp² -hybridized orbitals whereas in C — C.

The σ –bond is formed due to the overlapping of two even bigger sp³ -hybridized orbitals. Again, the multiplicity of the bond between two atoms increases, and the atoms come closer to each other leading to a decrease in bond length. So, bond length follows the order: C=C < C=C < C— C.

Question 16. Write structures and IUPAC names of the alkynes having molecular formula C₅H₈.
Answer:

Structures and IUPAC names of the alkynes having molecular formula C₅H_n are-

⇒ \(\stackrel{5}{\mathrm{C}} \mathrm{H}_3 \stackrel{4}{\mathrm{C}} \mathrm{H}_2 \stackrel{3}{\mathrm{C}} \mathrm{H}_2 \stackrel{2}{\mathrm{C}} \equiv \stackrel{1}{\mathrm{C}} \mathrm{H} \quad \text { (Pent-1-one) }\)

⇒ \(\stackrel{5}{\mathrm{C}} \mathrm{H}_3 \stackrel{4}{\mathrm{C}} \mathrm{H}_2 \stackrel{3}{\mathrm{C}} \equiv \stackrel{2}{\mathrm{C}} \stackrel{1}{\mathrm{C}} \mathrm{H}_3 \quad \text { (Pent-2-one) }\) Δ

Question 17. What happens when the ethanolic solution of 1,1,2,2-tetrabromoethane is heated with zinc dust?
Answer:

When the ethanolic solution of 1,1,2,2-tetrabromoethane is heated with Zn dust, acetylene is formed as the product.

Question 18. What happens when the gas obtained by the action of water on calcium carbide is passed through an ammoniacal AgNO₃ Solution? Identify the solution through which acetylene gas is passed to form a red precipitate.
Answer:

Acetylene is formed by die action of water on calcium carbide. When acetylene gas is passed through ammoniacal silver nitrate solution, a white precipitate of silver acetylide (Ag₂C₂) is obtained.

Question 19. Which gas is used in carbide lamps or Hawker’s lamps? How does the gas produce a bright illuminating flame in the lamp?
Answer:

Acetylene gas is used in carbide lamps or Hawker’s lamps for producing bright illuminating flame. The percentage of carbon in acetylene is greater than in saturated hydrocarbons having the same number of carbon atoms.

As a result, incomplete combustion of acetylene gas takes place and the heated carbon particles thus formed produce Hluminating flame and bright light.

Question 20. What is Lindlar’s catalyst? Mention its use.
Answer:

Use of Lindlar’s catalyst:

This catalyst is used to add 1 molecule H₂ i.e., for partial hydrogenation of an alkyne. As cis-hydrogenation takes place in this case, a cis-alkene can be prepared from a non-terminal alkyne by using this catalyst.

Question 21. Which out of ethylene & acetylene is more acidic and why?
Answer:

The greater the s -the character of a hybridized carbon atom, the greater will be its electronegativity. The s -character of sp hybridized carbon atom of acetylene is greater than that of the sp² -hybridised carbon atom of ethylene.

So the electronegativity of the carbon atom of acetylene (CH=CH) is greater than the carbon atom of ethylene (CH₂=CH₂).

The tendency of the H -atom attached to the more electro¬ negative carbon atom to be removed as a proton (H+) is relatively higher. So, the acidity of ethylene is less than that of acetylene, i.e., acetylene is more acidic than ethylene.

Question 22. How will you convert ethyne into ethanol?
Answer:

Question 23. Which out of ethyne and propyne is more acidic and why?
Answer:

In ethyne (HC=CH), there are two terminal hydrogen atoms whereas, in propyne (CH₃C=CH) there is only one. Apart from this, one electron-donating —CH₃ group is attached to the carbon atom present on the other side of the triple bond in the propyne molecule.

This decreases the acidity of the alkynyl hydrogen atom. So, ethyne is more acidic than propyne.

Question 24. How will you distinguish between but-1-yne and but-2-yne?
Answer:

But-1-yne (CH₃CH₂C=CH) being a terminal alkyne reacts with a solution of ammoniacal stiver nitrate (AgNO₃) to form a white precipitate of silver 1-butynide.

However, but-2-yne (CH₃C=CCH₃) being a non-terminal alkyne does not undergo this type of reaction. So, but-1-one and but-2-one can be distinguished by observing the result of the above-mentioned test using ammoniacal stiver nitrate solution.

Question 25. How will you convert propylene into propylene?
Answer:

Question 26. What is the expected shape ofa benzene molecule in the absence of resonance?
Answer:

In the absence of resonance, benzene will be considered as the compound, 1,3,5-cyclohexatriene. The structure of this hypothetical compound is—

The structure will be such because each carbon-carbon bond length will be unequal. As, C=C is shorter than C—C, the structure of the molecule will appear as an irregular hexagon instead ofa regular one.

Question 27. Write structures of two aromatic ions in which there is a p-orbital containing 2 electrons and the other in which there is a vacant orbital.
Answer:

Cyclopentadienyl anion has two electrons in one of its p -orbital whereas, cyclopropenyl cation has a vacant p -orbital.

Question 28. Write the name and structural formula of the dibromobenzene which forms three mononitro compounds.
Answer:

Question 29. What happens when each of the following compounds is heated with acidified K₂Cr₂O₇ solution—
Answer:

Ethylbenzene:

Ethylbenzene gets oxidised to benzoic acid.

Question 30. Benzene exhibits a greater tendency towards substitution reactions but a lesser tendency towards addition reactions — Explain.
Answer:

Benzene attains stability by resonance. If benzene undergoes addition reactions, then it no longer participates in resonance. The aromaticity of benzene is no longer retained due to loss of conjugation.

Consequently, the extra stability of benzene is lost. So, benzene has a lower tendency to undergo an addition reaction. However, during the formation of the substitution product, the aromaticity and stability of benzene remain intact. So, benzene has a higher tendency towards substitution reactions.

Question 31. Benzene burns with a luminous sooty flame but methane bums with a non-luminous flame with no black smoke. Why?
Answer:

Due to the high percentage of carbon, elementary carbon is produced during the burning of benzene. As a result, black smoke is formed. The presence of hot carbon particles in the flame makes the flame luminous. In methane, the percentage of carbon is low so no elementary carbon is produced during the burning of methane. Thus, methane burns with a non-luminous flame with no black smoke.

Question 32. Name the electrophiles that participate in the following reactions: nitration, chlorination, Frieclel-Crafts alkylation, Friedel-Crafts acylation, and sulphonation.
Answer:

Electrophiles which participate in nitration, chlorination, Friedel-Crafts alkylation, Friedel-Crafts acylation, and sulphonation reactions are nitronium ion (⁺NO₂), positively charged chlorine ion (Cl⁺) or chlorine-iron (III) Chloride complex (Cl — Cl⁺— ^–FeCl₃), carbocation (R), B: acylium ion (R⁺CO) and sulfur trioxide (SO₃) respectively.

Question 33. Why iodobenzene cannot be prepared directly by combining benzene and iodine in the presence of iron filings? Why is this reaction possible in the presence of nitric acid?
Answer:

Iodobenzene cannot be directly prepared by combining benzene and iodine because the reaction is reversible. However, in the presence of nitric acid, this reaction becomes possible because nitric acid oxidizes the hydrogen iodide as it is formed and so drives the reaction to the right.

Question 34. Write the structure of the compound 4-(1-isopropyl butyl)-3-propyl undecane.
Answer:

Question 35. Write the structure of 2,2,3-trimethylpentane and label the 1°, 2°, 3° and 4° carbon atoms.
Answer:

Question 36. Write the IUPAC names of two different optically active alkanes with the lowest molecular mass.
Answer:

2 optically active alkanes with the lowest molecular masses-

Question 37. What is the state of hybridization of the quaternary carbon atom present in the neopentane molecule? Write the IUPAC name and structure of the alkane formed by the combination of neopentyl and tert-butyl groups.
Answer:

The state of hybridization of the quaternary carbon atom In a nooporitanu molecule is sp³

The alkane formed by the combination of neopentyl and terf-butyl groups Is—

IUPAC name: 2,2,4,4-tetramethylpentane

Question 38. How many chain isomers will be obtained on replacement of different H-atoms of n-pentane? Write their structures and IUPAC names.
Answer:

Three non-equivalent H-atoms are present in a pentane (CH₃CH₂CH₂CH₂CH₃) molecule. Thus, the replacement of these H-atoms by —CH₃ groups results in three isomeric alkanes. These are as follows —

CH₃CH₂CH₂CH₂CH₂CH₃ (hexane)

CH₃CH₂CH₂CH(CH₃)₂(2-methylpentane)

CH₃CH₂CH(CH₃)CH₂CH₃(3-methylpentane)

Question 39. Write the trivial and IUPAC names of the branched chain alkane with the lowest molecular mass.
Answer:

The trivial name of the branched chain alkane with the lowest molecular mass is isobutane and its IUPAC name is 2-methylpropane.

Question 40. Write the IUPAC name and structure of the alkane having formula C₈H₁₈ and containing a maximum number of methyl groups.
Answer:

The alkane of formula C₈H₁₈ containing the maximum number of methyl groups is

IUPAC name: 2,2,3,3-tetramethylbutane

Question 41. Benzene cannot be used as a solvent in ozonolysis of unsaturated hydrocarbons—why?
Answer:

Benzene itself reacts with ozone to form a triozonide. So, benzene cannot be used as a solvent in ozonolysis of unsaturated hydrocarbons

Question 42. Which one of the following three alkyl halides does not undergo Wurtz reaction and why?

1. CH₃CH₂Br
2. CH₃I
3. (CH₃)₃CB

Answer:

1. The second step of the Wurtz reaction is an S_N2 reaction.
2. An S_N2 reaction is very susceptible to steric effect so, a 3° alkyl halide does not take part in an S_N2 reaction.
3. So, (CH₃)₃CBr being a 3° alkyl halide does not undergo Wurtz reaction

Question 43. Which compound is the strongest acid and why?

1. HC ≡CH
2. C₆H₆
3. C₂H₆
4. CH₃OH

Answer:

CH₃OH is the strongest acid because here the H-atom is bonded to oxygen which is more electronegative titan carbon irrespective of its state of hybridisation

Question 44. Give two equations for the preparation of propane from the Grignard reagent.
Answer:

Question 45. How can CH₃D be prepared from CH₄?
Answer:

Question 46. The preparation of which of the following alkanes by Wurtz reaction is not practicable?

1. (CH₃)₃C-C(CH₃)₃
2. CH₃—CH(CH₃)—CH₂CH₃
3. (CH₃)₂CHCH₂CH₂CH(CH₃)₂

Answer:

1. Not practicable. Although it is a symmetrical alkane, prepared from a carboxylic acid? its preparation requires a f-alkyl halide, (CH₃)₃CX, which does not participate in Wurtz reaction.
2. Not practicable. It is an unsymmetrical alkane.
3. Practicable, because it is a symmetrical alkane

Question 47. When a concentrated aqueous solution of a mixture of sodium salts of two monocarboxylic acids is subjected to electrolysis, ethane, propane, and butane are liberated at the anode. Write the structures and names of the two starting sodium salts
Answer:

Ethane may be produced from two CH₃COONa molecules, propane from 1 CH₃COONa molecule and one CH₃CH₂COONa molecule, and butane (CH₂CH₂CH₂CH₃) from 2CH₃CH₂COONa molecules. Evidentlyethane, propane & butane may be obtained from the mixture of sodium ethanoate (CH₃COONa) & sodium propionate (CH₃CH₂COONa)

Question 48. How can an alkane having one carbon atom less be prepared from a carboxylic acid?
Answer:

Question 49. Arrange the following compounds in the increasing order of their acidic character.

1. H₂O, CH₂=CH₂, NH₃
2. HC≡CH, CH₃OH, CH₃CH₃

Answer:

1. Increasing order of acidic character: CH₂=CH₂ < NH₃ < H₂O [Because, the increasing order of electronegativity: C₂ < N < O ]

2. Increasing order of acidic character: CH₃CH₃ < HC = CH < CH₃OH [increasing order of electronegativity: c ,<C. <ol

Question 50. An alkane (molecular mass 72) produces only one monochloro derivative. Give IIJPAC the name of the compound. Give reasons.
Ana.

As the alkane (C_nH_2n+2) on monochlorination produces only one monochloride derivative, so all of its hydrogens are equivalent. The molecular mass of the alkane is 72, i.e.,12 × n +{2n + 2) = 72, or n = 5. Hence, the alkane contains a carbon atom, l.e., it is one of the isomeric pentanes. The pentane in which all the H-atoms are equivalent is neopentane, (CH₃)₄ C. The IUPAC name of the compound is 2,2-dimethylpropane.

Question 51. 
Answer:

A = CH₃CHO (Acetaldehyde);

B= CHCHOH

(Ethyl alcohol); C = CHCl₃

D = HCOONa (Sodium formate)

Question 52. Acetylene does not react with NaOH or KOH, even though it possesses an acidic character —why?
Answer:

Acetylene (HC = CH) is a weaker acid than water (H₂O) and OH- is a weak base than HC=C^–. As a result, weak acid HC=CH does not react with weak base OH^– to form strong acid H₂O and strong base HC≡C. So acetylene fails to react with NaOH or KOH.

Question 53.  Write the products obtained when propyne ion (CH₃C = ^–C:) is allowed to react with

1. H₂O,
2. CH₃OH
3. H₃ (liquid)
4. 1-hexene and
5. Hexane.

Also mention, where no reaction occurs
Answer:

1. CH₃C = CH + OH^–
2. CH₃C = CH + CH₃O^–
3. No reaction
4. No reaction
5. No reaction.

(Propyne is a weaker acid than H₂O and CH₃OH but it is a stronger acid than, NH3 1-hexene and hexane.]

Question 54. Write the names of the products obtained in the following cases
Answer:

Answer:

1. Br CH₂CH₂ COOH (3-bromopropanoic acid).
2. C₆H₅ COCH₃ (Acetophenone).

Question 55. How can allyl chloride be prepared from 1-propanol?
Answer:

Question 56. trans-pent-2-ene is polar but Frans-but-2-ene is non-polar —why
Answer:

In a trans-but-2-ene molecule, the two C-—CH₃ bond moments, oriented in opposite directions cancel each other and so the net dipole moment in a trans-but-2-ene molecule is zero. On the other hand, in the trans-pent-2-ene molecule, although the C-*-CH₃ and C—C₂H₅ bond moments act in opposite directions, they are not equal in magnitude so they cannot cancel each other. Hence, the molecule possesses a net dipole moment.

Question 57. Convert acetylene into but-2-one.
Answer:

Question 58. Ozonolysis of an alkene leads to the formation of an aldehyde and an isomeric ketone having the molecular formula, C₃H₆O. Identify the alkene.
Answer:

The aldehyde and the ketone having molecular formula, C₃H₆O are CH₃CH₂CHO and CH₃COCH₃ respectively.

Therefore, the starting alkene is 2-methylpent-2-ene. (CH₃)₂C= O + O=CHCH₂CH₃-(CH₃)₂C =CHCH₂CH

Question 59. An alkene having molecular formula, C₄H₈ reacts with HBr to form a tertiary alkyl bromide. Identify the alkene and the alkyl bromide.
Answer:

Question 60. Carry out the following transformation (in two steps): Methyl acetylene →1-bromopropan
Answer:

Question 61. H How can butan-2-one be prepared from acetylene?
Answer:

Question 62. Write the IUPAC names of the products obtained when buta-1,3-diene reacts with bromine in1:1 molar ratio
Answer:

Question 63. Distinguish between buta-1,3-diene and but-1-one.
Answer:

But-1-one (CH₃CH₂C = CH) reacts with ammoniacal cuprous chloride solution to give a red precipitate. Buta-1,3- diene, however, does not respond to this test.

Question 64. Prove that benzene is insoluble in water and is lighter than water.
Answer:

Benzene is taken in a separating flask and water is added to it. The mixture is shaken and then allowed to stand until the two layers are separated. The upper layer contains benzene while water settles at the bottom. This observation proves that benzene is insoluble in water and is lighter than water.

Question 65. Why does benzene burn with a sooty flame?
Answer:

As the carbon content in benzene molecule (C₆H₆, C: H = 1: 1 ) is relatively higher as compared to the saturated hydrocarbon, hexane (C₆H₁₄ C: H = 1:2.3), elementary carbon is formed during burning of benzene. So, benzene bums with a sooty flame

Question 66. Nitration of aniline with 98% H₂SO₄ and cone. [Anilinium sulfate (water soluble)] HNO₃ occurs very slowly and mainly metasubstitution occurs —why?
Answer:

In the presence of 98% H₂SO₄, the — NH₂ group of aniline takes up a proton (H) and is converted to an electron attracting (-1) and meta-directing — NH₃ group. Electron electrophilic attack of the —NO₂ group occurs very slowly. This accounts for the extremely slow rate of nitration of aniline and the formation of the mainly mega-substituted products under strong acidic conditions.

Question 67. Activating groups are ortho-/para- directing, while; the deactivating groups are meta- directing—why?
An8.

The activating groups by exerting their electron-donating +1 and/or +R effects increase electron densities at the ortho and para- positions to a larger extent than the meta-position. So, the electrophiles (E+) enter preferably the ortho- and parapositions. On the other hand, the deactivating groups by exerting their electron-withdrawing -I and I or -R effects decrease electron densities of meta-positions to a lesser extent than the ortho- and para-positions. So the electrophile enters preferably the relatively more electron-rich meta-position.

Question 68. How will you remove traces of aniline present in benzene?
Answer:

If benzene containing traces of aniline (basic) is shaken with a cone. H₂SO₄ then aniline dissolves in the acid forming a salt.- The acid layer is thus removed. In this way, traces of aniline present in benzene can be removed

Question 69. How can traces of phenol present in a sample of benzene be removed?
Answer:

If the sample of benzene containing traces of phenol (acidic) is shaken with 10% NaOH solution then phenol reacts with NaOH to form sodium phenoxide which dissolves in the aqueous layer. The aqueous layer is then removed. In this way, traces of phenol present in a sample of benzene can be removed.

Question 70. How will you distinguish between benzene and hexane by a simple test in the laboratory?
Answer:

The percentage of carbon in benzene is much higher than the corresponding alkane, hexane. So benzene bums with the formation of elementary carbon and as a result, a sooty flame is produced. During the burning of hexane no such sooty flame is produced. Thus, by observing the nature of the flame produced, the two compounds can be easily distinguished.

Question 71. Aniline does not undergo Friedel-Crafts reaction, though it contains an electron-donating group—why?
Answer:

Aniline (Lewis base) combines with AlCl₃ (Lewis acid) by donating the unshared electron pair on nitrogen, to form a complex. As a result, the N-atom of the — NH₂ group acquires a positive charge and this, being converted into an electron attracting group, decreases the electron density of the ring to such an extent that the Friedel-Crafts reaction does not occur.

Question 72.  How much trisubstituted benzene may be obtained from o-, m-, and p-chlorotoluene and why?
Answer:

Each of o – and m -chlorotoluene will give 4 trisubstituted benzenes, while p -chlorotoluene will give two trisubstituted benzenes. This is because there are 4 types of non-equivalent hydrogens (or positions) in each of the ortho- and meta¬ isomers and two types of non-equivalent hydrogens (or positions) in the para-isomer

Question 73. C₅H₁₂ and C₈H₁₈ are two alkanes that form one monochloride each reacting with chlorine. Write the names of the alkanes and structural formulas of the chlorides.
Answer:

As both the alkanes form one monochloride, it can be said that all H-atoms in these two alkanes are equivalent. Therefore, the alkane with molecular formula C₆H₁₂ is (CH₃)₄C and the alkane with molecular formula C₈H₁₈ is (CH₃)₃C-C(CH₃)₃

Question 74. How can you convert methyl acetylene to acetone?
Answer:

At 60-80°C temperature, when methyl acetylene or propyne is passed through a dilute (20%) solution of sulphuric acid containing 1% H₂SO₄, it combines with one molecule of water to form the unstable compound, 2-propanol. Addition of water molecule to unsymmetrical alkyne (propyne in this case) through Markownikoff’s rule. 2-propenol! Rapidly tautomerizes to acetone

Question 75. Mention two reactions of benzene to show its behavior is different from that of the open-chain unsaturated hydrocarbons.
Answer:

1. Benzene does not decolorise bromine water.
2. Benzene does not react with halogen acids such as HCl, HBr, etc

Question 76. Is it possible to isolate pure staggered ethane or pure eclipsed ethane at room temperature? Explain.
Answer:

The energy difference between eclipsed and staggered conformations of ethane is only 2.8kcal .mol^-1, which is easily achieved by collisions among the molecules at room temperature. So it is not possible to isolate either pure staggered or pure eclipsed form of ethane at room temperature.

Question 77. Explain why rotation about carbon-carbon double bond is hindered?.
Answer:

Carbon-carbon double bond consists of a σ -and a π bond. The π-bond is formed by lateral overlap of unhybridised p -p-orbitals of two carbon atoms above and below the plane of the carbon atoms. If an attempt is made to rotate one of the atoms of the double bond concerning the other, the p orbitals will no longer overlap, thereby causing the fission of the n bond. Since breaking of a π -bond requires a considerable amount of energy, which is not available at room temperature, rotation about the carbon-carbon double bond is hindered.

Question 78. Arrange the following carbanions in order of their Hg+ decreasing stability

Answer:

Due to greater s -character sp -hybridized carbon is more electronegative than sp³ -hybridized carbon and hence can accommodate the -ve charge more effectively. So 1 and 2 are more stable than 3. Again —CH group has an electron-donating +1 effect, therefore it interacts with the -ve charge on carbanion carbon and hedestabilizesises 1 relative to 2. Thus, the stability of the given carbanions decreases in the sequence: B> 1> 3

Question 79. Explain why Tert butylbenzene cannot be oxidized to benzoic acid by treatment with alkaline KMnO₄
Answer:

Alkylbenzenes can be oxidized to benzoic acid provided that the side chain contains one benzylic {hydrogen atom. Since tert-butylbenzene does not contain any benzylic hydrogen, so the alkyl chain cannot be oxidized to the — COOH group.

Question 80. Explain why HF forms hydrogen bonding with acetylene even though it is non-polar
Answer:

Due to the sp -hybridization of carbon, the electrons of the C —H bond of acetylene are attracted considerably toward carbon. Consequently, each carbon carries a partial negative charge and each hydrogen carries a partial positive charge. Owing to the presence of a partial positive charge on hydrogen, acetylene forms an H -bond with the F -atom of the HF molecule

⇒ \(\begin{array}{r}
\delta+\delta-\delta-\delta+\delta-\delta+ \\
\cdots \mathrm{H}-\mathrm{C} \equiv \stackrel{\delta}{\mathrm{C}}-\mathrm{H} \cdots \mathrm{F}-\mathrm{H}
\end{array}\)

Question 81. One mole ofa symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44u. Identify the alkene
Answer:

Let, the aldehyde be C_nH_2n+1CHO

The molecular mass of this aldehyde

=[12n + (2n +1) + (12 +1 + 16)]u

= (14n + 30)u

Thus, 14n + 30 = 44

n = 1

So the aldehyde is CH₃CHO.

The alkene is CH₃CH=CH— CH..

Question 82. The addition of HBr to 1 -butene gives a mixture of mechanisms: A and B as the main products together with a small amount of another compound C. Identify A, B, and C
Answer:

Question 83. An alkene, C₆H₁₂, reacts with HBr in the absence as well as in the presence of peroxide to give the same product. Find its structure.
Answer:

Symmetrical alkenes react with HBr in the presence or absence of peroxide to give the same product. Hence the given alkene may have the structure (1) or (2)

Question 84. How will you prepare a sample of propane free from other alkanes using ethyl bromide, methyl bromide, and diethyl ether as the organic compounds, together? The product‘T is iodobenzene. With other inorganic materials
Answer:

It can be prepared by Corey-House synthesis:

Question 85. The catalytic hydrogenation of which of the following is most exothermic?

Answer:

The least substituted alkene, having the lowest number of hyperconjugative structures, has the least thermodynamic stability (i.e., highest energy content) and so it has the highest heat of hydrogenation. Now, out of the given compounds, (C) is the least substituted alkene and so it has the highest heat of hydrogenation.

Question 86. Identify A, B, C & D in the following reaction—

Answer:

1. H —C≡C —H
2.  H —C ≡ CNa
3. H—C ≡ C—CH₃
4. H₂C=CH—CH₃

Question 87. Identify A, B, C Si D in the following reaction

Answer:

1.  CH₂=CH₂
2. CH= CH
3. CH₃CHO
4. CH₃—CH₃

Question 88. What organic compound is obtained when

1. Ethyl iodide is subjected to react with Zn -Cu couple/aqueous ethanol and 
2. Iodoform is heated with Ag powder?

Answer:

1. CH₃CH₃
2. C₂H₂
   

Question 89. Write structures of A and B in the following reactions.

Question 90.  Which one of these two reactions can be used for the identification of ethylenic unsaturation? Why? ½ + ½+½ + 1
Answer:

R-CH(Br)—CH₃

R—CH(Br) —CH₂Br

Question 91.  A hydrocarbon (A) is obtained when 1,2-dibromoethane reacts with alcoholic KOH. (A) decolourises alkaline KMnO₄ solution. (A) contains acidic hydrogen. Identify (A) with reasons.
Answer:

Question 92. Writes structures of A, B, and C 

Answer:

Question 93. How will you convert?

1. HC = CH→H₃C-CH₂-CH₃

Answer:

Question 94. An organic compound (A) composed of C and H contains 85.71 %C. It shows M⁺ at mlz = 42 in the mass spectrum. The compound reacts with HBr in the absence of peroxide to yield an organic compound (B) which is isomeric with the compound (C) obtained when the compound reacts with HBr in the presence of peroxide. Identify A, B, and C.
Answer:

⇒ \(C: H=\frac{85.71}{12}: \frac{14.29}{1}=7.14: 14.29=1: 2\)

Emperical formula = CH₂, Molar formula = (CH₂)_n

According to the problem, n × (12 + 2) = 42.

n = 3. So, the actual molecular formula is C₃H₆. DBE = 1. So, probable structures of CH₆ are—

Structure A is accepted as it can undergo given reactions:

Question 95.

1. Write equations of all the steps of the reaction of methane with chlorine in the presence of diffused sunlight

2. Identify A and B

Answer:

⇒ \(\mathrm{A} \Rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OSO}_3 \mathrm{H}, \mathrm{B} \Rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}\)

Question 18. Convert benzene into aniline by using the following reagents in the correct order: alkaline KMnO₄ and then HCl NH₃, heat; Br₂/KOH; CH₃Cl /anhydrous AlCl₃.
Answer:

Question 96. Write structures of the organic products obtained in the following reactions :

Answer:

Question 97. Identify (X) and (Y) in the following reactions:

Answer:

Question 98. Identify(M) and (N)in the following reactions

Answer:

Question 99. Write the product of the following reaction:

Answer:

Question 100. Benzene in reaction with NOCl in the presence of acid produces an organic compound (1).

1. On treatment with NaNH₂/Liq.NH₃ furnishes another organic compound (15).
2. Treatment with NaNO₂/HBF₄ affords an organic compound
3. Which on heating gives an organic compound.
4. Identify(1), (2), (3) and (4).

Answer:

Question 101. Two different compounds produce only acetaldehyde on ozonolysis. Draw the structures of the two compounds.
Answer:

Question 102. Write the name and the structural formula of the product obtained when hydrogen bromide reacts with propene in the presence of benzoyl peroxide.
Answer:

Question 103. Identify the compound in the following reaction

Answer:

Identify the compound A in the following reaction:

Question 104.

1. Among benzene and toluene, which one will undergo nitration reaction easily and why?

2. Identify A, B, C and D

Answer:

If the electron density of the benzene ring increases, then the reactivity of the ring towards electrophilic substitution also increases. In toluene, the —CH₃ group increases the electron density of the ring and as a result, the reactivity of the ring also increases due to +1 and the hyperconjugation effect of the —CH₃ group. So, nitration occurs more easily for toluene than unsubstituted benzene.

Question 105. Write the name and structural formula of A in the following reaction.

Answer:

Question 106. What happens when What happens ether?
Answer:

Question 107. Two isomeric compounds A and B have the molecular formula C₃H.Br forms the same compound C on dehydrobromination. C on ozonolysis produces acetaldehyde and formaldehyde. Identify A, B, and C
Answer:

⇒ \(A \Rightarrow \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}\)

⇒ \(B \Rightarrow \mathrm{CH}_3 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3\)

⇒ \(C \Rightarrow \mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2\)

Question 108. An alkene A on ozonolysis gives a mixture of ethanal & pentan-3-one. Write the structure & IUPAC name of A.
Answer:

Question 109. An alkene ‘A’ contains three C —C, eight C —H crbonds and one C — C π -bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write the IUPAC name of ‘A’.
Answer:

An aldehyde with molar mass 44u is ethanal (CH₃CH=:O). The formula of the alkene ‘A’ which on ozonolysis gives two moles of ethanol can be determined as follows—

⇒ \(\underset{\text { Ethanal }}{\mathrm{CH}_3 \mathrm{CH}}=\underset{\text { Ethanal }}{\mathrm{O}}+\underset{\text { But-2-ene (A) }}{\mathrm{O}}=\underset{\mathrm{CHCH}_3}{\mathrm{CH}_3-\mathrm{CH}}=\underset{\mathrm{CH}}{\mathrm{CH}}-\mathrm{CH}_3\)

There are three C — C cr -bonds, eight C —H tr -bonds, and one C —C 7t -bond in but-2-one.

Question 110. Propanal and pentane-3-one are the ozonolysis products of an alkene. What Is the structural formula of the alkene?
Answer:

Question 111. Write chemical equations of the combustion reaction of the following hydrocarbons: Butane Pantene Hexyne Toluene
Answer:

Question 112. Draw The Cis- and trans-structure of hex- 2 ene which isomer will have higher B>P and Why?
Answer:

The general formula of hex-2-ene is CH₃—CH₂—CH₂—CH=CH—CH₃. Structural formulas of cis-and trans-isomers of this compound are-

The ds-isomer being more polar than the trans-isomer has a higher value of dipole moment than that of the trans-isomer. Intermolecular dipole-dipole interaction in the case of cis-isomers is more than that in trans-isomers. So, the boiling point of the customer is higher.

Question 113. Why is benzene extraordinarily stable though it contains three double bonds?
Answer:

There are (4n + 2) delocalized 7r -electrons (n = 1) in the planar benzene molecule.

Consequently, it attains stability due to aromaticity. So, benzene is extraordinarily stable despite having three double bonds.

Question 114. Out of benzene, m-dinitrobenzene, and toluene which will undergo nitration most easily, and why
Answer:

Nitration of the benzene ring is an electrophilic substitution reaction. In this reaction, the presence of an activating group ( —CH₃) increases the reactivity of the benzene ring, whereas the presence of a deactivating group (—NO₂) decreases the reactivity of the benzene ring. Therefore, order of nitration is toluene > benzene > m-dinitrobenzene.

Question 115. Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Answer:

Ethylation of benzene means the introduction of an ethyl group in the benzene ring. This reaction is carried out by Friedel-Crafts reaction of benzene with ethyl halide (chloride or bromide), ethene, or ethanol. Lewis acids, other than anhydrous AlCl₃, that can be used in this reaction are anhydrous FeCl₃, SnCl₄, BF₃, HF, etc.

Question 116. Why does an iline not participate in the Friedel-Crafts reaction?
Answer:

Aniline reacts with AlCl₃ complexing \(\mathrm{C}_6 \mathrm{H}_5-\stackrel{\oplus}{\mathrm{N}} \mathrm{H}_2-\stackrel{\ominus}{\mathrm{AlCl}} \mathrm{Cl}_3\) which makes —NH₂ group electron-withdrawing nature. Consequently, the benzene ring becomes highly deactivated so aniline does not participate in the Friedel-Crafts reaction;

Question 117. The reaction of CH₂=CH—N(CH₃)₃I takes place contrary to Markownikoff’s rule—why?
Answer: 

The carbocation formed due to the addition of H+ to the carbon atom containing a higher number of hydrogen atoms becomes unstable because of the electron-attracting —NMe₃ group.

So, the reaction takes place contrary’ to Markownikoff’s rule;

Question 118. Mention the limitations of the Wurtzreaction.
Answer:

Limitations:

• Tertiary alkyl halides do not respond to this reaction,
• Methane cannot be prepared by this reaction and
• Preparation of unsymmetrical alkanes cannot be done by this method;

Question 119. What product is formed when the given compound reacts with HBr and why?
Answer: The product obtained

 is because the carbocation formed in the first step

Is, resonance stabilized due to —OCH₃ group

NCERT Solutions For Class 11 Chemistry Chapter 13 Hydrocarbons  Warm-Up Exercise Question And Answers

Question 1. What are the chief constituents of LPG?
Answer: The chief constituents of LPG are n-butane and isobutane.

Question 2. Why do C—C bonds instead of C—H bonds of alkanes dissociate due to the effect of heat?
Answer: The bond energy of the C— C bond (ΔH = 83 kcal. mol^-1) is less than that of the C—H bond (ΔH= 99 kcal. mol^-1). So, the C—C bond dissociates more easily than the C—H bond.

Question 3. Write the IUPAC name of the straight-chain hydrocarbon consisting of 20 carbon atoms.
Answer: IUPAC’s name of the straight-chain hydrocarbon consisting of 20 carbon atoms is eicosane.

Question 4. Give the structures of the isomers of molecular formula C₅H₁₂–
Answer: CH₃CH₂CH₂CH₂CH₃ (n -pentane) CH₃CH(CH₃)CH,CH₃ (isopentane) and (CH₃)₄C (neopentane)

Question 5. Explain why dry ether is used in the Wurtzreaction.
Answer:

Dry ether is used because it is present in ether, then it may react with metallic sodium thereby rendering it ineffective

2Na + 2H₂O→2NaOH + H₂

Question 6. Predict whether Me3CBr will take part in the Wurtz reaction or not
Answer:  Wurtz reaction proceeds through the S_N2 pathway. As tertiary alkyl halides do not participate in the S_N2 reaction (due to steric effect), Me₃CBr does not participate in the Wurtz reaction

Question 7. Explain why methane does not react with chlorine in the dark.
Answer:  The reaction does not take place because in the dark Cl —Cl bond does not dissociate to form Cl free radical;

Question 8. One molecule of a hydrocarbon produces one molecule each of acetone, methyl glyoxal, and formaldehyde on ozonolysis. Identify the hydrocarbon.
Answer:

The hydrocarbon is 3, 4-dimethylpenta-l, 3-diene [CH₃—C(CH₃)=C(CH₃)—CH=CH₂] or, 2,4-dimethylpenta-1,3-diene [CH₂=C(CH₃)—CH=C(CH₃) —CH₃];

Question 9. Explain why 1-butyne reacts with ammoniacal silver nitrate to produce a white precipitate, but 2-butyne does not
Answer:

1-butyne (CH₃CH₂C=CH) being a terminal alkyne reacts with ammoniacal AgNO₃  solution to produce a white precipitate but 2-butyne (CH₃C=CCH₃) being a non-terminal alkyne does not react with ammoniacal AgNO₃ solution;

Question 10. How will you detect the presence of acetylene in a gas mixture?
Answer:

If the gas mixture when passed through ammoniacal AgNO₃ solution or ammoniacal Cu₂Cl₂ solution forms a white or red precipitate, then the gas mixture contains acetylene.

Question 11. Explain why the carbon-carbon bond in acetylene is shorter than the carbon-carbon bond in ethylene.
Answer:

Cr -bond in acetylene (HC=CH) is formed due to the overlapping of two small sp-hybridized orbitals whereas in ethylene (H₂C=CH₂) it is formed by overlapping of two bigger sp² hybridized orbitals. So, the bond length of HC=CH <H₂C=CH₂;

Question 12. How will you distinguish between ethylene and acetylene?
Answer:

Acetylene reacts with ammoniacal AgNO₂ solution to form a white precipitate of silver acetylide (AgC=CAg) but ethylene does not give a similar reaction with ammoniacal AgNO₃ solution.

Question 13. The population of which conformation increases with the temperature rise?
Answer: The population of the less stable conformation Increases with the increase in temperature.

Question 14. What are the carbides that react with water to form methane commonly known as?
Answer: The carbides which react with water to form methane are commonly known as methanldes.

Question 15.

Question 16. Why are hydrocarbons insoluble in water but highly soluble in solvents like petroleum ether, benzene, carbon tetrachloride, etc?
Answer:

An important principle regarding dissolution is ‘like dissolves like’. It means that polar molecules dissolve in polar solvents while non-polar molecules dissolve in nonpolar solvents. This dissolution process is thermodynamically favorable. Water is a highly polar solvent whereas, petroleum, ether, benzene, and carbon tetrachloride are non-polar solvents. As hydrocarbons are non-polar compounds, they are insoluble in water but soluble in petroleum ether, benzene, and carbon tetrachloride.

Question 17. Why are the alkanes called paraffins?
Answer: Alkanes are called paraffin as their chemical reactivity is quite low (Latin: parum = little, affinis = affinity).

Question 18. What are the typical reactions of alkanes?
Answer: Typical reactions of alkanes are substitution reactions.

Question 19. Mention the type of mechanism through which halogenation of alkanes occurs.
Answer: Free-radical mechanism.

Question 20. What happens when methane is heated at 1000°C in the absence of air?
Answer:

Methane when heated at 1000°C in the absence of air, decomposes to form a fine powder of carbon which is known as carbon black:

⇒ \(\mathrm{CH}_4 →{1000^{\circ} \mathrm{C}} \mathrm{C}+2 \mathrm{H}_2 \uparrow\)

Question 21. What is the main constituent of natural gas which is used as a fuel?
Answer: The main constituent of natural gas which is used as a fuel is methane (90%).

Question 22. Why is light or heat essential for the chlorination of alkanes?
Answer:

Cl free radical is required for the initiation of the reaction between an alkane and chlorine, i.e., homolysis of the Cl—Cl bond is necessary. The energy required for this hemolysis is derived from light or heat. So, light or heat is essential for the chlorination of alkanes.

Question 23. Which gas is responsible for explosions in coal mines?
Answer: Methane is responsible for explosions in coal mines.

Question 24. Write the IUPAC name of freon – 113.
Answer: IUPAC name off neon-113, i.e., Cl₂FC— CClF₂ is 1,1,2-trichloro-1,2,2-trifluoroethane

Question 25. Which reaction helps locate the position of the double bond in alkenes?
Answer:

The reaction which helps locate the position of the double bonds in alkenes is ozonolysis.

Question 26. An alkene (C₄Hg) reacts with HBr in the presence or in the absence of peroxide to give the same compound. Identify the alkene.
Answer:

As tire given alkene (molecular formula: C₄HO) reacts with HBr to give the same product in the presence and absence of peroxide, the alkene is symmetrical. So, a symmetrical alkene with molecular formula C₄H₈ is but-2-ene (CH₃CH=CHCH₃).

Question 27. Calculate the number of sigma (or) and pi (n) bonds in methyl acetylene.
Answer:

In methyl acetylene (CH₃-C CH), there are 6 a-bonds and 2 bonds.

Question 28. Which of the following compounds will react with metallic sodium to produce H₂ gas?

1. C₂H₄
2. C₆H₆
3. C₂H₂
4. CH₃CH₂CH₃

Answer: 3. Acetylene (C₂H₂) reacts with metallic sodium to produce H2 gas:

⇒ \(\mathrm{HC} \equiv \mathrm{CH}+2 \mathrm{Na} \rightarrow \mathrm{NaC} \equiv \mathrm{CNa}+\mathrm{H}_2 \uparrow\)

Question 29. The C₂ — C₃ bond 1,3-butadiene possesses some double bond characteristics.
Answer:

The C₂—C₃ bond in 1,3-butadiene possesses some double bond character because of the delocalization of n -electrons.

Question 30. An arena when oxidised forms 1,3-dicarboxylic acid. Write the numbers of side chains and their position in the arena.
Answer:

As the arena gets oxidized to a dicarboxylic acid, it has two side chains. It can be said that the two side chains are at 1,3- or meta-position of each other because a 1,3-dicarboxylic acid forms in the oxidation.

Question 32. Distinguish between benzene and toluene with the help of a chemical reaction.
Answer:

Toluene on oxidation by alkaline KMn04 and subsequent acidification produces shining white crystals of benzoic acid. Benzene, on the other hand, does not undergo oxidation with alkaline KMnO₄ to form any white precipitate.

Question 33. Between — NH₂ and —NO₂, which group facilitates nucleophilic substitution reaction in the benzene ring?
Answer:

The group that facilitates nucleophilic substitution reaction in the benzene ring is — NO₂ because it decreases the electron density of the benzene ring.

Question 34. Arrange in order of increasing reactivity towards electrophilic substitution: benzene, nitrobenzene, toluene, chlorobenzene.
Answer:

The order of increasing reactivity towards electrophilic substitution of the compounds is :

Nitrobenzene< Chloro¬ benzene < Benzene < Toluene.

Question 35. Name the halogen carrier in the chlorination of benzene.
Answer: The compound that acts as the halogen carrier in the chlorination of benzene is either AlCl₃ or FeCl₃.

Question 36. Benzene undergoes de-sulphonation but not denitration. Why?
Answer:

Since sulphonation is a reversible reaction, benzene can undergo a desulphonation reaction. However, nitration is an irreversible reaction. So, benzene cannot undergo a nitration reaction.

Question 37. If the calculated and the experimental heats of combustion of benzene are 824.1 and 789.1 kcal mol^-1 respectively, then calculate the value of resonance energy of benzene.
Answer:

Resonance energy = calculated heat of combustion experimental heat of combustion = (824.1 – 789.1)kcal. mol^-1= 35 kcal .mol^-1