CLASS 11 · CHEMISTRY · CHAPTER 9

Hydrocarbons

Simple, point-wise answers to all 14 numbered in-text problems and all 25 end-of-chapter exercises.

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In-text Problems

Textbook pages 297–321

Question Problem 9.1

List the chain isomers of C₆H₁₄ and give their IUPAC names.

Steps only

n-Hexane; 2-methylpentane; 3-methylpentane; 2,2-dimethylbutane; 2,3-dimethylbutane.

Answer

n-Hexane; 2-methylpentane; 3-methylpentane; 2,2-dimethylbutane; 2,3-dimethylbutane.

Question Problem 9.2

List the eight different C₅H₁₁ alkyl groups and give the IUPAC name of the alcohol formed when –OH replaces the marked bond.

Steps only

The eight distinct C₅H₁₁ attachment groups give: pentan-1-, pentan-2-, pentan-3-, 3-methylbutan-1-, 2-methylbutan-1-, 2-methylbutan-2-, 3-methylbutan-2- and 2,2-dimethylpropan-1-ol.

Answer

Eight alcohols: pentan-1-ol, pentan-2-ol, pentan-3-ol, 3-methylbutan-1-ol, 2-methylbutan-1-ol, 2-methylbutan-2-ol, 3-methylbutan-2-ol and 2,2-dimethylpropan-1-ol.

Question Problem 9.3

Give IUPAC names for: (i) (CH₃)₃CCH₂C(CH₃)₃; (ii) (CH₃)₂C(C₂H₅)₂; (iii) tetra-tert-butylmethane.

Steps only

(i) 2,2,4,4-tetramethylpentane; (ii) 3,3-dimethylpentane; (iii) 3,3-di-tert-butyl-2,2,4,4-tetramethylpentane.

Answer

(i) 2,2,4,4-tetramethylpentane;
(ii) 3,3-dimethylpentane;
(iii) 3,3-di-tert-butyl-2,2,4,4-tetramethylpentane.

Question Problem 9.4

Write condensed structures for (i) 3,4,4,5-tetramethylheptane and (ii) 2,5-dimethylhexane.

Steps only

(i) CH₃CH₂CH(CH₃)C(CH₃)₂CH(CH₃)CH₂CH₃. (ii) CH₃CH(CH₃)CH₂CH₂CH(CH₃)CH₃.

Answer

(i) CH₃CH₂CH(CH₃)C(CH₃)₂CH(CH₃)CH₂CH₃.
(ii) CH₃CH(CH₃)CH₂CH₂CH(CH₃)CH₃.

Question Problem 9.5

For “2-ethylpentane” and “5-ethyl-3-methylheptane”, give structures and corrected IUPAC names.

Steps only

(i) CH₃CH₂CH(CH₃)CH₂CH₂CH₃; 3-methylhexane. (ii) Keep the same heptane skeleton; renumber to give ethyl locant 3 and methyl locant 5: 3-ethyl-5-methylheptane.

Answer

(i) CH₃CH₂CH(CH₃)CH₂CH₂CH₃; 3-methylhexane.
(ii) Keep the same heptane skeleton; renumber to give ethyl locant 3 and methyl locant 5: 3-ethyl-5-methylheptane.

Question Problem 9.6

Which sodium carboxylate is needed to prepare propane by heating with soda lime? Write the reaction.

Steps only

Sodium butanoate: CH₃CH₂CH₂COONa + NaOH —(CaO, Δ)→ CH₃CH₂CH₃ + Na₂CO₃.

Answer

Sodium butanoate: CH₃CH₂CH₂COONa + NaOH —(CaO, Δ)→ CH₃CH₂CH₃ + Na₂CO₃.

Question Problem 9.7

Give IUPAC names for the four alkene structures shown in the textbook problem.

Steps only

(i) 2,8-dimethyldeca-3,6-diene; (ii) octa-1,3,5,7-tetraene; (iii) 2-propylpent-1-ene; (iv) 4-ethyl-2,6-dimethyldec-4-ene.

Answer

(i) 2,8-dimethyldeca-3,6-diene;
(ii) octa-1,3,5,7-tetraene;
(iii) 2-propylpent-1-ene;
(iv) 4-ethyl-2,6-dimethyldec-4-ene.

Question Problem 9.8

Count the σ and π bonds in the four structures from Problem 9.7.

Steps only

(i) 33 σ, 2 π; (ii) 17 σ, 4 π; (iii) 23 σ, 1 π; (iv) 41 σ, 1 π.

Answer

(i) 33 σ, 2 π;
(ii) 17 σ, 4 π;
(iii) 23 σ, 1 π;
(iv) 41 σ, 1 π.

Question Problem 9.9

Write the open-chain alkene structural isomers of C₅H₁₀ and name them.

Steps only

Pent-1-ene (CH₂=CHCH₂CH₂CH₃); pent-2-ene (CH₃CH=CHCH₂CH₃); 2-methylbut-1-ene (CH₂=C(CH₃)CH₂CH₃); 3-methylbut-1-ene (CH₂=CHCH(CH₃)CH₃); 2-methylbut-2-ene (CH₃C(CH₃)=CHCH₃).

Answer

Question Problem 9.10

Draw the cis and trans forms of (i) CHCl=CHCl and (ii) C₂H₅C(CH₃)=C(CH₃)C₂H₅, and name them.

Steps only

(i) cis- and trans-1,2-dichloroethene. (ii) cis- and trans-3,4-dimethylhex-3-ene.

Answer

(i) cis- and trans-1,2-dichloroethene.
(ii) cis- and trans-3,4-dimethylhex-3-ene.

Question Problem 9.11

Which of these alkenes show cis-trans isomerism: (i) (CH₃)₂C=CHC₂H₅; (ii) CH₂=CBr₂; (iii) C₆H₅CH=CHCH₃; (iv) CH₃CH=C(Cl)CH₃?

Steps only

Only (iii) and (iv).

Answer

Only (iii) and (iv).

Question Problem 9.12

Name the main products when HBr adds to hex-1-ene (i) without peroxide and (ii) with peroxide.

Steps only

(i) 2-bromohexane. (ii) 1-bromohexane.

Answer

(i) 2-bromohexane.
(ii) 1-bromohexane.

Question Problem 9.13

List the structural isomers of the fifth alkyne, C₆H₁₀, and identify the types of isomerism.

Steps only

Hex-1-yne, hex-2-yne, hex-3-yne, 3-methylpent-1-yne, 4-methylpent-1-yne, 4-methylpent-2-yne and 3,3-dimethylbut-1-yne. They show position and chain isomerism.

Answer

Hex-1-yne, hex-2-yne, hex-3-yne, 3-methylpent-1-yne, 4-methylpent-1-yne, 4-methylpent-2-yne and 3,3-dimethylbut-1-yne. They show position and chain isomerism.

Question Problem 9.14

Show a reaction route to convert ethanoic acid into benzene.

Steps only

CH₃COOH → CH₃COONa; CH₃COONa + NaOH —(CaO, Δ)→ CH₄ + Na₂CO₃; 2CH₄ —(≈1500 K)→ C₂H₂ + 3H₂; 3C₂H₂ —(red-hot Fe, 873 K)→ C₆H₆.

Answer

CH₃COOH → CH₃COONa; CH₃COONa + NaOH —(CaO, Δ)→ CH₄ + Na₂CO₃; 2CH₄ —(≈1500 K)→ C₂H₂ + 3H₂; 3C₂H₂ —(red-hot Fe, 873 K)→ C₆H₆.

Chapter 9 Exercises

Textbook pages 252–257

Question 9.1

Why does chlorination of methane produce ethane?

Steps only

Chlorine radicals help form CH₃•; two CH₃• radicals join.

Answer

CH₃• + •CH₃ → C₂H₆.

Question 9.2

Give the IUPAC names of all seven structures (a)–(g), including the bond-line and ring structures shown in the textbook.

Steps only

(a) 2-Methylbut-2-ene. (b) Pent-1-en-3-yne. (c) Penta-1,4-diene. (d) 4-Phenylbut-1-ene. (e) 2-Methylphenol. (f) 5-(2-Methylpropyl)decane. (g) 4-Ethylnona-1,5,8-triene.

Answer

(a) 2-Methylbut-2-ene. (b) Pent-1-en-3-yne. (c) Penta-1,4-diene. (d) 4-Phenylbut-1-ene. (e) 2-Methylphenol. (f) 5-(2-Methylpropyl)decane. (g) 4-Ethylnona-1,5,8-triene.

Question 9.3

Write the structures of all isomers of (a) C₄H₈ with one double bond and (b) C₅H₈ with one triple bond.

Steps only

Move the multiple bond along each valid chain and include branched chains.

Answer

Three constitutional alkenes for C₄H₈; three constitutional alkynes for C₅H₈ (structures listed above).

Question 9.4

Write the products formed on ozonolysis of (a) pent-2-ene, (b) 3,4-dimethylhept-3-ene, (c) 2-ethylbut-1-ene and (d) 1-phenylbut-1-ene.

Steps only

Cut at C=C and turn each of its carbon atoms into C=O.

Answer

(a) ethanal + propanal; (b) butan-2-one + pentan-2-one; (c) methanal + pentan-3-one; (d) benzaldehyde + propanal.

Question 9.5

An alkene gives ethanal and pentan-3-one on ozonolysis. Identify the alkene.

Steps only

Join the carbonyl carbons from the two products using C=C.

Answer

3-Ethylpent-2-ene: CH₃CH=C(CH₂CH₃)₂.

Question 9.6

An alkene has three C–C σ bonds and eight C–H σ bonds, and its ozonolysis gives two moles of an aldehyde of molar mass 44 g mol⁻¹. Find its structure.

Steps only

Two ethanal molecules point to identical CH₃–CH= groups on either side of C=C.

Answer

But-2-ene, CH₃CH=CHCH₃.

Question 9.7

An alkene gives propanal and pentan-3-one on ozonolysis. Identify the alkene.

Steps only

Reconnect the two carbonyl carbons with a double bond.

Answer

3-Ethylhex-3-ene.

Question 9.8

Write balanced equations for the complete combustion of butane, pentene, hexyne and toluene.

Steps only

Balance C and H to make CO₂ and H₂O; then balance O₂.

Answer

Balanced equations are given for all four hydrocarbons.

Question 9.9

Draw cis- and trans-hex-2-ene. Which has the higher boiling point?

Steps only

Same side = cis; opposite sides = trans. Greater polarity raises the boiling point.

Answer

Cis-hex-2-ene has the higher boiling point.

Question 9.10

Benzene has three double bonds in its Kekulé structures. Why is it much more stable than a typical alkene?

Steps only

Six π electrons are shared around the entire flat ring.

Answer

Delocalisation gives benzene extra stability.

Question 9.11

State the conditions a molecule must meet to be aromatic.

Steps only

Check for a planar conjugated ring and count π electrons using 4n+2.

Answer

Cyclic + planar + conjugated + (4n+2) π electrons.

Question 9.12

Why are the three ring systems shown in the exercise not aromatic?

Steps only

(i) and (ii) are non-aromatic because an sp³ ring carbon interrupts cyclic conjugation. (iii) is non-aromatic because cyclooctatetraene is non-planar and has 8 π electrons, not a 4n+2 count.

Answer

(i) and
(ii) are non-aromatic because an sp³ ring carbon interrupts cyclic conjugation.
(iii) is non-aromatic because cyclooctatetraene is non-planar and has 8 π electrons, not a 4n+2 count.

Question 9.13

Show how to prepare (a) p-nitrobromobenzene, (b) m-nitrochlorobenzene, (c) p-nitrotoluene and (d) acetophenone from benzene.

Steps only

Install an ortho/para director before nitration for para products; nitrate first for a meta product.

Answer

(a) Br then NO₂; (b) NO₂ then Cl; (c) CH₃ then NO₂; (d) CH₃COCl/AlCl₃.

Question 9.14

For CH₃–CH₂–C(CH₃)₂–CH₂–CH(CH₃)₂, count the 1°, 2° and 3° carbon atoms and their attached hydrogens.

Steps only

Count carbon neighbours: one, two, three or four.

Answer

1°: 5 carbons/15 H; 2°: 2 carbons/4 H; 3°: 1 carbon/1 H; 4°: 1 carbon/0 H.

Question 9.15

How does branching affect the boiling point of an alkane?

Steps only

More branching → smaller contact area → weaker attractions.

Answer

More branching usually lowers the boiling point.

Question 9.16

What happens when HBr is added to propene with and without benzoyl peroxide? Give the mechanisms.

Steps only

No peroxide: Markovnikov addition. Peroxide: radical anti-Markovnikov addition.

Answer

Without peroxide gives 2-bromopropane; with peroxide gives 1-bromopropane.

Question 9.17

What products form on ozonolysis of o-xylene? How does this support the Kekulé structures of benzene?

Steps only

Break the three C=C bonds in each Kekulé form and compare the resulting carbonyl fragments.

Answer

Ozonolysis gives carbonyl fragments predicted by both Kekulé contributors, supporting resonance.

Question 9.18

Arrange benzene, n-hexane and ethyne in increasing order of acidity.

Steps only

Compare C–H acidity: sp³ < sp² < sp.

Answer

n-Hexane < benzene < ethyne.

Question 9.19

Why does benzene undergo electrophilic substitution more readily than nucleophilic substitution?

Steps only

The ring is electron-rich and substitution preserves aromaticity.

Answer

Benzene favours electrophilic substitution.

Question 9.20

How can benzene be prepared from (a) ethyne, (b) ethene and (c) hexane?

Steps only

Ethyne trimerises; ethene is converted to ethyne; hexane cyclises and loses hydrogen.

Answer

Use ethyne trimerisation, or catalytic dehydrocyclisation of n-hexane.

Question 9.21

Write the possible alkenes that give 2-methylbutane on hydrogenation.

Steps only

Put the double bond at each non-equivalent C–C position.

Answer

2-Methylbut-1-ene, 2-methylbut-2-ene and 3-methylbut-1-ene.

Question 9.22

Arrange the compounds in decreasing reactivity towards electrophilic substitution: (a) chlorobenzene, p-nitrochlorobenzene, dinitrochlorobenzene; (b) toluene, p-nitrotoluene, p-dinitrobenzene.

Steps only

More electron donation means faster electrophilic substitution; each NO₂ group strongly slows it.

Answer

(a) chlorobenzene > mono-nitro > di-nitro; (b) toluene > nitrotoluene > dinitrobenzene.

Question 9.23

Which compound is nitrated most easily: benzene, m-dinitrobenzene or toluene?

Steps only

The methyl group activates the ring; nitro groups deactivate it.

Answer

Toluene.

Question 9.24

Name a Lewis acid other than AlCl₃ that can be used to ethylate benzene by Friedel–Crafts reaction.

Steps only

Use another anhydrous Lewis acid to activate the ethyl halide.

Answer

Anhydrous FeCl₃.

Question 9.25

Why is the Wurtz reaction not preferred for making odd-carbon alkanes? Give an example.

Steps only

Two different halides create multiple coupling products.

Answer

It gives a mixture; e.g. methyl + ethyl halides form ethane, propane and butane.