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CLASS 11 · CHEMISTRY · CHAPTER 8

Organic Chemistry – Some Basic Principles and Techniques

Clear, point-wise answers to all 24 numbered in-text problems and all 40 end-of-chapter exercises.

64 worked answersComplete chapter answers

64 worked answers · short steps first, explanations when you need them

In-text Problems

Textbook pages 256–292

Question Problem 8.1

Count σ and π bonds in (a) HC≡CCH=CHCH₃ and (b) CH₂=C=CHCH₃.

Steps only

(a) Four C–C σ bonds and six C–H σ bonds: 10σ total; one π bond from C=C and two from C≡C: 3π total. (b) Three C–C σ bonds and six C–H σ bonds: 9σ total; two π bonds: 2π total.

Answer

(a) Four C–C σ bonds and six C–H σ bonds: 10σ total; one π bond from C=C and two from C≡C: 3π total. (b) Three C–C σ bonds and six C–H σ bonds: 9σ total; two π bonds: 2π total.

Question Problem 8.2

State the hybridisation of each carbon in (a) CH₃Cl, (b) (CH₃)₂CO, (c) CH₃CN, (d) HCONH₂ and (e) CH₃CH=CHCN.

Steps only

(a) sp³. (b) The two methyl carbons are sp³; the carbonyl carbon is sp². (c) CH₃ carbon is sp³; nitrile carbon is sp. (d) Carbonyl carbon is sp². (e) In order along the chain: sp³, sp², sp², sp.

Answer

(a) sp³. (b) The two methyl carbons are sp³; the carbonyl carbon is sp². (c) CH₃ carbon is sp³; nitrile carbon is sp. (d) Carbonyl carbon is sp². (e) In order along the chain: sp³, sp², sp², sp.

Question Problem 8.3

Give the hybridisation of carbon and molecular shape in (a) H₂C=O, (b) CH₃F and (c) HCN.

Steps only

(a) Carbon is sp²; the atoms around it are trigonal planar. (b) Carbon is sp³; the shape around carbon is tetrahedral. (c) Carbon is sp; the molecule is linear.

Answer

(a) Carbon is sp²; the atoms around it are trigonal planar. (b) Carbon is sp³; the shape around carbon is tetrahedral. (c) Carbon is sp; the molecule is linear.

Question Problem 8.4

Expand into complete structural formulas: (a) CH₃CH₂COCH₂CH₃; (b) CH₃CH=CH(CH₂)₃CH₃.

Steps only

(a) CH₃–CH₂–C(=O)–CH₂–CH₃. (b) CH₃–CH=CH–CH₂–CH₂–CH₂–CH₃.

Answer

(a) CH₃–CH₂–C(=O)–CH₂–CH₃. (b) CH₃–CH=CH–CH₂–CH₂–CH₂–CH₃.

Question Problem 8.5

Write condensed structures for the printed structures and a bond-line formula for (a) HOCH₂CH₂CH₂CH(CH₃)CH(CH₃)CH₃ and (b) the figure-based structure.

Steps only

(a) HOCH₂CH₂CH₂CH(CH₃)CH(CH₃)CH₃. (b) HOCH(CN)₂, also written NC–CH(OH)–CN.

Answer

(a) HOCH₂CH₂CH₂CH(CH₃)CH(CH₃)CH₃. (b) HOCH(CN)₂, also written NC–CH(OH)–CN.

Question Problem 8.6

Expand each of the four bond-line formulas shown in the textbook into a complete structural formula.

Steps only

(a) C₆H₁₁CH(CH₃)₂ (propan-2-ylcyclohexane). (b) CH₃(CH₂)₆CH₃ (octane). (c) HC≡CCH₂CH(OH)CH₃. (d) CH₃CH(CH₃)CH(CH₃)CH₃.

Answer

(a) C₆H₁₁CH(CH₃)₂ (propan-2-ylcyclohexane). (b) CH₃(CH₂)₆CH₃ (octane). (c) HC≡CCH₂CH(OH)CH₃. (d) CH₃CH(CH₃)CH(CH₃)CH₃.

Question Problem 8.7

Write the condensed and bond-line structures for (a) HO(CH₂)₃CH(CH₃)CH(CH₃)₂ and (b) HOCH(CN)₂.

Steps only

(a) HO–CH₂–CH₂–CH₂–CH(CH₃)–CH(CH₃)₂. (b) HO–CH(C≡N)₂: the central carbon is attached to –OH, H and two cyano groups.

Answer

(a) HO–CH₂–CH₂–CH₂–CH(CH₃)–CH(CH₃)₂. (b) HO–CH(C≡N)₂: the central carbon is attached to –OH, H and two cyano groups.

Question Problem 8.8

Give the IUPAC names of the four structures shown in the textbook.

Steps only

(i) 6-Methyloctan-3-ol. (ii) Hexane-2,4-dione. (iii) 5-Oxohexanoic acid. (iv) Hexa-1,3-dien-5-yne.

Answer

  • (i) 6-Methyloctan-3-ol.
  • (ii) Hexane-2,4-dione.
  • (iii) 5-Oxohexanoic acid.
  • (iv) Hexa-1,3-dien-5-yne.

Question Problem 8.9

Draw structures for (i) 2-chlorohexane, (ii) pent-4-en-2-ol, (iii) 3-nitrocyclohexene, (iv) cyclohex-2-en-1-ol and (v) 6-hydroxyheptanal.

Steps only

(i) CH₃–CH(Cl)–CH₂–CH₂–CH₂–CH₃. (ii) CH₃–CH(OH)–CH₂–CH=CH₂. (iii) Cyclohex-1-ene with –NO₂ on carbon 3. (iv) Cyclohex-2-en-1-ol: a six-membered ring with OH at C1 and C2=C3. (v) OHC–CH₂–CH₂–CH₂–CH₂–CH(OH)–CH₃.

Answer

  • (i) CH₃–CH(Cl)–CH₂–CH₂–CH₂–CH₃.
  • (ii) CH₃–CH(OH)–CH₂–CH=CH₂.
  • (iii) Cyclohex-1-ene with –NO₂ on carbon 3.
  • (iv) Cyclohex-2-en-1-ol: a six-membered ring with OH at C1 and C2=C3.
  • (v) OHC–CH₂–CH₂–CH₂–CH₂–CH(OH)–CH₃.

Question Problem 8.10

Draw structures for (a) o-ethylanisole, (b) p-nitroaniline, (c) 2,3-dibromo-1-phenylpentane and (d) 4-ethyl-1-fluoro-2-nitrobenzene.

Steps only

Place ring groups at the named positions; for (c), use a five-carbon chain with phenyl at C1 and Br at C2 and C3.

Answer

(a) Adjacent –OCH₃/–C₂H₅; (b) opposite –NH₂/–NO₂; (c) C₆H₅CH₂CHBrCHBrCH₂CH₃; (d) benzene with F-1, NO₂-2 and ethyl-4.

Question Problem 8.11

Show the products of heterolytic cleavage of (a) CH₃–SCH₃, (b) CH₃–CN and (c) CH₃–Cu.

Steps only

(a) The S–CH₃ bond breaks with the pair going to sulfur: CH₃S⁻ + CH₃⁺. (b) The C–CN bond breaks with the pair going to CN: CH₃⁺ + CN⁻. (c) The C–Cu bond breaks with the pair going to carbon: CH₃⁻ + Cu⁺.

Answer

(a) The S–CH₃ bond breaks with the pair going to sulfur: CH₃S⁻ + CH₃⁺. (b) The C–CN bond breaks with the pair going to CN: CH₃⁺ + CN⁻. (c) The C–Cu bond breaks with the pair going to carbon: CH₃⁻ + Cu⁺.

Question Problem 8.12

Classify the listed species HS⁻, C₂H₅O⁻, (CH₃)₃N, H₂N⁻, BF₃, CH₃⁺, CH₃CO⁺ and NO₂⁺ as nucleophiles or electrophiles.

Steps only

Nucleophiles: HS⁻, C₂H₅O⁻, (CH₃)₃N and H₂N⁻; they can donate an electron pair. Electrophiles: BF₃, CH₃⁺, CH₃CO⁺ and NO₂⁺; they can accept an electron pair.

Answer

Nucleophiles: HS⁻, C₂H₅O⁻, (CH₃)₃N and H₂N⁻; they can donate an electron pair. Electrophiles: BF₃, CH₃⁺, CH₃CO⁺ and NO₂⁺; they can accept an electron pair.

Question Problem 8.13

Identify the electrophilic carbon in CH₃CHO, CH₃CN and CH₃I.

Steps only

In each molecule, the carbon attached to the more electronegative atom is electrophilic: the carbonyl carbon in CH₃CHO, the nitrile carbon in CH₃CN and the carbon bonded to iodine in CH₃I.

Answer

In each molecule, the carbon attached to the more electronegative atom is electrophilic: the carbonyl carbon in CH₃CHO, the nitrile carbon in CH₃CN and the carbon bonded to iodine in CH₃I.

Question Problem 8.14

Which bond is more polar in each pair: (a) CH₃–H or CH₃–Br; (b) CH₃–NH₂ or CH₃–OH; (c) CH₃–OH or CH₃–SH?

Steps only

(a) C–Br. (b) C–O. (c) C–O. In each pair, the bond to the more electronegative atom is more polar.

Answer

(a) C–Br. (b) C–O. (c) C–O. In each pair, the bond to the more electronegative atom is more polar.

Question Problem 8.15

In which C–C bond of CH₃CH₂CH₂Br is the inductive effect least?

Steps only

It is least in the C₂–C₃ bond, farthest from bromine, because the inductive effect weakens as it travels through more bonds.

Answer

It is least in the C₂–C₃ bond, farthest from bromine, because the inductive effect weakens as it travels through more bonds.

Question Problem 8.16

Write the resonance structures of acetate ion, CH₃COO⁻, and show the electron movement.

Steps only

Two equivalent forms share the negative charge over both oxygens.

Answer

Two equivalent forms share the negative charge over both oxygens.

Question Problem 8.17

Write resonance forms of CH₂=CH–CHO and compare their stability.

Steps only

The neutral structure is the major contributor; charge-separated forms are minor.

Answer

The neutral structure is the major contributor; charge-separated forms are minor.

Question Problem 8.18

Why can the two printed structures labelled I and II not be major contributors to methyl acetate, CH₃COOCH₃?

Steps only

Neither I nor II is a major contributor: both have charge separation, and structure I also contains a carbon with an incomplete octet. The neutral structure of methyl acetate is more stable.

Answer

Neither I nor II is a major contributor: both have charge separation, and structure I also contains a carbon with an incomplete octet. The neutral structure of methyl acetate is more stable.

Question Problem 8.19

Why is (CH₃)₃C⁺ more stable than CH₃CH₂⁺, while CH₃⁺ is least stable?

Steps only

Methyl groups stabilise a nearby positive carbon by donating electron density through hyperconjugation. (CH₃)₃C⁺ has nine adjacent C–H bonds available for this effect, CH₃CH₂⁺ has three, and CH₃⁺ has none. So (CH₃)₃C⁺ is most stable and CH₃⁺ least stable.

Answer

Methyl groups stabilise a nearby positive carbon by donating electron density through hyperconjugation. (CH₃)₃C⁺ has nine adjacent C–H bonds available for this effect, CH₃CH₂⁺ has three, and CH₃⁺ has none. So (CH₃)₃C⁺ is most stable and CH₃⁺ least stable.

Question Problem 8.20

On complete combustion, 0.246 g of an organic compound gives 0.198 g CO₂ and 0.1014 g H₂O. Find the percentage of carbon and hydrogen.

Steps only

Carbon mass=(12/44)(0.198)=0.0540 g, so %C=(0.0540/0.246)×100≈21.95%. Hydrogen mass=(2/18)(0.1014)=0.01127 g, so %H=(0.01127/0.246)×100≈4.58%.

Answer

Carbon mass=(12/44)(0.198)=0.0540 g, so %C=(0.0540/0.246)×100≈21.95%. Hydrogen mass=(2/18)(0.1014)=0.01127 g, so %H=(0.01127/0.246)×100≈4.58%.

Question Problem 8.21

In Dumas’ method, 0.30 g of a compound gives 50 mL nitrogen at 300 K and 715 mm pressure. Aqueous tension is 15 mm. Find nitrogen percentage.

Steps only

Dry N₂ pressure=715−15=700 mm. Volume at STP=50×(700/760)×(273/300)=41.9 mL. Nitrogen mass=(28/22400)×41.9=0.0524 g. Percentage N=(0.0524/0.30)×100≈17.5%.

Answer

Dry N₂ pressure=715−15=700 mm. Volume at STP=50×(700/760)×(273/300)=41.9 mL. Nitrogen mass=(28/22400)×41.9=0.0524 g. Percentage N=(0.0524/0.30)×100≈17.5%.

Question Problem 8.22

In Kjeldahl’s method, ammonia from a 0.5 g compound neutralises 10 mL of 1 M H₂SO₄. Find the nitrogen percentage.

Steps only

10 mL of 1 M H₂SO₄ neutralises 20 mL of 1 M NH₃; this is 0.28 g N.

Answer

Nitrogen percentage = (0.28/0.50)×100 = 56%.

Question Problem 8.23

In Carius’ method, 0.15 g of a compound gives 0.12 g AgBr. Find the bromine percentage.

Steps only

Mass of Br=(80/188)×0.12=0.0511 g. Percentage Br=(0.0511/0.15)×100≈34.0%.

Answer

Mass of Br=(80/188)×0.12=0.0511 g. Percentage Br=(0.0511/0.15)×100≈34.0%.

Question Problem 8.24

In sulfur estimation, 0.157 g of a compound gives 0.4813 g BaSO₄. Find the sulfur percentage.

Steps only

Mass of S=(32/233)×0.4813=0.0661 g. Percentage S=(0.0661/0.157)×100≈42.1%.

Answer

Mass of S=(32/233)×0.4813=0.0661 g. Percentage S=(0.0661/0.157)×100≈42.1%.

Chapter 8 Exercises

Textbook pages 282–288

Question 8.1

State the hybridisation of each carbon atom in ketene (CH₂=C=O), propene (CH₃CH=CH₂), propanone ((CH₃)₂CO), acrylonitrile (CH₂=CHCN) and benzene (C₆H₆).

Steps only

Single-bond carbon: sp³; alkene/carbonyl/aromatic carbon: sp²; nitrile/linear carbon: sp.

Answer

Ketene: sp², sp; propene: sp³, sp², sp²; propanone: sp³, sp³, sp²; acrylonitrile: sp², sp², sp; benzene: all sp².

Question 8.2

Count σ and π bonds in benzene, cyclohexane, CH₂Cl₂, allene (CH₂=C=CH₂), nitromethane and HCONHCH₃.

Steps only

Count each single bond as σ; each double bond contributes one additional π.

Answer

Benzene 12σ/3π; cyclohexane 18σ/0π; CH₂Cl₂ 4σ/0π; allene 6σ/2π; nitromethane 6σ/1π; HCONHCH₃ 8σ/1π.

Question 8.3

Write bond-line structures for propan-2-ol, 2,3-dimethylbutanal and heptan-4-one.

Steps only

Place the functional group at the numbered carbon, then add the correct carbon chain and branches.

Answer

CH₃CH(OH)CH₃; OHCCH(CH₃)CH(CH₃)CH₃; CH₃CH₂CH₂COCH₂CH₂CH₃.

Question 8.6

Write the first five members of the homologous series beginning with (a) HCOOH, (b) CH₃COCH₃ and (c) CH₂=CH₂.

Steps only

Add one –CH₂– unit at a time and keep the functional group unchanged.

Answer

Each next member differs by –CH₂–: acids, ketones, and alkenes retain their group or double bond.

Question 8.7

Give condensed structures and identify the functional groups in 2,2,4-trimethylpentane, 2-hydroxypropane-1,2,3-tricarboxylic acid and hexanedial.

Steps only

Build the named parent structure and mark each suffix group.

Answer

(a) alkane; (b) three –COOH and one –OH; (c) two –CHO groups.

Question 8.9

Which is more stable: ⁻OCH₂CH₂NO₂ or ⁻OCH₂CH₃? Explain.

Steps only

The –NO₂ group has a strong electron-withdrawing (−I) effect.

Answer

⁻OCH₂CH₂NO₂ is more stable.

Question 8.10

Why do alkyl groups donate electron density when attached to a π system?

Steps only

Adjacent C–H σ electrons delocalise into the π system (hyperconjugation).

Answer

Alkyl groups donate by hyperconjugation.

Question 8.12

What are electrophiles and nucleophiles? Give examples.

Steps only

Nucleophile donates an electron pair; electrophile accepts one.

Answer

Examples: OH⁻ is a nucleophile; H⁺ is an electrophile.

Question 8.13

Classify the bold reagents in the three given reaction equations as nucleophiles or electrophiles.

Steps only

OH⁻ and CN⁻ donate electron pairs; CH₃CO⁺ accepts one.

Answer

(a) Nucleophile; (b) nucleophile; (c) electrophile.

Question 8.14

Classify the four reactions: bromoethane with HS⁻; a haloalkane with HI; bromoethane with OH⁻ giving an alkene; and an alcohol with HBr giving a bromoalkane.

Steps only

Check whether a group is replaced, added across a bond, or removed to form a multiple bond.

Answer

(a) Substitution; (b) addition; (c) elimination; (d) substitution.

Question 8.17

Define inductive and electromeric effects. Explain why (a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH and (b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH in acidity.

Steps only

Electron-withdrawing groups stabilise the carboxylate conjugate base; the inductive effect weakens with distance.

Answer

Chlorine’s −I effect strengthens acidity as more Cl atoms are added. Alkyl groups’ +I effect weakens acidity as branching increases.

Question 8.18

Briefly describe crystallisation, distillation and chromatography, giving an example of each.

Steps only

Crystallisation uses solubility; distillation uses boiling point; chromatography uses different phase affinities.

Answer

These methods separate by solubility, volatility and movement between phases, respectively.

Question 8.19

Which method separates two compounds that have different solubilities in a solvent?

Steps only

Use an immiscible solvent and a separating funnel.

Answer

Differential extraction.

Question 8.20

Differentiate ordinary distillation, distillation under reduced pressure and steam distillation.

Steps only

Normal pressure; lowered pressure for heat-sensitive/high-boiling liquids; steam for volatile water-immiscible compounds.

Answer

They differ in pressure and whether steam is used as a co-distilling vapour.

Question 8.21

Explain the chemistry of Lassaigne’s test.

Steps only

Sodium fusion changes covalently bonded elements into water-soluble ions for testing.

Answer

N→CN⁻; S→S²⁻; halogen→X⁻.

Question 8.22

How do Dumas and Kjeldahl methods differ in estimating nitrogen in an organic compound?

Steps only

Dumas measures N₂ gas; Kjeldahl measures ammonia released from ammonium salt.

Answer

Dumas: N₂ volume. Kjeldahl: NH₃ titration.

Question 8.23

State the principle used to estimate halogens, sulfur and phosphorus in an organic compound.

Steps only

Convert each element into a measurable stable precipitate.

Answer

Halogen→AgX; sulfur→BaSO₄; phosphorus→phosphate precipitate.

Question 8.24

Explain the principle of paper chromatography.

Steps only

Components partition differently between stationary and moving phases.

Answer

Different partitioning separates the spots.

Question 8.25

Why is nitric acid added to the sodium extract before silver nitrate when testing for halogens?

Steps only

Remove CN⁻ and S²⁻, which would also react with Ag⁺.

Answer

To prevent false precipitates in the halogen test.

Question 8.26

Why is an organic compound fused with sodium when testing for nitrogen, sulfur and halogens?

Steps only

Fusion changes covalent elements into detectable water-soluble ions.

Answer

To form ionic sodium salts for qualitative tests.

Question 8.27

Name a suitable method to separate calcium sulfate and camphor.

Steps only

Camphor sublimes; calcium sulfate remains solid.

Answer

Sublimation.

Question 8.28

Why can an organic liquid vaporise below its normal boiling point during steam distillation?

Steps only

Water and organic vapour pressures add to reach atmospheric pressure sooner.

Answer

It co-distils with steam at a lower temperature.

Question 8.29

Will CCl₄ give a white AgCl precipitate when heated with silver nitrate? Explain.

Steps only

CCl₄ contains covalent C–Cl bonds, not free Cl⁻ ions.

Answer

No white AgCl precipitate.

Question 8.30

Why is KOH used to absorb CO₂ in carbon estimation?

Steps only

KOH removes CO₂ by converting it into carbonate.

Answer

To absorb and measure the CO₂ formed.

Question 8.31

Why is acetic acid used instead of sulfuric acid before the lead acetate test for sulfur?

Steps only

Avoid adding sulfate, which precipitates lead as PbSO₄.

Answer

Use acetic acid to avoid interfering PbSO₄.

Question 8.32

An organic compound is 69% carbon and 4.8% hydrogen by mass, with the rest oxygen. Find the masses of CO₂ and H₂O formed from 0.20 g on complete combustion.

Steps only

CO₂: 0.20×0.69×44/12=0.506 g; H₂O: 0.20×0.048×9=0.0864 g.

Answer

0.506 g CO₂ and 0.0864 g H₂O.

Question 8.33

A 0.50 g sample is treated with 50 mL of 0.5 M H₂SO₄. The remaining acid needs 60 mL of 0.5 M NaOH. Find the nitrogen percentage.

Steps only

n(N)=0.050−0.030=0.020 mol; m(N)=0.28 g; %N=56.0%.

Answer

56.0% nitrogen.

Question 8.34

A 0.3780 g chlorinated compound gives 0.5740 g AgCl. Find the chlorine percentage.

Steps only

%Cl=0.5740×35.45/143.32/0.3780×100≈37.6%.

Answer

37.6% chlorine.

Question 8.35

A 0.468 g sulfur compound gives 0.668 g BaSO₄. Calculate the sulfur percentage.

Steps only

%S=0.668×32.06/233.39/0.468×100≈19.6%.

Answer

19.6% sulfur.

Question 8.36

For CH₂=CH–CH₂–CH₂–C≡CH, identify the hybrid orbitals forming the C₂–C₃ bond.

Steps only

C₂ is sp²; C₃ is sp³.

Answer

sp²–sp³ overlap.

Question 8.37

Which compound gives the Prussian blue colour in Lassaigne’s test for nitrogen?

Steps only

The blue complex is ferric ferrocyanide.

Answer

Fe₄[Fe(CN)₆]₃.

Question 8.39

Which is a particularly useful modern technique for separating and purifying organic compounds: crystallisation, distillation, sublimation or chromatography?

Steps only

Chromatography separates compounds by their different phase affinities.

Answer

Chromatography.

Question 8.40

How is CH₃CH₂I+KOH(aq)→CH₃CH₂OH+KI classified?

Steps only

OH⁻ replaces the leaving group I⁻.

Answer

Nucleophilic substitution.

Question 8.4

Give IUPAC names for the six structures shown in Exercise 8.4.

Steps only

(a) 2,5-Dimethylheptane. (b) 1-Bromo-3-methylpentane. (c) 3-Chloro-4-methylhexane. (d) 2-Methylbutan-1-ol. (e) 3,3-Dimethylbutanal. (f) 2,2-Dichloroethan-1-ol.

Answer

(a) 2,5-Dimethylheptane. (b) 1-Bromo-3-methylpentane. (c) 3-Chloro-4-methylhexane. (d) 2-Methylbutan-1-ol. (e) 3,3-Dimethylbutanal. (f) 2,2-Dichloroethan-1-ol.

Question 8.5

Choose the correct name in each pair: (i) 2,2-dimethylpentane / 2-dimethylpentane; (ii) 2,4,7-trimethyloctane / 2,5,7-trimethyloctane; (iii) 2-chloro-4-methylpentane / 4-chloro-2-methylpentane; (iv) but-3-yn-1-ol / but-4-ol-1-yne.

Steps only

(i) 2,2-Dimethylpentane. (ii) 2,4,7-Trimethyloctane. (iii) 2-Chloro-4-methylpentane. (iv) But-3-yn-1-ol.

Answer

  • (i) 2,2-Dimethylpentane.
  • (ii) 2,4,7-Trimethyloctane.
  • (iii) 2-Chloro-4-methylpentane.
  • (iv) But-3-yn-1-ol.

Question 8.8

Identify the functional groups in the three structures shown in Exercise 8.8.

Steps only

(a) Vanillin: aldehyde, ether (methoxy) and phenolic –OH. (b) The amino-aromatic ester: primary aromatic amine, ester and tertiary amine. (c) The unsaturated nitro compound: alkene and nitro group.

Answer

(a) Vanillin: aldehyde, ether (methoxy) and phenolic –OH. (b) The amino-aromatic ester: primary aromatic amine, ester and tertiary amine. (c) The unsaturated nitro compound: alkene and nitro group.

Question 8.11

Describe the important resonance contributors for (a) phenol, (b) nitrobenzene, (c) but-2-enal, (d) benzaldehyde, (e) benzyl cation and (f) allyl cation.

Steps only

(a) Phenol: an oxygen lone pair is donated into the ring, placing negative charge at ortho and para positions while oxygen becomes positive. (b) Nitrobenzene: the nitro group withdraws ring electron density, placing positive charge at ortho/para positions. (c) CH3CH=CHCHO: conjugation shares electron density across C=C–C=O; a contributor has O− and positive charge at the β-carbon. (d) Benzaldehyde: ring donation toward –CHO gives O− and positive charge at ortho/para ring positions. (e) Benzyl cation: positive charge is shared between benzylic carbon and ortho/para ring positions. (f) Allyl cation: positive charge is shared between the two terminal carbons.

Answer

(a) Phenol: an oxygen lone pair is donated into the ring, placing negative charge at ortho and para positions while oxygen becomes positive. (b) Nitrobenzene: the nitro group withdraws ring electron density, placing positive charge at ortho/para positions. (c) CH3CH=CHCHO: conjugation shares electron density across C=C–C=O; a contributor has O− and positive charge at the β-carbon. (d) Benzaldehyde: ring donation toward –CHO gives O− and positive charge at ortho/para ring positions. (e) Benzyl cation: positive charge is shared between benzylic carbon and ortho/para ring positions. (f) Allyl cation: positive charge is shared between the two terminal carbons.

Question 8.15

For the three pairs shown in Exercise 8.15, state whether each pair represents structural isomers, geometrical isomers or resonance contributors.

Steps only

(a) Structural isomers. (b) Geometrical (cis/trans) isomers. (c) Resonance contributors.

Answer

(a) Structural isomers. (b) Geometrical (cis/trans) isomers. (c) Resonance contributors.

Question 8.16

Classify the four bond cleavages in Exercise 8.16 as homolysis or heterolysis and name the intermediates.

Steps only

(a) C–C bond splits evenly: homolysis, giving two methyl radicals. (b) C–Cl bond electrons go to chlorine: heterolysis, giving CH3+ and Cl−. (c) tert-Butyl bromide splits with both electrons to Br: heterolysis, giving a tertiary carbocation and Br−. (d) The C–Mg bond pair stays with carbon: heterolysis, giving a carbanion-like CH3− and MgBr+.

Answer

(a) C–C bond splits evenly: homolysis, giving two methyl radicals. (b) C–Cl bond electrons go to chlorine: heterolysis, giving CH3+ and Cl−. (c) tert-Butyl bromide splits with both electrons to Br: heterolysis, giving a tertiary carbocation and Br−. (d) The C–Mg bond pair stays with carbon: heterolysis, giving a carbanion-like CH3− and MgBr+.

Question 8.38

Which of the four carbocations shown in Exercise 8.38 is most stable?

Steps only

The tert-butyl carbocation, (CH3)3C+, is most stable because it is tertiary and is stabilised by electron donation from three alkyl groups.

Answer

The tert-butyl carbocation, (CH3)3C+, is most stable because it is tertiary and is stabilised by electron donation from three alkyl groups.

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