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.
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(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.
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(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.
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(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.
Write the condensed and bond-line structures for (a) HO(CH₂)₃CH(CH₃)CH(CH₃)₂ and (b) HOCH(CN)₂.
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(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.
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(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.
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(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.
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(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.
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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?
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(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?
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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.
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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.
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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?
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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.
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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.
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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.
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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.
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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.
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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).
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.
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Check whether a group is replaced, added across a bond, or removed to form a multiple bond.
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?
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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.
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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?
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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.
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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.
A 0.3780 g chlorinated compound gives 0.5740 g AgCl. Find the chlorine percentage.
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%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.
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%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.
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.
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(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.
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(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.
Classify the four bond cleavages in Exercise 8.16 as homolysis or heterolysis and name the intermediates.
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(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.