Build the named parent first, then place each halogen and branch at its numbered carbon.
Answer
(i) CH₃CH(Cl)CH(CH₃)CH₂CH₃.
(ii) p-BrC₆H₄Cl.
(iii) Cyclohexane ring with Cl on C1 and ethyl on C4.
(iv) ICH₂CH(o-ClC₆H₄)(CH₂)₅CH₃.
(v) CH₃CH(Br)CH₂CH₃.
(vi) CH₃CH₂CH(I)CH[C(CH₃)₃]CH₂CH₂CH₃.
(vii) 1-Br-4-[CH(CH₃)CH₂CH₃]-2-methylbenzene.
(viii) BrCH₂CH=CHCH₂Br.
Question 6.4
Which has the greatest dipole moment: CH2Cl2, CHCl3 or CCl4?
Steps only
CH2Cl2. Its bond dipoles do not cancel as much as in the other molecules; CCl4 is symmetrical and has zero net dipole.
Answer
CH2Cl2. Its bond dipoles do not cancel as much as in the other molecules; CCl4 is symmetrical and has zero net dipole.
Question 6.5
A hydrocarbon C5H10 does not react with chlorine in the dark but forms one monochloro product in sunlight. Identify it.
Steps only
Cyclopentane. It is a saturated ring, so chlorine substitution needs light; its five hydrogen positions are equivalent by symmetry, giving one monochloro product.
Answer
Cyclopentane. It is a saturated ring, so chlorine substitution needs light; its five hydrogen positions are equivalent by symmetry, giving one monochloro product.
Question 6.6
Write all isomers with formula C4H9Br.
Steps only
1-Bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane.
Answer
1-Bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane.
Question 6.7
Give equations for preparing 1-iodobutane from (i) butan-1-ol, (ii) 1-chlorobutane and (iii) but-1-ene.
Steps only
(i) CH3CH2CH2CH2OH + HI → CH3CH2CH2CH2I + H2O. (ii) CH3CH2CH2CH2Cl + NaI (acetone) → CH3CH2CH2CH2I + NaCl↓. (iii) CH2=CHCH2CH3 + HBr (peroxide) → CH3CH2CH2CH2Br; then NaI/acetone → CH3CH2CH2CH2I.
Answer
(i) CH3CH2CH2CH2OH + HI → CH3CH2CH2CH2I + H2O.
(ii) CH3CH2CH2CH2Cl + NaI (acetone) → CH3CH2CH2CH2I + NaCl↓.
A nucleophile that can attack through either of two different atoms is ambident. CN⁻ can attack through carbon to form a nitrile with KCN, or through nitrogen with AgCN to form an isocyanide.
Answer
A nucleophile that can attack through either of two different atoms is ambident. CN⁻ can attack through carbon to form a nitrile with KCN, or through nitrogen with AgCN to form an isocyanide.
Question 6.9
In each pair, which reacts faster by SN2 with OH⁻: (i) CH3Br or CH3I, (ii) (CH3)3CCl or CH3Cl?
Steps only
(i) CH3I, because I⁻ is the better leaving group. (ii) CH3Cl, because the small methyl group allows easier backside attack; the tertiary centre is strongly hindered.
Answer
(i) CH3I, because I⁻ is the better leaving group.
(ii) CH3Cl, because the small methyl group allows easier backside attack; the tertiary centre is strongly hindered.
Question 6.10
List alkenes from elimination of (i) 1-bromo-1-methylcyclohexane, (ii) 2-chloro-2-methylbutane, (iii) 2,2,3-trimethyl-3-bromopentane; identify the major product.
Steps only
(i) 1-Methylcyclohexene (major) and methylenecyclohexane. (ii) 2-Methylbut-2-ene (major) and 2-methylbut-1-ene. (iii) 3,4,4-trimethylpent-2-ene (major, E/Z forms possible) and 2-ethyl-3,3-dimethylbut-1-ene.
Answer
(i) 1-Methylcyclohexene (major) and methylenecyclohexane.
(ii) 2-Methylbut-2-ene (major) and 2-methylbut-1-ene.
(iii) 3,4,4-trimethylpent-2-ene (major, E/Z forms possible) and 2-ethyl-3,3-dimethylbut-1-ene.
Question 6.11
Give reagent routes for the ten conversions in Exercise 6.11.
Steps only
(i) Convert some ethanol to bromoethane (HBr). Convert another portion to ethyne: dehydrate to ethene, add Br2 to make 1,2-dibromoethane, then eliminate twice with NaNH2. Deprotonate ethyne with NaNH2 and alkylate with bromoethane to get but-1-yne. (ii) Ethane → bromoethane (Br2/hν) → ethene (alc KOH) → 1,2-dibromoethane (Br2) → bromoethene (one dehydrohalogenation). (iii) Propene → 1-bromopropane (HBr/peroxide) → 1-nitropropane (AgNO2). (iv) Toluene → benzyl chloride (Cl2/hν) → benzyl alcohol (aq KOH). (v) Propene → 1,2-dibromopropane (Br2) → propyne (excess NaNH2, then water). (vi) Ethanol → chloroethane (SOCl2) → fluoroethane (AgF). (vii) CH3Br → CH3CN (KCN); CH3CN + CH3MgBr, then hydrolysis → propanone. (viii) But-1-ene → butan-2-ol (H2O/H+) → but-2-ene (dehydration). (ix) 2 CH3CH2CH2CH2Cl + 2 Na (dry ether) → n-octane + 2 NaCl. (x) Bromobenzene + Na/dry ether → biphenyl.
Answer
(i) Convert some ethanol to bromoethane (HBr). Convert another portion to ethyne: dehydrate to ethene, add Br2 to make 1,2-dibromoethane, then eliminate twice with NaNH2. Deprotonate ethyne with NaNH2 and alkylate with bromoethane to get but-1-yne.
Explain (i) lower dipole moment of chlorobenzene than cyclohexyl chloride, (ii) why polar alkyl halides do not mix with water, (iii) why Grignard reagents need dry conditions.
Steps only
(i) In chlorobenzene, resonance gives the C–Cl bond partial double-bond character and reduces its polarity; ring geometry also affects dipole addition. (ii) Their interactions with water are not strong enough to replace water–water hydrogen bonds. (iii) Water protonates and destroys a Grignard reagent: RMgX + H2O → RH + Mg(OH)X.
Answer
(i) In chlorobenzene, resonance gives the C–Cl bond partial double-bond character and reduces its polarity; ring geometry also affects dipole addition.
(ii) Their interactions with water are not strong enough to replace water–water hydrogen bonds.
(iii) Water protonates and destroys a Grignard reagent: RMgX + H2O → RH + Mg(OH)X.
Question 6.13
Give uses of Freon-12, DDT, carbon tetrachloride and iodoform.
Steps only
Freon-12 was used as a refrigerant; DDT as an insecticide; carbon tetrachloride as a solvent and in fire extinguishers; iodoform as an antiseptic. Freon-12 and DDT are now restricted or phased out in many uses because of serious environmental harm; carbon tetrachloride is toxic.
Answer
Freon-12 was used as a refrigerant; DDT as an insecticide; carbon tetrachloride as a solvent and in fire extinguishers; iodoform as an antiseptic. Freon-12 and DDT are now restricted or phased out in many uses because of serious environmental harm; carbon tetrachloride is toxic.
Question 6.14
Find the main organic product in each reaction: (i) CH₃CH₂CH₂Cl + NaI (acetone, heat); (ii) (CH₃)₃CBr + KOH (ethanolic or aqueous); (iii) CH₃CH(Br)CH₂CH₃ + aqueous NaOH; (iv) CH₃CH₂Br + KCN; (v) C₆H₅ONa + C₂H₅Cl; (vi) CH₃CH₂CH₂OH + SOCl₂; (vii) CH₃CH₂CH=CH₂ + HBr; (viii) CH₃CH=C(CH₃)₂ + HBr.
Steps only
Identify substitution, elimination, or addition; for KOH, check whether the solvent is water or alcohol.
Answer
(i) CH₃CH₂CH₂I (1-iodopropane).
(ii) Ethanolic KOH: 2-methylpropene; aqueous KOH: 2-methylpropan-2-ol.
Show the mechanism for n-butyl bromide reacting with KCN to give n-butyl cyanide.
Steps only
CN⁻ attacks the carbon bonded to Br from the opposite side in one step; the C–Br bond breaks at the same time. CH3CH2CH2CH2Br + CN⁻ → CH3CH2CH2CH2CN + Br⁻. The product is pentanenitrile.
Answer
CN⁻ attacks the carbon bonded to Br from the opposite side in one step; the C–Br bond breaks at the same time. CH3CH2CH2CH2Br + CN⁻ → CH3CH2CH2CH2CN + Br⁻. The product is pentanenitrile.
Question 6.16
Arrange each set from slowest to fastest in an SN2 reaction: (i) 2-bromo-2-methylbutane, 1-bromopentane, 2-bromopentane; (ii) 1-bromo-3-methylbutane, 2-bromo-2-methylbutane, 2-bromo-3-methylbutane; (iii) 1-bromobutane, 1-bromo-2,2-dimethylpropane, 1-bromo-2-methylbutane, 1-bromo-3-methylbutane.
Steps only
Less crowding near the carbon bearing Br means a faster SN2 reaction; branching next door also slows it.
Which is hydrolysed faster by aqueous KOH: benzyl chloride or benzhydryl chloride?
Steps only
Benzhydryl chloride, C6H5CHClC6H5. Its leaving group forms a carbocation stabilised by two phenyl rings, so hydrolysis is faster.
Answer
Benzhydryl chloride, C6H5CHClC6H5. Its leaving group forms a carbocation stabilised by two phenyl rings, so hydrolysis is faster.
Question 6.18
Why does p-dichlorobenzene have a higher melting point than the ortho and meta isomers?
Steps only
The para isomer is more symmetrical and packs more efficiently in a crystal. Better packing strengthens the crystal lattice and raises its melting point.
Answer
The para isomer is more symmetrical and packs more efficiently in a crystal. Better packing strengthens the crystal lattice and raises its melting point.
Question 6.19
Give suitable reagent sequences for all twenty conversions listed in Exercise 6.19.
Steps only
(i) Propene → 1-bromopropane (HBr/peroxide) → propan-1-ol (aq KOH).
(ii) Make ethyne from ethanol: ethanol → ethene → 1,2-dibromoethane → ethyne. Make bromoethane from another portion of ethanol. Convert ethyne to sodium acetylide, then react it with bromoethane to form but-1-yne.
(iii) 1-Bromopropane → propene (alcoholic KOH) → 2-bromopropane (HBr).
(iv) Toluene → benzyl chloride (Cl2, light) → benzyl alcohol (aqueous KOH).
(v) Benzene → bromobenzene (Br2/FeBr3) → nitration (conc. HNO3/H2SO4); separate the para isomer to obtain 4-bromonitrobenzene.
(vi) Benzyl alcohol → benzyl bromide (HBr) → phenylacetonitrile (KCN) → hydrolysis → 2-phenylethanoic acid.
(vii) Ethanol → bromoethane (HBr) → propanenitrile (KCN).
(viii) Aniline → diazonium salt (NaNO2/HCl, 273–278 K) → chlorobenzene (CuCl).
(ix) 2-Chlorobutane + Na (dry ether) → 3,4-dimethylhexane (Wurtz reaction).
(x) 2-Methylpropene + HCl → 2-chloro-2-methylpropane.
(xi) Chloroethane → propanenitrile (KCN) → hydrolysis → propanoic acid.
(xii) But-1-ene → 1-bromobutane (HBr/peroxide) → 1-iodobutane (NaI/acetone).
(xiii) 2-Chloropropane → propene (alcoholic KOH) → 1-bromopropane (HBr/peroxide) → propan-1-ol (aqueous KOH).
(xiv) Propan-2-ol + I2/NaOH → iodoform (CHI3) + propanone.
(xv) Chlorobenzene → p-nitrochlorobenzene (nitration; separate para isomer) → p-nitrophenol (aq. NaOH, 443 K; then acidify).
(xvi) 2-Bromopropane → propene (alcoholic KOH) → 1-bromopropane (HBr/peroxide).
(xvii) 2 molecules of chloroethane + 2 Na (dry ether) → butane + 2 NaCl.
(xviii) Benzene → bromobenzene (Br2/FeBr3); then 2 C6H5Br + 2 Na (dry ether) → biphenyl + 2 NaBr.
(xix) tert-Butyl bromide → 2-methylpropene (alcoholic KOH) → isobutyl bromide (HBr/peroxide).
(xx) Aniline + CHCl3/alcoholic KOH → phenyl isocyanide.
(ii) Make ethyne from ethanol: ethanol → ethene → 1,2-dibromoethane → ethyne. Make bromoethane from another portion of ethanol. Convert ethyne to sodium acetylide, then react it with bromoethane to form but-1-yne.
Why do alkyl chlorides form alcohols with aqueous KOH but alkenes with alcoholic KOH?
Steps only
In water, OH⁻ acts mainly as a nucleophile and replaces Cl, forming an alcohol. In alcohol, the basic conditions favour removal of a neighbouring H together with Cl, forming an alkene by elimination.
Answer
In water, OH⁻ acts mainly as a nucleophile and replaces Cl, forming an alcohol. In alcohol, the basic conditions favour removal of a neighbouring H together with Cl, forming an alkene by elimination.
Question 6.21
A primary C₄H₉Br isomer (a) reacts with alcoholic KOH to form alkene (b). Adding HBr to (b) gives another C₄H₉Br isomer (c). Wurtz reaction of (a) forms a branched C₈H₁₈ (d), not n-octane. Identify a–d and write the three reaction equations.
Steps only
The starting halide is isobutyl bromide; elimination gives 2-methylpropene, HBr gives tert-butyl bromide, and Wurtz coupling gives 2,5-dimethylhexane.
State the product(s): (i) n-butyl chloride + alcoholic KOH; (ii) bromobenzene + Mg in dry ether; (iii) chlorobenzene + hot aqueous NaOH, high pressure, then acidify; (iv) ethyl chloride + aqueous KOH; (v) methyl bromide + Na in dry ether; (vi) methyl chloride + KCN.
Steps only
Check whether each condition causes elimination, substitution, Grignard formation, aryl hydrolysis, coupling, or nitrile formation.