Intext Questions
1. Write structures of the following compounds: (i) 2-Chloro-3-methylpentane (ii) 1-Chloro-4-ethylcyclohexane (iii) 4-tert. Butyl-3-iodoheptane (iv) 1,4-Dibromobut-2-ene (v) 1-Bromo-4-sec. butyl-2-methylbenzene.
Answer
(i) CH₃–CH(Cl)–CH(CH₃)–CH₂–CH₃
(ii) A cyclohexane ring carrying Cl on C-1 and –CH₂CH₃ on C-4 (para-like positions of the ring): Cl–C₆H₁₀–C₂H₅ (1,4).
(iii) CH₃CH₂–CH(I)–CH[C(CH₃)₃]–CH₂CH₂CH₃
(iv) BrCH₂–CH=CH–CH₂Br
(v) A benzene ring with Br at C-1, CH₃ at C-2 and –CH(CH₃)CH₂CH₃ (sec-butyl) at C-4.
2. Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer
H₂SO₄ reacts with KI to give HI, but being an oxidising agent it then oxidises HI to I₂: 2KI + H₂SO₄
⇒ 2HI + K₂SO₄
⇒ 2HI + H₂SO₄
⇒ I₂ + SO₂ + 2H₂O. So HI is not available to convert the alcohol to an alkyl iodide. A non-oxidising acid such as H₃PO₄ is used instead.
3. Write structures of different dihalogen derivatives of propane.
Answer
Taking dichloro derivatives (C₃H₆Cl₂), four isomers are possible:
(i) ClCH₂CH₂CH₂Cl – 1,3-dichloropropane
(ii) ClCH₂CHClCH₃ – 1,2-dichloropropane
(iii) Cl₂CHCH₂CH₃ – 1,1-dichloropropane
(iv) CH₃CCl₂CH₃ – 2,2-dichloropropane
4. Among the isomeric alkanes of molecular formula C₅H₁₂, identify the one that on photochemical chlorination yields (i) A single monochloride. (ii) Three isomeric monochlorides. (iii) Four isomeric monochlorides.
Answer
The number of monochlorides equals the number of sets of equivalent hydrogens.
(i) Neopentane, C(CH₃)₄ (2,2-dimethylpropane) – all 12 H are equivalent
⇒ one monochloride.
(ii) n-Pentane, CH₃CH₂CH₂CH₂CH₃ – three kinds of H (C-1/C-5, C-2/C-4, C-3)
⇒ three monochlorides.
(iii) Isopentane, (CH₃)₂CHCH₂CH₃ (2-methylbutane) – four kinds of H
⇒ four monochlorides.
5. Draw the structures of major monohalo products in each of the following reactions: (i) cyclohexanol + SOCl₂ → (ii) 4-ethylnitrobenzene + Br₂ (heat or UV light) → (iii) 4-(hydroxymethyl)phenol (HO–C₆H₄–CH₂OH) + HCl, heat → (iv) 1-methylcyclohexene + HI → (v) CH₃CH₂Br + NaI → (vi) cyclohexene + Br₂, heat/UV light →
Answer
(i) C₆H₁₁OH + SOCl₂
⇒ C₆H₁₁Cl (chlorocyclohexane) + SO₂ + HCl
(ii) O₂N–C₆H₄–CH₂CH₃ + Br₂
⇒ O₂N–C₆H₄–CH(Br)CH₃ (1-bromo-1-(4-nitrophenyl)ethane); free-radical substitution occurs at the benzylic carbon.
(iii) HO–C₆H₄–CH₂OH + HCl
⇒ HO–C₆H₄–CH₂Cl (4-(chloromethyl)phenol); only the alcoholic –OH is replaced, the phenolic –OH is not.
(iv) 1-Methylcyclohexene + HI
⇒ 1-iodo-1-methylcyclohexane (Markovnikov addition; I goes to the ring carbon bearing CH₃).
(v) CH₃CH₂Br + NaI
⇒ CH₃CH₂I + NaBr (Finkelstein reaction, in dry acetone).
(vi) Cyclohexene + Br₂ (heat/UV)
⇒ 3-bromocyclohexene (allylic substitution).
6. Arrange each set of compounds in order of increasing boiling points. (i) Bromomethane, Bromoform, Chloromethane, Dibromomethane. (ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Answer
(i) Chloromethane < Bromomethane < Dibromomethane < Bromoform. Boiling point rises with molecular mass (stronger van der Waals forces).
(ii) Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane. Branching lowers surface area and b.p.; a longer chain raises it.
7. Which alkyl halide from the following pairs would you expect to react more rapidly by an SN2 mechanism? Explain your answer. (i) CH₃CH₂CH₂CH₂Br or CH₃CH₂CH(Br)CH₃ (ii) CH₃CH₂CH(Br)CH₃ or (CH₃)₃CBr (iii) CH₃CH(CH₃)CH₂CH₂Br or CH₃CH₂CH(CH₃)CH₂Br
Answer
SN2 rate decreases with steric crowding at (and near) the carbon bearing the halogen.
(i) CH₃CH₂CH₂CH₂Br – it is a primary halide, least hindered.
(ii) CH₃CH₂CH(Br)CH₃ – a secondary halide reacts faster than a tertiary one.
(iii) CH₃CH(CH₃)CH₂CH₂Br – the methyl branch is farther from the C–Br carbon; in the other compound the CH₃ on the adjacent carbon hinders the backside attack.
8. In the following pairs of halogen compounds, which compound undergoes faster SN1 reaction? (i) 2-chloro-2-methylpropane and 2-chlorobutane (ii) 2-chloroheptane and 1-chlorohexane
Answer
(i) 2-Chloro-2-methylpropane, (CH₃)₃CCl – it forms a tertiary carbocation, which is more stable than the secondary one.
(ii) 2-Chloroheptane – it forms a secondary carbocation, more stable than the primary carbocation from 1-chlorohexane.
9. Identify A, B, C, D, E, R and R¹ in the following: Cyclohexyl bromide + Mg (dry ether) → A —H₂O→ B; R–Br + Mg (dry ether) → C —D₂O→ CH₃CH(D)CH₃; R¹–X —Na/ether→ (CH₃)₃C–C(CH₃)₃; R¹–X —Mg→ D —H₂O→ E.
Answer
A = C₆H₁₁MgBr (cyclohexylmagnesium bromide); B = C₆H₁₂ (cyclohexane), since C₆H₁₁MgBr + H₂O
⇒ C₆H₁₂ + Mg(OH)Br.
C gives CH₃CHDCH₃ with D₂O, so C = (CH₃)₂CHMgBr and R = (CH₃)₂CH– (isopropyl).
Wurtz coupling of R¹X gives 2,2,3,3-tetramethylbutane, so R¹ = (CH₃)₃C– (tert-butyl).
D = (CH₃)₃CMgX (tert-butylmagnesium halide); E = (CH₃)₃CH (2-methylpropane).
Exercises
1. Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: (i) (CH₃)₂CHCH(Cl)CH₃ (ii) CH₃CH₂CH(CH₃)CH(C₂H₅)Cl (iii) CH₃CH₂C(CH₃)₂CH₂I (iv) (CH₃)₃CCH₂CH(Br)C₆H₅ (v) CH₃CH(CH₃)CH(Br)CH₃ (vi) CH₃C(C₂H₅)₂CH₂Br (vii) CH₃C(Cl)(C₂H₅)CH₂CH₃ (viii) CH₃CH=C(Cl)CH₂CH(CH₃)₂ (ix) CH₃CH=CHC(Br)(CH₃)₂ (x) p-ClC₆H₄CH₂CH(CH₃)₂ (xi) m-ClCH₂C₆H₄CH₂C(CH₃)₃ (xii) o-Br-C₆H₄CH(CH₃)CH₂CH₃
Answer
(i) 2-Chloro-3-methylbutane – secondary alkyl halide.
(ii) 3-Chloro-4-methylhexane – secondary alkyl halide.
(iii) 1-Iodo-2,2-dimethylbutane – primary alkyl halide.
(iv) 1-Bromo-3,3-dimethyl-1-phenylbutane – secondary benzylic halide.
(v) 2-Bromo-3-methylbutane – secondary alkyl halide.
(vi) 1-Bromo-2-ethyl-2-methylbutane – primary alkyl halide.
(vii) 3-Chloro-3-methylpentane – tertiary alkyl halide.
(viii) 3-Chloro-5-methylhex-2-ene – vinylic halide.
(ix) 4-Bromo-4-methylpent-2-ene – tertiary allylic halide.
(x) 1-Chloro-4-(2-methylpropyl)benzene – aryl halide.
(xi) 1-Chloromethyl-3-(2,2-dimethylpropyl)benzene – primary benzylic halide.
(xii) 1-Bromo-2-(1-methylpropyl)benzene – aryl halide.
2. Give the IUPAC names of the following compounds: (i) CH₃CH(Cl)CH(Br)CH₃ (ii) CHF₂CBrClF (iii) ClCH₂C≡CCH₂Br (iv) (CCl₃)₃CCl (v) CH₃C(p-ClC₆H₄)₂CH(Br)CH₃ (vi) (CH₃)₃CCH=CClC₆H₄I-p
Answer
(i) 2-Bromo-3-chlorobutane
(ii) 1-Bromo-1-chloro-1,2,2-trifluoroethane
(iii) 1-Bromo-4-chlorobut-2-yne
(iv) 2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane
(v) 2-Bromo-3,3-bis(4-chlorophenyl)butane
(vi) 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene
3. Write the structures of the following organic halogen compounds. (i) 2-Chloro-3-methylpentane (ii) p-Bromochlorobenzene (iii) 1-Chloro-4-ethylcyclohexane (iv) 2-(2-Chlorophenyl)-1-iodooctane (v) 2-Bromobutane (vi) 4-tert-Butyl-3-iodoheptane (vii) 1-Bromo-4-sec-butyl-2-methylbenzene (viii) 1,4-Dibromobut-2-ene
Answer
(i) CH₃CH(Cl)CH(CH₃)CH₂CH₃
(ii) Br–C₆H₄–Cl (Br and Cl para, 1,4 on the benzene ring)
(iii) Cyclohexane ring with Cl at C-1 and C₂H₅ at C-4
(iv) ICH₂–CH(C₆H₄Cl-o)–CH₂CH₂CH₂CH₂CH₂CH₃, i.e. the C-2 of octane carries a 2-chlorophenyl group and C-1 carries I
(v) CH₃CH(Br)CH₂CH₃
(vi) CH₃CH₂CH(I)CH[C(CH₃)₃]CH₂CH₂CH₃
(vii) Benzene ring: Br at C-1, CH₃ at C-2, CH(CH₃)CH₂CH₃ at C-4
(viii) BrCH₂CH=CHCH₂Br
4. Which one of the following has the highest dipole moment? (i) CH₂Cl₂ (ii) CHCl₃ (iii) CCl₄
Answer
CH₂Cl₂ (about 1.60 D). CCl₄ is symmetrical tetrahedral, so its bond dipoles cancel (μ = 0). In CHCl₃ (μ ≈ 1.03 D) the resultant of two C–Cl dipoles is partly opposed by the third, whereas in CH₂Cl₂ the two C–Cl dipoles reinforce each other along with the C–H dipoles.
5. A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Answer
C₅H₁₀ with no reaction with Cl₂ in the dark has no C=C bond, so it is a cycloalkane. A single monochloride means all 10 H are equivalent
⇒ the hydrocarbon is cyclopentane. C₅H₁₀ + Cl₂ —hν
⇒ C₅H₉Cl (chlorocyclopentane) + HCl.
6. Write the isomers of the compound having formula C₄H₉Br.
Answer
(i) CH₃CH₂CH₂CH₂Br – 1-bromobutane
(ii) CH₃CH₂CH(Br)CH₃ – 2-bromobutane (exists as a pair of enantiomers, being chiral)
(iii) (CH₃)₂CHCH₂Br – 1-bromo-2-methylpropane
(iv) (CH₃)₃CBr – 2-bromo-2-methylpropane
7. Write the equations for the preparation of 1-iodobutane from (i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene.
Answer
(i) CH₃CH₂CH₂CH₂OH + KI + H₃PO₄ (heat)
⇒ CH₃CH₂CH₂CH₂I + KH₂PO₄ + H₂O (or using red P/I₂).
(ii) CH₃CH₂CH₂CH₂Cl + NaI —dry acetone
⇒ CH₃CH₂CH₂CH₂I + NaCl (Finkelstein reaction; NaCl precipitates and drives it forward).
(iii) CH₃CH₂CH=CH₂ + HBr —peroxide
⇒ CH₃CH₂CH₂CH₂Br (anti-Markovnikov); then CH₃CH₂CH₂CH₂Br + NaI —dry acetone
⇒ CH₃CH₂CH₂CH₂I + NaBr.
8. What are ambident nucleophiles? Explain with an example.
Answer
Nucleophiles that have two different nucleophilic sites and can attack through either of them are called ambident nucleophiles. Example: cyanide ion, [:C≡N:]⁻. Attack through carbon gives alkyl cyanides (R–CN, with KCN, which is ionic), and attack through nitrogen gives isocyanides (R–NC, with AgCN, which is covalent). Similarly the nitrite ion (–O–N=O) gives alkyl nitrites R–O–N=O (via O, with KNO₂) or nitroalkanes R–NO₂ (via N, with AgNO₂).
9. Which compound in each of the following pairs will react faster in SN2 reaction with –OH? (i) CH₃Br or CH₃I (ii) (CH₃)₃CCl or CH₃Cl
Answer
(i) CH₃I – I⁻ is a better leaving group than Br⁻ because the C–I bond is weaker.
(ii) CH₃Cl – the methyl carbon is far less hindered than the tertiary carbon, so backside attack is easy.
10. Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene: (i) 1-Bromo-1-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 2,2,3-Trimethyl-3-bromopentane.
Answer
According to Saytzeff (Zaitsev) rule, the more substituted alkene is the major product.
(i) 1-Methylcyclohexene (major, trisubstituted, endocyclic C=C) and methylenecyclohexane, C₆H₁₀=CH₂ (minor).
(ii) (CH₃)₂C=CHCH₃, 2-methylbut-2-ene (major) and CH₂=C(CH₃)CH₂CH₃, 2-methylbut-1-ene (minor).
(iii) Substrate: (CH₃)₃C–C(Br)(CH₃)–CH₂CH₃. β-Hydrogens are on the CH₃ and CH₂ groups only. Products: (CH₃)₃C–C(CH₃)=CHCH₃, 3,4,4-trimethylpent-2-ene (major, more substituted) and (CH₃)₃C–C(=CH₂)–CH₂CH₃, 2-ethyl-3,3-dimethylbut-1-ene (minor). (3,4,4-Trimethylpent-2-ene exists as cis and trans forms.)
11. How will you bring about the following conversions? (i) Ethanol to but-1-yne (ii) Ethane to bromoethene (iii) Propene to 1-nitropropane (iv) Toluene to benzyl alcohol (v) Propene to propyne (vi) Ethanol to ethyl fluoride (vii) Bromomethane to propanone (viii) But-1-ene to but-2-ene (ix) 1-Chlorobutane to n-octane (x) Benzene to biphenyl.
Answer
(i) CH₃CH₂OH + SOCl₂
⇒ CH₃CH₂Cl
⇒ then HC≡CH + NaNH₂
⇒ HC≡C⁻Na⁺; HC≡CNa + CH₃CH₂Cl
⇒ CH₃CH₂C≡CH (but-1-yne) + NaCl.
(ii) CH₃CH₃ + Br₂ —hν
⇒ CH₃CH₂Br
⇒ alc. KOH
⇒ CH₂=CH₂
⇒ Br₂/CCl₄
⇒ BrCH₂CH₂Br
⇒ alc. KOH (1 mol)
⇒ CH₂=CHBr (bromoethene).
(iii) CH₃CH=CH₂ + HBr —peroxide
⇒ CH₃CH₂CH₂Br
⇒ AgNO₂ (ethanol)
⇒ CH₃CH₂CH₂NO₂ + AgBr.
(iv) C₆H₅CH₃ + Cl₂ —hν
⇒ C₆H₅CH₂Cl
⇒ aq. KOH
⇒ C₆H₅CH₂OH + KCl.
(v) CH₃CH=CH₂ + Br₂ (CCl₄)
⇒ CH₃CHBrCH₂Br
⇒ alc. KOH, then NaNH₂
⇒ CH₃C≡CH (propyne).
(vi) CH₃CH₂OH + PBr₃
⇒ CH₃CH₂Br
⇒ AgF (Swarts reaction)
⇒ CH₃CH₂F + AgBr.
(vii) CH₃Br + Mg (dry ether)
⇒ CH₃MgBr; CH₃MgBr + CH₃CHO, then H₂O
⇒ CH₃CH(OH)CH₃
⇒ oxidation (K₂Cr₂O₇/H⁺ or Cu, 573 K)
⇒ CH₃COCH₃ (propanone).
(viii) CH₃CH₂CH=CH₂ + HBr
⇒ CH₃CH₂CH(Br)CH₃ (Markovnikov)
⇒ alc. KOH
⇒ CH₃CH=CHCH₃ (major, Saytzeff product).
(ix) 2 CH₃CH₂CH₂CH₂Cl + 2Na —dry ether
⇒ CH₃(CH₂)₆CH₃ (n-octane) + 2NaCl (Wurtz reaction).
(x) C₆H₆ + Br₂ —FeBr₃
⇒ C₆H₅Br; 2 C₆H₅Br + 2Na —dry ether
⇒ C₆H₅–C₆H₅ (biphenyl) + 2NaBr (Fittig reaction).
12. Explain why (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride? (ii) alkyl halides, though polar, are immiscible with water? (iii) Grignard reagents should be prepared under anhydrous conditions?
Answer
(i) In chlorobenzene the C–Cl carbon is sp² hybridised (more electronegative than sp³ carbon), so the C–Cl bond is less polar. Also, resonance gives the C–Cl bond partial double-bond character and pushes electron density from Cl into the ring, opposing the bond dipole. In cyclohexyl chloride the carbon is sp³ and no resonance occurs, so its dipole moment is larger.
(ii) To dissolve, an alkyl halide would have to break the strong hydrogen bonds between water molecules. The new attractions between haloalkane and water are weaker than these, so the energy released is too small and they do not mix.
(iii) Grignard reagents are very reactive; the carbon attached to Mg behaves as a carbanion and grabs a proton from any source. Even traces of water destroy them: RMgX + H₂O
⇒ RH + Mg(OH)X. Hence dry (anhydrous) conditions are needed.
13. Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Answer
Freon-12 (CCl₂F₂): refrigerant in refrigerators and air conditioners, propellant in aerosol sprays.
DDT: insecticide, widely used to control mosquitoes that spread malaria and against insects attacking crops (now restricted because it is non-biodegradable and accumulates in food chains).
Carbon tetrachloride: manufacture of refrigerants and aerosol propellants; feedstock for CFCs and some pharmaceuticals; earlier used as a cleaning fluid, spot remover and fire extinguisher.
Iodoform (CHI₃): used earlier as an antiseptic (its action is due to liberation of free iodine); replaced now because of its unpleasant smell.
14. Write the structure of the major organic product in each of the following reactions: (i) CH₃CH₂CH₂Cl + NaI (acetone, heat) → (ii) (CH₃)₃CBr + KOH (ethanol, heat) → (iii) CH₃CH(Br)CH₂CH₃ + NaOH (water) → (iv) CH₃CH₂Br + KCN (aq. ethanol) → (v) C₆H₅ONa + C₂H₅Cl → (vi) CH₃CH₂CH₂OH + SOCl₂ → (vii) CH₃CH₂CH=CH₂ + HBr (peroxide) → (viii) CH₃CH=C(CH₃)₂ + HBr →
Answer
(i) CH₃CH₂CH₂I (1-iodopropane)
(ii) (CH₃)₂C=CH₂ (2-methylpropene) – elimination
(iii) CH₃CH(OH)CH₂CH₃ (butan-2-ol)
(iv) CH₃CH₂CN (propanenitrile)
(v) C₆H₅OC₂H₅ (ethoxybenzene, phenetole) – Williamson synthesis
(vi) CH₃CH₂CH₂Cl (1-chloropropane) + SO₂ + HCl
(vii) CH₃CH₂CH₂CH₂Br (1-bromobutane) – anti-Markovnikov addition
(viii) CH₃CH₂C(Br)(CH₃)₂ (2-bromo-2-methylbutane) – Markovnikov addition
15. Write the mechanism of the following reaction: nBuBr + KCN —EtOH–H₂O→ nBuCN
Answer
KCN is ionic and gives CN⁻, an ambident nucleophile. Since C–C bond is stronger than C–N, attack occurs mainly through carbon. n-Butyl bromide is a primary halide, so the reaction follows the SN2 path in one step:
N≡C⁻ + CH₃CH₂CH₂CH₂–Br
⇒ [N≡C···CH₂(C₃H₇)···Br]⁻ (transition state: C partly bonded to both CN and Br, other three groups in a plane)
⇒ N≡C–CH₂CH₂CH₂CH₃ + Br⁻
CN⁻ attacks from the side opposite to Br; the C–CN bond forms as the C–Br bond breaks, the configuration at carbon is inverted, and rate = k[nBuBr][CN⁻].
16. Arrange the compounds of each set in order of reactivity towards SN2 displacement: (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.
Answer
(i) 1-Bromopentane (1°) > 2-Bromopentane (2°) > 2-Bromo-2-methylbutane (3°)
(ii) 1-Bromo-3-methylbutane (1°) > 2-Bromo-3-methylbutane (2°) > 2-Bromo-2-methylbutane (3°)
(iii) 1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2,2-dimethylpropane. All are primary; reactivity falls as branching moves closer to the C–Br carbon (more steric hindrance).
17. Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH.
Answer
C₆H₅CHClC₆H₅. Hydrolysis proceeds by SN1; the carbocation (C₆H₅)₂CH⁺ is stabilised by resonance with two phenyl rings, while C₆H₅CH₂⁺ has only one. The more stable carbocation forms faster, so C₆H₅CHClC₆H₅ hydrolyses more easily.
18. p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.
Answer
The para isomer is symmetrical and fits more closely and regularly in the crystal lattice than the ortho and meta isomers. Stronger intermolecular forces in the lattice need more energy to break, so p-dichlorobenzene melts at a higher temperature (about 323 K versus 256 K and 249 K).
19. How the following conversions can be carried out? (i) Propene to propan-1-ol (ii) Ethanol to but-1-yne (iii) 1-Bromopropane to 2-bromopropane (iv) Toluene to benzyl alcohol (v) Benzene to 4-bromonitrobenzene (vi) Benzyl alcohol to 2-phenylethanoic acid (vii) Ethanol to propanenitrile (viii) Aniline to chlorobenzene (ix) 2-Chlorobutane to 3,4-dimethylhexane (x) 2-Methyl-1-propene to 2-chloro-2-methylpropane (xi) Ethyl chloride to propanoic acid (xii) But-1-ene to n-butyliodide (xiii) 2-Chloropropane to 1-propanol (xiv) Isopropyl alcohol to iodoform (xv) Chlorobenzene to p-nitrophenol (xvi) 2-Bromopropane to 1-bromopropane (xvii) Chloroethane to butane (xviii) Benzene to diphenyl (xix) tert-Butyl bromide to isobutyl bromide (xx) Aniline to phenylisocyanide
Answer
(i) CH₃CH=CH₂ + HBr —peroxide
⇒ CH₃CH₂CH₂Br
⇒ aq. KOH
⇒ CH₃CH₂CH₂OH
(ii) CH₃CH₂OH + SOCl₂
⇒ CH₃CH₂Cl; HC≡CH + NaNH₂
⇒ HC≡CNa; HC≡CNa + CH₃CH₂Cl
⇒ CH₃CH₂C≡CH + NaCl
(iii) CH₃CH₂CH₂Br —alc. KOH, heat
⇒ CH₃CH=CH₂ —HBr (no peroxide)
⇒ CH₃CHBrCH₃
(iv) C₆H₅CH₃ + Cl₂ —hν
⇒ C₆H₅CH₂Cl
⇒ aq. KOH
⇒ C₆H₅CH₂OH
(v) C₆H₆ + Br₂ —FeBr₃
⇒ C₆H₅Br
⇒ conc. HNO₃ + conc. H₂SO₄
⇒ p-BrC₆H₄NO₂ (major, Br is o/p directing; o-isomer separated)
(vi) C₆H₅CH₂OH + SOCl₂
⇒ C₆H₅CH₂Cl
⇒ KCN (alc.)
⇒ C₆H₅CH₂CN
⇒ H₃O⁺, heat
⇒ C₆H₅CH₂COOH
(vii) CH₃CH₂OH + PBr₃
⇒ CH₃CH₂Br
⇒ KCN (aq. ethanol)
⇒ CH₃CH₂CN
(viii) C₆H₅NH₂ + NaNO₂ + HCl (273–278 K)
⇒ C₆H₅N₂⁺Cl⁻
⇒ Cu₂Cl₂/HCl (Sandmeyer)
⇒ C₆H₅Cl + N₂
(ix) 2 CH₃CH₂CH(Cl)CH₃ + 2Na —dry ether
⇒ CH₃CH₂CH(CH₃)–CH(CH₃)CH₂CH₃ (3,4-dimethylhexane) + 2NaCl
(x) (CH₃)₂C=CH₂ + HCl
⇒ (CH₃)₃CCl (Markovnikov addition)
(xi) CH₃CH₂Cl + KCN (alc.)
⇒ CH₃CH₂CN
⇒ H₃O⁺, heat
⇒ CH₃CH₂COOH
(xii) CH₃CH₂CH=CH₂ + HBr —peroxide
⇒ CH₃CH₂CH₂CH₂Br
⇒ NaI/dry acetone
⇒ CH₃CH₂CH₂CH₂I
(xiii) CH₃CHClCH₃ —alc. KOH
⇒ CH₃CH=CH₂ —HBr, peroxide
⇒ CH₃CH₂CH₂Br —aq. KOH
⇒ CH₃CH₂CH₂OH
(xiv) CH₃CH(OH)CH₃ + I₂ + NaOH (warm)
⇒ CH₃COCH₃ (oxidation)
⇒ CHI₃ (yellow ppt) + CH₃COONa (iodoform reaction)
(xv) C₆H₅Cl + conc. HNO₃/conc. H₂SO₄
⇒ p-ClC₆H₄NO₂
⇒ NaOH (aq), 443 K, then H⁺
⇒ p-HOC₆H₄NO₂ (the –NO₂ group at para position activates Cl towards nucleophilic substitution)
(xvi) CH₃CHBrCH₃ —alc. KOH
⇒ CH₃CH=CH₂ —HBr, peroxide
⇒ CH₃CH₂CH₂Br
(xvii) 2 CH₃CH₂Cl + 2Na —dry ether
⇒ CH₃CH₂CH₂CH₃ + 2NaCl (Wurtz reaction)
(xviii) C₆H₆ + Br₂ —FeBr₃
⇒ C₆H₅Br; 2 C₆H₅Br + 2Na —dry ether
⇒ C₆H₅–C₆H₅ + 2NaBr (Fittig reaction)
(xix) (CH₃)₃CBr —alc. KOH
⇒ (CH₃)₂C=CH₂ —HBr, peroxide
⇒ (CH₃)₂CHCH₂Br (isobutyl bromide)
(xx) C₆H₅NH₂ + CHCl₃ + 3KOH (alc.), heat
⇒ C₆H₅NC + 3KCl + 3H₂O (carbylamine reaction)
20. The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Answer
In water, KOH is fully ionised and the OH⁻ ions are heavily hydrated; they act mainly as nucleophiles and substitute Cl to give alcohols: R–Cl + OH⁻
⇒ R–OH + Cl⁻. In alcohol, KOH gives alkoxide ions (RO⁻ from C₂H₅OH + OH⁻ ⇌ C₂H₅O⁻ + H₂O), which are stronger bases than OH⁻ and less solvated. They pull off a β-hydrogen, causing dehydrohalogenation (β-elimination) to give alkenes: CH₃CH₂Cl + C₂H₅O⁻
⇒ CH₂=CH₂ + C₂H₅OH + Cl⁻.
21. Primary alkyl halide C₄H₉Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
Answer
Since (a) is primary but is not n-butyl bromide, (a) = (CH₃)₂CHCH₂Br (1-bromo-2-methylpropane).
(CH₃)₂CHCH₂Br + KOH (alc.)
⇒ (CH₃)₂C=CH₂ (b, 2-methylpropene) + KBr + H₂O
(CH₃)₂C=CH₂ + HBr
⇒ (CH₃)₃CBr (c, 2-bromo-2-methylpropane, an isomer of a)
2 (CH₃)₂CHCH₂Br + 2Na —dry ether
⇒ (CH₃)₂CHCH₂–CH₂CH(CH₃)₂ (d, 2,5-dimethylhexane, C₈H₁₈) + 2NaBr
(n-Butyl bromide would give n-octane instead.)
22. What happens when (i) n-butyl chloride is treated with alcoholic KOH, (ii) bromobenzene is treated with Mg in the presence of dry ether, (iii) chlorobenzene is subjected to hydrolysis, (iv) ethyl chloride is treated with aqueous KOH, (v) methyl bromide is treated with sodium in the presence of dry ether, (vi) methyl chloride is treated with KCN?
Answer
(i) Dehydrohalogenation gives but-1-ene: CH₃CH₂CH₂CH₂Cl + KOH (alc.)
⇒ CH₃CH₂CH=CH₂ + KCl + H₂O
(ii) A Grignard reagent is formed: C₆H₅Br + Mg —dry ether
⇒ C₆H₅MgBr (phenylmagnesium bromide)
(iii) Chlorobenzene is hydrolysed only under drastic conditions: C₆H₅Cl + NaOH (aq) —623 K, 300 atm
⇒ C₆H₅ONa —H⁺
⇒ C₆H₅OH (phenol)
(iv) Ethanol is formed by substitution: C₂H₅Cl + KOH (aq)
⇒ C₂H₅OH + KCl
(v) Ethane is formed (Wurtz reaction): 2CH₃Br + 2Na —dry ether
⇒ CH₃–CH₃ + 2NaBr
(vi) Methyl cyanide (ethanenitrile) is formed: CH₃Cl + KCN
⇒ CH₃CN + KCl
Important Extra Questions
Why are aryl halides less reactive towards nucleophilic substitution than alkyl halides?
Answer
Resonance gives the C–X bond partial double-bond character; the carbon is sp² (shorter, stronger bond); the phenyl cation is unstable so SN1 is impossible; and the electron-rich ring repels the incoming nucleophile.
What is the effect of an –NO₂ group at the ortho/para position on the reactivity of chlorobenzene?
Answer
It withdraws electrons and stabilises the carbanion intermediate, so substitution of Cl by OH⁻ becomes easier; 2,4,6-trinitrochlorobenzene is hydrolysed even by warm water.
Why is chloroform stored in dark coloured bottles filled to the brim?
Answer
In light and air chloroform is oxidised to the very poisonous phosgene: 2CHCl₃ + O₂
⇒ 2COCl₂ + 2HCl. Dark bottles cut off light and full bottles keep air out.
What is a racemic mixture? Why does SN1 hydrolysis of an optically active halide give one?
Answer
An equimolar mixture of two enantiomers, optically inactive. In SN1 the planar carbocation is attacked equally from both faces, so both configurations form.
Write the Swarts reaction.
Answer
Heating an alkyl chloride or bromide with a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃ gives the alkyl fluoride: CH₃Br + AgF
⇒ CH₃F + AgBr.