Answer to Some Selected Problems UNIT 8 8.25 15 g UNIT 12 12.32 Mass of carbon dioxide formed = 0.505 g Mass of water formed = 0.0864 g 12.33 % fo nitrogen = 56 12.34 % of chlorine = 37.57 12.35 % of sulphur = 19.66 UNIT 13 • 13.1 Due to the side reaction in termination step by the combination of two CH3 free radicals. 13.2 (a) 2-Methyl-but-2-ene (b) Pent-1-ene-3-yne (c) Buta-1, 3-diene (d) 4-Phenylbut-1-ene (e) 2-Methylphenol (f) 5-(2-Methylpropyl)-decane (g) 4-Ethyldeca –1,5,8- triene 13.3 (a) (i) CH2 = CH – CH2 – CH3 But-1-ene (ii) CH3 – CH2 = CH – CH3 But-2-ene (iii) CH2 = C – CH3 2-Methylpropene|CH3 (b) (i) HC ≡C – CH2 – CH2 – CH3 Pent-1-yne (ii) CH3 – C ≡C – CH2 – CH3 Pent-2-yne (iii) CH3 – CH – C ≡CH 3-Methylbut-1-yne| CH3 13.4 (i) Ethanal and propanal (ii) Butan-2-one and pentan-2-one (iii) Methanal and pentan-3-one (iv) Propanal and benzaldehyde 13.5 3-Ethylpent-2-ene 13.6 But-2-ene 13.7 4-Ethylhex-3-ene CH3 – CH2– C = CH – CH2–CH3|CH2–CH3 13.8 (a) 4 10 2C H (g) 13/2O (g) + Δ⎯→ 2 24CO (g) 5H O(g) + (b) 5 10 2C H (g) 15/2 O (g) + Δ⎯→ 2 25CO (g) 5H O(g) + (c) 6 10 2C H (g) 17/2 O (g) + Δ⎯→ 2 26CO (g) 5H O(g) + (d) 7 8 2C H (g) 9O (g) + Δ⎯→ 2 27CO (g) 4H O(g) + cis-Hex-2-ene trans-Hex-2-ene The cis form will have higher boiling point due to more polar nature leading to stronger intermolecular dipole–dipole interaction, thus requiring more heat energy to separate them. 13.10 Due to resonance 13.11 Planar, conjugated ring system with delocalisation of (4n+2) π electrons, where, n is an integer 13.12 Lack of delocalisation of (4n +2) π electrons in the cyclic system. 13.13 (i) (ii) (iii) 13.14 15 H attached to 1° carbons4 H attached to 2° carbons1 H attached to 3° carbons 13.15 More the branching in alkane, lower will be the boiling point. 13.16 Refer to addition reaction of HBr to unsymmetrical alkenes in the text. All the three products cannot be obtained by any one of the Kekulé’s structures. This shows that benzene is a resonance hybrid of the two resonating structures. 13.18 H – C ≡C – H > CH > CH. Due to maximum s orbital character in66614enthyne (50 per cent) as compared to 33 per cent in benzene and 25 per cent in n-hexane. 13.19 Due to the presence of 6 π electrons, benzene behaves as a rich source of electrons thus being easily attacked by reagents deficient in electrons. 13.20 (i) Br2 alc. KOH NaNH 2CH ⎯⎯→CH −CH ⎯⎯⎯⎯→CH =CHBr ⎯⎯⎯→ (ii) 2422 2 || Br Br (iii)CH3 | 13.21 CH2 = C – CH2 – CH3 2-Methylbut-1-ene CH3| CH3 – C = CH – CH3 2-Methylbut-2-ene CH3| CH3 – CH –CH = CH2 3-Methylbut-1-ene 13.22 (a) Chlorobenzene>p-nitrochlorobenzene> 2,4 – dinitrochlorobenzene (b) Toluene > p-CH-CH-NO > p-ON–CH–NO36422642 13.23 Toleune undergoes nitration most easily due to electron releasing nature of the methyl group. 13.24 FeCl3 13.25 Due to the formation of side products. For example, by starting with 1-bromopropane and 1-bromobutane, hexane and octane are the side products besides heptane.