CBSE · NCERT · Class 12 Chemistry · Chapter 6

NCERT Solutions: Class 12 Chemistry Chapter 6 - Haloalkanes and Haloarenes

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Chapter-wise NCERT intext questions and exercise answers for Haloalkanes and Haloarenes, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Exercises 15
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1Exercises15 questions
Q.6.4Which one of the following has the highest dipole moment? (i) CH2Cl2 (ii) CHCl3(iii) CCl4v
  1. i. CH2Cl2
  2. ii. CHCl3
  3. iii. CCl4
Solution

CCl4 has zero dipole moment because its tetrahedral bond dipoles cancel. In CHCl3, partial cancellation is greater than in CH2Cl2. Therefore CH2Cl2 has the highest net dipole moment among the three.

Answer:

(i) CH2Cl2.

Q.6.5A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.v
Solution

The formula C5H10 suggests either an alkene or cycloalkane. Since it does not react with chlorine in the dark, it is not an alkene. Cyclopentane has all hydrogens equivalent, so photochemical chlorination gives only one monochloro product, chlorocyclopentane.

Answer:

Cyclopentane.

Q.6.6Write the isomers of the compound having formula C4H9Br.v
Solution

The four carbon skeletons/positions possible are n-butyl bromide, sec-butyl bromide, isobutyl bromide and tert-butyl bromide. Their condensed structures are CH3CH2CH2CH2Br, CH3CHBrCH2CH3, (CH3)2CHCH2Br and (CH3)3CBr.

Answer:

1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane and 2-bromo-2-methylpropane.

Q.6.7Write the equations for the preparation of 1-iodobutane from (i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene.v
Solution

From 1-butanol, use HI or PI3 generated in situ. From 1-chlorobutane, use Finkelstein reaction with NaI in dry acetone. From but-1-ene, first add HBr in the presence of peroxide to form 1-bromobutane, then replace Br by I using NaI in acetone.

Answer:

(i) CH3CH2CH2CH2OH + HI → CH3CH2CH2CH2I + H2O. (ii) CH3CH2CH2CH2Cl + NaI → CH3CH2CH2CH2I + NaCl. (iii) Convert but-1-ene to 1-bromobutane by anti-Markovnikov HBr addition, then use NaI in acetone.

Q.6.8What are ambident nucleophiles? Explain with an example.v
Solution

Cyanide ion is ambident because it can attack through carbon to give alkyl cyanide, R-CN, or through nitrogen to give isocyanide, R-NC. Nitrite ion is also ambident; it can give R-ONO or R-NO2 depending on the attacking atom.

Answer:

Ambident nucleophiles can attack through two different donor atoms.

Q.6.9Which compound in each of the following pairs will react faster in SN2 reaction with –OH? (i) CH3Br or CH3I (ii) (CH3)3CCl or CH3Clv
Solution

SN2 reactions are favoured by better leaving groups and low steric hindrance. I- is a better leaving group than Br-, so CH3I reacts faster than CH3Br. Methyl chloride is far less hindered than tert-butyl chloride, so CH3Cl reacts faster than (CH3)3CCl in SN2 reaction.

Answer:

(i) CH3I; (ii) CH3Cl.

Q.6.12Explain 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?v
Solution

In chlorobenzene, resonance gives the C-Cl bond partial double-bond character and the sp2 carbon is more electronegative, reducing the bond moment compared with cyclohexyl chloride. Alkyl halides are polar but do not compensate for breaking water-water hydrogen bonds, so they are immiscible. Grignard reagents are destroyed by moisture: RMgX + H2O → RH + Mg(OH)X.

Answer:

(i) C-Cl bond in chlorobenzene has partial double-bond character and lower polarity. (ii) Alkyl halides cannot form strong hydrogen bonds with water. (iii) Grignard reagents react rapidly with water.

Q.6.13Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.v
Solution

Freon-12, CCl2F2, was widely used in refrigeration and aerosol propellants. DDT was used as an insecticide, especially against malaria-spreading mosquitoes. Carbon tetrachloride was used as solvent, cleaning fluid and fire extinguisher fluid, though it is toxic and ozone-depleting. Iodoform was used as an antiseptic because it releases iodine.

Answer:

Freon-12 is used as refrigerant/aerosol propellant; DDT as insecticide; carbon tetrachloride as solvent/fire extinguisher feedstock though restricted; iodoform as an antiseptic due to liberation of iodine.

Q.6.15Write the mechanism of the following reaction: nBuBr + KCN nBuCNv
Solution

Cyanide ion attacks the primary carbon of n-butyl bromide from the backside while the C-Br bond breaks in the same step: CH3CH2CH2CH2Br + CN- → CH3CH2CH2CH2CN + Br-. Because n-butyl bromide is primary, steric hindrance is low and SN2 substitution is favoured.

Answer:

The reaction proceeds by an SN2 mechanism.

Q.6.16Arrange 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.v
Solution

SN2 reactivity decreases with steric hindrance: methyl > primary > secondary >> tertiary. Branching near the reacting carbon also slows SN2 attack. The listed orders follow increasing steric crowding around the carbon bearing bromine.

Answer:

(i) 1-bromopentane > 2-bromopentane > 2-bromo-2-methylbutane. (ii) 1-bromo-3-methylbutane > 2-bromo-3-methylbutane > 2-bromo-2-methylbutane. (iii) 1-bromobutane > 1-bromo-3-methylbutane > 1-bromo-2-methylbutane > 1-bromo-2,2-dimethylpropane.

Q.6.17Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH.v
Solution

Hydrolysis in aqueous KOH can proceed through carbocation formation for benzylic halides. C6H5CHClC6H5 forms a diphenylmethyl carbocation, stabilised by resonance with two phenyl rings, whereas C6H5CH2Cl forms a benzyl carbocation stabilised by only one phenyl ring. Greater carbocation stability gives faster hydrolysis.

Answer:

C6H5CHClC6H5 is hydrolysed more easily.

Q.6.18p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.v
Solution

p-Dichlorobenzene has a symmetrical structure, so its molecules fit more closely and regularly in the crystal lattice. Stronger packing increases lattice energy and therefore melting point compared with the less symmetrical ortho and meta isomers.

Answer:

The para isomer is more symmetrical and packs better in the crystal lattice.

Q.6.20The 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.v
Solution

In aqueous solution, OH- is strongly solvated and acts mainly as a nucleophile, replacing Cl- to form alcohol. In alcoholic KOH, ethoxide/alkoxide character and heating favour abstraction of beta hydrogen, elimination of HCl and formation of alkene.

Answer:

Aqueous KOH favours nucleophilic substitution, while alcoholic KOH favours beta-elimination.

Q.6.21Primary 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.v
Solution

(CH3)2CHCH2Br + alc. KOH → (CH3)2C=CH2 + KBr + H2O. Then (CH3)2C=CH2 + HBr → (CH3)3CBr. Wurtz reaction: 2(CH3)2CHCH2Br + 2Na → (CH3)2CHCH2CH2CH(CH3)2 + 2NaBr. This C8H18 product differs from n-octane formed from n-butyl bromide.

Answer:

(a) is 1-bromo-2-methylpropane, (CH3)2CHCH2Br. (b) is 2-methylpropene. (c) is 2-bromo-2-methylpropane. (d) is 2,5-dimethylhexane.

Q.6.22What 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?v
Solution

(i) CH3CH2CH2CH2Cl + alc. KOH → CH3CH2CH=CH2 + KCl + H2O. (ii) C6H5Br + Mg → C6H5MgBr in dry ether. (iii) C6H5Cl + NaOH at high temperature/pressure gives C6H5ONa, which on acidification gives phenol. (iv) C2H5Cl + KOH(aq) → C2H5OH + KCl. (v) 2CH3Br + 2Na → C2H6 + 2NaBr. (vi) CH3Cl + KCN → CH3CN + KCl.

Answer:

(i) But-1-ene forms. (ii) Phenylmagnesium bromide forms. (iii) Phenol forms only under drastic conditions. (iv) Ethanol forms. (v) Ethane forms. (vi) Ethanenitrile forms.