JEE Chemistry Haloalkanes and Haloarenes MCQs Set 23

Answer: (b) 3-Hydroxycyclohexan-1-one

Explanation:
1. 3-Hydroxycyclohexanone is a beta-hydroxy ketone: the alpha hydrogen is acidic and loss of water gives a conjugated enone (cyclohex-2-en-1-one).
2. The conjugated product is especially stable, so dehydration is easiest.

Teacher's Note:
1. Beta-hydroxy carbonyl compounds dehydrate readily.

Answer: (c) \( E2 \) mechanism

Explanation:
1. Methoxide is a strong base and heat favours elimination.
2. A secondary halide with a strong base undergoes bimolecular elimination to give cyclohexene.

Teacher's Note:
1. Strong base plus heat on a secondary halide gives E2.

Answer: (d) (2-Bromophenyl)methanethiol

Explanation:
1. The benzylic C-Br bond is very reactive towards nucleophilic substitution (\( S_N1 \)/\( S_N2 \)).
2. The aryl C-Br bond has partial double bond character and is inert under these conditions.
3. So only \( \ce{CH2Br} \) is converted to \( \ce{CH2SH} \).

Teacher's Note:
1. Aryl halides do not undergo ordinary nucleophilic substitution.

Answer: (d) 4-Nitrophenyl benzoate

Explanation:
1. The ring attached to O (the phenoxy ring) is activated by the lone pair on O, though weaker than in a phenol.
2. The benzoyl ring is deactivated by the C=O group, so nitration occurs on the phenoxy ring, para to the O.

Teacher's Note:
1. The ring bearing the donor atom is the more reactive ring.

Answer: (c) 2-(3-Iodophenyl)acetonitrile

Explanation:
1. The benzylic C-Cl bond undergoes \( S_N2 \) substitution by cyanide in the polar aprotic solvent DMF.
2. The aryl C-I bond is unreactive towards \( S_N2 \), so iodine stays on the ring.

Teacher's Note:
1. Substitution occurs at the benzylic halide only.

Answer: (c) \( \ce{C6H5-CHCl-OCH3} \)

Explanation:
1. The reaction proceeds by \( S_N1 \) through a carbocation.
2. In the first compound the cation is stabilised by both the phenyl ring and the lone pair of oxygen (oxocarbenium ion), making it the most stable.

Teacher's Note:
1. Lone pair donation from O gives the greatest stabilisation.

Answer: (d) 2-Bromo-2-methylpentane with \( \ce{CH3ONa/CH3OH} \), heat

Explanation:
1. With a good leaving group (Br) and a small base, E2 gives the more stable, more substituted alkene (Saytzeff).
2. Quaternary ammonium hydroxides, alkyl fluorides and ester pyrolysis give the Hofmann (less substituted) alkene as the major product.

Teacher's Note:
1. Poor leaving groups and bulky bases favour the Hofmann product.

Answer: (c) c > d > b > a

Explanation:
1. Toluene is activated by the +I and hyperconjugation effect of CH3.
2. Chlorobenzene is deactivated (-I outweighs +M), and nitrobenzene is strongly deactivated (-M, -I).

Teacher's Note:
1. Activating groups speed up EAS; deactivating groups slow it down.

Answer: (a) Butanoyl chloride + \( \ce{AlCl3} \), followed by Zn-Hg/HCl

Explanation:
1. Acylation followed by Clemmensen reduction gives n-butylbenzene (no rearrangement).
2. The other three routes generate the secondary butyl carbocation (by rearrangement of the n-butyl cation or by protonation) and give 2-phenylbutane.

Teacher's Note:
1. Acylation then reduction gives straight chain alkylbenzenes.

Answer: (d) X = concentrated alcoholic NaOH, 80 \(^\circ\)C; Y = HBr/\( \ce{CHCl3} \), 20 \(^\circ\)C

Explanation:
1. Step 1 is dehydrohalogenation to propene, needing concentrated alcoholic NaOH and heat.
2. Step 2 adds HBr to propene by Markovnikov addition to give 2-bromopropane.
3. \( \ce{Br2} \) would give 1,2-dibromopropane, and dilute aqueous NaOH gives an alcohol.

Teacher's Note:
1. Alcoholic alkali eliminates; aqueous alkali substitutes.

Answer: (b) position isomers

Explanation:
1. A comes from nitration of chlorobenzene (o/p directing), giving mainly 4-chloronitrobenzene.
2. B comes from chlorination of nitrobenzene (m directing), giving 3-chloronitrobenzene.
3. Both are \( \ce{C6H4ClNO2} \) differing in the position of substituents.

Teacher's Note:
1. Cl directs o/p and NO2 directs meta.

Answer: (c) I > II > IV > III

Explanation:
1. Good leaving groups are weak bases (conjugate bases of strong acids).
2. Acid strengths: HBr > acetic acid > phenol > ethanol, so leaving ability is Br- > acetate > phenoxide > ethoxide.

Teacher's Note:
1. Weaker base means better leaving group.

Answer: (b) \( S_N2 \)

Explanation:
1. Pyridine takes up HCl and frees Cl- as a nucleophile.
2. Cl- attacks the alkyl chlorosulfite from the back side, giving inversion: an \( S_N2 \) pathway (without pyridine the \( S_Ni \) pathway gives retention).

Teacher's Note:
1. Pyridine converts the reaction to \( S_N2 \).

Answer: (b) (b) and (d)

Explanation:
1. \( S_N1 \) gives racemisation (a mixture of retention and inversion), not complete inversion, so (b) is false.
2. The rate-determining step is the ionisation of the alkyl halide, so the nucleophile's strength does not matter, and (d) is false.
3. All other statements are true.

Teacher's Note:
1. Inversion is characteristic of \( S_N2 \), not \( S_N1 \).

Answer: (b) 4-Hydroxy-3-nitrobenzoic acid

Explanation:
1. The OH group is a strong activating ortho-para director and COOH is a deactivating meta director.
2. Both effects direct the incoming NO2 to the position ortho to OH (meta to COOH): C3.

Teacher's Note:
1. The stronger activating group controls the position.

Answer: (a) \( \ce{CH3CHCl2} \)

Explanation:
1. \( \ce{CH3CHCl2} \) is a geminal dihalide; hydrolysis gives \( \ce{CH3CH(OH)2} \), which loses water to give \( \ce{CH3CHO} \).
2. 1,2-Dichloroethane gives ethylene glycol and acetyl chloride gives acetic acid.

Teacher's Note:
1. Geminal dihalides give carbonyl compounds on hydrolysis.

Answer: (d) (a), (c), (d) and (e)

Explanation:
1. HCl with ZnCl2 (Lucas reagent), PCl3, PCl5 and SOCl2 all replace OH by Cl.
2. NaCl is not acidic and cannot protonate the OH, so it cannot convert an alcohol to an alkyl chloride.

Teacher's Note:
1. The OH must be converted into a good leaving group first.

Answer: (c) II > I > III

Explanation:
1. II ionises to the tropylium ion, which is aromatic (6 pi electrons) and very stable.
2. I gives an ordinary secondary carbocation.
3. III would give the cyclopentadienyl cation (4 pi electrons), which is anti-aromatic and extremely unstable.

Teacher's Note:
1. Aromatic cations form easily, anti-aromatic cations do not.