JEE Chemistry Haloalkanes and Haloarenes MCQs Set 20

Answer: (b) \( \ce{CH3OCH2CH2CH=CHCH2CH3} \) with the Z (cis) double bond

Explanation:
1. Excess sodamide makes the dianion; the acetylide carbanion (more nucleophilic) is ethylated by 1 equivalent of ethyl iodide, giving \( \ce{HOCH2CH2C#CCH2CH3} \) (as the alkoxide).
2. The remaining alkoxide is then methylated by methyl iodide: \( \ce{CH3OCH2CH2C#CCH2CH3} \).
3. Lindlar hydrogenation gives the cis (Z) alkene.

Teacher's Note:
1. Terminal-alkyne carbanion is alkylated before the alkoxide; Lindlar gives cis.

Answer: (c) (2S)-2-hydroxy-4-methylpentanoic acid (retention of configuration)

Explanation:
1. Nitrous acid converts the primary amino group to a diazonium ion, which is displaced by water.
2. In alpha-amino acids the neighbouring carboxylate attacks the carbon (forming an alpha-lactone) and is then opened by water: two inversions give net retention of configuration.
3. The OH replaces NH2 with the same priority order (OH > COOH > CH2 > H), so the descriptor stays S.

Teacher's Note:
1. Double inversion via neighbouring-group participation gives retention.

Answer: (c) \( \ce{(CH3)3C-Cl} \)

Explanation:
1. SN1 needs a stable carbocation; tert-butyl chloride ionises to the tertiary cation.
2. n-Propyl chloride and 2-phenylethyl chloride are primary (would give SN2), and chlorobenzene has a strong sp2 C-Cl bond.

Teacher's Note:
1. Tertiary, allylic and benzylic halides favour SN1.

Answer: (d) A was \( \ce{C6H5I} \)

Explanation:
1. Benzyl iodide is hydrolysed by NaOH to iodide ion, which gives yellow AgI with silver nitrate. Iodobenzene is not hydrolysed (strong aryl C-I bond), so no precipitate forms.
2. Hence B (precipitate) is benzyl iodide and A is iodobenzene. The nitric acid is needed to neutralise excess NaOH, which would otherwise precipitate \( \ce{Ag2O} \).

Teacher's Note:
1. Aryl halides do not hydrolyse under ordinary conditions.

Answer: (b) steric hindrance

Explanation:
1. In SN2 the nucleophile must approach the carbon from the back side; three alkyl groups block this approach.
2. Hence tertiary halides react by SN1 or eliminate instead.

Teacher's Note:
1. SN2 order: methyl > primary > secondary >> tertiary.

Answer: (d) D > C > A > B

Explanation:
1. For nucleophiles with the same attacking atom (oxygen), nucleophilicity parallels basicity.
2. Conjugate acid pKa values: methanol 15.5, water 15.7, phenol 10, acetic acid 4.8. Hence basicity and nucleophilicity: \( \ce{CH3O-} \) > \( \ce{HO-} \) > \( \ce{PhO-} \) > \( \ce{AcO-} \).

Teacher's Note:
1. Same attacking atom: stronger base = better nucleophile.

Answer: (a) \( \ce{RCH2X > R2CHX > R3CX} \)

Explanation:
1. SN2 is governed by steric hindrance to back-side attack, so reactivity falls as alkyl substitution on the carbon increases: primary > secondary > tertiary.

Teacher's Note:
1. Primary halides are the best SN2 substrates.

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

Explanation:
1. A geminal dihalide is hydrolysed to a gem-diol, \( \ce{CH3CH(OH)2} \), which loses water to give the aldehyde \( \ce{CH3CHO} \).
2. Ethyl chloride gives ethanol, 1,2-dichloroethane gives ethylene glycol and acetyl chloride gives acetate.

Teacher's Note:
1. Gem-dihalides at the end of a chain give aldehydes; in the middle, ketones.

Answer: (b) 2-methylpropan-2-ol

Explanation:
1. The Lucas reaction proceeds by SN1; the rate depends on carbocation stability.
2. The tertiary alcohol forms the stable tertiary carbocation fastest: tertiary > secondary > primary.

Teacher's Note:
1. Lucas test: tertiary alcohols give turbidity immediately.

Answer: (c) a carbocation

Explanation:
1. \( \ce{SbCl5} \) is a Lewis acid and removes chloride, giving the planar benzylic carbocation \( \ce{PhCH+CH3} \).
2. The planar cation is recaptured from either face, so the optical activity is lost.

Teacher's Note:
1. Planar carbocations cause racemisation.

Answer: (c) a tertiary alcohol, by SN1

Explanation:
1. Lucas reagent (conc. HCl + ZnCl2) converts alcohols to chlorides; the fastest alcohol gives the most stable carbocation.
2. Tertiary alcohols react immediately by SN1.

Teacher's Note:
1. Tertiary: immediate turbidity; secondary: 5-10 min; primary: only on heating.

Answer: (d) the Swarts reaction

Explanation:
1. Direct fluorination is explosive and unselective, and the Sandmeyer reaction does not introduce fluorine.
2. The Swarts reaction heats an alkyl chloride or bromide with a metallic fluoride (\( \ce{AgF} \), \( \ce{Hg2F2} \), \( \ce{SbF3} \)) to exchange the halogen; the Finkelstein reaction gives iodides.

Teacher's Note:
1. Swarts: R-Cl/Br to R-F; Finkelstein: to R-I.

Answer: (b) 5-tert-butylcyclohex-2-en-1-ol (the OH on the allylic carbon remote from the tert-butyl group)

Explanation:
1. NBS with light brominates the allylic position by a radical chain mechanism; the bulky tert-butyl group makes the allylic CH2 remote from it the more accessible position.
2. Aqueous potassium carbonate then replaces the allylic bromide by OH (SN1/SN2), giving the allylic alcohol. No oxidising agent is present, so no ketone or acid forms and elimination is a minor path.
3. Product: an allylic alcohol.

Teacher's Note:
1. NBS/hv = allylic bromination; then hydrolysis gives the allylic alcohol.

Answer: (d) 1-(2-hydroxyphenyl)-1-iodoethane, \( \ce{HOC6H4-CHICH3} \)

Explanation:
1. The aryl ethyl ether (Ar-O-Et) cleaves at the alkyl-oxygen bond: iodide attacks the ethyl group, giving the phenol (Ar-OH) and ethyl iodide; aryl C-O bonds never cleave.
2. The benzylic ether cleaves at the benzylic carbon (via the stable benzylic cation), which captures iodide: \( \ce{ArCH(I)CH3} \) and methanol (or methyl iodide).
3. Product: o-hydroxy compound with the benzylic iodide.

Teacher's Note:
1. Ar-O-R cleaves to ArOH + RI; benzylic ethers cleave at the benzylic carbon.

Answer: (a) \( \ce{CH2=CH-CH2-Br + C6H5ONa} \)

Explanation:
1. Allyl phenyl ether is \( \ce{C6H5-O-CH2-CH=CH2} \). By Williamson synthesis, the aryl oxygen must be the nucleophile (phenoxide) and the alkyl part must come from a good SN2 halide (allyl bromide).
2. Aryl bromide (alkoxide as nucleophile) cannot react, since aryl halides do not undergo SN2.

Teacher's Note:
1. Williamson synthesis for alkyl aryl ethers: phenoxide + alkyl halide.

Answer: (d) methyl alcohol

Explanation:
1. Primary aliphatic amines give unstable diazonium salts \( \ce{CH3N2+Cl-} \) with nitrous acid; these lose \( \ce{N2} \) at once to give a carbocation, which is captured by water (mainly) to give methanol (with some methyl chloride).

Teacher's Note:
1. Aliphatic primary amine + HNO2 gives alcohol + N2.

Answer: (b) nucleophilic substitution

Explanation:
1. An alkoxide ion (nucleophile) replaces the halide of an alkyl halide in an SN2 reaction.

Teacher's Note:
1. Williamson = SN2 of alkoxide on alkyl halide.

Answer: (c) 2,2-dichloropropane

Explanation:
1. Immediate turbidity with the Lucas reagent means III is a tertiary alcohol.
2. III is formed from a ketone II by addition of \( \ce{CH3MgBr} \): II = acetone and III = 2-methylpropan-2-ol.
3. Acetone arises from hydrolysis of the gem-dichloride \( \ce{CH3CCl2CH3} \), 2,2-dichloropropane.

Teacher's Note:
1. Gem-dihalide in the middle of a chain gives a ketone on hydrolysis.

Answer: (c) 1-chloro-1-(4-methoxyphenyl)propane

Explanation:
1. Grignard addition to the aldehyde gives the secondary benzylic alcohol \( \ce{ArCH(OH)CH2CH3} \) (A).
2. With HCl, the benzylic alcohol ionises to a benzylic cation (stabilised by the para-OMe group) that takes up chloride (SN1): \( \ce{ArCH(Cl)CH2CH3} \). The aryl OCH3 is not affected by HCl.

Teacher's Note:
1. Benzylic alcohols + HX give benzylic halides.

Answer: (d) A = benzyl chloride, B = benzyl isocyanide

Explanation:
1. Chloromethylation (Blanc reaction) puts \( \ce{CH2Cl} \) on benzene: A = benzyl chloride.
2. Silver cyanide is largely covalent, so the nitrogen end of the ambident cyanide attacks the carbon and the product is the isocyanide \( \ce{PhCH2-NC} \) (KCN would give the cyanide).

Teacher's Note:
1. AgCN gives isocyanides; KCN gives nitriles.

Answer: (b) 1-methoxy-2-methylbenzene (2-methylanisole)

Explanation:
1. Sodium hydroxide converts the phenol to the phenoxide; methyl iodide then attacks through oxygen (Williamson synthesis, SN2) to give the ether.
2. Ring methylation (the other products) is a minor side reaction.

Teacher's Note:
1. Williamson synthesis with a phenoxide gives the aryl alkyl ether.

Answer: (d) addition-elimination (SNAr)

Explanation:
1. The two nitro groups, ortho to chlorine, stabilise the negative charge of the intermediate (Meisenheimer complex) formed when methoxide adds to the ring carbon bearing Cl.
2. Loss of chloride then restores aromaticity: addition followed by elimination. A benzyne mechanism needs a very strong base and is not needed here.

Teacher's Note:
1. Ortho/para nitro groups activate halobenzenes for addition-elimination.