JEE Chemistry Haloalkanes and Haloarenes MCQs Set 27

Multiple Choice Questions (MCQs) for JEE Chemistry: Haloalkanes and Haloarenes

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Multiple Choice Questions

Question 1: Optically active 2-bromopentane reacts with aqueous OH⁻ to give 2-pentanol with inverted configuration (a single enantiomer). Which mechanism operates?
(a) SN1
(b) SN2
(c) SNi
(d) SNAE
Show Answer & Explanation

Answer: (b) SN2

Explanation:
1. Complete inversion of configuration with a single product is the sign of a one-step back-side attack, i.e. SN2.
2. SN1 would give a racemic mixture, and SNi gives retention.
3. Answer: (b)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 2: Which of the following reactions occurs most rapidly?
(a) (CH3)2CHOH + HBr
(b) (CH3)2CHOH + HI
(c) (CH3)3C–OH + HBr
(d) (CH3)3C–OH + HI
Show Answer & Explanation

Answer: (d) (CH3)3C–OH + HI

Explanation:
1. The order of reactivity of alcohols is 3° > 2° > 1° (stability of the carbocation), and of hydrogen halides HI > HBr > HCl.
2. The combination of the tertiary alcohol and HI is the fastest.
3. Answer: (d)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 3: In SN2 reactions, the correct order of reactivity for CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is:
(a) CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl
(b) (CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl
(c) CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl
(d) CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl
Show Answer & Explanation

Answer: (d) CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl

Explanation:
1. SN2 is slowed by steric hindrance at the carbon being attacked.
2. Reactivity falls as more alkyl groups replace H: methyl > primary > secondary > tertiary.
3. Answer: (d)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 4: Which of the following undergoes the fastest reaction with water?
(a) C6H5CH(Cl)OCH3
(b) C6H5CH(Cl)CH3
(c) C6H5CH(Cl)CH2CH3
(d) C6H5CH(Cl)C6H5
Show Answer & Explanation

Answer: (a) C6H5CH(Cl)OCH3

Explanation:
1. Hydrolysis by water is an SN1 reaction, so the stability of the carbocation decides.
2. In (a) the OCH3 group donates a lone pair directly to the cationic carbon (+M), besides the phenyl ring.
3. That makes the cation more stable than the alkyl-substituted or diphenyl cations.
4. Answer: (a)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 5: Which electrophile is formed in the Reimer-Tiemann reaction?
(a) Carbocation
(b) Carbanion
(c) Carbene
(d) Free radical
Show Answer & Explanation

Answer: (c) Carbene

Explanation:
1. Chloroform with a strong base loses HCl (alpha elimination) to give dichlorocarbene, :CCl2.
2. Dichlorocarbene is the electrophile that attacks the phenoxide ring.
3. Answer: (c)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 6: Which of the following statements is not correct for nucleophilic substitution reactions?
(a) The correct decreasing order of SN1 reactivity is (CH3)3C–Cl > (CH3)2CHCl > CH3CH2Cl
(b) Stereochemical inversion may occur in an SN2 reaction
(c) Racemisation may occur in an SN1 reaction
(d) A carbocation is formed during an SN2 reaction
Show Answer & Explanation

Answer: (d) A carbocation is formed during an SN2 reaction

Explanation:
1. Statements (a), (b) and (c) are correct.
2. In SN2 the bond to the nucleophile forms while the C–X bond breaks, in a single step through a transition state; no carbocation intermediate forms.
3. A carbocation is an intermediate of SN1 only.
4. Answer: (d)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 7: The reactivity order of the following benzyl chlorides for the SN1 reaction is:
I
II
III
IV
(a) I > II > III > IV
(b) II > III > IV > I
(c) IV > III > II > I
(d) III > IV > II > I
Show Answer & Explanation

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

Explanation:
1. The rate depends on the benzylic carbocation.
2. A para alkyl group stabilises it by hyperconjugation; the effect is larger when the alkyl group has more alpha C–H bonds: CH3 (3) > CH2CH3 (2) > CH(CH3)2 (1).
3. Benzyl chloride with no alkyl group is the least reactive.
4. Order: II > III > IV > I.
5. Answer: (b)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 8: Which of the following reactions is the least feasible?
(a) Chlorobenzene + NaOH at 360°C → phenol
(b) 4-Chloronitrobenzene + NaOH, heat → 4-nitrophenol
(c) 2-Chloroanisole + NaOH → 2-methoxyphenol
(d) 2,4-Dinitrochlorobenzene + NaOH, heat → 2,4-dinitrophenol
Show Answer & Explanation

Answer: (c) 2-Chloroanisole + NaOH → 2-methoxyphenol

Explanation:
1. Replacement of Cl on a benzene ring by OH needs either very severe conditions (chlorobenzene, 623 K) or electron-withdrawing groups ortho or para to Cl (reactions b and d).
2. In 2-chloroanisole the OCH3 group is an electron donor and gives no activation, and no severe conditions are given, so it is the least feasible.
3. Answer: (c)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 9: In which reaction is a halide not obtained?
(a) R–OH + NaBr
(b) R–OH + HBr
(c) R–OH + SOCl2
(d) R–OH + PCl5
Show Answer & Explanation

Answer: (a) R–OH + NaBr

Explanation:
1. HBr, SOCl2 and PCl5 all convert an alcohol into the alkyl halide.
2. NaBr alone does not protonate the OH group, which is a very poor leaving group, so no reaction occurs (it works only with concentrated H2SO4).
3. Answer: (a)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 10: p-Hydroxybenzyl alcohol on reaction with HCl gives:
(a)
(b)
(c)
(d)
Show Answer & Explanation

Answer: (b)

Explanation:
1. HCl converts the benzylic alcohol OH into Cl through the stable benzylic carbocation.
2. The phenolic C–OH bond has partial double-bond character and does not react.
3. The product is 4-(chloromethyl)phenol.
4. Answer: (b)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 11: CH3–CHCl–CH(CH3)–CH3 is heated with alcoholic KOH to give A (major) and B (minor). A and B respectively are:
(a) 3-methylbut-1-ene, 2-methylbut-2-ene
(b) 2-methylbut-2-ene, 3-methylbut-1-ene
(c) 2-methylbut-1-ene, 3-methylbut-1-ene
(d) 2-methylbut-2-ene, 2-methylbut-1-ene
Show Answer & Explanation

Answer: (b) 2-methylbut-2-ene, 3-methylbut-1-ene

Explanation:
1. Dehydrohalogenation follows Saytzeff's rule: the more substituted alkene is the major product.
2. Loss of H from C-3 (tertiary) gives (CH3)2C=CH–CH3, 2-methylbut-2-ene (major).
3. Loss of H from C-1 gives CH2=CH–CH(CH3)2, 3-methylbut-1-ene (minor).
4. Answer: (b)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 12: Chlorobenzene on reaction with conc. HNO3/conc. H2SO4 gives A, and A with NaOH at 443 K gives B. B as the major product is:
(a)
(b)
(c)
(d)
Show Answer & Explanation

Answer: (c)

Explanation:
1. Nitration of chlorobenzene gives mainly p-chloronitrobenzene (A), as Cl directs ortho and para and the para product is favoured.
2. The Cl in A is activated by the para NO2 group, so NaOH at 443 K replaces it by OH.
3. B is p-nitrophenol.
4. Answer: (c)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

Question 13: Which of the following is insoluble in water?
(a) D.D.T.
(b) CHI3
(c) CH3CH2Cl
(d) All of these
Show Answer & Explanation

Answer: (d) All of these

Explanation:
1. Haloalkanes cannot form hydrogen bonds with water, and the energy needed to break water's hydrogen bonds is not repaid.
2. D.D.T., iodoform and chloroethane are all only sparingly soluble in water; DDT and iodoform are essentially insoluble.
3. Answer: (d)

Teacher's Note:
1. Decide first whether the substrate and conditions favour SN1 (stable carbocation, polar protic solvent) or SN2 (unhindered carbon, strong nucleophile, polar aprotic solvent).

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Haloalkanes and Haloarenes Objective Questions & Solutions for JEE Chemistry

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