JEE Chemistry Haloalkanes and Haloarenes MCQs Set 12

Answer: (a) inversion more than retention leading to partial racemisation

Explanation:
1. The planar sp\(^2\) carbocation can be attacked from either face, giving both retained and inverted products.
2. The inverted product is slightly in excess because the leaving group partly shields the front side, so there is partial racemisation.

Teacher's Note:
1. \(S_N1\) gives racemisation with slight inversion; \(S_N2\) gives complete inversion.

Answer: (c) \( \ce{C6H5CHBrCH2CH3} \)

Explanation:
1. Protonation of the double bond occurs so as to give the more stable benzylic carbocation \( \ce{C6H5CH^+CH2CH3} \).
2. Bromide then attacks this carbocation to give \( \ce{C6H5CHBrCH2CH3} \).

Teacher's Note:
1. Markovnikov addition goes through the most stable carbocation; benzylic is the most stable.

Answer: (c) Rb

Explanation:
1. Fluoride is more nucleophilic when it is poorly ion-paired with the metal ion.
2. The larger the cation, the weaker the ion pairing with F\(^-\), so Rb is the best choice.

Teacher's Note:
1. Larger cation means a freer, more nucleophilic anion.

Answer: (c) \( \ce{C6H5C(CH3)(C6H5)Br} \)

Explanation:
1. \(S_N1\) reactivity depends on the stability of the carbocation formed.
2. \( \ce{C6H5C^+(CH3)(C6H5)} \) is stabilised by two phenyl rings (resonance) and a methyl group (inductive effect), so it is the most stable.

Teacher's Note:
1. More resonance and hyperconjugation means a faster \(S_N1\) reaction.

Answer: (c) \( \ce{C6H5CH3} \)

Explanation:
1. Benzyl bromide forms the Grignard reagent \( \ce{C6H5CH2MgBr} \).
2. Hydrolysis with H\(_3\)O\(^+\) gives the hydrocarbon toluene.

Teacher's Note:
1. Grignard reagents are converted to hydrocarbons by water or acid.

Answer: (d) \( \ce{RX + KOH -> ROH + KX} \)

Explanation:
1. In \( \ce{RX + KOH -> ROH + KX} \) the nucleophile OH\(^-\) replaces the halide ion.
2. The other reactions are Wurtz coupling, reduction and Grignard formation.

Teacher's Note:
1. Nucleophile in, leaving group out.

Answer: (c) 4

Explanation:
1. The double bond (with different groups on each carbon) shows cis-trans isomerism, giving 2 forms.
2. The carbon bearing Br, H, CH\(_3\) and the chain is a chiral centre, giving 2 forms.
3. Total number of stereoisomers = 2 x 2 = 4.

Teacher's Note:
1. Stereoisomers = 2 to the power of the number of stereo elements.

Answer: (a) 1, 2 and 3

Explanation:
1. Lactic acid, 2-butanol and 2,3-dibromopentane each have a carbon bonded to four different groups.
2. 3-Bromopentane has two identical ethyl groups on the carbon bearing Br, so it is achiral.

Teacher's Note:
1. Look for a carbon with four different groups.

Answer: (b) racemic mixture

Explanation:
1. Chlorination of n-butane is a free radical reaction through a planar secondary radical.
2. Chlorine can attack from either face with equal probability, giving equal amounts of d and l forms: a racemic mixture.

Teacher's Note:
1. Achiral reactants give racemic products without a chiral influence.

Answer: (a) \( \ce{Cl2} \)/UV light

Explanation:
1. Alkanes undergo free radical chlorination in the presence of ultraviolet light.
2. This replaces H by Cl at both the primary and secondary positions.

Teacher's Note:
1. Light starts the radical chain; iron promotes ring substitution in arenes.

Answer: (c) III < I < II

Explanation:
1. Among isomeric alkyl halides, branching reduces the surface area.
2. Dispersion forces become weaker, so the boiling point decreases with branching: tert-butyl (73 degrees C) < isobutyl (91 degrees C) < n-butyl (102 degrees C).

Teacher's Note:
1. More branching means a lower boiling point.

Answer: (b) gem-dihalide

Explanation:
1. Ethylidene chloride is \( \ce{CH3CHCl2} \).
2. Both chlorine atoms are on the same carbon, so it is a geminal (gem) dihalide.

Teacher's Note:
1. Ethylene dichloride (\( \ce{ClCH2CH2Cl} \)) is the vicinal dihalide.

Answer: (d) \( \ce{(CH3)3C-I} \)

Explanation:
1. All have the same tertiary carbocation, so reactivity depends on the leaving group.
2. The C-I bond is the weakest and I\(^-\) is the best leaving group, so the iodide reacts fastest.

Teacher's Note:
1. Leaving group ability: I\(^-\) > Br\(^-\) > Cl\(^-\) > F\(^-\).

Answer: (d) \( \ce{C6H5CHClCH2CH3} \)

Explanation:
1. Addition of H\(^+\) gives the secondary carbocation \( \ce{C6H5CH2CH^+CH3} \).
2. A hydride shift converts it to the more stable benzylic carbocation \( \ce{C6H5CH^+CH2CH3} \), which captures Cl\(^-\) to give \( \ce{C6H5CHClCH2CH3} \).

Teacher's Note:
1. Carbocations rearrange to become more stable.

Answer: (c) 1-bromo-2-methylbutane

Explanation:
1. The longest chain containing the carbon with Br has four carbons (butane), with Br at C1 and a methyl group at C2.
2. The name is 1-bromo-2-methylbutane (substituents in alphabetical order).

Teacher's Note:
1. Choose the longest chain and give the lowest locants.

Answer: (b) 3-bromopentane

Explanation:
1. Diethylbromomethane is \( \ce{(C2H5)2CHBr} \).
2. The longest chain is five carbons with Br on the middle carbon, so it is 3-bromopentane.

Teacher's Note:
1. Number from either end; Br lies on C3.

Answer: (b) 2

Explanation:
1. The two middle carbons each carry H, OH, and two different carbon groups (CHO and the neighbouring CH(OH)COOH side), so they are asymmetric.
2. The CHO and COOH carbons are sp\(^2\) and cannot be asymmetric.

Teacher's Note:
1. Only sp\(^3\) carbons can be chiral centres.

Answer: (a) I only

Explanation:
1. A racemic mixture arises when a chiral halide reacts by \(S_N1\).
2. Compound I (2-bromobutane, secondary) is chiral and can ionise.
3. II is tertiary but achiral, and III is chiral but primary, so it reacts by \(S_N2\) with inversion only.

Teacher's Note:
1. Chiral and \(S_N1\) capable are both needed.

Answer: (c) I < III < II

Explanation:
1. A nitro group at the ortho or para position activates the ring towards nucleophilic substitution by withdrawing electrons (resonance).
2. At the meta position it can act only by induction, so the activation is much smaller.
3. So I < III < II.

Teacher's Note:
1. o/p nitro groups activate, m nitro group only weakly.

Answer: (d) I < II < III

Explanation:
1. Each o/p nitro group stabilises the negatively charged Meisenheimer intermediate.
2. So the reactivity increases with the number of such groups: I < II < III.

Teacher's Note:
1. More electron withdrawing groups give easier substitution.