JEE Chemistry Haloalkanes and Haloarenes MCQs Set 17

Answer: (c) a carbocation

Explanation:
1. The slow first step is ionisation of the C-X bond, giving a planar carbocation, which the nucleophile attacks in a fast second step.

Teacher's Note:
1. SN1: two steps with a carbocation intermediate, first order.

Answer: (d) 2

Explanation:
1. SN1 is first order in the alkyl halide only: rate = k[RX], independent of the nucleophile (solvent) concentration.
2. Doubling [RX] doubles the rate.

Teacher's Note:
1. Rate of SN1 does not depend on [Nu].

Answer: (d) \( \ce{PhCH(Cl)CH2CH3} \)

Explanation:
1. The secondary alcohol ionises (Lucas conditions) to a secondary carbocation at C-2.
2. A 1,2-hydride shift from the benzylic carbon gives the benzylic cation \( \ce{PhCH+CH2CH3} \), stabilised by the ring.
3. Chloride attacks it to give 1-chloro-1-phenylpropane.

Teacher's Note:
1. Carbocations rearrange to the most stable ion; benzylic is more stable than secondary alkyl.

Answer: (c) racemisation

Explanation:
1. A tertiary benzylic alcohol is protonated and loses water to give a planar, delocalised carbocation (SN1).
2. Bromide attacks either face, so the product is racemic.

Teacher's Note:
1. Tertiary and benzylic substrates react by SN1.

Answer: (b) 4

Explanation:
1. Rate = k[RX][Nu]; doubling both gives \( 2\times 2 = 4 \).

Teacher's Note:
1. SN2 is second order overall.

Answer: (c) \( \ce{CH3CH2CH2Cl} \) and \( \ce{CH3(CH2)4C#CH} \)

Explanation:
1. 4-Decyne is \( \ce{CH3CH2CH2-C#C-CH2CH2CH2CH2CH3} \): a propyl group on one side and a pentyl group on the other of the triple bond, with the carbons numbered so that the triple bond is C4-C5.
2. Propyl chloride (3 C) plus 1-heptyne (7 C) gives the 10-carbon product with the triple bond at C4-C5; butyl chloride with 1-hexyne would give 5-decyne, and the other pairs give 9 carbons.

Teacher's Note:
1. The acetylide ion is a carbon nucleophile in SN2 reactions with primary halides.

Answer: (a) 1-bromobutane by SN2

Explanation:
1. A primary alcohol cannot form a stable primary carbocation, so after protonation of OH, bromide attacks by SN2 and displaces water.

Teacher's Note:
1. Primary alcohols react with HX by SN2; tertiary by SN1.

Answer: (b)

Explanation:
1. The strong base removes the OH proton to give an alkoxide, which attacks the other end of the chain intramolecularly (SN2, Williamson-type).
2. Formation of the six-membered ring is favourable, giving 2-methyltetrahydropyran.

Teacher's Note:
1. Intramolecular Williamson synthesis forms 5- and 6-membered cyclic ethers readily.

Answer: (d) retention of configuration (SNi)

Explanation:
1. In the absence of pyridine the alkyl chlorosulphite decomposes by an internal return mechanism (SNi): chloride is delivered from the same face, so the configuration is retained.

Teacher's Note:
1. SNi: retention. SOCl2 with pyridine: inversion.

Answer: (c)

Explanation:
1. Each NO2 group ortho or para to Cl stabilises the Meisenheimer (anionic sigma) complex.
2. 2,4,6-Trinitrochlorobenzene has three activating groups, so it reacts fastest.

Teacher's Note:
1. More ortho/para nitro groups give faster SNAr.

Answer: (b)

Explanation:
1. The cyano group (-M, -I) activates the position para to it for SNAr, and fluoride is a good leaving group in this mechanism.
2. Dimethylamine replaces fluorine only; the cyano group stays.

Teacher's Note:
1. SNAr replaces the halogen, not the activating group.

Answer: (c) The reaction in DMSO is faster

Explanation:
1. Ethanol is protic and solvates hydroxide by hydrogen bonding, lowering its nucleophilicity.
2. DMSO is polar aprotic: it solvates the cation but leaves the anion bare and very reactive.

Teacher's Note:
1. Polar aprotic solvents accelerate SN2; polar protic solvents accelerate SN1.

Answer: (a) two transition states and one intermediate, the first step (ionisation) having the larger activation energy

Explanation:
1. SN1 is a two-step process: ionisation (slow, high barrier) gives a carbocation intermediate, followed by a fast capture, so there are two transition states with a dip between them.
2. SN2 has a single peak.

Teacher's Note:
1. Intermediate = energy minimum between two peaks.

Answer: (b) the allylic chloride solvolyses faster than a simple primary alkyl chloride, and crotyl alcohol gives a mixture of 1-bromobut-2-ene and 3-bromobut-1-ene

Explanation:
1. Allylic halides ionise to delocalised allylic cations, so they solvolyse far faster than simple primary halides.
2. The allylic cation from crotyl alcohol has positive charge shared by C-1 and C-3, so bromide attacks at both to give a mixture.

Teacher's Note:
1. Delocalised allylic cations lead to allylic rearrangement products.

Answer: (d) Hydroxide is a poor leaving group because it is a strong base, while chloride is a weak base

Explanation:
1. A good leaving group is a weak base. \( \ce{OH-} \) is a strong base (conjugate acid water, pKa 15.7), so it will not leave unless the OH is first protonated or converted into a better leaving group.
2. Chloride is a weak base (conjugate acid HCl), so the equilibrium lies to the left.

Teacher's Note:
1. Protonating the OH (HX) or using SOCl2 or PCl5 makes it a good leaving group.

Answer: (c) m-hydroxybenzyl bromide and resorcinol (benzene-1,3-diol)

Explanation:
1. The benzylic OH is protonated and replaced by Br (SN1, stable benzylic cation) while the phenolic OH, whose C-O bond has partial double-bond character, is untouched.
2. In the aryl methyl ether, HBr cleaves the O-CH3 bond (bromide attacks the methyl carbon, SN2), giving the phenol and methyl bromide; the aryl-O bond does not break.

Teacher's Note:
1. Aryl alkyl ethers with HX give phenol + alkyl halide.
2. C(sp2)-O bonds are not cleaved by nucleophilic substitution.

Answer: (b) benzyl bromide and 4-bromophenol

Explanation:
1. The phenoxy ring is activated by the ether oxygen, so bromine enters para to it.
2. HBr then cleaves the ether at the benzylic C-O bond (the aryl C-O bond cannot be broken by nucleophilic substitution), giving benzyl bromide and 4-bromophenol.

Teacher's Note:
1. Aryl benzyl ether + HBr gives phenol + benzyl bromide.

Answer: (c) tert-butyl iodide and methanol, by SN1

Explanation:
1. The ether oxygen is protonated, and the C-O bond to the tertiary carbon breaks to give the stable tert-butyl cation and methanol (SN1).
2. Iodide captures the cation, giving tert-butyl iodide; the methyl group is not attacked.

Teacher's Note:
1. Ethers with a tertiary alkyl group cleave by SN1; with primary and methyl groups by SN2.

Answer: (a) the cyclopropenyl cation is aromatic (2 pi electrons) whereas the cyclopentadienyl cation is antiaromatic (4 pi electrons)

Explanation:
1. Ionisation of 3-chlorocyclopropene gives the cyclopropenyl cation, a planar cyclic conjugated system with \( 4n+2 = 2 \) pi electrons, which is aromatic and strongly stabilised.
2. Ionisation of 5-chlorocyclopenta-1,3-diene gives the cyclopentadienyl cation with 4 pi electrons: antiaromatic, strongly destabilised.

Teacher's Note:
1. Aromatic cations form easily; antiaromatic ones do not.

Answer: (b) 2-bromopentane and 2-bromo-2-methylbutane

Explanation:
1. SN1 rate depends on carbocation stability: secondary beats primary, and tertiary beats primary.

Teacher's Note:
1. Order: 3 degree > 2 degree > 1 degree.

Answer: (a) NaCN

Explanation:
1. Cyanide displaces chloride by SN2 to give pentanenitrile.
2. AlCl3 is a Lewis acid, Mg/ether gives the Grignard reagent and alcoholic KOH gives elimination.

Teacher's Note:
1. SN2 at a primary halide with a good nucleophile.