JEE Chemistry Haloalkanes and Haloarenes MCQs Set 22

Answer: (b) hex-3-yne

Explanation:
1. Sodamide removes the acidic terminal alkyne proton to give \( \ce{CH3CH2C#C- Na+} \).
2. The acetylide ion is a good nucleophile and displaces bromide from the primary ethyl bromide (SN2), giving \( \ce{CH3CH2C#CCH2CH3} \), hex-3-yne.

Teacher's Note:
1. Acetylide + primary alkyl halide gives a higher alkyne.

Answer: (d) elimination followed by addition

Explanation:
1. A very strong base (such as \( \ce{NH2-} \)) removes a proton ortho to the halogen and halide is lost (elimination), giving the benzyne intermediate.
2. The nucleophile then adds to the triple bond and the anion is protonated (addition).

Teacher's Note:
1. Benzyne route: elimination-addition. SNAr: addition-elimination.

Answer: (a) the SN1 mechanism is encouraged by polar solvents

Explanation:
1. SN1 involves ionisation to a carbocation and a halide ion; polar (especially protic) solvents solvate and stabilise these ions and so speed up SN1 reactions.
2. SN2 with a chiral substrate gives inversion (a single stereoisomer), not a racemic mixture, and the solvent does influence the mechanism.

Teacher's Note:
1. Polar protic solvents favour SN1.

Answer: (d) ethyne with \( \ce{NaNH2} \), then \( \ce{n-C5H11Br} \), then \( \ce{H2O, Hg^{2+}, H2SO4} \)

Explanation:
1. Mercury-catalysed hydration gives a single methyl ketone only from a terminal alkyne (Markovnikov hydration gives \( \ce{R-CO-CH3} \)); an unsymmetrical internal alkyne gives a mixture of two ketones.
2. Ethyne is alkylated once with n-pentyl bromide to give hept-1-yne, which on hydration gives only 2-heptanone. The other routes give internal alkynes (hept-2-yne or hept-3-yne), i.e. mixtures of ketones or 3-heptanone.

Teacher's Note:
1. Terminal alkyne hydration gives methyl ketones cleanly.

Answer: (b) Q

Explanation:
1. M is 4-nitrochlorobenzene (major), N is 3-chloronitrobenzene (nitro meta to Cl, no activation), P is 4-nitroanisole (no leaving group).
2. Q is 2,4-dinitrochlorobenzene: two nitro groups ortho and para to chlorine, so SNAr is fastest.

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

Answer: (c) 4-chloroanisole

Explanation:
1. 2,4-Dinitrochlorobenzene undergoes SNAr readily. 4-Methoxybenzaldehyde (no alpha hydrogen) undergoes the Cannizzaro reaction and the arylacetaldehyde undergoes aldol-type reactions with hot concentrated alkali.
2. 4-Chloroanisole has an unactivated aryl chloride (the OCH3 group deactivates SNAr), so it does not react under these conditions.

Teacher's Note:
1. Unactivated aryl halides need very drastic conditions.

Answer: (c) 4-bromobenzyl bromide

Explanation:
1. Methylamine reacts by SN2 on the halide carbon; the best site is an unhindered primary benzylic bromide, because the benzene ring stabilises the SN2 transition state.
2. The ring bromine of the first compound is unreactive. The 1,4-dibromo compound has secondary benzylic bromines (more hindered), the 2,3-dibromo compound has secondary non-benzylic bromines, and the 7-bromonorbornane bromine is on a hindered bridge.

Teacher's Note:
1. Primary benzylic bromides are among the best SN2 substrates.

Answer: (d) tert-butyl phenyl ether

Explanation:
1. tert-Butyl phenyl ether would need either phenoxide + tert-butyl halide (E2 elimination to isobutene) or tert-butoxide + a halobenzene (aryl halides do not undergo SN2), so it cannot be made this way.
2. THF (intramolecular, from 4-halobutanol), anisole (phenoxide + CH3I) and cyclopentyl methyl ether (cyclopentoxide + CH3I) can all be made.

Teacher's Note:
1. Williamson fails with tertiary halides and with aryl halides.

Answer: (b) benzyl bromide

Explanation:
1. SN1 needs a halide that gives a stabilised carbocation: benzyl bromide gives the resonance-stabilised benzyl cation.
2. Bromobenzene and vinyl bromide have sp2 C-Br bonds (the cations are very unstable), and 4-bromobut-1-ene is a primary non-allylic bromide that would react by SN2.

Teacher's Note:
1. Benzylic and allylic halides can react by SN1.

Answer: (d) (i) and (iii)

Explanation:
1. (i) Protonation of the alkene gives the tertiary carbocation, which captures iodide.
2. (iii) A tertiary alcohol with HCl is protonated, loses water to the tertiary cation (SN1), and this cation captures chloride.
3. (ii) Chlorobenzene with sodamide reacts through benzyne (elimination-addition), with no carbocation.

Teacher's Note:
1. Alkene + HX and tertiary alcohol + HX proceed through carbocations; benzyne has none.

Answer: (a) retention of configuration

Explanation:
1. With thionyl chloride alone the intermediate alkyl chlorosulphite collapses by an internal return (SNi) in which chlorine is delivered to the carbon from the same side as the departing group.
2. This gives retention of configuration.

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

Answer: (d) the ethyl ether, \( \ce{PhCH2CH(OC2H5)CH3} \), with retention of configuration

Explanation:
1. Sodium forms the sodium alkoxide; the oxygen attacks ethyl bromide (SN2, Williamson).
2. The C-O bond at the stereocentre is never broken, so the configuration at the stereocentre is retained.

Teacher's Note:
1. Williamson synthesis with an alkoxide retains stereochemistry at the alkoxide carbon.

Answer: (b) enantiomers

Explanation:
1. NaH/CH3I (Williamson): the stereocentre's C-O bond is retained, giving the methyl ether M with the same configuration as X.
2. TsCl converts OH to the tosylate with retention; then \( \ce{CH3O-} \) displaces tosylate by SN2 with inversion, giving the methyl ether M'.
3. M and M' have the same constitution but opposite configuration at the single stereocentre: enantiomers.

Teacher's Note:
1. Retention route and inversion route give enantiomers.

Answer: (c) \( \ce{AgNO3} \) increases the rate of solvolysis in an SN1 reaction of an alkyl halide

Explanation:
1. Silver ion removes halide as insoluble AgX, so ionisation to the carbocation is faster.
2. SN1 gives racemisation, aryl halides are much less reactive than alkyl halides, and nucleophiles are electron-pair donors that attack electron-poor species.

Teacher's Note:
1. Ag+ assists the departure of the halide.

Answer: (c) the cyclohex-2-en-1-yl (allylic) cation

Explanation:
1. A cyclopropyl cation is destabilised and opens as the leaving group departs, a ring-opening that is disrotatory and that cleaves the bond common to both rings.
2. This opens the three-membered ring and enlarges the five-membered ring to a six-membered ring, giving the allylic cyclohexenyl cation, stabilised by delocalisation of the positive charge over two carbons.

Teacher's Note:
1. Cyclopropyl cations ring-open to allyl cations.

Answer: (a) X = 1-ethoxy-2-methylpropan-2-ol; Y = 2-ethoxy-2-methylpropan-1-ol

Explanation:
1. Under basic conditions the strong nucleophile EtO- attacks the less hindered CH2 carbon (SN2): \( \ce{(CH3)2C(OH)CH2OEt} \).
2. Under acidic conditions the protonated epoxide opens with SN1-like character, and ethanol attacks the more substituted (tertiary) carbon, which bears more positive charge: \( \ce{(CH3)2C(OEt)CH2OH} \).

Teacher's Note:
1. Base: less hindered carbon; acid: more substituted carbon.

Answer: (d) Statements B and D only are correct

Explanation:
1. (A) is the Williamson synthesis (SN2, methyl bromide) and (C) is an SN1 solvolysis of a tertiary halide in an alcohol, giving the ether: both are feasible.
2. (B) the strongly basic alkoxide with a tertiary halide gives mostly the alkene by E2, so the ether is not obtained; (D) aryl halides do not undergo substitution under ordinary conditions. So B and D are not feasible.

Teacher's Note:
1. Tertiary halide + strong base: elimination; aryl halide + nucleophile: no reaction.

Answer: (b) Statements A and D only are correct

Explanation:
1. All four are primary bromides, which cannot form stable carbocations, so all react by SN2 (A true, B and C false).
2. The rate falls sharply as branching next to the reacting carbon increases (beta-methyl groups hinder back-side attack, neopentyl bromide almost not at all): the steric effect (D true).

Teacher's Note:
1. Branching at the beta carbon slows SN2 dramatically (neopentyl halides are practically inert).

Answer: (b) Statements A, B and D only are correct

Explanation:
1. The Grignard reagent adds to the carbonyl group (nucleophilic addition, D) to give the alkoxide.
2. The alkoxide attacks the carbon bearing chlorine within the same molecule (intramolecular nucleophilic attack, B) displacing chloride, which is a nucleophilic substitution (A).
3. No water is lost, so dehydration (C) is not involved.

Teacher's Note:
1. Addition then intramolecular Williamson-type ring closure.

Answer: (d) 75 %

Explanation:
1. The first term (second order, depends on [OH-]) is the SN2 contribution; the second (first order) is the SN1 contribution.
2. SN2 term = 6.0 x 10^-5 x 0.01 [R-X] = 6.0 x 10^-7 [R-X]; SN1 term = 2 x 10^-7 [R-X].
3. Percentage SN2 = 6.0/(6.0 + 2) x 100 = 75 %.

Teacher's Note:
1. Compare the two rate terms.

Answer: (a) 4

Explanation:
1. The allylic bromide ionises to an allylic cation \( \ce{CH3CH2C(CH3)=CH-CH2+ <-> CH3CH2C+(CH3)-CH=CH2} \), with positive charge on both ends.
2. Water captures the primary end to give the allylic alcohol \( \ce{CH3CH2C(CH3)=CHCH2OH} \) as E and Z isomers (2 products), and captures the tertiary end to give \( \ce{CH3CH2C(OH)(CH3)CH=CH2} \), which has a stereocentre and forms as R and S (2 products).
3. Total = 4.

Teacher's Note:
1. Allylic cations are captured at both ends; count geometric and optical isomers.