JEE Chemistry Haloalkanes and Haloarenes MCQs Set 19

Answer: (c) 15

Explanation:
1. SN2 rate = k[RX][OH-], so the new rate / old rate = 3 x (1/2) = 1.5.
2. Hence X = 1.5 and 10X = 15.

Teacher's Note:
1. SN2 is first order in each reactant.

Answer: (d) 4

Explanation:
1. A white precipitate of AgCl needs easy ionisation of the C-Cl bond to a stable carbocation.
2. Cyclopropenyl cation (ii), tropylium cation (from the cycloheptatrienyl chloride), the 9-xanthenyl cation (stabilised by two benzene rings and oxygen) and the diphenyl-stabilised cation of (vi) are all stable, so (ii), (iii), (v) and (vi) ionise.
3. Cyclopentadienyl cation is antiaromatic (i) and chlorobenzene (iv) has a strong sp2 C-Cl bond, so these give no precipitate. Total = 4.

Teacher's Note:
1. Aromatic carbocations (3-, 7-membered rings with 2 or 6 pi electrons) are unusually stable; antiaromatic ones are very unstable.

Answer: (b) 22

Explanation:
1. Both reactions proceed by SN1 through the planar 1-phenylethyl cation.
2. Water adds to either face to give the (R)- and (S)-alcohols, so X = 2; chloride adds to either face to give the (R)- and (S)-chlorides, so Y = 2.
3. The number XY is therefore 22.

Teacher's Note:
1. Benzylic secondary substrates racemise through SN1.

Answer: (d) 4

Explanation:
1. The protonated alcohol loses water to give the secondary cation \( \ce{PhCH2-CH+-CH3} \); this can be attacked directly or rearrange by a hydride shift to the more stable benzylic cation \( \ce{PhCH+-CH2CH3} \).
2. 2-Bromo-1-phenylpropane has one stereocentre (2 isomers) and 1-bromo-1-phenylpropane has one stereocentre (2 isomers).
3. Total = 4.

Teacher's Note:
1. Secondary cations next to a benzylic CH2 rearrange to the benzylic cation.

Answer: (a) Statements A and B only are correct

Explanation:
1. Quinuclidine has its alkyl groups tied back in a ring, so its nitrogen is less hindered and more nucleophilic than that of triethylamine: k1 > k1' and k2 > k2'. Statement C (k2' > k2) is therefore false.
2. Methyl iodide reacts much faster than 2-iodopropane in SN2 for the same amine: k1 > k2 (B true) and k1' > k2', so D is false.
3. Correct statements: A and B.

Teacher's Note:
1. Less hindered amines are better SN2 nucleophiles; methyl halides beat secondary halides.

Answer: (d) All four

Explanation:
1. (A) PCl5 replaces the OH of the alcohol by Cl; (C) thionyl chloride likewise replaces OH by Cl; (D) HI replaces the OH of the tertiary alcohol by I through SN1.
2. (B) The benzylthiolate replaces chloride on the activated ring (SNAr).
3. Every reaction is a substitution at carbon by a nucleophile, so all four are correct.

Teacher's Note:
1. Replacement of OH by halogen and SNAr are both nucleophilic substitutions.

Answer: (b) Statements B and C only are correct

Explanation:
1. In SN1 the rate-determining step is the slow ionisation of the protonated ether to a stable (tertiary) carbocation and alcohol (B).
2. The carbocation is then captured by iodide in a fast step (C).
3. Step A is an SN2 displacement by iodide on a primary carbon, and capture of a cation by iodide is never rate determining (D).

Teacher's Note:
1. SN1: slow ionisation, fast capture.

Answer: (c) Statements A, B and C only are correct

Explanation:
1. Cyclopentadiene is unusually acidic (the anion is aromatic, 6 pi electrons), so BuLi removes a proton: n-butane + lithium cyclopentadienide (A and C true).
2. The cyclopentadienide anion then attacks the CH2 of chloromethyl methyl ether and displaces chloride by SN2 (B true).
3. The nucleophile in step 2 is the cyclopentadienyl anion, not Bu- (D false).

Teacher's Note:
1. Cyclopentadienyl anion is aromatic, hence the CH acidity of cyclopentadiene (pKa about 16).

Answer: (c) SN2, SN1, SNi, SN2

Explanation:
1. Inversion means SN2 (I and IV); racemisation means SN1 (II); retention in an alcohol-to-chloride change means SNi, as with thionyl chloride (III).
2. The order is SN2, SN1, SNi, SN2.

Teacher's Note:
1. Match the stereochemical outcome to the mechanism.

Answer: (d) the sodium salt of a chiral secondary alcohol, R-ONa, with \( \ce{CH3I} \)

Explanation:
1. In the Williamson reaction the alkoxide oxygen attacks \( \ce{CH3I} \); the bond to the stereocentre is never broken, so the configuration is retained.
2. A secondary bromide or tosylate with iodide in acetone is SN2 (inversion), and a tertiary bromide in water is SN1 (racemisation).

Teacher's Note:
1. Retention is expected whenever the stereocentre's C-O or C-X bond is not broken.

Answer: (b) 5-bromopent-1-ene, \( \ce{BrCH2CH2CH2CH=CH2} \)

Explanation:
1. Tetrahydropyran is a six-membered ring containing O and five carbons, so the dibromide must be 1,5-dibromopentane.
2. Anti-Markovnikov addition of HBr (peroxide) to a terminal alkene puts Br on the terminal carbon, so A is 5-bromopent-1-ene. It has no stereocentre and is optically inactive.
3. The other isomers would give 1,4-, 1,3- or 1,2-dibromides, which cannot give the six-membered cyclic ether.

Teacher's Note:
1. Anti-Markovnikov HBr (peroxide effect) adds Br to the terminal carbon.

Answer: (d) (II), (i), (P)

Explanation:
1. A secondary benzylic bromide with NaOH in aqueous medium reacts partly by SN2 (inversion) and partly by SN1 (racemisation), giving a largely racemic alcohol (P).
2. (I), (iii), (P) is impossible (a primary bromide cannot give the secondary alcohol); (III), (iv), (Q) and (IV), (ii), (S) cannot be correct since SOCl2 and HI convert alcohols to halides, not to alcohols of different configuration.

Teacher's Note:
1. Aqueous medium plus a secondary benzylic halide means mixed SN1/SN2.

Answer: (a) (III), (ii), (P)

Explanation:
1. The chiral alcohol, protonated in acidic aqueous acetone, ionises to the planar benzylic cation, which is captured by water to give the racemic alcohol P (SN1 only).
2. (II), (iii), (P) is an SN2 case (strong nucleophile in DMSO would give inversion, not racemisation) and (I), (i), (R) is a primary SN2 case. (IV), (ii), (S) is not possible.

Teacher's Note:
1. Strongly acidic aqueous conditions and a stable benzylic cation favour pure SN1.

Answer: (d) a single stereoisomer

Explanation:
1. SN2 proceeds with complete inversion of configuration, so only one stereoisomer of the product forms.
2. It is not necessarily the enantiomer of the substrate (the nucleophile differs, so the product is a different compound) and the sign of rotation need not reverse.

Teacher's Note:
1. Inversion of configuration is not the same as a change in sign of rotation.

Answer: (b) \( \ce{(CH3)3S+ I-} \)

Explanation:
1. Hydroxide attacks the methyl group by SN2; the best leaving group wins.
2. The sulphonium ion \( \ce{(CH3)3S+} \) loses neutral \( \ce{(CH3)2S} \), an excellent leaving group (better than \( \ce{(CH3)3N} \) from the ammonium salt). Dimethyl ether would have to lose \( \ce{CH3O-} \), a poor leaving group, and tert-butyl chloride gives mainly isobutene or tert-butanol, not methanol.

Teacher's Note:
1. Neutral, weakly basic leaving groups (R2S, R3N) are good; RO- is poor.

Answer: (c) 1-bromo-1-methylcyclopentane

Explanation:
1. A = cyclobutylmagnesium bromide; B = 1-cyclobutylethanol, \( \ce{CH3CH(OH)-C4H7} \).
2. HBr protonates B and water leaves to give the secondary cation next to the strained four-membered ring; ring expansion (a C-C shift) relieves strain and gives the tertiary 1-methylcyclopentyl cation.
3. Bromide adds to it, giving 1-bromo-1-methylcyclopentane.

Teacher's Note:
1. Ring expansion of cyclobutylcarbinyl cations to cyclopentyl cations is favoured by strain relief.

Answer: (c) \( \ce{C6H5OC2H5} \)

Explanation:
1. Ethoxide converts phenol to the phenoxide ion (phenol is far more acidic than ethanol).
2. Phenoxide attacks ethyl iodide (Williamson synthesis, SN2) to give ethyl phenyl ether.

Teacher's Note:
1. Williamson synthesis: alkoxide/phenoxide + primary alkyl halide.

Answer: (d) the benzylic bromide replaced by SPh with inversion of configuration, fluorine retained

Explanation:
1. Thiophenoxide is an excellent nucleophile and DMF is a polar aprotic solvent: the secondary benzylic bromide reacts by SN2, with inversion.
2. The fluorine is meta to the nitro group (nitro is at C4, F at C2), so it is not activated for SNAr and is retained.

Teacher's Note:
1. Benzylic halide SN2 competes best when the ring halogen is not activated (meta to NO2).

Answer: (b) phenol and \( \ce{CH3Br} \)

Explanation:
1. The ether oxygen is protonated; bromide attacks the methyl carbon (SN2).
2. The aryl-oxygen bond is never cleaved because attack at an sp2 carbon is impossible, so the products are phenol and methyl bromide.

Teacher's Note:
1. Alkyl aryl ethers always cleave at the alkyl-O bond.

Answer: (d) N-(4-bromobenzyl)phthalimide

Explanation:
1. The phthalimide anion is a nitrogen nucleophile and attacks the more reactive benzylic C-Cl carbon by SN2 (displacing chloride).
2. The aryl C-Br bond (sp2) is inert to SN2, so the bromine stays on the ring.

Teacher's Note:
1. Benzylic halides undergo SN2; aryl halides do not.

Answer: (a) S > P > R > Q

Explanation:
1. Phenacyl chloride is activated by the adjacent carbonyl group (the transition state is stabilised by overlap), so S is fastest by a large margin.
2. Methyl chloride (unhindered) is slightly faster than allyl chloride, and the secondary 2-chloropropane is the slowest.
3. Order: S > P > R > Q.

Teacher's Note:
1. alpha-Halo ketones are exceptionally reactive in SN2.

Answer: (d) two phenyl groups are replaced by two para-methoxyphenyl groups

Explanation:
1. The reaction goes by SN1: protonated ether loses ROH to give the triarylmethyl cation.
2. Para-methoxy groups donate electrons by resonance and stabilise the cation, so the more of them, the faster the ionisation. Replacing a phenyl by methyl removes a stabilising ring and slows it.

Teacher's Note:
1. Electron-donating para substituents speed up SN1 for aryl cations.