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Which of the following statements about SN2 reactions is/are correct?

1. CH3CH2CH2I reacts more readily than (CH3)2CHI in SN2 reactions
2. CH3CH2CH2Cl reacts more readily than CH3CH2CH2Br in SN2 reactions
3. CH3CH2CH2CH2Br reacts more readily than (CH3)3CCH2Br in SN2 reactions
4. CH3-O-C6H4-CH2Br reacts more readily than O2N-C6H4-CH2Br in SN2 reactions

Correct answer: CH3CH2CH2I reacts more readily than (CH3)2CHI in SN2 reactions

Solution

Option A: n-propyl iodide (primary) reacts faster than isopropyl iodide (secondary) in SN2 — correct. Option B: Cl is a worse leaving group than Br, so propyl chloride reacts more slowly than propyl bromide — incorrect. Option C: n-butyl bromide is primary and less hindered than neopentyl-type (CH3)3CCH2Br (which has severe 1,2-steric strain) — n-butyl bromide reacts faster — correct. Option D: methoxy group (OCH3) is electron-donating, which destabilises the transition state for SN2 at the benzylic position relative to the electron-withdrawing nitro group (NO2 activates benzylic SN2 via stabilisation of partial negative charge in TS) — so NO2-C6H4-CH2Br reacts faster — D is incorrect.

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