See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: A pair of enantiomers is produced when an SN1 reaction occurs at a chiral center, generating a carbocation intermediate that is planar (sp2), allowing nucleophilic attack from both faces to give equal amounts of R and S products (a racemic mixture = pair of enantiomers). Alternatively, if the substrate undergoes SN1 with complete racemization, a pair of enantiomers results. Step 1 - Identify the reaction mechanism for each option: (a) CN- in DMSO: CN- is a strong nucleophile, DMSO is a polar aprotic solvent. This favors SN2. SN2 on a chiral center gives inversion (Walden inversion), producing a single enantiomer (the inverted product), NOT a pair of enantiomers. (b) H2O as nucleophile/solvent: H2O is a weak nucleophile and a polar protic solvent. The substrate appears to be a secondary alkyl iodide (2-iodo-4-methylpentane type). Weak nucleophile + polar protic solvent + secondary substrate favors SN1. SN1 proceeds through a planar carbocation intermediate, allowing attack from both faces, producing a racemic mixture — a pair of enantiomers. (c) OH- in DMF: OH- is a strong nucleophile, DMF is a polar aprotic solvent. This strongly favors SN2. The substrate is a disubstituted cyclohexane (trans or cis relationship). SN2 gives inversion at the reaction center, producing a single stereochemical outcome (the other diastereomer), not a pair of enantiomers. (d) CH3OH as nucleophile: CH3OH is a weak nucleophile and polar protic solvent — conditions that could favor SN1. However, looking at the structure, the carbon bearing I appears to be a secondary carbon (3-iodopentane), but CH3OH acting as nucleophile in an SN1 would give racemization. However, 3-iodopentane has a stereocenter, but the two groups on either side are both ethyl groups — making the product (3-methoxypentane) achiral (the carbon bearing OCH3 with two ethyl groups is not a stereocenter). Therefore, no pair of enantiomers is formed; the product is achiral. Step 2 - Conclusion: Only option (b) — SN1 reaction of a secondary chiral alkyl iodide with H2O (polar protic, weak nucleophile) — produces a planar carbocation intermediate followed by attack from both faces, yielding a racemic mixture (pair of enantiomers). Why others fail: - (a) SN2 with strong nucleophile/polar aprotic solvent → single inverted product - (c) SN2 with strong nucleophile/polar aprotic solvent → single inverted diastereomer - (d) SN1 possible but product carbon is not a stereocenter (two identical ethyl groups) → achiral product, no enantiomers Therefore, the correct answer is B.