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Which of the following reactions proceeds with inversion of configuration? (A) (S)-2-butanol + Na gives the sodium alkoxide, then reacts with CH3Br via SN2. (B) (S)-2-butanol + TsCl (tosyl chloride) gives the tosylate, then tosylate reacts with CH3ONa via SN2. (C) (S)-2-butanol + PCl5 via SNi mechanism gives 2-chlorobutane.

1. Reaction A only
2. Reaction B only
3. Reaction C only
4. Reactions A and B

Correct answer: Reaction B only

Solution

Reaction A: (S)-2-butanol reacts with Na to form sodium (S)-2-butoxide. This breaks only the O-H bond, so configuration at chiral center is retained. The sodium butoxide then reacts with CH3Br: the oxygen (nucleophile) attacks the methyl carbon of CH3Br in SN2. The chiral center (C2 of butoxide) is NOT attacked. Configuration at C2 is retained. No inversion at chiral center. Reaction B: (S)-2-butanol + TsCl: the O-H bond is replaced by O-Ts, C-O bond is retained. Configuration at chiral center is retained (still S). The tosylate is a good leaving group. CH3ONa then attacks... wait, if CH3ONa attacks the chiral carbon in SN2, there IS inversion. But the product would be (R)-2-methoxylbutane (ether). Actually the reaction shown has the tosylate oxygen leaving (the whole OTs group leaves from chiral carbon), and OCH3 attacks chiral carbon via SN2 -> inversion of configuration. Product is (R)-2-methoxybutane. So reaction B proceeds with inversion. Reaction C: SNi (substitution nucleophilic internal) is a concerted intramolecular mechanism where the chloride ion from PCl5 attacks from the same face as the departing group -> RETENTION of configuration. Answer: Reaction B only.

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