The enantiopure alcohol shown is subjected to four successive steps: (1) TsCl, pyridine; (2) NaBr, DMF; (3) NaOAc, DMF; (4) K2CO3 in methanol (cleaves the acetate by attack at the acyl carbon). What is the final alcohol?

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JEE Advanced levelSingle correct optionDifficulty 3 of 3o3244

The enantiopure alcohol shown is subjected to four successive steps: (1) TsCl, pyridine; (2) $NaBr$ , DMF; (3) $NaOAc$ , DMF; (4) $K X_{2} CO X_{3}$ in methanol (cleaves the acetate by attack at the acyl carbon). What is the final alcohol?

A(S)-butan-2-ol
B(R)-butan-2-ol
CRacemic butan-2-ol
D(R)-2-bromobutane

Answer: A

The compound is (S)-butan-2-ol. Step 1 retains configuration (O-H bond cleaved). Step 2 is

S X_{N} 2

with inversion, giving (R)-2-bromobutane. Step 3 is another inversion, giving (S)-2-butyl acetate. Step 4 cleaves the acyl C-O bond, so the stereocentre is untouched. Two inversions: net retention, (S)-butan-2-ol.

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The enantiopure alcohol shown is subjected to four successive steps: (1) TsCl, pyridine; (2) NaBr, DMF; (3) NaOAc, DMF; (4) KX2​COX3​ in methanol (cleaves the acetate by attack at the acyl carbon). What is the final alcohol?

Answer: A

More questions on Stereochemical outcome of substitution on drawn chiral substrates

The enantiopure alcohol shown is subjected to four successive steps: (1) TsCl, pyridine; (2) $NaBr$ , DMF; (3) $NaOAc$ , DMF; (4) $K X_{2} CO X_{3}$ in methanol (cleaves the acetate by attack at the acyl carbon). What is the final alcohol?