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The major product in the following reaction is 2-butyne + Na/liqNH_3/ followed by reaction with OsO₄ /NaHSO₃

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racemic mixture of butane – 2, 3 – diol

a pure enantiomers of butane – 2, 3 – diol

pair of enantiomers of butane – 2, 3 – diol but in unequal amount

meso butane – 2, 3 – diol

answer is C.

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Detailed Solution

$2 - butyne + Na / {liqNH}_{3} \to trans alkene \to_{Syn - hydro x ylation}^{{OsO}_{4} / {NaHSO}_{3}} racemic mixture of butane - 2, 3 - diol$

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