$2 - b u t y n e + N a / N H_{3} (l)$ then $O s O_{4} / N a H S O_{3} ⟶$
The major product in the above reaction is

a pure enantiomers of

b u t a n e - 2, 3 - d i o l

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m e s o

b u t a n e - 2, 3 - d i o l

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racemic mixture of

b u t a n e - 2, 3 - d i o l

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pair of enantiomers of

b u t a n e - 2, 3 - d i o l

but in unequal amount

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Solution

The correct option is C racemic mixture of $b u t a n e - 2, 3 - d i o l$
The major product of the given reaction is represented by option (C). The reaction is as shown above.

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M e s o - b u t a n - 2, 3 - d i o l

2, 3

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