Consider following reaction and find incorrect option(s) regarding product(s)

Number of optically active compound formed is 10

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Number of pair of enantiomers is 5

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Number of pair of diastereomers formed is 4

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Number of optically inactive compounds formed is 2

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Solution

The correct option is D Number of optically inactive compounds formed is 2

There are 8 compoundws formed by monochlorination, out of which there are four pairs of enantiomers.

There are 24 diastereomers.

Since the starting compound was symmetric to begin with, all the 8 compounds are optically active. That is, there are no meso copounds, neither are there any optically inactive compounds.

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C H_{3} O H H B r - -- \to A K C N - -- \to B H_{2} / N i - --- \to C H N O_{2} - --- \to D

C H_{2} = C H_{2} + C l_{2} / C C l_{4} - ----- \to Product A

C H_{2} = C H_{2} + H_{2} N i / P t / P d - ----- \to Product B

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