An organic compound A has molecular formula $C_{8} H_{12}$ and is optically active It reacts with $H_{2} / P t$ to give $B (C_{8} H_{18})$ which is optically inactive Careful hydrogenation of A with $P d / B a S O_{4}$ gives $C (C_{8} H_{14})$ which is still optically active Compound A reacts with Na in liquid ammonia to give Diastereomer of C, but it is optically inactive Deduce structures of A to D.

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Solution

The structures of products A to D are shown in the figure. A contains a carbon carbon triple bond and a carbon carbon double bond.In the conversion of A to B, both double and triple bonds are hydrogenated to single bonds. In the conversion A to C and A to D, the triple bond is selectively reduced to double bond. However, the newly formed double bond has cis orintation in C and trans orientation in D.

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