An acyclic hydrocarbon P with molecular formu | vedclass

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(B) Step $1$: Analyze the final product. The final product is $2$ moles of acetone $(CH_3COCH_3)$. This implies that the intermediate $Q$ must be $2,3

Vedclass · Accepted Answer

$(D, B)$

Vedclass · Answer

(B) Step $1$: Analyze the final product. The final product is $2$ moles of acetone $(CH_3COCH_3)$. This implies that the intermediate $Q$ must be $2,3$-dimethylbut-$2$-ene,which upon ozonolysis gives $2$ moles of acetone.Step $2$: Analyze the conversion of $Q$ to acetone. $Q$ undergoes acid-catalyzed dehydration followed by ozonolysis. $2,3$-dimethylbut-$2$-ene is formed by the dehydration of $2,3$-dimethylbutan-$2$-ol. Thus,$Q$ is $(CH_3)_2CH-CH(OH)-CH_3$ (Option $B$).Step $3$: Analyze the formation of $Q$ from $P$. $P$ $(C_6H_{10})$ reacts with $dil. H_2SO_4/HgSO_4$ to form a ketone,which is then reduced by $NaBH_4$ to an alcohol $Q$. For $Q$ to be $2,3$-dimethylbutan-$2$-ol,the ketone must be $3,3$-dimethylbutan-$2$-one. This ketone is formed from the hydration of $3,3$-dimethylbut-$1$-yne $(P)$. Thus,$P$ is $(CH_3)_2CH-C \equiv C-H$ (Option $D$).Therefore,$P$ is $D$ and $Q$ is $B$.

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