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(B) $1$. The molecular formula $C_6H_{12}O$ corresponds to the general formula $C_nH_{2n}O$,which indicates that the compound $P$ is either an aldehyd

Vedclass · Accepted Answer

$CH_3-CO-CH(CH_3)-CH_2-CH_3$

Vedclass · Answer

(B) $1$. The molecular formula $C_6H_{12}O$ corresponds to the general formula $C_nH_{2n}O$,which indicates that the compound $P$ is either an aldehyde or a ketone.$2$. $P$ gives a positive test with $2,4-$dinitrophenylhydrazine $(2,4-DNP)$,which confirms the presence of a carbonyl group $(C=O)$.$3$. $P$ gives a negative test with Tollens reagent,which indicates that $P$ is a ketone,not an aldehyde.$4$. $P$ is optically active,meaning it must contain at least one chiral carbon atom.$5$. Let's examine the options:- Option $A$: $CH_3-CO-CH_2-CH_2-CH_2-CH_3$ (Hexan-$2$-one) is not optically active.- Option $B$: $CH_3-CO-CH(CH_3)-CH_2-CH_3$ ($3-$methylpentan-$2$-one) has a chiral carbon at the $C-3$ position. It is a ketone and gives a positive $2,4-DNP$ test and a negative Tollens test.- Option $C$: This is an aldehyde,which would give a positive Tollens test.- Option $D$: $CH_3-CO-CH_2-CH(CH_3)_2$ ($4-$methylpentan-$2$-one) is not optically active.$6$. Therefore,the correct structure is $3-$methylpentan-$2$-one.

$P$ is an optically active compound with molecular formula $C_6H_{12}O$. When $P$ is treated with $2,4-$dinitrophenylhydrazine,it gives a positive test. However,in the presence of Tollens reagent,$P$ gives a negative test. Predict the structure of $P$.

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