An optically active compound C_6H_{12}O gives | vedclass

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(C) $1$. The compound gives a positive $2,4-DNP$ test,which indicates the presence of a carbonyl group (aldehyde or ketone).$2$. It gives a negative T

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

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(C) $1$. The compound gives a positive $2,4-DNP$ test,which indicates the presence of a carbonyl group (aldehyde or ketone).$2$. It gives a negative Tollens' reagent test,which indicates that it is a ketone,not an aldehyde.$3$. The molecular formula is $C_6H_{12}O$. $4$. The compound must be optically active,meaning it must contain a chiral center (a carbon atom bonded to four different groups).$5$. Option $C$ is $CH_3-CO-CH(CH_3)-CH_2-CH_3$ ($3$-methylpentan$-2-$one). The carbon at the $3^{rd}$ position is chiral because it is attached to $-H$,$-CH_3$,$-COCH_3$,and $-CH_2CH_3$ groups. Thus,it is optically active.

An optically active compound $C_6H_{12}O$ gives a positive test with $2,4-$dinitrophenylhydrazine,but a negative test with Tollens' reagent. What is the structure of the compound?

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