Compound (A) with molecular formula C_6H_{12} | vedclass

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(B) The molecular formula $C_6H_{12}O$ corresponds to a degree of unsaturation of $1$,indicating either an aldehyde or a ketone.Since the compound giv

Vedclass · Accepted Answer

$CH_3-C(=O)-CH(CH_3)-CH_2-CH_3$

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(B) The molecular formula $C_6H_{12}O$ corresponds to a degree of unsaturation of $1$,indicating either an aldehyde or a ketone.Since the compound gives a positive $2,4$-dinitrophenylhydrazine test,it contains a carbonyl group.$A$ negative Tollens test confirms that the compound is a ketone,not an aldehyde.For the compound to be optically active,it must possess a chiral center.In option $(B)$,$CH_3-C(=O)-CH(CH_3)-CH_2-CH_3$,the carbon atom at the $3^{rd}$ position is bonded to four different groups: $-H$,$-CH_3$,$-CH_2CH_3$,and $-COCH_3$,making it a chiral center.

Compound $(A)$ with molecular formula $C_6H_{12}O$ is optically active. Compound $(A)$ gives a negative Tollens test and a positive test with $2,4$-dinitrophenylhydrazine. Identify $(A)$.

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