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

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(C) $1$. Compound $(A)$ has the molecular formula $C_6H_{12}O_3$.$2$. It gives a positive iodoform test,which indicates the presence of a $CH_3-C(=O)-

Vedclass · Accepted Answer

$CH_3-C(=O)-CH_2-CH(OCH_3)_2$

Vedclass · Answer

(C) $1$. Compound $(A)$ has the molecular formula $C_6H_{12}O_3$.$2$. It gives a positive iodoform test,which indicates the presence of a $CH_3-C(=O)-$ group.$3$. It gives a negative Tollens' test,meaning it does not contain an aldehyde group.$4$. Upon acid-catalyzed hydrolysis,it gives a product that shows a positive Tollens' test,indicating the formation of an aldehyde group.$5$. The structure $CH_3-C(=O)-CH_2-CH(OCH_3)_2$ contains a $CH_3-C(=O)-$ group (positive iodoform test) and an acetal group (negative Tollens' test).$6$. Hydrolysis of the acetal group $-CH(OCH_3)_2$ in the presence of $H^+$ yields an aldehyde group $-CHO$,which gives a positive Tollens' test.$7$. Therefore,the compound is $CH_3-C(=O)-CH_2-CH(OCH_3)_2$.

Compound $(A)$ with molecular formula $C_6H_{12}O_3$ gives a positive iodoform test and a negative Tollens' test. Upon hydrolysis with a drop of $H_2SO_4$,it gives a product that shows a positive Tollens' test. Compound $(A)$ is:

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