Compound (A),C_6H_{12}O_3,when treated with I | vedclass

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(C) $1$. The compound $(A)$ gives a yellow precipitate with $I_2/NaOH$,indicating the presence of a methyl ketone group $(CH_3CO-)$.$2$. $(A)$ does no

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$CH_3-CO-CH_2-CH(OCH_3)_2$

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(C) $1$. The compound $(A)$ gives a yellow precipitate with $I_2/NaOH$,indicating the presence of a methyl ketone group $(CH_3CO-)$.$2$. $(A)$ does not react with Tollens' reagent,meaning it does not contain a free aldehyde group.$3$. Upon hydrolysis,$(A)$ gives a positive Tollens' test,indicating the formation of an aldehyde group.$4$. Option $(C)$,$CH_3-CO-CH_2-CH(OCH_3)_2$,is an acetal. It contains a methyl ketone group,so it gives the iodoform test. It does not have a free aldehyde group,so it does not react with Tollens' reagent. Upon hydrolysis,the acetal group $-CH(OCH_3)_2$ converts to an aldehyde group $-CHO$,which gives a positive Tollens' test.

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