I. (CH_3)_2 C=CH_2 xrightarrow{KMnO_4 / H^{+} | vedclass

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(A) The oxidative cleavage of alkenes using hot acidic $KMnO_4$ results in the breaking of the $C=C$ double bond.For reaction $I$: $(CH_3)_2 C=CH_2$ u

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Ketone and Carboxylic acid

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(A) The oxidative cleavage of alkenes using hot acidic $KMnO_4$ results in the breaking of the $C=C$ double bond.For reaction $I$: $(CH_3)_2 C=CH_2$ undergoes cleavage to form $(CH_3)_2 C=O$ (acetone,a ketone) and $CO_2 + H_2 O$. Thus,$X$ is a ketone.For reaction $II$: $CH_3-CH=CH-CH_3$ undergoes cleavage to form $2CH_3COOH$ (acetic acid,a carboxylic acid). However,looking at the provided options and the structure of $Y$ in the image,the product is $CH_3COOH$. Given the options,the correct functional groups are Ketone $(X)$ and Carboxylic acid $(Y)$.

$I$. $(CH_3)_2 C=CH_2 \xrightarrow{KMnO_4 / H^{+}} X + CO_2 + H_2 O$
$II$. $CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4 / H^{+}} Y$
The functional groups in $X$ and $Y$ are respectively

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$I$. $(CH_3)_2 C=CH_2 \xrightarrow{KMnO_4 / H^{+}} X + CO_2 + H_2 O$$II$. $CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4 / H^{+}} Y$The functional groups in $X$ and $Y$ are respectively

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$I$. $(CH_3)_2 C=CH_2 \xrightarrow{KMnO_4 / H^{+}} X + CO_2 + H_2 O$
$II$. $CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4 / H^{+}} Y$
The functional groups in $X$ and $Y$ are respectively