In the following sequence of reactions,C_{3}H | vedclass

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(A) Step $1$: Hydration of propene $(C_{3}H_{6})$ with $H^{+}/H_{2}O$ follows Markovnikov's rule to form propan$-2-$ol $(A)$: $CH_{3}-CH=CH_{2} + H_{2

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$CHI_{3}, CH_{3}COOK$

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(A) Step $1$: Hydration of propene $(C_{3}H_{6})$ with $H^{+}/H_{2}O$ follows Markovnikov's rule to form propan$-2-$ol $(A)$: $CH_{3}-CH=CH_{2} + H_{2}O \xrightarrow{H^{+}} CH_{3}-CH(OH)-CH_{3} (A)$Step $2$: Propan$-2-$ol $(A)$ is a secondary alcohol with a methyl group attached to the carbinol carbon,which undergoes the iodoform reaction with $KIO/dil. KOH$ to form iodoform $(B)$ and potassium acetate $(C)$: $CH_{3}-CH(OH)-CH_{3} + 4KIO \rightarrow CHI_{3} (B) + CH_{3}COOK (C) + 3KOH + H_{2}O$Thus,$B$ is $CHI_{3}$ and $C$ is $CH_{3}COOK$.

In the following sequence of reactions,
$C_{3}H_{6}$ $\xrightarrow{H^{+}/H_{2}O} A$ $\xrightarrow[dil. KOH]{KIO} B + C$
The compounds $B$ and $C$ respectively are :

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In the following sequence of reactions,$C_{3}H_{6}$ $\xrightarrow{H^{+}/H_{2}O} A$ $\xrightarrow[dil. KOH]{KIO} B + C$The compounds $B$ and $C$ respectively are :