In the following sequence of reactions,CH_3CH | vedclass

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(D) Step $1$: $CH_3CH_2CH_2Br$ reacts with alcoholic $KOH$ (dehydrohalogenation) to form propene $(A)$ $(CH_3CH=CH_2)$.Step $2$: Propene $(A)$ reacts

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Propan$-2-$ol

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(D) Step $1$: $CH_3CH_2CH_2Br$ reacts with alcoholic $KOH$ (dehydrohalogenation) to form propene $(A)$ $(CH_3CH=CH_2)$.Step $2$: Propene $(A)$ reacts with $HBr$ following Markovnikov's rule to form $2$-bromopropane $(B)$ $(CH_3CH(Br)CH_3)$.Step $3$: $2$-bromopropane $(B)$ reacts with aqueous $KOH$ (nucleophilic substitution) to form propan$-2-$ol $(C)$ $(CH_3CH(OH)CH_3)$.

In the following sequence of reactions,
$CH_3CH_2CH_2Br$ $\xrightarrow{KOH(alc.)}$ $(A)$ $\xrightarrow{HBr}$ $(B)$ $\xrightarrow{KOH(aq.)}$ $(C)$
The product $(C)$ is

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In the following sequence of reactions,$CH_3CH_2CH_2Br$ $\xrightarrow{KOH(alc.)}$ $(A)$ $\xrightarrow{HBr}$ $(B)$ $\xrightarrow{KOH(aq.)}$ $(C)$The product $(C)$ is