The product (C) in the below mentioned reacti | vedclass

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(D) Step $1$: Dehydrohalogenation of $CH_3-CH_2-CH_2-Br$ with $KOH_{(alc)}$ gives propene $(A)$ as $CH_3-CH=CH_2$.Step $2$: Electrophilic addition of

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Propan$-2-$ol

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(D) Step $1$: Dehydrohalogenation of $CH_3-CH_2-CH_2-Br$ with $KOH_{(alc)}$ gives propene $(A)$ as $CH_3-CH=CH_2$.Step $2$: Electrophilic addition of $HBr$ to propene follows Markovnikov's rule to give $2-$bromopropane $(B)$ as $CH_3-CH(Br)-CH_3$.Step $3$: Nucleophilic substitution of $2-$bromopropane with $KOH_{(aq)}$ gives propan$-2-$ol $(C)$ as $CH_3-CH(OH)-CH_3$.

The product $(C)$ in the below mentioned reaction is:
$CH_3-CH_2-CH_2-Br$ $\xrightarrow[\Delta]{KOH_{(alc)}} A$ $\xrightarrow{HBr} B$ $\xrightarrow[KOH_{(aq)}]{\Delta} C$

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The product $(C)$ in the below mentioned reaction is:$CH_3-CH_2-CH_2-Br$ $\xrightarrow[\Delta]{KOH_{(alc)}} A$ $\xrightarrow{HBr} B$ $\xrightarrow[KOH_{(aq)}]{\Delta} C$