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(A) $CH_3-CH_2-CH_2-Br$ undergoes dehydrohalogenation with alcoholic $KOH$ to form propene $(A)$.Propene then reacts with $HBr$ via Markovnikov additi

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$A.$ Propan$-2-$ol

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(A) $CH_3-CH_2-CH_2-Br$ undergoes dehydrohalogenation with alcoholic $KOH$ to form propene $(A)$.Propene then reacts with $HBr$ via Markovnikov addition to form $2$-bromopropane $(B)$.Finally,$2$-bromopropane reacts with aqueous $KOH$ via nucleophilic substitution to form propan-$2$-ol $(C)$.$CH_3-CH_2-CH_2-Br$ $\xrightarrow{KOH(alc)} CH_3-CH=CH_2 (A)$ $\xrightarrow{HBr} CH_3-CH(Br)-CH_3 (B)$ $\xrightarrow{KOH(aq.)} CH_3-CH(OH)-CH_3 (C)$

In the following sequence of reactions,the product $C$ is: $CH_3-CH_2-CH_2-Br$ $\xrightarrow{KOH(alc)} (A)$ $\xrightarrow{HBr} (B)$ $\xrightarrow{KOH(aq.)} (C)$

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