n-Propanol on treatment with concentrated HBr | vedclass

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Question

(C) $n$-Propanol $(CH_3CH_2CH_2OH)$ reacts with concentrated $HBr$ to form $n$-propyl bromide $(CH_3CH_2CH_2Br)$ as product $P$.$n$-Propyl bromide rea

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Butanoic acid

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(C) $n$-Propanol $(CH_3CH_2CH_2OH)$ reacts with concentrated $HBr$ to form $n$-propyl bromide $(CH_3CH_2CH_2Br)$ as product $P$.$n$-Propyl bromide reacts with $KCN$ via nucleophilic substitution to form butanenitrile $(CH_3CH_2CH_2CN)$ as product $Q$.Butanenitrile on heating with an aqueous acidic solution undergoes hydrolysis to form butanoic acid $(CH_3CH_2CH_2COOH)$ as product $R$.$CH_3CH_2CH_2OH$ $\xrightarrow{HBr} CH_3CH_2CH_2Br$ $\xrightarrow{KCN} CH_3CH_2CH_2CN$ $\xrightarrow{H_3O^{+}} CH_3CH_2CH_2COOH$

$n$-Propanol on treatment with concentrated $HBr$ gives $P$. The product $P$ on reaction with $KCN$ gave the product $Q$. The product $Q$ on heating with aqueous acidic solution,furnished the product $R$. The product $R$ is

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