n-Propanol on treatment with concentrated HBr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Vedclass · Accepted Answer

Butanoic acid

Vedclass · Answer

(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

Similar Questions

The major products obtained in the reaction of oxalic acid with conc. $H_2SO_4$ upon heating are

The reaction $CH_3COOAg + Br_2 \xrightarrow{CS_2}$ is known as the Hunsdiecker reaction. What is the main product of this reaction?

Identify the major product formed in the following reaction sequence:

What is obtained when oxalic acid is dehydrated in the presence of concentrated $H_2SO_4$?

Correct order of $pK_{a}$ of given compounds is $:-$
$(P) \ CH_{3}COOH$
$(Q) \ FCH_{2}COOH$
$(R) \ ClCH_{2}COOH$
$(S) \ C_{2}H_{5}COOH$

Vedclass Test Series

Exam Paper Generator

Online Exam Module