C_4H_7OCl (x) xrightarrow{NH_3} C_4H_9NO xrig | vedclass

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

Question

(A) The reaction sequence is: $C_4H_7OCl$ $(x)$ $\xrightarrow{NH_3} C_4H_9NO$ (amide) $\xrightarrow{Br_2 + KOH} CH_3-CH_2-CH_2-NH_2$ (propylamine).Thi

Vedclass · Accepted Answer

Butanoyl chloride $(CH_3CH_2CH_2COCl)$

Vedclass · Answer

(A) The reaction sequence is: $C_4H_7OCl$ $(x)$ $\xrightarrow{NH_3} C_4H_9NO$ (amide) $\xrightarrow{Br_2 + KOH} CH_3-CH_2-CH_2-NH_2$ (propylamine).This is a Hofmann bromamide degradation reaction,which converts an amide $(R-CONH_2)$ into a primary amine $(R-NH_2)$ with one carbon atom less.Since the product is propylamine $(CH_3CH_2CH_2NH_2)$,the amide must be butanamide $(CH_3CH_2CH_2CONH_2)$.Butanamide is formed by the reaction of butanoyl chloride $(CH_3CH_2CH_2COCl)$ with ammonia $(NH_3)$.Therefore,compound $x$ is butanoyl chloride.

$C_4H_7OCl$ $(x)$ $\xrightarrow{NH_3} C_4H_9NO$ $\xrightarrow{Br_2 + KOH} CH_3-CH_2-CH_2-NH_2$. Compound $x$ is:

Similar Questions

The strongest acid amongst the following compounds is:

In the esterification reaction of alcohols with carboxylic acids,which of the following occurs?

What sized ring is formed when two carboxylic acids form a dimeric structure through intermolecular hydrogen bonding? (Consider the hydrogen bond as a bond in the ring structure.)

In the given reaction,the reactant is $\gamma$-valerolactone. Predict the product $[X]$ for the reaction: $\text{Reactant} \xrightarrow{H_2O/H^{+}} [X]$.

Oxalic acid may be distinguished from tartaric acid by

Vedclass Test Series

Exam Paper Generator

Online Exam Module