An organic compound A on treatment with NH_3 | vedclass

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

Question

(D) The reaction sequence is as follows: $CH_3CH_2COOH (A)$ $\xrightarrow{NH_3} CH_3CH_2COONH_4 (B)$ $\xrightarrow{\Delta} CH_3CH_2CONH_2 (C)$ $\xrigh

Vedclass · Accepted Answer

$CH_3CH_2COOH$

Vedclass · Answer

(D) The reaction sequence is as follows: $CH_3CH_2COOH (A)$ $\xrightarrow{NH_3} CH_3CH_2COONH_4 (B)$ $\xrightarrow{\Delta} CH_3CH_2CONH_2 (C)$ $\xrightarrow{Br_2/KOH} CH_3CH_2NH_2$ (Ethylamine). The last step is the Hoffmann Bromamide degradation reaction,where an amide is converted into an amine with one less carbon atom. Since the product is ethylamine $(CH_3CH_2NH_2)$,which has $2$ carbon atoms,the amide $(C)$ must have $3$ carbon atoms. Therefore,compound $A$ is propanoic acid $(CH_3CH_2COOH)$.

An organic compound $A$ on treatment with $NH_3$ gives $B$,which on heating gives $C$. $C$ when treated with $Br_2$ in the presence of $KOH$ produces ethyl amine. Compound $A$ is:

Similar Questions

How can ethylamine be prepared?

Which of the following amines cannot be prepared by Gabriel phthalimide synthesis?

Identify product $B$ in the following reaction: $CH_3Br + AgNO_2$ $\longrightarrow A$ $\xrightarrow{Sn / HCl} B$

Write chemical equations for the following reactions:
$(i)$ Reaction of ethanolic $NH_3$ with $C_2H_5-Cl$.
$(ii)$ Ammonolysis of benzyl chloride and reaction of amine so formed with two moles of $CH_3-Cl$.

$CH_3-CH_2-I$ $\xrightarrow{NaCN} A$ $\xrightarrow[Partial \ hydrolysis]{\Theta OH} B$ $\xrightarrow[Br_2]{NaOH} C$. Product $C$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module