When propionamide reacts with Br_2 in the pre | vedclass

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

Question

(B) The reaction of an amide with $Br_2$ in the presence of an alkali ($KOH$ or $NaOH$) is known as the Hofmann bromamide degradation reaction.In this

Vedclass · Accepted Answer

$CH_3CH_2NH_2$

Vedclass · Answer

(B) The reaction of an amide with $Br_2$ in the presence of an alkali ($KOH$ or $NaOH$) is known as the Hofmann bromamide degradation reaction.In this reaction,the amide group $(-CONH_2)$ is converted into a primary amine $(-NH_2)$ with the loss of one carbon atom.Propionamide is $CH_3CH_2CONH_2$.According to the Hofmann bromamide degradation reaction: $CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$.The product formed is ethylamine $(CH_3CH_2NH_2)$.

When propionamide reacts with $Br_2$ in the presence of alkali the product is :

Similar Questions

What is the product formed when nitrobenzene undergoes nitration?

Reduction of nitrobenzene with zinc $(Zn)$ and ammonium chloride $(NH_4Cl)$ gives:

What product will be formed from the Hofmann exhaustive methylation of the following compound?
$CH_3-CH(CH_3)-CH_2-NH-CH_2-CH_2-CH_3 \xrightarrow[(i) CH_3I (excess), (ii) Ag_2O, (iii) \Delta]{} \text{Product}$

$A$ primary amine,$RNH_2$ can be reacted with $CH_3-X$ to get a secondary amine,$RNHCH_3$,but the disadvantage is that $3^o$ amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where $RNH_2$ forms only $2^o$ amine?

Identify $X$ and $Y$ in the following reactions:
$Nitrobenzene \xrightarrow{Zn/NH_4Cl} X$
$Nitrobenzene \xrightarrow{Zn + KOH/C_2H_5OH} Y$

Vedclass Test Series

Exam Paper Generator

Online Exam Module