Predict the product of the following reaction | vedclass

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

Question

(B) The reaction $Br_2/KOH$ is the Hofmann bromamide degradation reaction.In this reaction,an amide $(-CONH_2)$ is converted into a primary amine $(-N

Vedclass · Accepted Answer

Vedclass · Answer

(B) The reaction $Br_2/KOH$ is the Hofmann bromamide degradation reaction.In this reaction,an amide $(-CONH_2)$ is converted into a primary amine $(-NH_2)$ with the loss of one carbon atom.The starting material is $2	ext{-bromo-6-(methoxycarbonyl)phenylacetamide}$.The $-CONH_2$ group is converted to $-NH_2$.The $-Br$ and $-COOCH_3$ groups remain unaffected.Thus,the product is $2	ext{-bromo-6-(methoxycarbonyl)benzylamine}$.

Predict the product of the following reaction:
$2\text{-bromo-6-(methoxycarbonyl)phenylacetamide} \xrightarrow{Br_2/KOH} ?$

Similar Questions

Aniline on treatment with a mixture of conc. $HNO_3$ and conc. $H_2SO_4$ yields:

Which of the following is the strongest base?

$A$ solution contains $1 \ g \ mol$ each of $p-$toluene diazonium chloride and $p-$nitrophenyl diazonium chloride. To this,$1 \ g \ mol$ of alkaline solution of phenol is added. Predict the major product and explain your answer.

Primary nitro compounds react with $HNO_2$ to form crystalline solids which,on treatment with $NaOH$,give:

The correct statements of the following are:
$A$. Aniline forms a stable benzene diazonium chloride at $285 \ K$.
$B$. $N$-Phenylethanamide is less reactive towards nitration than aniline.
$C$. $p-CH_3C_6H_4COCl$ is Hinsberg reagent.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Predict the product of the following reaction:$2\text{-bromo-6-(methoxycarbonyl)phenylacetamide} \xrightarrow{Br_2/KOH} ?$