The Conversion of an amide by action of NaOH and $B r_{2}$ to primary amine that has one carbon less than the starting amide is known as Hofmann-Bromamide reaction.

$R - O | | C - N H_{2} B r_{2} + N a O H - ------ \to (o r) N a o B r R - N H_{2} + N a B r + N a_{2} C O_{3}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B

The Conversion of an amide by action of NaOH and $B r_{2}$ to primary amine that has one carbon less than the starting amide is known as Hofmann-Bromamide reaction.

$R - O | | C - N H_{2} B r_{2} + N a O H - ------ \to (o r) N a o B r R - N H_{2} + N a B r + N a_{2} C O_{3}$

Hofmann Bromamide reaction is Intra molecular

Suggest Corrections

Similar questions

The Conversion of an amide by action of NaOH and $B r_{2}$ to primary amine that has one carbon less than the starting amide is known as Hofmann-Bromamide reaction.

$R - O | | C - N H_{2} B r_{2} + N a O H - ------ \to (o r) N a o B r R - N H_{2} + N a B r + N a_{2} C O_{3}$

Q. (1) How will you convert an amide into an amine having one carbon less than the starting compound
(2) Name the reaction

Q. The reaction of an amide with bromine and alkali to form a primary amine is called

Q. Assertion :In Hofmann bromide reaction, the amine formed has one carbon atom less than the parent

1^{o}

amide. Reason: N-methyl acetamide undergoes Hofimann bromamide reaction.

Q. Assertion :In Hoffmann bromide reaction, the amine formed has one carbon atom less than the parent

1^{o}

amide. Because Reason:

N -

methyl acetamide undergoes Hoffmann bromamide reaction.

Join BYJU'S Learning Program

The Conversion of an amide by action of NaOH and Br2 to primary amine that has one carbon less than the starting amide is known as Hofmann-Bromamide reaction. R−O||C−NH2Br2+NaOH−−−−−−−→(or)NaoBr R−NH2+NaBr+Na2CO3

The Conversion of an amide by action of NaOH and $B r_{2}$ to primary amine that has one carbon less than the starting amide is known as Hofmann-Bromamide reaction.

$R - O | | C - N H_{2} B r_{2} + N a O H - ------ \to (o r) N a o B r R - N H_{2} + N a B r + N a_{2} C O_{3}$