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

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Solution

$R - O | | C - N H_{2} B r_{2} - --- \to 4 K O H R - N H_{2} + K_{2} C O_{3} + 2 K B r + 2 H_{2} O$

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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.

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

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. 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. Write and name the reaction to prepare amine containing less number of carbon atoms from acid-amide.

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.

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

R−O||C−NH2Br2−−−−→4KOHR−NH2+K2CO3+2KBr+2H2O

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

$R - O | | C - N H_{2} B r_{2} - --- \to 4 K O H R - N H_{2} + K_{2} C O_{3} + 2 K B r + 2 H_{2} O$