Hoffmann Bromamide Degradation: Mechanism, One Carbon Fewer and Exam Traps

ChemistryClass 12, AminesFor JEE and NEET aspirants

Hoffmann bromamide degradation turns a primary amide into a primary amine with one carbon fewer. The amide is warmed with bromine and an aqueous or ethanolic solution of sodium (or potassium) hydroxide. The alkyl or aryl group moves from the carbonyl carbon to the nitrogen, and the carbonyl carbon leaves as carbonate. The reaction is named after A. W. von Hofmann, so most chemistry books spell it Hofmann; NCERT writes Hoffmann. It is in the Class 12 NCERT chapter on amines, and JEE Main and NEET ask it in reaction sequences.

At a glance

General:
$RCON H_{2} + B r_{2} + 4NaOH RN H_{2} + N a_{2} C O_{3} + 2NaBr + 2 H_{2} O$
Benzamide to aniline:
$C_{6} H_{5} CON H_{2} + B r_{2} + 4NaOH C_{6} H_{5} N H_{2} + N a_{2} C O_{3} + 2NaBr + 2 H_{2} O$
Needs: a primary amide, $RCON H_{2}$ , with two N–H bonds.
Product: a primary amine, $RN H_{2}$ , with one carbon fewer than the amide.
Per mole of amide: one mole of $B r_{2}$ and four of NaOH; by-products $N a_{2} C O_{3}$ , NaBr and water.
Intermediate: the isocyanate, R–N=C=O.

1. Mechanism

Bromine and hydroxide first replace one H on the nitrogen by Br. The second N–H is now more acidic, and once it is removed the molecule rearranges: R moves from carbon to nitrogen just as bromide leaves.

Figure 1: The key step is the shift of R from C to N while bromide leaves. The carbonyl carbon ends as carbonate, so the amine has one carbon fewer than the amide.

N-Bromination. $O H^{-}$ removes one N–H proton and the anion takes Br from $B r_{2}$ : $RCONHBr$ .
Second deprotonation. $O H^{-}$ removes the remaining N–H proton, which the Br has made more acidic.
Rearrangement. R moves from C to N with its pair of electrons while $B r^{-}$ leaves, giving the isocyanate R–N=C=O. This is the key step.
Hydrolysis. Water and $O H^{-}$ add to the isocyanate; the carbamic acid formed, $RNHCOOH$ , loses $C O_{2}$ (as carbonate in the alkali), leaving $RN H_{2}$ .

Exam Trick

"Hoffmann takes a carbon away." Count the carbons: the amine always has one fewer than the amide. Benzamide ( $C_{6} H_{5} CON H_{2}$ , 7 C) gives aniline ( $C_{6} H_{5} N H_{2}$ , 6 C), not benzylamine.

2. Which amide gives which amine

Amide	Amine formed
Ethanamide, $C H_{3} CON H_{2}$	Methanamine, $C H_{3} N H_{2}$
Propanamide, $C H_{3} C H_{2} CON H_{2}$	Ethanamine, $C H_{3} C H_{2} N H_{2}$
Butanamide, $C H_{3} C H_{2} C H_{2} CON H_{2}$	Propan-1-amine, $C H_{3} C H_{2} C H_{2} N H_{2}$
Benzamide, $C_{6} H_{5} CON H_{2}$	Aniline, $C_{6} H_{5} N H_{2}$
2-Phenylethanamide, $C_{6} H_{5} C H_{2} CON H_{2}$	Benzylamine, $C_{6} H_{5} C H_{2} N H_{2}$
N-Methylethanamide, $C H_{3} CONHC H_{3}$	No amine: the N has only one H

Three things make the reaction useful. It gives a primary amine only, without the secondary and tertiary amines that ammonolysis of alkyl halides produces. And it shortens the chain by one carbon, so an acid can be stepped down: $RCOOH$ with $N H_{3}$ on heating gives $RCON H_{2}$ , and the degradation then gives $RN H_{2}$ . And it makes aromatic amines such as aniline, which Gabriel phthalimide synthesis cannot.

JEE Advanced

The group keeps its configuration. If the migrating carbon is a stereocentre, it keeps its configuration (retention), because R moves within the molecule and is never free. Its R/S label can still change if the priorities of its groups change.

Why the second N–H matters. Only an amide with two N–H bonds can lose one H to bromination and a second to the base. An N-substituted amide, RCONHR′, stops after bromination and gives no amine. Phthalimide has only one N–H, yet with $B r_{2}$ and KOH it gives anthranilic acid (2-aminobenzoic acid): the alkali first opens the ring to a primary amide, which then degrades.

Other halogen sources. Sodium hypobromite, NaOBr, does the same job, since $B r_{2}$ and NaOH form NaOBr in the first place. Sodium hypochlorite with NaOH also works, through the N-chloroamide instead of the N-bromoamide.

Not the Hofmann elimination. The same chemist's name is on a different reaction: heating a quaternary ammonium hydroxide gives an alkene, mainly the less substituted one.

3. Worked examples

Solved Example 1

C H_{3} C H_{2} C H_{2} I

is treated with NaCN to give A; partial hydrolysis of A gives B, and B with NaOH and

B r_{2}

gives C. Identify A, B and C.

Solution:

A is butanenitrile, $C H_{3} C H_{2} C H_{2} CN$ (the chain gains a carbon). B is butanamide, $C H_{3} C H_{2} C H_{2} CON H_{2}$ . C is propan-1-amine, $C H_{3} C H_{2} C H_{2} N H_{2}$ : the Hoffmann degradation takes the carbon away again.

Solved Example 2

Benzamide is heated with bromine in ethanolic NaOH. Is the product benzylamine?

Solution:

No. The carbonyl carbon is lost, so the product is aniline, $C_{6} H_{5} N H_{2}$ ( $C_{6} H_{7} N$ ). Benzylamine, $C_{6} H_{5} C H_{2} N H_{2}$ , comes from 2-phenylethanamide, $C_{6} H_{5} C H_{2} CON H_{2}$ .

Solved Example 3

How many moles of NaOH and of

B r_{2}

are used per mole of amide, and what are the by-products?

Solution:

One mole of $B r_{2}$ and four moles of NaOH. The by-products are $N a_{2} C O_{3}$ , NaBr and water:

RCON H_{2} + B r_{2} + 4NaOH RN H_{2} + N a_{2} C O_{3} + 2NaBr + 2 H_{2} O

4. Exam traps

Watch out

The amine has one carbon fewer: benzamide gives aniline, not benzylamine.
Only primary amides, $RCON H_{2}$ , react; N-substituted amides give no amine.
The product is always a primary amine, never secondary or tertiary.
One $B r_{2}$ and four NaOH per amide; the by-products are $N a_{2} C O_{3}$ , NaBr and water.
$B r_{2}$ and NaOH do not act on a nitrile (they are not a reducing agent); to use this route, first partly hydrolyse the nitrile to the amide.
The intermediate is an isocyanate, R–N=C=O. An isocyanide (R–N≡C, foul smell) comes from the carbylamine reaction, not from this one.
Hoffmann (NCERT) and Hofmann are the same chemist; the Hofmann elimination is a different reaction.
The migrating group keeps its configuration.

Past-paper questions on Hoffmann bromamide degradation

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 15 questions in them that test this:

JEE Main 11 (2025-2026)
JEE Advanced 1 (2025)
NEET 3 (2023-2024, 2026)

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