(a) Write the structures of A, B and C in the following reactions : (i) Benzoic acid [$C_6H_5COOH$] $\xrightarrow{NH_3, \Delta}$ A $\xrightarrow{Br_2 + NaOH}$ B $\xrightarrow{NaNO_2 + HCl, 0^\circ C}$ C (ii) $CH_3CH_2Br \xrightarrow{KCN}$ A $\xrightarrow{LiAlH_4}$ B $\xrightarrow{HNO_2, 0^\circ C}$ C OR (b) How will you convert the following : (i) Aniline to p-bromoaniline (ii) Ethanoic acid to methanamine (iii) Butanenitrile to 1-aminobutane

(a) Write the structures of A, B and C in the following reactions : (i) Benzoic acid [C₆H₅COOH] xrightarrowNH₃

Question

Q. (a) Write the structures of A, B and C in the following reactions :

(i) Benzoic acid [

C_6H_5COOH

]

\xrightarrow{NH_3, \Delta}

\xrightarrow{Br_2 + NaOH}

\xrightarrow{NaNO_2 + HCl, 0^\circ C}

(ii)

CH_3CH_2Br \xrightarrow{KCN}

\xrightarrow{LiAlH_4}

\xrightarrow{HNO_2, 0^\circ C}

(OR)

(b) How will you convert the following :

(i) Aniline to p-bromoaniline

(ii) Ethanoic acid to methanamine

(iii) Butanenitrile to 1-aminobutane

CBSECBSE Class XII Board 2023Subjective· 3mImportance★★★★★

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Concept understanding — Hofmann Bromamide Reaction

Hofmann Bromamide Degradation

Imagine you have an amide — a molecule with a carbonyl group ( $\ce{-CO-}$ ) attached to a nitrogen. You want to turn it into a primary amine, but you also want to chop off one carbon from the chain. That is exactly what the Hofmann bromamide reaction does: it shortens the carbon skeleton by one carbon and gives you an amine.

The Intuition

The reaction uses bromine ( $\ce{Br2}$ ) in the presence of a strong alkali (like $\ce{NaOH}$ or $\ce{KOH}$ ). The alkali first deprotonates the amide nitrogen, making it a strong nucleophile. This nucleophile attacks bromine, forming an N-bromoamide. Under the strongly basic conditions, this intermediate loses a bromide ion and undergoes a rearrangement — the alkyl group attached to the carbonyl carbon migrates from carbon to nitrogen. The result is an isocyanate intermediate ( $\ce{R-N=C=O}$ ). Finally, the isocyanate is hydrolysed by the aqueous alkali to give a primary amine and carbon dioxide.

The net effect: the carbonyl carbon is lost as $\ce{CO2}$ , and the alkyl group ends up attached to the nitrogen.

Important

The product amine has one fewer carbon than the starting amide. The lost carbon is the carbonyl carbon.

The Precise Statement

Hofmann bromamide degradation (also called Hofmann rearrangement) is the conversion of a primary amide to a primary amine with one fewer carbon atom, using bromine and an aqueous alkali (usually $\ce{NaOH}$ or $\ce{KOH}$ ).

The general reaction is:

\ce{R-CONH2 + Br2 + 4 NaOH -> R-NH2 + 2 NaBr + Na2CO3 + 2 H2O}

Or, in a more compact form:

\ce{R-CONH2 ->[Br2, NaOH] R-NH2 + CO2}

Step-by-Step Mechanism

Deprotonation: The amide nitrogen is deprotonated by the strong base, forming an amide anion.

\ce{R-CONH2 + OH- -> R-CONH- + H2O}

Bromination: The amide anion attacks bromine, forming an N-bromoamide.

\ce{R-CONH- + Br2 -> R-CONHBr + Br-}

Second deprotonation: The N-bromoamide is deprotonated again by the base.

\ce{R-CONHBr + OH- -> R-CONBr- + H2O}

Rearrangement: The alkyl group migrates from the carbonyl carbon to the nitrogen, with simultaneous loss of bromide ion. This forms an isocyanate.

\ce{R-CONBr- -> R-N=C=O + Br-}

Hydrolysis: The isocyanate reacts with water to form a carbamic acid, which spontaneously decarboxylates (loses $\ce{CO2}$ ) to give the primary amine.

\ce{R-N=C=O + H2O -> R-NH-COOH -> R-NH2 + CO2}

Tip

The rearrangement step (step 4) is the key. The alkyl group migrates with its bonding electrons — it is a 1,2-shift from carbon to the electron-deficient nitrogen. This is why the carbon skeleton shortens by one carbon.

Key Points for Exams

Starting material: Primary amide ( $\ce{R-CONH2}$ ) only. Secondary or tertiary amides do not undergo this reaction.
Reagents: $\ce{Br2}$ and $\ce{NaOH}$ (or $\ce{KOH}$ ). Sometimes $\ce{Cl2}$ can be used instead of $\ce{Br2}$ , but bromine is more common.
Product: Primary amine with one fewer carbon.
By-products: $\ce{NaBr}$ , $\ce{Na2CO3}$ , $\ce{H2O}$ (or $\ce{CO2}$ if written in the simplified form). …

(a) (i) benzamide → aniline (Hofmann) → benzenediazonium chloride; (ii) propanenitrile → propan-1-amine → propan-1-ol.

(b) (i) protect-brominate-deprotect aniline → p-bromoaniline; (ii) acetic acid → acetamide → methanamine (Hofmann); (iii) $LiAlH_4$ reduces butanenitrile → 1-aminobutane.

Reaction sequences

(i) Benzoic acid → A → B → C

C_6H_5COOH + NH_3 \xrightarrow{\Delta} \underset{A}{C_6H_5CONH_2} + H_2O

C_6H_5CONH_2 + Br_2 + 4NaOH \rightarrow \underset{B}{C_6H_5NH_2} + Na_2CO_3 + 2NaBr + 2H_2O \quad(\text{Hofmann; one C lost})

C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0^\circ C} \underset{C}{C_6H_5N_2^+Cl^-} + NaCl + 2H_2O

So A = benzamide, B = aniline, C = benzenediazonium chloride.

(ii) Ethyl bromide → A → B → C

CH_3CH_2Br + KCN \rightarrow \underset{A}{CH_3CH_2CN} + KBr

CH_3CH_2CN \xrightarrow{LiAlH_4} \underset{B}{CH_3CH_2CH_2NH_2}

CH_3CH_2CH_2NH_2 + HNO_2 \xrightarrow{0^\circ C} \underset{C}{CH_3CH_2CH_2OH} + N_2 + H_2O

Aliphatic diazonium salts are unstable and decompose at once, giving the alcohol. So A = propanenitrile, B = propan-1-amine, C = propan-1-ol.

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

CBSE 2026Set A1 markMCQ
$Q. By which of the following processes, is methyl amine prepared ?$
(a) Wurtz reaction
(b) Hoffmann Bromamide reaction
(c) Friedel-Crafts reaction
(d) Kolbe's reaction
›Reveal solutionSolution

The Hoffmann bromamide degradation converts an amide to a primary amine with one fewer carbon; acetamide -> methylamine.

In the Hoffmann bromamide (degradation) reaction, an amide is treated with bromine and alkali (KOH), losing the carbonyl carbon and forming a primary amine with one carbon atom fewer:

CH3-CONH2 + Br2 + 4 KOH --> CH3-NH2 + K2CO3 + 2 KBr + 2 H2O

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. C6H5CONH2 + Br2 + 4NaOH -----> Product, Product is$
(a) C6H5COOH
(b) C6H5NC
(c) C6H5NH2
(d) C6H6
›Reveal solutionSolution

Treating a primary amide with Br2 and excess NaOH is the Hofmann bromamide degradation, which shortens the carbon chain by one and gives a primary amine.

$\text{C}_6\text{H}_5\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + 2\text{NaBr} + \text{Na}_2\text{CO}_3 + 2\text{H}_2\text{O}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Hofmann bromamide degradation reaction is shown by:$
(a) CH3CH2NH2
(b) CH3CH2NO2
(c) CH3CONH2
(d) CH3CH2CN
›Reveal solutionSolution

The Hofmann bromamide degradation converts a primary AMIDE (RCONH2) into a primary amine with one carbon less, using Br2/KOH — only a compound with the amide (-CONH2) functional group can undergo it.

RCONH2 + Br2 + 4KOH → RNH2 + K2CO3 + 2KBr + 2H2O.

…

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CBSE 2025Set ANNUAL1 mark
$Q. Write the structure and IUPAC name of the amine produced by the Hofmann degradation of benzamide.$
›Reveal solutionSolution

Hofmann degradation of benzamide (Br $_2$ /KOH) removes the carbonyl carbon as $CO_2$ (via $K_2CO_3$ ) and converts the amide nitrogen directly into the ring's amino group, giving aniline.

The Hofmann bromamide degradation converts an amide, $R-CONH_2$ , into a primary amine, $R-NH_2$ , with the loss of one carbon atom (as carbonate), when treated with bromine in aqueous/alcoholic potassium hydroxide. Mechanistically it proceeds through an N-bromoamide, then a nitrene/isocyanate intermediate, and finally hydrolysis and decarboxylation of the resulting carbamic acid.

For benzamide, $R = C_6H_5-$ :

…

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CBSE 2024Set B1 mark
$Q. Write True or False: Primary amines is prepared by Hoffmann bromamide reaction.$
›Reveal solutionSolution

Hofmann bromamide degradation is indeed a standard method for preparing primary amines from amides.

In the Hofmann bromamide degradation reaction, an amide is treated with bromine and concentrated aqueous/ethanolic sodium hydroxide (or KOH). This converts the amide (RCONH2) into a primary amine (RNH2) with the loss of one carbon atom (as CO2, via an isocyanate intermediate):

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. In Hofmann-bromamide reaction an amide is converted to :$
(a) Primary amine
(b) Secondary amine
(c) Tertiary amine
(d) Aldehyde
›Reveal solutionSolution

The Hofmann bromamide reaction converts an amide into a primary amine with one carbon atom less than the starting amide.

When an amide (RCONH2) is treated with bromine in aqueous or ethanolic sodium hydroxide, it is converted to a primary amine (RNH2) containing one carbon atom less than the amide. This is called the Hofmann bromamide degradation reaction.

RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O

…

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CBSE 2023Set F1 markMCQ
$Q. Which of the following is formed when acetamide reacts with Br2/KOH?$
(a) Acetone
(b) Methyl amine
(c) Acetaldehyde
(d) Ammonia
›Reveal solutionSolution

Hofmann bromamide degradation converts acetamide (CH3CONH2) to methylamine (CH3NH2), a primary amine with one fewer carbon.

An amide treated with bromine and aqueous KOH (or NaOH) undergoes the Hofmann bromamide (degradation) reaction, giving a primary amine that has one carbon fewer than the amide:

CH3CONH2 + Br2 + 4KOH → CH3NH2 + K2CO3 + 2KBr + 2H2O

…

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CBSE 2023Set A1 mark
$Q. Match the following. Column A item: 'Hoffmann bromide'. Choose its correct match from Column B:$
(a) Ether
(b) Primary amine
(c) Lactose
(d) C12H22O11
(e) Glucose
(f) Negative ions
(g) C6H5SO2Cl
(h) +7
›Reveal solutionSolution

The Hofmann bromamide (Hoffmann bromide) degradation reaction converts an amide into a primary amine with one less carbon atom.

$RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O$

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. When acetamide is heated with an aqueous solution of sodium hydroxide and bromine, the product formed is$
(a) Methylamine
(b) Ethylamine
(c) Propylamine
(d) None of these
›Reveal solutionSolution

The Hofmann bromamide degradation converts an amide R-CONH2 into the primary amine R-NH2, removing the carbonyl carbon entirely.

CH3CONH2 (acetamide) + Br2 + 4NaOH --> CH3NH2 (methylamine) + 2NaBr + Na2CO3 + 2H2O …

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CBSE 2023Set ANNUAL1 mark
$Q. What is Hoffmann's bromamide reaction? Give an example with a suitable chemical reaction. (½+½=1)$
›Reveal solutionSolution

Hofmann's bromamide degradation converts an amide $RCONH_2$ into a primary amine $RNH_2$ having one fewer carbon atom, by treating it with bromine in concentrated aqueous/alcoholic sodium hydroxide.

Reaction ( $\tfrac12$ mark): When an amide is treated with bromine in the presence of concentrated $NaOH$ solution, it degrades to give a primary amine containing one carbon atom less than the amide, along with the loss of the carbonyl carbon as carbonate:

$R-CO-NH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + 2NaBr + Na_2CO_3 + 2H_2O$

(Mechanistically, bromine first brominates the amide N-H to give an $N$ -bromoamide, which loses $HBr$ under base to form a nitrene/isocyanate intermediate that rearranges and hydrolyses, expelling $CO_2$ (trapped as carbonate) to give the amine — this is why the product amine has one carbon less than the starting amide.)

Example ( $\tfrac12$ mark): Acetamide is converted to methylamine:

…

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CBSE 2022Set ANNUAL1 mark
$C_6H_5CONH_2$
›Reveal solutionSolution

The two routes give different amines: LiAlH₄ reduction keeps the carbon skeleton (benzylamine), while the Hofmann degradation strips out one carbon as CO₂ (aniline).

Route 1 — $LiAlH_4$ reduction: Reduction of the amide carbonyl converts $-CONH_2$ directly into $-CH_2NH_2$ , without loss of any carbon:

$C_6H_5CONH_2 \xrightarrow[\text{ii) }H_2O]{\text{i) }LiAlH_4} C_6H_5CH_2NH_2 \ (\text{benzylamine})$

Route 2 — Hofmann bromamide degradation: Treating the amide with $Br_2/KOH$ converts it to an isocyanate intermediate which hydrolyses with loss of the carbonyl carbon as $CO_2$ , giving an amine with one carbon less than the starting amide, and with the $-NH_2$ attached directly to the ring:

$C_6H_5CONH_2 \xrightarrow{Br_2/KOH} C_6H_5NH_2 \ (\text{aniline}) + CO_2$

…

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CBSE 2022Set ANNUAL1 mark
$Q. Write the reagent required (denoted '?') for the following reaction: CH3CONH2 --?--> CH3NH2.$
›Reveal solutionSolution

Amides are converted to amines with one fewer carbon by the Hofmann bromamide degradation, using bromine in aqueous alkali.

Identifying the reaction: CH3CONH2 (acetamide, 2 carbons) converting to CH3NH2 (methylamine, 1 carbon) is a degradation that removes the carbonyl carbon — the Hofmann bromamide (Hofmann rearrangement) reaction.

Reagent and conditions: Bromine (Br2) in aqueous sodium hydroxide (NaOH).

…

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Q.(a) Write the structures of A, B and C in the following reactions :

Concept understanding — Hofmann Bromamide Reaction

Hofmann Bromamide Degradation

The Intuition

The Precise Statement

Step-by-Step Mechanism

Key Points for Exams

Reaction sequences

(i) Benzoic acid → A → B → C

(ii) Ethyl bromide → A → B → C

Conversions

(i) Aniline → p-bromoaniline

Q.(a) Write the structures of A, B and C in the following reactions :

Concept understanding — Hofmann Bromamide Reaction

Hofmann Bromamide Degradation

The Intuition

The Precise Statement

Step-by-Step Mechanism

Key Points for Exams

Reaction sequences

(i) Benzoic acid → A → B → C

(ii) Ethyl bromide → A → B → C

Conversions

(i) Aniline → p-bromoaniline

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