Write the structures of A, B and C in the following reactions : (a) CH₃-CH₂-Br xrightarrowKCN A xrightarrow[hy

Question

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

(a)

CH_3-CH_2-Br \xrightarrow{KCN} A \xrightarrow[\text{hydrolysis}]{\text{Partial}} B \xrightarrow{NaOH + Br_2} C

(b)

C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{NaNO_2/HCl} B \xrightarrow{C_2H_5OH} C

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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). …

Start your 14-day free trial to unlock the full solution →

Part (a) involves nucleophilic substitution, partial nitrile hydrolysis, and Hofmann degradation to yield ethylamine. Part (b) involves reduction of a nitro group, diazotisation, and reduction of the diazonium salt to yield benzene.

Let's break down each reaction sequence step by step, understanding the underlying organic chemistry principles.

Part (a): $CH_3-CH_2-Br \xrightarrow{KCN} A \xrightarrow[\text{hydrolysis}]{\text{Partial}} B \xrightarrow{NaOH + Br_2} C$

This sequence demonstrates the chain lengthening of an alkyl halide, followed by conversion to an amide, and then a carbon-shortened amine.

Step 1: $CH_3-CH_2-Br \xrightarrow{KCN} A$
- Concept: This is a classic nucleophilic substitution reaction. Bromoethane is an alkyl halide, and the cyanide ion ( $CN^-$ ) from KCN is a strong nucleophile. The bromine atom, being a good leaving group, is replaced by the cyanide group. This reaction typically proceeds via an $S_N2$ mechanism for primary alkyl halides.
- Reaction: $CH_3-CH_2-Br + KCN \longrightarrow CH_3-CH_2-CN + KBr$
- Product A: Propanenitrile (or ethyl cyanide). This reaction is significant because it adds a carbon atom to the chain, converting a two-carbon compound into a three-carbon compound.
Step 2: $A \xrightarrow[\text{hydrolysis}]{\text{Partial}} B$
- Concept: Nitriles can undergo hydrolysis. Complete hydrolysis (with strong acid or base and heat) converts a nitrile into a carboxylic acid. However, under controlled or partial hydrolysis conditions (e.g., with dilute acid or base, or specific catalysts), nitriles are converted into amides. The triple bond of the nitrile is attacked by water, eventually forming a carbonyl group and an amino group.
- Reaction: $CH_3-CH_2-CN \xrightarrow[\text{hydrolysis}]{\text{Partial}} CH_3-CH_2-CONH_2$
- Product B: Propanamide.
Watch out

Be careful to distinguish between partial and complete hydrolysis of nitriles. Partial hydrolysis yields amides, while complete hydrolysis yields carboxylic acids.
Step 3: $B \xrightarrow{NaOH + Br_2} C$
- Concept: This is the Hofmann bromamide degradation (or Hofmann rearrangement) reaction. It is a very important named reaction used to convert a primary amide into a primary amine with one less carbon atom. The reaction involves the formation of an intermediate isocyanate, which then hydrolyzes to the amine. The carbon atom of the carbonyl group is lost as carbon dioxide.
- Reaction: $CH_3-CH_2-CONH_2 \xrightarrow{NaOH + Br_2} CH_3-CH_2-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$
- Product C: Ethylamine.
The general form of the Hofmann bromamide degradation is:

$R-CONH_2 + Br_2 + 4NaOH \longrightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$

Part (b): $C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{NaNO_2/HCl} B \xrightarrow{C_2H_5OH} C$

This sequence illustrates the reduction of a nitro group, followed by diazotisation, and then reduction of the diazonium salt.

Step 1: $C_6H_5NO_2 \xrightarrow{Fe/HCl} A$
- Concept: The nitro group ( $-\text{NO}_2$ ) is readily reduced to an amino group ( $-\text{NH}_2$ ) under various conditions. $Fe/HCl$ (or $Sn/HCl$ , or catalytic hydrogenation) is a common and effective reducing agent for this transformation, especially for aromatic nitro compounds. The acidic conditions protonate the amine, so a subsequent base treatment (like $NaOH$ ) is often used to liberate the free amine, but for simplicity, we often write the amine directly.
- Reaction: $C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2$
- Product A: Aniline (aminobenzene).
Step 2: $A \xrightarrow{NaNO_2/HCl} B$ …

Start your 14-day free trial to unlock the full solution →

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

…

Start your 14-day free trial to see the full solution →
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.

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

…

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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, $R = C_6H_5-$ :

…

Start your 14-day free trial to see the full solution →
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):

…

Start your 14-day free trial to see the full solution →
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

…

Start your 14-day free trial to see the full solution →
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

…

Start your 14-day free trial to see the full solution →
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.

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

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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:

…

Start your 14-day free trial to see the full solution →
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:

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

…

Start your 14-day free trial to see the full solution →
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).

…

Start your 14-day free trial to see the full solution →

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

Part (a): CH3−CH2−Br→KCNA→hydrolysisPartialB→NaOH+Br2CCH_3-CH_2-Br \xrightarrow{KCN} A \xrightarrow[\text{hydrolysis}]{\text{Partial}} B \xrightarrow{NaOH + Br_2} CCH3​−CH2​−BrKCN​APartialhydrolysis​BNaOH+Br2​​C

Part (b): C6H5NO2→Fe/HClA→NaNO2/HClB→C2H5OHCC_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{NaNO_2/HCl} B \xrightarrow{C_2H_5OH} CC6​H5​NO2​Fe/HCl​ANaNO2​/HCl​BC2​H5​OH​C

🎓Unlock everything free for 14 days

Part (a): $CH_3-CH_2-Br \xrightarrow{KCN} A \xrightarrow[\text{hydrolysis}]{\text{Partial}} B \xrightarrow{NaOH + Br_2} C$

Part (b): $C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{NaNO_2/HCl} B \xrightarrow{C_2H_5OH} C$

Unlock everything free for 14 days