Q.Suggest a route by which the following conversion can be accomplished: cyclohexanecarboxamide (a cyclohexane ring bearing a –C(=O)NH2 group) is to be converted into N-methylcyclohexanamine (a cyclohexane ring bearing a –NH–CH3 group).

Concept and Intuition

In the Hoffmann bromamide degradation, a primary amide $R{-}CO{-}NH_2$ reacts with

$Br_2/NaOH$ to give the amine $R{-}NH_2$ — the alkyl/aryl group $R$ migrates directly

onto nitrogen while the carbonyl carbon is expelled as carbon dioxide (via an isocyanate

intermediate). The crucial consequence: the product amine has one carbon less than the starting amide, and the carbon skeleton of $R$ is otherwise carried over unchanged

(no rearrangement of $R$ itself).

Step-by-Step Solution

1. Target product: propan-1-amine, $CH_3CH_2CH_2{-}NH_2$ — a straight 3-carbon chain with the amine on the terminal carbon.
2. Since Hoffmann degradation removes exactly the carbonyl carbon, the parent amide must have $R = CH_3CH_2CH_2{-}$ (propyl), i.e. the amide is $CH_3CH_2CH_2{-}C(=O){-}NH_2$ = butanamide (4 carbons total).
3. Check each option:
   • (A) $CH_3CH_2C(=O)NH_2$ (propanamide, $R=$ ethyl) → gives ethanamine ($CH_3CH_2NH_2$), not propan-1-amine.
   • (B) $CH_3CH(CH_3)C(=O)NH_2$ (2-methylpropanamide, $R=$ isopropyl) → gives isopropylamine ($(CH_3)_2CHNH_2$), a branched 3-carbon amine, not propan-1-amine.
   • (C) $CH_3CH_2CH_2C(=O)NH_2$ (butanamide, $R=$ n-propyl) → gives …

Concept and Intuition

Benzamide, $C_6H_5CONH_2$, can be converted to an amine in two very different ways, and the key distinguishing feature is carbon count:

• $Br_2/NaOH$ (Hofmann bromamide degradation) removes the carbonyl carbon entirely (as $CO_2$ after rearrangement), so the amine formed has one carbon fewer than the amide.
• $LiAlH_4$ is a simple reducing agent: it reduces the $C=O$ of the amide down to a $CH_2$ group, keeping all the original carbons.

Step-by-Step Solution

1. Right-hand arrow: $C_6H_5CONH_2 \xrightarrow{Br_2/NaOH} X$. This is the classic Hofmann bromamide degradation: the amide is converted (via an isocyanate intermediate) into a primary amine with one carbon less than the starting amide — the phenyl group is retained but the carbonyl carbon is lost as carbonate/$CO_2$. So $X = C_6H_5NH_2$ (aniline). …

Concept and Intuition

The Hofmann bromamide degradation is a name reaction that converts an amide $\mathrm{R{-}CONH_2}$ directly into a primary amine $\mathrm{R{-}NH_2}$ using bromine and concentrated NaOH — crucially, the carbonyl carbon is lost in the process (extruded as carbonate/CO₂ via an isocyanate intermediate), so the product amine has exactly one carbon fewer than the parent amide, and it is always a primary amine regardless of the amide's structure.

Step-by-Step Solution

1. $\mathrm{CH_3COOH + NH_3} \to \mathrm{CH_3COONH_4}$ (ammonium acetate, an acid-base salt).
2. On heating ($\Delta$), ammonium acetate loses water (dehydration) to give acetamide, $\mathrm{CH_3CONH_2}$ — this is P (2 carbons).
3. P undergoes the Hofmann bromamide reaction with $\mathrm{Br_2/NaOH}$: mechanistically, this proceeds via an N-bromoamide → nitrene/isocyanate rearrangement (the alkyl/aryl group migrates from carbon to nitrogen) → hydrolysis of the isocyanate to a carbamic acid → spontaneous decarboxylation, releasing $\mathrm{CO_2}$ and leaving the amine.
4. Net result: $\mathrm{CH_3CONH_2} \to \mathrm{CH_3NH_2}$ (methylamine) + $\mathrm{CO_2}$ + salts — Y is methylamine, a primary amine with one carbon less than P (P has 2 C, Y has 1 C). …

Concept and Intuition

The Hofmann bromamide (Hofmann degradation) reaction converts a primary amide, $\mathrm{R-CONH_2}$, into a primary amine, $\mathrm{R-NH_2}$, with the loss of the carbonyl carbon as carbon dioxide/carbonate. Mechanistically: Br$_2$/NaOH first brominates the amide nitrogen (N-bromoamide), base-induced rearrangement then migrates the R group from carbon to nitrogen with simultaneous loss of bromide, forming an isocyanate (R$-$N=C=O), which is rapidly hydrolysed under the basic aqueous conditions to the amine plus carbonate. The key structural consequence is that the amine formed has one fewer carbon than the starting amide, and the R group (here, phenyl) ends up directly attached to nitrogen.

Step-by-Step Solution

1. Start: $\mathrm{PhCONH_2}$ (benzamide), where Ph is attached to the carbonyl carbon.
2. Br$_2$/NaOH brominates nitrogen, then base-promoted rearrangement migrates the phenyl group from the carbonyl carbon directly to nitrogen (with loss of the carbonyl carbon as isocyanate then carbonate/CO$_2$).
3. Net result: the phenyl group is now bonded straight to $-$NH$_2$, i.e. $\mathrm{Ph-NH_2}$ (aniline) — one carbon shorter than the starting amide. …

Concept and Intuition

The transformation given — an amide, $R-CONH_2$, converting into an amine with one less carbon, $R-NH_2$ — is the signature outcome of the Hofmann bromamide degradation reaction:

Mechanistically, $Br_2$/NaOH first brominates the amide nitrogen to give an N-bromoamide; base-mediated rearrangement (loss of bromide, migration of the R group from carbon to nitrogen) generates an isocyanate ($R-N=C=O$), which is then hydrolyzed under the basic aqueous conditions to give the primary amine $R-NH_2$ directly (with the original carbonyl carbon lost as carbonate).

The other reagents don't fit this transformation:

• $Zn-Hg/HCl$ (Clemmensen reduction) reduces a ketone/aldehyde carbonyl to $-CH_2-$, not applicable to an amide going to an amine with carbon loss.
• $HCl/SnCl_2$ (Stephen reduction) reduces nitriles to aldehydes.
• $ZnCl_2/HCl$ (Lucas reagent) is a test/reaction for alcohols, not related here.

Step-by-Step Solution …