Explain the mechanism of the Hofmann rearrangement when cyclohexanecarboxamide ( $C_{6} H_{11} CON H_{2}$ ) is treated with sodium hypochlorite (NaOCl) and sodium hydroxide (NaOH). Write the balanced overall equation and give the name of the organic product formed.

hard 5 marks Class 12 · Chemistry · Amines

Answer

Product: cyclohexylamine (

C_{6} H_{11} N H_{2}

); Overall equation:

C_{6} H_{11} CON H_{2} + NaOCl + 2 NaOH \to C_{6} H_{11} N H_{2} + NaCl + N a_{2} C O_{3} + H_{2} O

Step-by-step solution

Step 1 – Formation of N‑chloroamide: $RCON H_{2} + OC l^{-} \to RCONHCl + O H^{-}$ . The amide nitrogen is chlorinated by hypochlorite, giving the N‑chloroamide.
Step 2 – Deprotonation: In the strongly basic medium, the N‑chloroamide loses a proton to give the anionic N‑chloroamide, $RCONHC l^{-}$ , which is more nucleophilic at nitrogen.
Step 3 – Rearrangement (migration): The electron‑rich nitrogen attacks the carbonyl carbon, causing the R‑group (cyclohexyl) to migrate from carbon to nitrogen while the carbonyl carbon–nitrogen bond breaks, releasing $C O_{2}$ as carbonate later. This step converts the N‑chloroamide into an isocyanate intermediate, $R - NCO$ .
Step 4 – Hydrolysis of isocyanate: The isocyanate reacts with water (present from NaOH) to give the amine and carbonic acid, which rapidly converts to $C O_{2}$ and $H_{2} O$ : $R - NCO + 2 H_{2} O \to R - N H_{2} + H_{2} C O_{3}$ .
Step 5 – Neutralisation and formation of by‑products: $H_{2} C O_{3}$ reacts with excess NaOH to give $N a_{2} C O_{3}$ , and the chlorine from hypochlorite ends up as $NaCl$ .
Combine all steps to obtain the overall balanced equation: $C_{6} H_{11} CON H_{2} + NaOCl + 2 NaOH \to C_{6} H_{11} N H_{2} + NaCl + N a_{2} C O_{3} + H_{2} O$ .
Thus, the organic product formed is cyclohexylamine.

Common mistake: Assuming the carbon skeleton is retained; forgetting the loss of carbon as

C O_{2}

Concept tested: Hofmann rearrangement mechanism