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Chapter 29: Problem 15

Caprolactam, the monomer from which nylon 6 is synthesized, is prepared from cyclohexanone in two steps. In Step 1, cyclohexanone is treated with hydroxylamine to form cyclohexanone oxime. Treatment of the oxime with concentrated sulfuric acid in Step 2 gives caprolactam by a reaction called a Beckmann rearrangement. Propose a mechanism for the conversion of cyclohexanone oxime to caprolactam.

Short Answer

Expert verified
Answer: The proposed mechanism involves four major steps: 1. Protonation of cyclohexanone oxime by concentrated sulfuric acid 2. Nucleophilic attack of water on the carbonyl carbon 3. Rearrangements through the migration of the alkyl group, leading to a nitrilium ion 4. De-protonation of the nitrilium ion by the bisulfate anion and the formation of caprolactam.

Step by step solution

01

Protonation of the oxime

The first step involves the protonation of the nitrogen atom in cyclohexanone oxime by concentrated sulfuric acid (H2SO4). This increases the electrophilicity of the oxime and makes it more susceptible to nucleophilic attack. Cyclohexanone oxime + H2SO4 -> Protonated cyclohexanone oxime + HSO4^- #Step 2: Nucleophilic attack of water#
02

Attack by water on the carbonyl carbon

After protonation, a water molecule (present in the reaction mixture) acts as a nucleophile and attacks the carbonyl carbon of the protonated cyclohexanone oxime. This forms a tetrahedral intermediate with the nitrogen atom still bearing a positive charge. Protonated cyclohexanone oxime + H2O -> Tetrahedral intermediate #Step 3: Rearrangement of the intermediate#
03

Beckmann rearrangement

In the Beckmann rearrangement step, the migration of the alkyl group (R) linked to the nitrogen atom of the oxime occurs. This migration results in the breaking of the C-N bond and the formation of a new C-N bond. The intermediate formed at this stage is called a nitrilium ion, having a positive charge on the nitrogen atom and a double bond with a carbon atom. Tetrahedral intermediate -> Nitrilium ion #Step 4: Proton loss and caprolactam formation#
04

De-protonation and caprolactam formation

The final step is the de-protonation of the nitrilium ion by the bisulfate anion (HSO4^-) which leads to the formation of a cyclic amide, caprolactam. Nitrilium ion + HSO4^- -> Caprolactam + H2SO4 Thus, the proposed mechanism for the conversion of cyclohexanone oxime to caprolactam via Beckmann rearrangement involves protonation of the oxime, nucleophilic attack by water, rearrangement of the intermediate, and de-protonation to form the final product, caprolactam.

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