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Question 46

The correct structure of product 'A' formed in the following reaction,

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Solution

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Cannizzaro Reaction Mechanism Analysis

Overall Reaction Context:
2 Ph-CHO + NaOD (in D2O) → Ph-CH2OD (Product 'A') + Ph-COO-Na+

This reaction is a textbook Cannizzaro Reaction (self oxidation-reduction of aldehydes lacking α-hydrogens). Here is the precise step-by-step pathway showing why the deuterium ends up on the oxygen and not the carbon atom.

1Nucleophilic Attack by OD¯

The deuterated hydroxide ion (OD¯) acts as a nucleophile and attacks the electrophilic carbonyl carbon of the first benzaldehyde molecule. This pushes the π-electrons up onto the carbonyl oxygen, forming a tetrahedral intermediate.

Ph C O H OD¯ → Ph C O¯ H OD
2Hydride Transfer (The Crucial Step)

The tetrahedral intermediate reformulates its carbonyl stable double bond. In doing so, it expels a hydride ion (H¯). This hydride shifts directly to a second molecule of benzaldehyde.

Note: Because this is a direct intermolecular transfer, the hydride never mixes with the solvent, meaning no deuterium can bind to this central carbon.

Ph C O¯ OD H Ph C O H → Ph-C(=O)-OD (Benzoic acid-d) + O¯ | Ph-C-H | H
3Proton / Deuteron Exchange with Solvent

The generated carboxylic acid is highly acidic, and the alkoxide intermediate is a strong base. Rapid acid-base proton/deuteron exchange occurs instantly with the deuterated solvent (D2O) or directly between them to form stable final products.

Ph-CH2-O¯ + D-OD → Ph-CH2-OD (Benzyl alcohol-d) [Product 'A']

Conclusion Structure

As demonstrated by the electron tracking arrows:

  • The carbon atom receives its second hydrogen strictly from the aldehyde group of the neighboring partner molecule (H transfer).
  • The oxygen anion pulls its tracking item from the deuterated workup environment (D pick up).

Therefore, the carbon maintains two normal H atoms, while the oxygen captures the OD variant.

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