Question

Identify the products of following reaction: Formaldehyde + Benzaldehyde $\xrightarrow{i. \text{conc. NaOH } ii. \text{H}_3\text{O}^+}$ Products.

• A. Phenylmethanol and methanol
• B. Methanol and benzoic acid
• C. Methanoic acid and phenylmethanol
• D. Methanoic acid and benzoic acid

Hint:

Crossed Cannizzaro Rule of Thumb: If Formaldehyde is one of the reactants, it is ALWAYS the sacrificial lamb! Formaldehyde will always be oxidized to formic acid, forcing the other bulky aldehyde to be reduced to the alcohol.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a Crossed Cannizzaro Reaction. When two aldehydes with no $\alpha$-hydrogen are treated with a strong base, one is oxidized to a carboxylic acid and the other is reduced to an alcohol.

Step 2: Formula Application:
In a crossed Cannizzaro involving Formaldehyde ($HCHO$), the Formaldehyde is always the one that gets oxidized because it is more reactive towards nucleophilic attack.

Step 3: Explanation:
1. Formaldehyde ($HCHO$) $\to$ Oxidized to Methanoic acid (Formic acid). 2. Benzaldehyde ($C_6H_5CHO$) $\to$ Reduced to Phenylmethanol (Benzyl alcohol).

Step 4: Final Answer:
The products are methanoic acid and phenylmethanol.