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 Question:
The reaction involves two distinct aldehydes (formaldehyde and benzaldehyde), both of which completely lack $\alpha$-hydrogens, reacting in the presence of concentrated strong base (NaOH). This is a classic textbook Crossed Cannizzaro Reaction. We must identify the specific oxidation and reduction products formed after acidification.

Step 2: Detailed Explanation:
The Cannizzaro reaction is a disproportionation (self-oxidation-reduction) reaction. One molecule of aldehyde is oxidized to a carboxylic acid salt, while the other is simultaneously reduced to a primary alcohol.
In a Crossed Cannizzaro reaction between two different non-enolizable aldehydes, the distribution of products is not random. It is strictly dictated by the relative electrophilicity (reactivity) of their carbonyl carbons.
Formaldehyde ($\text{HCHO}$): It has two hydrogen atoms, making its carbonyl carbon highly unhindered sterically and intensely electrophilic due to a lack of electron-donating alkyl groups.
Benzaldehyde ($\text{C}_6\text{H}_5\text{CHO}$): The bulky phenyl ring provides steric hindrance and donates electron density via resonance, significantly lowering the electrophilicity of the carbonyl carbon.
Reaction Mechanism:
1. The $\text{OH}^-$ nucleophile from the concentrated base will preferentially and rapidly attack the more reactive aldehyde, which is formaldehyde.
2. This attack turns formaldehyde into the intermediate that ultimately donates a hydride ion, resulting in its oxidation to a formate salt ($\text{HCOO}^- \text{Na}^+$).
3. The donated hydride ion forces the other, less reactive aldehyde (benzaldehyde) to be reduced to an alcohol ($\text{C}_6\text{H}_5\text{CH}_2\text{OH}$, benzyl alcohol or phenylmethanol).
4. Upon subsequent acidification ($\text{H}_3\text{O}^+$), the sodium formate salt is protonated to form methanoic acid ($\text{HCOOH}$).
Therefore, the final stable products are Methanoic acid and Phenylmethanol.

Step 3: Final Answer:
The products are Methanoic acid and phenylmethanol, matching option (c).