Step 1: Understanding the Concept:
The reaction of aldehydes lacking \(\alpha\)-hydrogen atoms with concentrated alkali (\(\text{NaOH}\) or \(\text{KOH}\)) results in a disproportionation reaction called the Cannizzaro reaction. When two different aldehydes without \(\alpha\)-hydrogens are used, it is a crossed Cannizzaro reaction.
Step 2: Key Formula or Approach:
Identify the reaction type based on the reagents. In a crossed Cannizzaro reaction, the more reactive aldehyde (usually formaldehyde) gets oxidized to a carboxylic acid, and the less reactive one gets reduced to an alcohol.
Step 3: Detailed Explanation:
In a crossed Cannizzaro reaction involving formaldehyde (\(\text{HCHO}\)) and another non-enolizable aldehyde like benzaldehyde (\(\text{C}_6\text{H}_5\text{CHO}\)), the highly reactive formaldehyde is always oxidized to the corresponding carboxylic acid salt (formate), while the other aldehyde is reduced to the corresponding alcohol.
The reaction proceeds as follows:
\[ \text{HCHO} + \text{C}_6\text{H}_5\text{CHO} \xrightarrow{\text{conc. NaOH}} \text{HCOO}^-\text{Na}^+ + \text{C}_6\text{H}_5\text{CH}_2\text{OH} \]
The products formed in the basic medium are sodium formate and benzyl alcohol (phenyl methanol).
The subsequent step involves acidification (\(\text{H}_3\text{O}^+\)), which converts the formate salt into methanoic acid (formic acid):
\[ \text{HCOO}^-\text{Na}^+ \xrightarrow{\text{H}_3\text{O}^+} \text{HCOOH} \]
Therefore, the final isolated products are Methanoic acid (\(\text{HCOOH}\)) and Phenyl methanol (\(\text{C}_6\text{H}_5\text{CH}_2\text{OH}\)).
Step 4: Final Answer:
The products are methanoic acid and phenyl methanol.