Question

What will be produced in a reduction reaction of methanal with ${{H}_{2}}$ gas in the presence of a $Pd$ catalyst?
(A) $C{{H}_{3}}OH$
(B) ${{C}_{2}}{{H}_{5}}OH$
(C) ${{C}_{3}}{{H}_{7}}OH$
(D) ${{C}_{4}}{{H}_{10}}OH$

Answer 1

Start by writing the structure of methanal. Write the reaction which takes place when methanal is treated with hydrogen gas in the presence of palladium catalyst. Identify the product formed in the reaction and then select the correct answer.

Complete step by step solution:
- Methanal is the reactant which is undergoing a reduction reaction.
- Methanal is the simplest aldehyde. It is also known commonly as formaldehyde.
- We know that aldehydes easily undergo reduction in the presence of reducing agents to form primary alcohols.
- When methanal is treated with hydrogen gas, ${{H}_{2}}$in the presence of a palladium catalyst, methanal gets reduced to form methanol which is a primary alcohol. The reaction is represented as,
$HCHO + {{H}_{2}}\xrightarrow{Pd}C{{H}_{3}}OH$
- This is an example of an additional reaction and is also known as hydrogenation. Hydrogen attacks the carbonyl group from the same side and therefore, it is a syn addition.
- Therefore, $C{{H}_{3}}OH$ is produced in a reduction reaction of methanal with ${{H}_{2}}$ gas in the presence of a $Pd$ catalyst.

Therefore, the correct answer is option (A).

Note: Remember methanal is the simplest aldehyde having only one carbon atom that is the carbonyl carbon atom. Aldehydes easily undergo reduction reaction to form primary alcohols and ketones on reduction form secondary alcohols. Aldehydes on oxidation form carboxylic acid and ketones also form carboxylic acid having less number of carbon atoms.