Question

Write the name of the product formed by the action of \( \text{LiAlH}_4 \) ether on acetamide.

Hint:

Lithium aluminum hydride (\( \text{LiAlH}_4 \)) is a strong reducing agent that reduces amides to amines.

Answer 1

The action of lithium aluminum hydride (\( \text{LiAlH}_4 \)) in ether on acetamide reduces the amide group to the corresponding amine. In this case, acetamide (\( \text{CH}_3\text{CONH}_2 \)) is reduced to ethylamine (\( \text{CH}_3\text{NH}_2 \)). The reaction is as follows: \[ \text{CH}_3\text{CONH}_2 \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{NH}_2 \] Thus, the product formed is ethylamine.