Based on the information, we can identify compound (A) as butanamide (C\(_4\)H\(_9\)N) because it undergoes reduction with DIBAL-H to form an aldehyde (B). The fact that (B) gives a positive Tollens' test indicates that it is an aldehyde. The negative iodoform test suggests that (B) is not a methyl ketone. Based on the information that (B) can also be formed by treating ethanal (acetaldehyde) with NaOH, we can conclude that (B) is ethanal.
Reactions:
The reaction of butanamide (A) with DIBAL-H followed by hydrolysis can be represented as follows:
\[
\text{Butanamide} \xrightarrow{\text{DIBAL-H}} \text{Butanal} \xrightarrow{\text{H}_2\text{O}} \text{Ethanal (B)}
\]