An organic compound (A), $C_9H_{10}O$, gives a 2,4-DNP reaction, negative Tollens' test and iodoform test. On drastic oxidation with chromic acid, it forms B, having molecular formulae, $C_7H_6O_2$. Identify A and B.

A: Propiophenone, B: Benzoic acid

The correct answer is Option (2) → A: Propiophenone, B: Benzoic acid

Step 1: 2,4-DNP test

A positive 2,4-DNP test indicates the presence of a carbonyl group (either an aldehyde or a ketone).

Step 2: Negative Tollens’ test

Tollens’ test is positive for aldehydes. Since the test is negative, compound A is not an aldehyde.

Therefore, A must be a ketone.

Step 3: Iodoform test

The Iodoform test is positive for methyl ketones (containing the $-COCH_3$ group). Since A gives a positive iodoform test, it must contain a $-COCH_3$ group.

Thus, A must be an aromatic methyl ketone.

Among the options, Propiophenone ($C_6H_5-CO-CH_3$) fits:

• Molecular formula $C_9H_{10}O$
• Contains methyl ketone group
• Gives 2,4-DNP test
• Negative Tollens' test
• Positive iodoform test

So A is Propiophenone.

Step 4: Oxidation with chromic acid

Strong oxidation of alkyl side chain attached to benzene converts it to benzoic acid.

Propiophenone on drastic oxidation gives Benzoic acid ($C_7H_6O_2$).

Thus B is Benzoic acid.

Option-wise Explanation

Option 1: 3-Phenylpropanal is an aldehyde and would give positive Tollens' test. Incorrect.

Option 2: Propiophenone satisfies all given tests and oxidation product. Correct.

Option 3: 1-Phenylacetophenone has incorrect formula and structure. Incorrect.

Option 4: 3-Phenylpropanal is aldehyde and oxidation product would not match $C_7H_6O_2$. Incorrect.