See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the functional group requirements: The compound reacts with 2,4-DNP (2,4-dinitrophenylhydrazine), which means it must contain a carbonyl group (aldehyde or ketone). This eliminates options (c) and (d), which are alcohols (no carbonyl group to react with 2,4-DNP). Step 2 - Apply the haloform test condition: The haloform test is positive for methyl ketones (CH3COR) and acetaldehyde (CH3CHO), as well as secondary alcohols that can be oxidized to methyl ketones (like 2-butanol). The compound must give a NEGATIVE haloform test, meaning it is NOT a methyl ketone or acetaldehyde. Step 3 - Evaluate option (a): CH3—C(=O)—CH2—CH3 is methyl ethyl ketone (butanone, MEK). It contains a CH3CO— group, so it GIVES a POSITIVE haloform test. This eliminates option (a). Step 4 - Evaluate option (b): CH3—CH(CH3)—CHO is 2-methylpropanal (isobutyraldehyde). It is an aldehyde (reacts with 2,4-DNP ✓). It does NOT have a CH3CO— group; the carbonyl carbon is a —CHO at the end of a branched chain, not adjacent to a methyl group in the methyl ketone sense. Therefore, it gives a NEGATIVE haloform test ✓. Step 5 - Confirm molecular formula for (b): CH3—CH(CH3)—CHO → C4H8O ✓. Step 6 - Why (c) and (d) fail: Cyclobutanol and 2-butanol are alcohols; they do not react with 2,4-DNP (which requires a carbonyl compound). Therefore, the correct answer is B.