Nucleophiles such as ammonia and its derivatives, \(H_2N-Z\), add to the carbonyl group of aldehydes and ketones. The reaction is reversible and catalysed by acid. The equilibrium favours product formation due to rapid dehydration of the intermediate to form \(-C=N-Z\), where Z = alkyl, aryl, OH, NH2, C6H5NH, NHCONH2, etc.
What is formed when acetone reacts with semicarbazide?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Acetone semicarbazone
The passage states that a reagent \( \ce{H2N-Z} \) adds to the carbonyl group and the intermediate loses water to give >C=N–Z. Semicarbazide is \( \ce{H2N-NH-CO-NH2} \), so Z = –NHCONH2:
\( \displaystyle \ce{(CH3)2C=O + H2N-NH-CO-NH2} \xrightarrow{\ce{H^+}} \ce{(CH3)2C=N-NH-CO-NH2 + H2O} \)
The product of a carbonyl compound with semicarbazide is a semicarbazone; here it is acetone semicarbazone.
Why the other options are incorrect
Option (2) Acetone semicarbazide — Incorrect. Semicarbazide is the reagent; once it joins the carbonyl carbon through a C=N bond, the product is named a semicarbazone.
Option (3) Acetone hydrazone — Incorrect. A hydrazone, >C=N–NH2, forms with hydrazine (Z = –NH2).
Option (4) Acetone oxime — Incorrect. An oxime, >C=N–OH, forms with hydroxylamine (Z = –OH).
NCERT (2026-27 edition):
Table 8.2: Some N-Substituted Derivatives of Aldehydes and Ketones (>C=N-Z)
