Answer the question on basis of passage given below:
Organic compounds containing carbon-oxygen double bonds (C=O) are called carbonyl compounds. These are highly polar molecules. Hence, their boiling points are higher than hydrocarbons and ethers. Aldehydes can be prepared from dehydrogenation or oxidation of alcohols and by reduction of acyl halides. Ketones are prepared by oxidation of secondary alcohols or by hydration of alkynes etc. Aldehydes and ketones undergo nucleophilic addition reaction with a number of nucleophiles such as HCN, NaHSO3, ammonia derivatives. Carboxylic acids are prepared by the oxidation of primary alcohols, aldehydes and alkenes, by hydrolysis of nitriles. Carboxylic acids are more acidic than alcohols and phenols.
The product (X) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → R–CH=CH–CH=N–NH–CO–NH2
Aldehydes and ketones react with ammonia derivatives H2N–Z in weakly acidic medium by nucleophilic addition followed by loss of water; the C=O becomes C=N–Z. With semicarbazide (Z = \( \ce{NHCONH2} \)) the product is a semicarbazone.
Semicarbazide, \( \ce{H2N-NH-CO-NH2} \), has two \( \ce{NH2} \) groups, but only the one at the hydrazine end is nucleophilic; the lone pair of the other \( \ce{NH2} \) is delocalised onto the C=O (amide resonance). So the free \( \ce{NH2} \) of the hydrazine end attacks the CHO carbon, and the C=C double bond of the aldehyde is not touched:
R–CH=CH–CHO + \( \ce{H2N-NH-CO-NH2} \) → \( \ce{R-CH=CH-CH=N-NH-CO-NH2} \) + \( \ce{H2O} \)
This is the structure in Option (4).
Why the other options are incorrect
Option (1) — Incorrect. It joins the aldehyde carbon to a carbon (\( \ce{C=C(NH2)-NH-NH2} \)); the C=O of semicarbazide has disappeared and no C=N bond is formed, which is not what a condensation with an ammonia derivative gives.
Option (2) — Incorrect. Here the aldehyde carbon is bonded to the amide nitrogen (\( \ce{CH=N-CO-NH-NH2} \)); that nitrogen is not nucleophilic because its lone pair is delocalised onto the C=O, so it does not react.
Option (3) — Incorrect. It shows the C=C converted into \( \ce{CH2-CO} \), a change of the carbon chain that this reaction does not bring about; only the CHO group reacts.
NCERT (2026-27 edition):
Reaction with Ammonia and Its Derivatives – …
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 09-06 S3 07-06 S3 | – |