Question:
Identify the reagent ' \(R\) ' used in the following reaction.
Ketone \(\xrightarrow{\mathrm{R}}\) Semicarbazone
Identify the reagent ' \(R\) ' used in the following reaction.
Ketone \(\xrightarrow{\mathrm{R}}\) Semicarbazone
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Semicarbazones are condensation products of carbonyl compounds with semicarbazide.
Updated On: Nov 13, 2026
- \(\mathrm{NH}_{2}\mathrm{OH}\)
- \(\mathrm{NH}_{2}-\mathrm{NH}-\mathrm{CONH}_{2}\)
- \(\mathrm{NH}_{2}-\mathrm{NH}-\mathrm{C}_{6}\mathrm{H}_{5}\)
- \(\mathrm{NH}_{2}-\mathrm{NH}_{2}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Aldehydes and ketones react with ammonia derivatives \(\text{NH}_2-\text{Z}\) to form compounds with C=N-Z.
Step 2: Identify Z
For a semicarbazone, \(Z=-\text{NH}-\text{CO}-\text{NH}_2\), so the reagent is semicarbazide \(\text{NH}_2\text{NHCONH}_2\).
Step 3: Compare the options
(A) gives an oxime, (C) gives a phenylhydrazone and (D) gives a hydrazone. So (B) is correct.
Final Answer:
The reagent is semicarbazide. \[ \boxed{\text{(B) }\text{NH}_{2}-\text{NH}-\text{CONH}_{2}} \]
Aldehydes and ketones react with ammonia derivatives \(\text{NH}_2-\text{Z}\) to form compounds with C=N-Z.
Step 2: Identify Z
For a semicarbazone, \(Z=-\text{NH}-\text{CO}-\text{NH}_2\), so the reagent is semicarbazide \(\text{NH}_2\text{NHCONH}_2\).
Step 3: Compare the options
(A) gives an oxime, (C) gives a phenylhydrazone and (D) gives a hydrazone. So (B) is correct.
Final Answer:
The reagent is semicarbazide. \[ \boxed{\text{(B) }\text{NH}_{2}-\text{NH}-\text{CONH}_{2}} \]
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