Question

Identify the reagent \(R\) used in following reaction.
\(\text{Ketone} \xrightarrow{R} \text{semi carbazone}\)

• A. \(\text{NH}_2\text{OH}\)
• B. \(\text{NH}_2\text{NHCONH}_2\)
• C. \(\text{NH}_2\text{NHC}_6\text{H}_5\)
• D. \(\text{NH}_2 - \text{NH}_2\)

Hint:

Ketone derivatives: oxime (NH$_2$OH), hydrazone (NH$_2$NH$_2$), semicarbazone (NH$_2$NHCONH$_2$).

Answer 1

The Correct Option is B

The question asks us to identify the reagent \( R \) used to convert a ketone to its semicarbazone derivative.

The reaction of a ketone with semicarbazide forms a semicarbazone. The semicarbazone formation is a part of organic compound derivatization reactions, where the semicarbazide group reacts selectively with carbonyl compounds (like ketones) to form corresponding semicarbazone.

Step-by-step Explanation:

1. The reaction involves a ketone, which has the functional group \( \text{C}= \text{O} \).
2. A semicarbazone comes from the reaction of the ketone with semicarbazide. The general reaction is:
   \(\text{Ketone} + \text{Semicarbazide} \rightarrow \text{Semicarbazone}\)
3. The structure of semicarbazide is \( \text{NH}_2\text{NHCONH}_2 \).
4. The mechanism involves nucleophilic attack by the nitrogen atom of semicarbazide on the carbon of the carbonyl group in the ketone, followed by elimination of water.

Given the options:

• \(\text{NH}_2\text{OH}\): This is hydroxylamine, not involved in semicarbazone formation.
• \(\text{NH}_2\text{NHCONH}_2\): This is semicarbazide, which reacts with ketones to form semicarbazones. Thus, this is the correct answer.
• \(\text{NH}_2\text{NHC}_6\text{H}_5\): This is phenylhydrazine, not related to semicarbazone formation.
• \(\text{NH}_2 - \text{NH}_2\): This is hydrazine, which forms hydrazones, not semicarbazones.

Therefore, the correct reagent for the conversion of ketone to semicarbazone is \(\text{NH}_2\text{NHCONH}_2\).