Question

An organic compound gives a positive semicarbazide test and on reduction produces propan-2-ol. The compound is:

• A. Propanal
• B. Propanone
• C. Ethanal
• D. Methanal

Hint:

Remember: \[ \text{Aldehyde} \rightarrow \text{Primary Alcohol} \] \[ \text{Ketone} \rightarrow \text{Secondary Alcohol} \] A positive semicarbazide test confirms the presence of a carbonyl group.

Answer 1

The Correct Option is B

Concept: Semicarbazide reacts with aldehydes and ketones containing the carbonyl group. The reaction forms crystalline derivatives called semicarbazones. \[ RCHO + H_2NNHCONH_2 \rightarrow RCH=NNHCONH_2 \] \[ RCOR' + H_2NNHCONH_2 \rightarrow RC(=NNHCONH_2)R' \] Thus, a positive semicarbazide test confirms the presence of a carbonyl compound. The second clue given in the question is reduction. Reduction of:
• Aldehydes gives primary alcohols.
• Ketones gives secondary alcohols. Therefore, both pieces of information must be used simultaneously.

Step 1: Interpretation of semicarbazide test.
Since the compound gives a positive semicarbazide test, it must contain: \[ >C=O \] Therefore the compound is either:
• An aldehyde
• A ketone All four options except alcohols satisfy this condition.

Step 2: Use the reduction clue.
The reduction product is: \[ \text{Propan-2-ol} \] Structure: \[ CH_3CHOHCH_3 \] This is a secondary alcohol. Since ketones reduce to secondary alcohols, the original compound must be a ketone.

Step 3: Identify the ketone.
The ketone having three carbon atoms is: \[ CH_3COCH_3 \] which is propanone. Reduction: \[ CH_3COCH_3 \xrightarrow{[H]} CH_3CHOHCH_3 \] The product formed is exactly propan-2-ol.

Step 4: Verification of other options.
Propanal reduces to: \[ CH_3CH_2CH_2OH \] which is propan-1-ol. Ethanal reduces to ethanol. Methanal reduces to methanol. None of these gives propan-2-ol. Therefore: \[ \boxed{\text{Propanone}} \]