Question:

The structure of Q in the following reaction scheme is.
PhNHNH 2 (excess).

Updated On: Nov 17, 2025
  • PhNHNH 2 (excess).
  • PhNHNH 2 (excess).
  • PhNHNH 2 (excess).
  • PhNHNH 2 (excess).
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The Correct Option is B

Solution and Explanation

To determine the structure of compound Q in the reaction scheme, we need to analyze the given reaction and understand the role of the reagents. 

  • Starting with the compound \(M\) (an enantiopure sugar) which has the following structure:
Structure of M
  • The reagent used here is \(\text{PhNHNH}_2\) in excess. This reagent is commonly used in the Wolff-Kishner reduction, which reduces aldehydes and ketones to hydrocarbons.
  • In the first step, the carbonyl group (CHO) in \(M\) is expected to react with the \(\text{PhNHNH}_2\) excess to form an intermediate hydrazone (N). This results in the removal of the oxygen and creation of a double bond with nitrogen.
  • The intermediate hydrazone \(N\) then further reacts to eliminate nitrogen, reducing the carbonyl to a methylene group (CH2), resulting in compound \(P\), which is an alkane.
  • The structure of compound Q would therefore have the hydroxyl groups intact but the aldehyde group reduced to a CH2 group.
  • The correct representation of Q is thus the structure with the CHO group replaced by CH2. Among the options, the correct one corresponds with the following structure:

Therefore, the structure of Q is where the CHO group has been reduced to CH2, preserving the enantiopurity and positions of hydroxyl groups.

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