Question:

The major product in the following reaction is
NaH Raney-Ni/H2

Updated On: Nov 17, 2025
  • NaH Raney-Ni/H2
  • NaH Raney-Ni/H2
  • NaH Raney-Ni/H2
  • NaH Raney-Ni/H2
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The Correct Option is C

Solution and Explanation

The given reaction involves the conversion of a ketone into an alkane using sodium hydride (NaH) and Raney nickel (Raney-Ni) with hydrogen gas (H2). Let's break it down step-by-step:

  1. The initial compound is a cyclic ketone with a dimethylsulfide (SMe2) group. This group serves as a protective group for the ketone.
  2. Upon treatment with NaH, the deprotonation of the compound occurs. NaH acts as a strong base and is involved in the preliminary preparation step for the reduction.
  3. Raney-Ni and H2 are then used for the desulfurization process. Raney-Ni is a nickel-aluminium alloy catalyst commonly utilized for hydrogenation reactions.
  4. The reaction using Raney-Ni and H2 reduces both the SMe2 group and the ketone, ultimately producing an alkane by removing the sulfur and reducing the carbonyl group.

The major product here will be the alkane derivative of the original ketone, where the ketone and the thioether group are reduced completely.

From the given options, the correct product of this reaction sequence is the complete reduction of the ketone into an alkane structure:

NaH Raney-Ni/H2

This option shows the correct reduction of the ketone into an alkane, accurately reflecting the process using NaH and Raney-Ni/H2.

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