Question

Predict the correct intermediate and product in the following reaction :
$H_{3}C-C\equiv CH\underset{HgS{O}_4}{\overset{H_{2}O,H_{2}S{O}_4}{\to }}\underset{\text{(A)}}{\mathrm{Intermediate}}\to \underset{\text{(B)}}{\mathrm{product}}$

• A.

  

• B.

  

• C.

  

• D.

  

Answer 1

The Correct Option is C

The transformation in the given reaction is an example of acid-catalyzed hydration of an alkyne to form a ketone. The reaction sequence can be broken down into key steps as explained below:

1. Formation of the Intermediate:
   The given reaction involves the addition of water with concentrated sulfuric acid (\text{H}_2\text{SO}_4) in the presence of mercuric sulfate (\text{HgSO}_4) as a catalyst to a terminal alkyne. Initially, the alkyne produces an enol intermediate through nucleophilic addition of water. This enol is an unstable form known as vinyl alcohol.
2. Rearrangement to Form the Ketone:
   The enol tautomerizes to form a more stable keto form by a process of keto-enol tautomerism. In this case, the enol undergoes rearrangement to produce the corresponding ketone. It is crucial to note that the keto form is usually more thermodynamically stable compared to the enol.

The correct intermediate and product according to the reaction scheme are identified as follows:

• (A) Intermediate: Enol (\text{Vinyl Alcohol})
• (B) Product: Ketone (\text{e.g., Acetone})

The correct option depicting this intermediate and product transformation is:

This reaction is a classic example of Markovnikov's rule where the hydroxyl group attaches to the most substituted carbon in the alkyne. Thus, the given reaction sequence leads to the formation of a ketone, validating the choice of the correct image.