A in above reaction is : $CH_{3}-C\equiv CH+H_{2}O\xrightarrow[333K]{Hg^{2+}/H^{+}} A$

Question

Accepted Answer

In the given reaction, we are starting with propyne ($CH_{3}-C\equiv CH$) and reacting it with water in the presence of mercuric ion ($Hg^{2+}$) and an acidic medium ($H^{+}$) at 333K. This is a classic example of the hydration of an alkyne, which follows the Markovnikov's rule and involves the formation of an enol intermediate that tautomerizes to a ketone.

1. **Explanation for Option 3 ($CH_{3}-C(=O)-CH_{3}$):**
   - The reaction involves the addition of water across the triple bond of propyne. The mercuric ion acts as a catalyst to facilitate the addition of water.
   - According to Markovnikov's rule, the hydroxyl group ($OH$) from water will add to the more substituted carbon atom, which in this case is the middle carbon of the propyne.
   - This results in the formation of an enol intermediate ($CH_{3}-C(OH)=CH_{2}$), which is unstable and undergoes tautomerization.
   - Tautomerization is the process where the enol form rearranges to a more stable keto form. In this case, the enol tautomerizes to acetone ($CH_{3}-C(=O)-CH_{3}$), which is a ketone.
   - Therefore, the correct product of the reaction is acetone, which corresponds to Option 3.

2. **Explanation for why the other options are incorrect:**
   - **Option 1 ($CH_{3}-CH_{2}-CH_{2}OH$):** This structure represents an alcohol with the hydroxyl group on the terminal carbon. This product would result from the anti-Markovnikov addition of water, which is not favored in the presence of $Hg^{2+}$ and an acidic medium.
   - **Option 2 ($CH_{3}-CH_{2}-CHO$):** This structure is an aldehyde, which would not be formed from the hydration of propyne. The reaction conditions favor the formation of a ketone rather than an aldehyde.
   - **Option 4 ($CH_{3}-CH(OH)-CH_{3}$):** This structure is a secondary alcohol, which would result from the direct addition of water to the double bond without tautomerization. However, the reaction conditions specifically lead to the formation of a ketone through tautomerization, not an alcohol.

In conclusion, the correct product of the reaction is acetone ($CH_{3}-C(=O)-CH_{3}$), which is Option 3. The reaction mechanism and conditions support the formation of a ketone via enol-keto tautomerization, making Option 3 the correct answer.