`CH_(3)-C-=CHunderset(dil.H_(2)SO_(4))overset(HgSO_(4))(to)` product
Product of the above reaction is

Options

A. an aldehyde
B. a ketone
C. an alcohol
D. a carboxylic acid

Answer

To determine the product of the reaction given by the equation \( CH_3-C \equiv CH \) in the presence of dilute \( H_2SO_4 \) and \( HgSO_4 \), we will follow the steps of the reaction mechanism for the hydration of alkynes.

### Step-by-Step Solution:

1. **Identify the Reactant**: The starting material is an alkyne, specifically 1-butyne, represented as \( CH_3-C \equiv CH \).

2. **Electrophilic Addition Reaction**: The reaction involves the hydration of the alkyne, which is an electrophilic addition reaction. The reagents used are dilute sulfuric acid (\( H_2SO_4 \)) and mercuric sulfate (\( HgSO_4 \)).

3. **Formation of Mercuric Ion Complex**: The \( Hg^{2+} \) ion from \( HgSO_4 \) acts as an electrophile and attacks the triple bond of the alkyne. This forms a mercuric ion complex. The reaction can be represented as:
\[ CH_3-C \equiv CH + Hg^{2+} \rightarrow CH_3-C(=CH-Hg^+) \text{ (mercuric ion complex)} \]

4. **Nucleophilic Attack by Water**: In the next step, water acts as a nucleophile and attacks the carbon atom that is bonded to the \( Hg^+ \) ion. This leads to the formation of an alcohol intermediate:
\[ CH_3-C(=CH-Hg^+) + H_2O \rightarrow CH_3-C(OH)-CH_2-Hg^+ \]

5. **Deprotonation**: The alcohol intermediate undergoes deprotonation, resulting in the formation of an enol:
\[ CH_3-C(OH)-CH_2-Hg^+ \rightarrow CH_3-C(=CH_2) + Hg^{2+} + H^+ \]

6. **Tautomerization to Ketone**: The enol form is unstable and readily tautomerizes to form a more stable ketone. The tautomerization involves the migration of a hydrogen atom and the rearrangement of the double bond:
\[ CH_3-C(=CH_2) \rightarrow CH_3-C(=O)-CH_3 \text{ (butan-2-one)} \]

7. **Final Product**: The final product of the reaction is butan-2-one (a ketone):
\[ \text{Product} = CH_3-C(=O)-CH_3 \]

### Final Answer:
The product of the reaction is **butan-2-one**.

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