Addition of water to alkynes occurs in acidic medium and in the presence of `Hg^(2+)` ions as a catalyst. Which of the following products will be formed on addition of water to but-1-yne under these conditions ?

Options

A. `CH_(3)-CH_(2)-CH_(2)-overset(O)overset("||")C-H`
B. `CH_(3)-CH_(2)-overset(O)overset("||")C-CH_(3)`
C. `CH_(3)-CH_(2)-overset(O)overset("||")C-OH+CO_(2)`
D. `CH_(3)-overset(O)overset("||")C-OH+H-overset(O)overset("||")C-H`

Answer

To solve the problem of the addition of water to but-1-yne in acidic medium with mercury(II) ions as a catalyst, we can follow these steps:

### Step 1: Identify the Reactant
The reactant is but-1-yne, which has the structure:
\[ \text{CH}_3\text{C} \equiv \text{C} \text{H} \]

### Step 2: Understand the Reaction Conditions
The reaction occurs in an acidic medium with mercury(II) ions (Hg²⁺) as a catalyst. The presence of Hg²⁺ will facilitate the addition of water (H₂O) to the alkyne.

### Step 3: Electrophilic Attack
In the presence of Hg²⁺, the alkyne will undergo an electrophilic attack. The triple bond (C≡C) is electron-rich and will attract the electrophile (Hg²⁺). The reaction proceeds as follows:
1. The triple bond donates electrons to Hg²⁺, forming a vinyl cation:
\[ \text{CH}_3\text{C}^+ \text{C} \text{H} - \text{Hg} \]

### Step 4: Nucleophilic Attack by Water
Next, water acts as a nucleophile and attacks the more stable carbocation:
1. Water (H₂O) donates a lone pair to the carbocation, resulting in the formation of an organomercury intermediate:
\[ \text{CH}_3\text{C}(\text{OH})\text{C} - \text{Hg} \]

### Step 5: Proton Transfer
The oxygen in the hydroxyl group (OH) will have a positive charge after the nucleophilic attack. To neutralize this charge, a proton (H⁺) from the solution will be removed, leading to:
\[ \text{CH}_3\text{C}(\text{OH})\text{C} - \text{Hg} \rightarrow \text{CH}_3\text{C}(\text{OH})\text{C} + \text{Hg}^+ \]

### Step 6: Tautomerization
The product formed is an enol (with a hydroxyl group and a double bond). This enol can undergo tautomerization to form a more stable keto form:
1. The hydrogen from the hydroxyl group shifts to the adjacent carbon, and the double bond shifts to form a carbonyl group:
\[ \text{CH}_3\text{C}(\text{OH})\text{C} \rightarrow \text{CH}_3\text{C}(\text{O})\text{C} + \text{H}_2\text{O} \]

### Step 7: Final Product
The final product after tautomerization is:
\[ \text{CH}_3\text{C}(\text{O})\text{CH}_2\text{CH}_3 \]
This is butan-2-one (or methyl ethyl ketone).

### Conclusion
Thus, the product formed on the addition of water to but-1-yne under the given conditions is butan-2-one.

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