In the reaction
`H-C=C-H overset(NaNH_(2))to A overset(CH_(3)I)to B overset(Hg^(2+))underset(H_(3)O^(+))to C`
the final product 'C' is :

Options

A. `CH_(3)CHO`
B. `CH_(3)CH_(2)CHO`
C. `CH_(3)-overset(O)overset(|)C-CH_(3)`
D. `CH_(3)CH_(2)COOH`

Answer

To solve the given reaction sequence step by step, we will analyze each reaction and identify the products formed at each stage.

### Step 1: Identify the starting compound
The starting compound is acetylene, which is represented as:
\[ \text{H-C≡C-H} \]

### Step 2: Reaction with Sodium Amide (NaNH₂)
When acetylene reacts with sodium amide (NaNH₂), a deprotonation occurs. The hydrogen atom from one of the terminal carbons is removed, resulting in the formation of a sodium acetylide:
\[ \text{H-C≡C-H} + \text{NaNH}_2 \rightarrow \text{H-C≡C}^- \text{Na}^+ + \text{NH}_3 \]
This product can be labeled as **A**:
\[ \text{A} = \text{H-C≡C}^- \text{Na}^+ \]

### Step 3: Reaction with Methyl Iodide (CH₃I)
Next, the sodium acetylide reacts with methyl iodide (CH₃I). The sodium ion (Na⁺) is replaced by a methyl group (CH₃), leading to the formation of a new alkyne:
\[ \text{H-C≡C}^- \text{Na}^+ + \text{CH}_3\text{I} \rightarrow \text{H-C≡C-CH}_3 + \text{NaI} \]
This product can be labeled as **B**:
\[ \text{B} = \text{H-C≡C-CH}_3 \]

### Step 4: Reaction with Mercury(II) Ion (Hg²⁺) and Hydronium Ion (H₃O⁺)
The next step involves the reaction of product B with mercury(II) ions (Hg²⁺) in the presence of hydronium ions (H₃O⁺). This reaction leads to the addition of water across the triple bond according to Markovnikov's rule:
\[ \text{H-C≡C-CH}_3 + \text{Hg}^{2+} + \text{H}_3\text{O}^+ \rightarrow \text{CH}_3\text{C(OH)-C}^+ \text{H}_2 \]
This product can be labeled as **C**:
\[ \text{C} = \text{CH}_3\text{C(OH)-C}^+ \text{H}_2 \]

### Step 5: Tautomerization to Form the Final Product
The product C is an enol, which can undergo tautomerization to form a ketone. The hydrogen from the hydroxyl group (OH) shifts to the adjacent carbon, resulting in the formation of a ketone:
\[ \text{CH}_3\text{C(OH)-C}^+ \text{H}_2 \rightarrow \text{CH}_3\text{C=O-CH}_3 \]
This final product is acetone:
\[ \text{Final Product C} = \text{CH}_3\text{C(=O)CH}_3 \]

### Final Answer
The final product C is:
\[ \text{C} = \text{CH}_3\text{C(=O)CH}_3 \] (Acetone)

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