In the reaction :
`HC -= CH overset(NaNH_(2)) to overset(CH_(3)I)to Y underset(H^(+))overset(H_(2)O, Hg^(2+))to Z`,

Options

A. `CH_(3)CHO`
B. `CH_(3)CH_(2)CH=CH_(2)`
C. `CH_(3)COCH_(3)`
D. `CH_(3)CH_(2)CHO`

Explanation

To solve the reaction step by step, let's break down the transformations occurring in the reaction.

### Step 1: Identify the starting material
The starting material is **ethyne (HC≡CH)**.

### Step 2: Reaction with sodium amide (NaNH₂)
When ethyne reacts with sodium amide (NaNH₂), it undergoes deprotonation. The hydrogen atom attached to the carbon atom is replaced by a sodium ion (Na⁺), resulting in the formation of a **sodium acetylide**.

**Reaction:**
\[ HC≡CH + NaNH₂ \rightarrow HC≡C^{-}Na^{+} \]

**Product X:** Sodium acetylide (HC≡C⁻Na⁺)

### Step 3: Reaction with methyl iodide (CH₃I)
Next, sodium acetylide reacts with methyl iodide (CH₃I). In this step, the acetylide ion (HC≡C⁻) acts as a nucleophile and attacks the electrophilic carbon in methyl iodide, leading to the formation of a new carbon-carbon bond.

**Reaction:**
\[ HC≡C^{-}Na^{+} + CH₃I \rightarrow HC≡C-CH₃ + NaI \]

**Product Y:** Propyne (HC≡C-CH₃)

### Step 4: Reaction with water (H₂O) and mercuric ion (Hg²⁺)
The propyne (Y) then undergoes hydration in the presence of water and mercuric ion (Hg²⁺) in an acid-catalyzed reaction. This step converts the alkyne into an alcohol through the Markovnikov addition of water.

**Reaction:**
\[ HC≡C-CH₃ + H₂O \overset{Hg^{2+}}{\rightarrow} CH₃-CHOH-CH₃ \]

**Product Z:** 2-Propanol (CH₃-CHOH-CH₃)

### Summary of Products:
- **X:** Sodium acetylide (HC≡C⁻Na⁺)
- **Y:** Propyne (HC≡C-CH₃)
- **Z:** 2-Propanol (CH₃-CHOH-CH₃)