What is (Z) is the following sequence of reaction?
`HC-=CH overset((i)2NaNH_(2)(ii) 2CH_(3)I)rarr (X) overset(HgSO_(4),H_(2)SO_(4))rarr(Y) overset((i)NaOH+Br_(3)(ii)H_(3O^(+)))rarr (Z)`

Options

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

Answer

To determine the compound (Z) in the given sequence of reactions, we will break down each step systematically.

### Step 1: Starting Material
The starting material is **HC≡CH** (ethyne or acetylene).

### Step 2: Reaction with Sodium Amide (NaNH₂)
When ethyne reacts with sodium amide (NaNH₂), the acidic hydrogen from the terminal alkyne is abstracted. This results in the formation of the acetylide ion:

\[ \text{HC≡C}^- \text{Na}^+ \]

This reaction occurs twice, leading to the formation of **C≡C** (the acetylide ion) with two sodium ions:

\[ \text{2HC≡C}^- \text{Na}^+ \]

### Step 3: Reaction with Methyl Iodide (CH₃I)
Next, the acetylide ion reacts with two moles of methyl iodide (CH₃I). The nucleophilic acetylide ion attacks the carbon in methyl iodide, resulting in the formation of **1-butyne**:

\[ \text{C≡C}^- + 2 \text{CH}_3\text{I} \rightarrow \text{CH}_3\text{C≡CCH}_3 \]

This compound is **X**: **1-butyne** (C₄H₆).

### Step 4: Reaction with HgSO₄ and H₂SO₄
The next step involves the hydration of 1-butyne in the presence of mercury(II) sulfate (HgSO₄) and sulfuric acid (H₂SO₄). This reaction leads to the formation of an enol intermediate, which subsequently tautomerizes to form a ketone:

\[ \text{CH}_3\text{C≡CCH}_3 \xrightarrow{\text{HgSO}_4, \text{H}_2\text{SO}_4} \text{CH}_3\text{C(=O)CH}_2\text{CH}_3 \]

This compound is **Y**: **butan-2-one** (C₄H₈O).

### Step 5: Reaction with NaOH and Br₂
The next reaction involves butan-2-one (Y) reacting with sodium hydroxide (NaOH) and bromine (Br₂). This is a haloform reaction, where the methyl ketone reacts to form bromoform (CHBr₃) and a carboxylic acid:

\[ \text{CH}_3\text{C(=O)CH}_2\text{CH}_3 \xrightarrow{\text{NaOH, Br}_2} \text{CH}_3\text{CH}_2\text{COOH} \]

After hydrolysis, the final product **Z** is **propanoic acid** (C₃H₆O₂).

### Final Answer
Thus, the compound (Z) is **propanoic acid (CH₃CH₂COOH)**.

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