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 final product (Z) in the given sequence of reactions, we will break down the steps as follows:

### Step 1: Reaction of Acetylene with Sodium Amide and Methyl Iodide
1. **Starting Material**: Acetylene (HC≡CH)
2. **Reagents**: 2 moles of Sodium Amide (NaNH₂) and 2 moles of Methyl Iodide (CH₃I)
3. **Reaction**: Sodium amide is a strong base and will deprotonate the acetylene, removing both acidic hydrogen atoms. This forms disodium acetylene (Na₂C₂H₂).
4. **Substitution**: The disodium acetylene will then react with 2 moles of methyl iodide, leading to the substitution of the sodium ions with methyl groups.

**Product (X)**: The product after this reaction will be 1,2-dimethylacetylene (CH₃C≡CCH₃).

### Step 2: Hydration of 1,2-Dimethylacetylene
1. **Reagents**: Mercury(II) sulfate (HgSO₄) and sulfuric acid (H₂SO₄)
2. **Reaction**: The addition of water (hydration) occurs across the triple bond of the alkyne, forming an enol intermediate.
3. **Tautomerization**: The enol will tautomerize to form a more stable keto form.

**Product (Y)**: The product after this step will be 2-pentanone (CH₃C(=O)C₂H₅).

### Step 3: Haloform Reaction
1. **Reagents**: Bromine (Br₂) and Sodium Hydroxide (NaOH)
2. **Reaction**: The presence of a methyl ketone (2-pentanone) will undergo a haloform reaction. This involves the halogenation of the methyl group adjacent to the carbonyl, followed by hydrolysis.
3. **Final Step**: The addition of acid (H₃O⁺) will convert the resulting sodium salt into the corresponding carboxylic acid.

**Product (Z)**: The final product after this reaction will be propanoic acid (CH₃CH₂COOH).

### Final Answer
Thus, the final product (Z) is **Propanoic Acid (CH₃CH₂COOH)**.

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