`CH-=CH overset(H_(g)SO_(4))underset(H_(2)SO_(4))rarr (X) overset(LiAIH_(4))rarr (Y) overset(P_(4)//Br_(2))rarr (Z)`
In this sequence of reaction, the (Z) is:

Options

A. Ethylene bromide
B. Ethanol
C. Ethyl bromide
D. Ethylidene bromide

Explanation

To solve the given reaction sequence step by step, let's analyze each reaction carefully:

### Step 1: Reaction of Alkyne with H₂SO₄
The starting compound is an alkyne, represented as CH≡CH (ethyne or acetylene). When this alkyne reacts with sulfuric acid (H₂SO₄) in the presence of mercury(II) sulfate (HgSO₄), it undergoes hydration.

The reaction proceeds as follows:
- The alkyne reacts with H₂SO₄ to form an intermediate vinyl alcohol (enol), which is CH₂=CHOH (vinyl alcohol).
- This vinyl alcohol then undergoes tautomerization to form an aldehyde. In this case, the aldehyde formed is acetaldehyde (ethanal), represented as CH₃CHO.

Thus, the product of this step (X) is:
\[ X = CH₃CHO \]

### Step 2: Reduction with Lithium Aluminum Hydride (LiAlH₄)
The next step involves the reduction of the aldehyde (X) using lithium aluminum hydride (LiAlH₄).

The reaction is as follows:
- LiAlH₄ is a strong reducing agent that reduces aldehydes to primary alcohols.
- Therefore, the acetaldehyde (CH₃CHO) will be reduced to ethanol (CH₃CH₂OH).

Thus, the product of this step (Y) is:
\[ Y = CH₃CH₂OH \]

### Step 3: Reaction with Phosphorus and Bromine (P₄/Br₂)
In the final step, ethanol (Y) is reacted with phosphorus (P₄) and bromine (Br₂). This reaction is known as the Hell-Volhard-Zelinsky (HVZ) reaction.

The reaction proceeds as follows:
- Phosphorus reacts with bromine to form phosphorus tribromide (PBr₃).
- The hydroxyl group (-OH) of ethanol reacts with PBr₃, leading to the formation of an intermediate that ultimately results in the substitution of the hydroxyl group with a bromine atom.

As a result, we obtain bromoethane (ethyl bromide), represented as CH₃CH₂Br.

Thus, the final product (Z) is:
\[ Z = CH₃CH₂Br \]

### Final Answer
The compound Z is bromoethane (ethyl bromide).