`[P]overset(Br_(2))rarrC_(2)H_(4)Br_(2)overset(NaNH_(2))rarrQ overset(20%H_(2)SO_(4))rarr Roverset(ZnHg//HCl)rarrS`
The species P, Q, R and S respectively are

Options

A. ethene, ethyne, ethanal, ethane
B. ethane, ethyne, ethanal, ethene
C. ethene, ethyne, ethanal, ethanol
D. ethyne, ethane, ethene, ethanal

Answer

To solve the problem, we will analyze each step of the reaction sequence provided in the question.

### Step-by-Step Solution:

1. **Identify the starting compound (P)**:
The first reaction involves the addition of bromine (Br₂) to compound P, resulting in C₂H₄Br₂. Since bromine adds across a double bond, compound P must be an alkene. The simplest alkene that can yield C₂H₄Br₂ upon bromination is ethene (C₂H₄).

**Thus, P = C₂H₄ (Ethene)**.

2. **Determine compound Q**:
The next step involves treating C₂H₄Br₂ with sodium amide (NaNH₂). This reaction typically leads to the formation of an alkyne by dehydrohalogenation. In this case, the removal of two bromine atoms and the formation of a triple bond occurs, resulting in ethyne (C₂H₂).

**Thus, Q = C₂H₂ (Ethyne)**.

3. **Identify compound R**:
The next reaction involves treating ethyne (Q) with 20% H₂SO₄ in the presence of mercuric ion (Hg²⁺). This reaction is known as the hydration of alkynes, which leads to the formation of an aldehyde. The product formed from ethyne under these conditions is acetaldehyde (C₂H₄O).

**Thus, R = C₂H₄O (Acetaldehyde)**.

4. **Determine compound S**:
Finally, the acetaldehyde (R) is treated with zinc amalgam (Zn/Hg) and hydrochloric acid (HCl). This reaction is a reduction process, which will convert the aldehyde into an alkane. The product of this reduction will be ethane (C₂H₆).

**Thus, S = C₂H₆ (Ethane)**.

### Summary of the Species:
- P = C₂H₄ (Ethene)
- Q = C₂H₂ (Ethyne)
- R = C₂H₄O (Acetaldehyde)
- S = C₂H₆ (Ethane)

### Final Answer:
The species P, Q, R, and S respectively are:
- P: Ethene (C₂H₄)
- Q: Ethyne (C₂H₂)
- R: Acetaldehyde (C₂H₄O)
- S: Ethane (C₂H₆)