`[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 need to identify the species P, Q, R, and S based on the given reactions. Let's break it down step by step.

### Step 1: Identify P
The reaction starts with a compound P that reacts with Br₂ to form C₂H₄Br₂. Since C₂H₄Br₂ is a vicinal dibromide, P must be an alkyne (specifically ethyne, C₂H₂).

**P = C₂H₂ (ethyne)**

### Step 2: Identify Q
Next, we react C₂H₄Br₂ with NaNH₂. NaNH₂ is a strong base and will deprotonate the vicinal dibromide, leading to the formation of a triple bond. The resulting compound will be an alkyne.

C₂H₄Br₂ → C₂H₂ (ethyne) + 2Br⁻

**Q = C₂H₂ (ethyne)**

### Step 3: Identify R
Now, we react Q (C₂H₂) with 20% H₂SO₄. The hydration of ethyne in the presence of sulfuric acid leads to the formation of acetaldehyde (an aldehyde).

C₂H₂ + H₂O → CH₃CHO (acetaldehyde)

**R = CH₃CHO (acetaldehyde)**

### Step 4: Identify S
Finally, we react R (acetaldehyde) with Zn/Hg in HCl. This is a Clemensen reduction, which reduces the carbonyl group (C=O) to a methylene group (C-H), converting the aldehyde to an alkane.

CH₃CHO + Zn/Hg/HCl → C₂H₆ (ethane)

**S = C₂H₆ (ethane)**

### Summary of Results
- **P = C₂H₂ (ethyne)**
- **Q = C₂H₂ (ethyne)**
- **R = CH₃CHO (acetaldehyde)**
- **S = C₂H₆ (ethane)**

### Final Answer
The species P, Q, R, and S respectively are:
- P: Ethyne (C₂H₂)
- Q: Ethyne (C₂H₂)
- R: Acetaldehyde (CH₃CHO)
- S: Ethane (C₂H₆)