In the following reaction, A and B respectively are, `A overset(HBr)rarrC_(2)H_(5)Br overset(B)rarr A`

Options

A. `C_(2)H_(4)&" alcoholic "KOH`
B. `C_(2)H_(4)Cl&" aqueous "KOH`
C. `C_(2)H_(5)OH and "aq. "KOH`
D. `C_(2)H_(2)& Br_(2)`

Explanation

To solve the problem, we need to identify the compounds A and B based on the given reactions. Let's break down the steps:

### Step 1: Identify Compound A
The reaction starts with a compound A that reacts with HBr to form C2H5Br. Since C2H5Br is the product, we can deduce that A is likely an alkene, specifically ethylene (C2H4). This is because alkenes react with HBr to form alkyl halides.

**Hint for Step 1:** Look for a compound that can react with HBr to yield an alkyl halide.

### Step 2: Reaction of A with HBr
When ethylene (C2H4) reacts with HBr, it undergoes an electrophilic addition reaction. The double bond in ethylene attacks the HBr, leading to the formation of a carbocation intermediate. The bromide ion (Br-) then attacks the carbocation, resulting in the formation of bromoethane (C2H5Br).

**Hint for Step 2:** Remember that alkenes react with HBr through electrophilic addition, forming carbocation intermediates.

### Step 3: Identify Compound B
The next part of the reaction indicates that B is involved in converting C2H5Br back to A. To regenerate A (which we identified as C2H4), B must be a strong base that can facilitate the elimination of HBr from C2H5Br. A common reagent for this purpose is alcoholic KOH (potassium hydroxide in alcohol).

**Hint for Step 3:** Think about reagents that can eliminate HBr from alkyl halides to form alkenes.

### Step 4: Elimination Reaction
When C2H5Br is treated with alcoholic KOH, an elimination reaction occurs. The KOH abstracts a hydrogen atom (H) from the ethyl group, and the bromine atom (Br) is eliminated as HBr. This results in the formation of ethylene (C2H4), which is compound A.

**Hint for Step 4:** Elimination reactions often involve the removal of a small molecule (like HBr) to form a double bond.

### Conclusion
In summary, we have identified:
- Compound A: Ethylene (C2H4)
- Compound B: Alcoholic KOH (a strong base)

Thus, the complete reaction can be summarized as:
1. A (C2H4) + HBr → C2H5Br
2. C2H5Br + B (alcoholic KOH) → A (C2H4)

**Final Answer:**
- A = Ethylene (C2H4)
- B = Alcoholic KOH