A compound (A) is formed by the reaction of ethylene with bromine which on reacting with aqueous KOH gives a compound (B). The compound (B) can also be prepared by the reaction of ethylene with the following:

Options

A. Baeyer’s reagent
B. Oxygen in the presence of silver catalyst and then acidic hydrolysis
C. performic acid and the product undergoes acidic hydrolysis
D. All of the above

Answer

To solve the problem, we will break it down into steps:

### Step 1: Identify Compound A
- **Reaction**: Ethylene (C2H4) reacts with bromine (Br2).
- **Mechanism**: The double bond in ethylene attacks one of the bromine atoms, leading to the formation of a bromonium ion. The other bromine ion then attacks from the backside, resulting in a vicinal dibromide.
- **Product A**: The product formed is 1,2-dibromoethane (BrCH2-CHBr).

### Step 2: Identify Compound B
- **Reaction**: Compound A (1,2-dibromoethane) reacts with aqueous KOH.
- **Mechanism**: The hydroxide ions (OH-) from KOH perform a bimolecular nucleophilic substitution (SN2) reaction, replacing the bromine atoms with hydroxyl groups.
- **Product B**: The product formed is ethylene glycol (HO-CH2-CH2-OH), also known as ethane-1,2-diol.

### Step 3: Check Other Reagents for Formation of Compound B
We need to determine whether the following reagents can also yield compound B (ethylene glycol) when reacted with ethylene.

1. **Baeyer’s Reagent (KMnO4)**
- **Reaction**: Ethylene reacts with Baeyer’s reagent (acidified KMnO4).
- **Mechanism**: KMnO4 oxidizes ethylene to form a diol.
- **Product**: The product is again ethylene glycol (HO-CH2-CH2-OH).
- **Conclusion**: This reagent can form compound B.

2. **Oxygen in the Presence of Silver Catalyst**
- **Reaction**: Ethylene reacts with oxygen in the presence of a silver catalyst.
- **Mechanism**: This reaction leads to the formation of an epoxide (cyclic ether).
- **Hydrolysis**: The epoxide can then be hydrolyzed (using acidic conditions) to yield ethylene glycol.
- **Product**: The product is ethylene glycol (HO-CH2-CH2-OH).
- **Conclusion**: This reagent can also form compound B.

3. **Performic Acid**
- **Reaction**: Ethylene reacts with performic acid (HCO3H).
- **Mechanism**: Similar to the previous reactions, an epoxide is formed, which can be hydrolyzed.
- **Hydrolysis**: The hydrolysis of the epoxide yields ethylene glycol.
- **Product**: The product is ethylene glycol (HO-CH2-CH2-OH).
- **Conclusion**: This reagent can also form compound B.

### Final Conclusion
Since all the reagents (Baeyer’s reagent, oxygen with silver catalyst, and performic acid) lead to the formation of compound B (ethylene glycol), the correct answer is:
- **All of the above**.