Product – I `overset("aq KOH")larrC_2H_5Br overset("alc. KOH")rarr`
Product – II, the correct Statement is

Options

A. product - I is obtained by the elimination reaction
B. product - I is obtained by the substitution reaction
C. The molecular formula of Product - I is `C_2H_4`, while the molecular formula of Product - II is `C_2H_6O`
D. Product -I is the isomer of dimethyl ether, while product -II is the dehydrated compound of product-I

Explanation

To solve the question regarding the reactions of C2H5Br (ethyl bromide) with aqueous KOH and alcoholic KOH, we will analyze each reaction step by step.

### Step 1: Reaction with Aqueous KOH
When ethyl bromide (C2H5Br) reacts with aqueous KOH, the primary mechanism involved is nucleophilic substitution (S_N2 mechanism).

1. **Nucleophilic Attack**: The hydroxide ion (OH⁻) from KOH acts as a nucleophile and attacks the carbon atom bonded to the bromine atom (Br).
2. **Formation of Product**: The bromine atom is displaced, and the product formed is ethanol (C2H5OH).

**Chemical Equation**:
\[ \text{C2H5Br} + \text{KOH (aq)} \rightarrow \text{C2H5OH} + \text{KBr} \]

### Step 2: Reaction with Alcoholic KOH
When ethyl bromide reacts with alcoholic KOH, the primary mechanism involved is elimination (E2 mechanism).

1. **Elimination Reaction**: The alcoholic KOH promotes the removal of a hydrogen atom from the β-carbon (the carbon adjacent to the one bonded to Br) while simultaneously removing the bromine atom.
2. **Formation of Product**: The product formed is ethylene (C2H4) along with water.

**Chemical Equation**:
\[ \text{C2H5Br} + \text{KOH (alc)} \rightarrow \text{C2H4} + \text{HBr} + \text{KOH} \]

### Summary of Products
- **Product I (from aqueous KOH)**: Ethanol (C2H5OH)
- **Product II (from alcoholic KOH)**: Ethylene (C2H4)

### Conclusion
The correct statement regarding the products formed is:
1. Product I is obtained by a substitution reaction (nucleophilic substitution).
2. Product II is obtained by an elimination reaction.