`H_(3)C- overset(CH_(3))overset(|)underset(CH_(3)) underset(|)( C) - Br + KOH (aq.) to H_(3) C - overset(CH_(3)) overset(|) underset(CH_(3)) underset(|)( C) - OH + KBr`
above reaction is

Options

A. `S_(N^1)`
B. `S_(N^2)`
C. `E_(1)`
D. Both (a) and (b)

Explanation

To solve the question regarding the reaction of tert-butyl bromide with potassium hydroxide (KOH), we need to analyze the reaction mechanism and identify the type of reaction taking place.

### Step-by-Step Solution:

1. **Identify the Reactants and Products:**
- The reactant is tert-butyl bromide (H₃C-C(CH₃)₂-Br).
- The product is tert-butyl alcohol (H₃C-C(CH₃)₂-OH) along with KBr as a byproduct.

2. **Determine the Reaction Type:**
- The reaction involves the substitution of a bromine atom (Br) with a hydroxyl group (OH).
- Since tert-butyl bromide is a tertiary alkyl halide, it is likely to undergo an SN1 reaction due to the stability of the tertiary carbocation that can be formed.

3. **Mechanism of SN1 Reaction:**
- **Step 1: Formation of Carbocation**
- The bromine atom (Br) acts as a leaving group and departs from the tert-butyl bromide, forming a stable tertiary carbocation (H₃C-C(CH₃)₂⁺).
- **Step 2: Nucleophilic Attack**
- The hydroxide ion (OH⁻) from KOH attacks the carbocation, leading to the formation of tert-butyl alcohol (H₃C-C(CH₃)₂-OH).
- **Byproduct Formation:**
- The leaving bromine ion (Br⁻) combines with potassium (K⁺) from KOH to form KBr.

4. **Final Reaction Equation:**
- The overall reaction can be summarized as:
\[ \text{H}_3\text{C}-\overset{\text{(CH}_3)}{\underset{\text{(CH}_3)}{|}(C)-Br} + \text{KOH (aq)} \rightarrow \text{H}_3\text{C}-\overset{\text{(CH}_3)}{\underset{\text{(CH}_3)}{|}(C)-OH} + \text{KBr} \]

5. **Conclusion:**
- The reaction is classified as an SN1 reaction, as it involves the formation of a carbocation intermediate and follows a two-step mechanism.