Rate of `S_(N)1` & `S_(N)2` reactions for the isomers of `C_(4)H_(9)Br` is
i. n- Butylbromide
ii isobutylborimide
ii s-Butylbromide
iv. t-Butylbromide
a. `i gt iigt iii gt iv ` for `S_(N)2`
b. `i gt ii gt iii gt iv` for `S_(N)1`
c. `iv gt iii gt ii gt` for `S_(N)2`
d. `iv gt iii gt ii gt i` for `S_(N)1`

Options

A. a & b
B. c & d
C. a & d
D. b & c

Explanation

To determine the rates of \( S_N1 \) and \( S_N2 \) reactions for the isomers of \( C_4H_9Br \), we need to analyze the structures of the given compounds and their reactivity based on the mechanisms of these reactions.

### Step 1: Identify the Structures of the Isomers
1. **n-Butylbromide**: This is a straight-chain compound with the structure \( CH_3-CH_2-CH_2-CH_2Br \). It is a primary alkyl halide.
2. **Isobutylbromide**: This compound has the structure \( (CH_3)_2CH-CH_2Br \). It is a primary alkyl halide but has a branched structure.
3. **s-Butylbromide**: This compound has the structure \( CH_3-CH_2-CHBr-CH_3 \). It is a secondary alkyl halide.
4. **t-Butylbromide**: This compound has the structure \( (CH_3)_3CBr \). It is a tertiary alkyl halide.

### Step 2: Analyze \( S_N1 \) Mechanism
The \( S_N1 \) mechanism involves the formation of a carbocation intermediate. The stability of the carbocation determines the rate of the reaction:
- Tertiary carbocations are the most stable, followed by secondary, and then primary.

**Order of Stability for \( S_N1 \)**:
- t-Butylbromide (3°) > s-Butylbromide (2°) > Isobutylbromide (1°) > n-Butylbromide (1°)

Thus, the order for \( S_N1 \) is:
\[ \text{t-Butylbromide} > \text{s-Butylbromide} > \text{Isobutylbromide} > \text{n-Butylbromide} \]
This corresponds to option D: \( iv > iii > ii > i \).

### Step 3: Analyze \( S_N2 \) Mechanism
The \( S_N2 \) mechanism involves a one-step bimolecular reaction where steric hindrance plays a crucial role. The less hindered the carbon, the faster the reaction:
- Primary substrates react fastest, followed by secondary, and tertiary substrates react the slowest.

**Order of Reactivity for \( S_N2 \)**:
- n-Butylbromide (1°) > Isobutylbromide (1°) > s-Butylbromide (2°) > t-Butylbromide (3°)

Thus, the order for \( S_N2 \) is:
\[ \text{n-Butylbromide} > \text{Isobutylbromide} > \text{s-Butylbromide} > \text{t-Butylbromide} \]
This corresponds to option A: \( i > ii > iii > iv \).

### Final Answers
- For \( S_N1 \): \( iv > iii > ii > i \) (Option D)
- For \( S_N2 \): \( i > ii > iii > iv \) (Option A)