Alkyl halides undergo nucleophilic substitution reactions in which halogen atom is replaced by other atom.
`RX + Nu:^(-) to Nu-R + X^(-)`
These reactions follow `S_(N)1` and `S_(N)2` type mechanism, in which `S_(N)1` takes place in two steps while `S_(N)2` takes place in single step. Due to their tendency to undergo substitution by a large number of nucleophiles, they form a variety of products.
In which of the following pairs, the first compound is better `S_(N)2` substrate ?

Options

A. 1-bromo-1-methyl cyclohexane, cyclohexyl bromide
B. 1-iodo-2, 2-dimethyl propane, isopropyl iodide
C. 2, 2,-dimethyl-1-chlorobutane, 2-chloro butane
D. isopropyl bromide, 2-bromobutane

Answer

To determine which of the given pairs has the first compound as a better substrate for the \( S_N2 \) reaction, we need to analyze the steric hindrance around the carbon atom bonded to the halogen in each compound. The \( S_N2 \) mechanism involves a backside attack by the nucleophile, which is more favorable when there is less steric hindrance.

### Step-by-Step Solution:

1. **Understanding the \( S_N2 \) Mechanism**:
- The \( S_N2 \) reaction occurs in a single step where the nucleophile attacks the carbon atom from the opposite side of the leaving group (halogen). This results in the simultaneous formation of a new bond and the breaking of the old bond.

2. **Identifying Steric Hindrance**:
- The steric hindrance around the carbon atom attached to the halogen is crucial. More bulky groups around this carbon will hinder the nucleophile's approach, making the reaction slower.

3. **Analyzing the Given Compounds**:
- For each pair of compounds, we will evaluate the structure to determine which has less steric hindrance.

4. **Example Analysis**:
- **Pair 1**: 1-bromo-1-methylcyclohexane vs. Cyclohexyl bromide
- 1-bromo-1-methylcyclohexane has a tertiary carbon (due to the methyl group), which is more hindered.
- Cyclohexyl bromide has a primary carbon, which is less hindered.
- **Conclusion**: Cyclohexyl bromide is a better \( S_N2 \) substrate.

- **Pair 2**: 2-iodo-2,2-dimethylpropane vs. 1-chlorobutane
- 2-iodo-2,2-dimethylpropane is a tertiary carbon, which is highly hindered.
- 1-chlorobutane is a primary carbon, which is less hindered.
- **Conclusion**: 1-chlorobutane is a better \( S_N2 \) substrate.

5. **Final Evaluation**:
- After analyzing all pairs, we conclude that the first compound in each pair is not necessarily the better \( S_N2 \) substrate due to steric hindrance considerations.

### Final Answer:
The first compound in each pair is not a better \( S_N2 \) substrate compared to the second compound. The better \( S_N2 \) substrates are those with less steric hindrance.