In `S(N)^(2)` reaction the correct order of reactivity of the following compounds:
(i) `CH_(3)Cl` (ii) `(CH_(3))_(3) C Cl`
(iii) `(CH_(3))_(2) CHCl` (iv) `CH_(3)CH_(2) - Cl`

Options

A. i > ii > iii > iv
B. iv > iii > ii > i
C. i > iv > iii > ii
D. iv > i > ii > iii

Answer

To determine the correct order of reactivity of the given compounds in an \( S_N2 \) reaction, we need to consider the concept of steric hindrance. The \( S_N2 \) mechanism involves a nucleophile attacking the carbon atom that is bonded to the leaving group (in this case, chlorine). The more sterically hindered the carbon atom is, the slower the reaction will be due to the difficulty the nucleophile has in approaching the carbon.

### Step-by-Step Solution:

1. **Identify the Compounds:**
- (i) \( CH_3Cl \) (Methyl chloride)
- (ii) \( (CH_3)_3CCl \) (Tertiary butyl chloride)
- (iii) \( (CH_3)_2CHCl \) (Isopropyl chloride)
- (iv) \( CH_3CH_2Cl \) (Ethyl chloride)

2. **Analyze Steric Hindrance:**
- **Methyl chloride (\( CH_3Cl \))** has no steric hindrance because it is a primary carbon. This means it is very reactive in \( S_N2 \) reactions.
- **Ethyl chloride (\( CH_3CH_2Cl \))** is also a primary carbon, but it has slightly more steric hindrance than methyl chloride due to the presence of an additional carbon. However, it is still quite reactive.
- **Isopropyl chloride (\( (CH_3)_2CHCl \))** is a secondary carbon. It has more steric hindrance compared to the primary carbons, which will slow down the \( S_N2 \) reaction.
- **Tertiary butyl chloride (\( (CH_3)_3CCl \))** is a tertiary carbon, which has the most steric hindrance. This makes it the least reactive in \( S_N2 \) reactions.

3. **Order of Reactivity:**
Based on the analysis of steric hindrance:
- Most reactive: \( CH_3Cl \) (Methyl chloride)
- Next: \( CH_3CH_2Cl \) (Ethyl chloride)
- Then: \( (CH_3)_2CHCl \) (Isopropyl chloride)
- Least reactive: \( (CH_3)_3CCl \) (Tertiary butyl chloride)

4. **Final Order:**
Therefore, the correct order of reactivity in \( S_N2 \) reactions is:
\[ CH_3Cl > CH_3CH_2Cl > (CH_3)_2CHCl > (CH_3)_3CCl \]

### Conclusion:
The order of reactivity of the compounds in \( S_N2 \) reactions is:
1. \( CH_3Cl \) (most reactive)
2. \( CH_3CH_2Cl \)
3. \( (CH_3)_2CHCl \)
4. \( (CH_3)_3CCl \) (least reactive)