Arrange the following compounds in order of their reactivity towards `S_(N^2)` reaction.
(i) `CH_3(CH_2)_3CH_2Br`
(ii) `(CH_3)_2CHCH_2CH_2Br`
(iii) `CH_3CH_2-overset(CH_3)overset(|)(CH)-CH_2Br`
(iv) `CH_3-underset(CH_3)underset(|)overset(CH_3)overset(|)C-CH_2Br`

Options

A. (i)gt(ii)gt(iii)gt(iv)
B. (ii)gt(iii)gt(iv)gt(i)
C. (iii)gt(i)gt(ii)gt(iv)
D. (iv)gt(ii)gt(i)gt(iii)

Answer

To determine the order of reactivity of the given compounds towards the \( S_N2 \) reaction, we need to consider the steric hindrance around the carbon atom that is attached to the bromine (Br) atom. The \( S_N2 \) mechanism is characterized by a bimolecular nucleophilic substitution, which is inversely proportional to steric hindrance; that is, the less hindered the carbon, the more reactive it is towards \( S_N2 \).

Let's analyze the compounds one by one:

1. **Compound (i): \( CH_3(CH_2)_3CH_2Br \)**
This compound has a primary carbon (the carbon attached to Br is bonded to only one other carbon). It has the least steric hindrance, making it the most reactive towards \( S_N2 \).

2. **Compound (ii): \( (CH_3)_2CHCH_2CH_2Br \)**
This compound has a secondary carbon (the carbon attached to Br is bonded to two other carbons). It has more steric hindrance than compound (i), thus it will be less reactive than compound (i) but more reactive than the others.

3. **Compound (iii): \( CH_3CH_2-(CH_3)-C(CH_2Br) \)**
This compound also has a secondary carbon, but it is branched. The branching increases steric hindrance compared to compound (ii), making it less reactive than compound (ii).

4. **Compound (iv): \( CH_3-(CH_3)-C(CH_3)-CH_2Br \)**
This compound has a tertiary carbon (the carbon attached to Br is bonded to three other carbons). Tertiary carbons are highly hindered, making this compound the least reactive towards \( S_N2 \).

Now, we can summarize the order of reactivity based on the steric hindrance:

- **Most Reactive:** Compound (i) \( CH_3(CH_2)_3CH_2Br \) (Primary)
- **Second Reactive:** Compound (ii) \( (CH_3)_2CHCH_2CH_2Br \) (Secondary)
- **Third Reactive:** Compound (iii) \( CH_3CH_2-(CH_3)-C(CH_2Br) \) (Secondary, more hindered)
- **Least Reactive:** Compound (iv) \( CH_3-(CH_3)-C(CH_3)-CH_2Br \) (Tertiary)

Thus, the final order of reactivity towards \( S_N2 \) is:
\[ \text{(i) > (ii) > (iii) > (iv)} \]