`{:("Column I","Column II"),("Primary alkyl bromide","SN"_(2)"relative rate"),((A) CH_(3)— CH_(2)— Br,(P) 10^(-5)),((B) Me-CH_(2)-CH_(2)-Br,(Q)10^(-2)),((C)Me-underset(Me)underset(|)(CH)-CH_(2)-Br,(R) 0.8),((D)Me-underset(Me)underset(|)overset(Me)overset(|)(C)-CH_(2)-Br,(S)1):}`

Answer

To solve the problem, we need to determine the relative rates of the SN2 reactions for the given primary alkyl bromides listed in Column I, and match them with the corresponding rates in Column II.

### Step-by-Step Solution:

1. **Identify the Structure of Each Alkyl Bromide:**
- (A) CH₃—CH₂—Br: This is a straight-chain primary alkyl bromide.
- (B) Me-CH₂-CH₂-Br: This is also a primary alkyl bromide with a methyl group attached to the first carbon.
- (C) Me- (Me) (CH)-CH₂-Br: This is a branched primary alkyl bromide.
- (D) Me- (Me) (C)-CH₂-Br: This is a more branched primary alkyl bromide.

2. **Understand SN2 Mechanism:**
- The SN2 reaction mechanism involves a backside attack on the carbon atom bonded to the leaving group (bromine in this case). The rate of the reaction is influenced by steric hindrance around the carbon atom. The less sterically hindered the carbon, the faster the reaction.

3. **Analyze Each Compound:**
- **(A) CH₃—CH₂—Br**: This is a simple primary bromide with minimal steric hindrance. It will have a relatively high rate.
- **(B) Me-CH₂-CH₂-Br**: This compound has a methyl group but is still primarily linear, so it will have a high rate, but slightly less than (A) due to the presence of the methyl group.
- **(C) Me- (Me) (CH)-CH₂-Br**: This compound has more branching, which increases steric hindrance, leading to a lower rate compared to (A) and (B).
- **(D) Me- (Me) (C)-CH₂-Br**: This compound is the most branched, resulting in the highest steric hindrance and thus the slowest rate.

4. **Assign Relative Rates:**
- Based on the analysis:
- (A) CH₃—CH₂—Br: Highest rate (P) 10^(-5)
- (B) Me-CH₂-CH₂-Br: Moderate rate (Q) 10^(-2)
- (C) Me- (Me) (CH)-CH₂-Br: Lower rate (R) 0.8
- (D) Me- (Me) (C)-CH₂-Br: Lowest rate (S) 1

5. **Match with Column II:**
- Now we match the compounds with the rates:
- (A) → (P) 10^(-5)
- (B) → (Q) 10^(-2)
- (C) → (R) 0.8
- (D) → (S) 1

### Final Answer:
- The matches are:
- (A) → (P)
- (B) → (Q)
- (C) → (R)
- (D) → (S)