`{:("column I"," ""Column II"),("Alkyl-bromide ","Relative rate of SN"^(1)),((A) CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-Br," "(P)1),((B) CH_(3)-underset(CH_(3))underset(|)(C)-Br," "(Q)11.6),((C)CH_(3)-CH_(2)-Br," "(R)1200000):}`

Answer

To solve the problem of matching the alkyl bromides with their relative rates of SN1 reactions, we will follow these steps:

### Step 1: Understand the SN1 Reaction Mechanism
The SN1 reaction involves two main steps:
1. The leaving group (Br) departs, forming a carbocation intermediate.
2. The nucleophile (Y-) attacks the carbocation to form the product.

The rate of the SN1 reaction is determined by the stability of the carbocation formed in the first step.

### Step 2: Identify the Structure of Each Alkyl Bromide
We need to analyze the structures provided in Column I:
- (A) CH₃-CH(CH₃)-Br
- (B) CH₃-CH(CH₃)-Br (This seems to be the same as A, possibly a typo)
- (C) CH₃-CH₂-Br

### Step 3: Determine the Carbocation Stability
The stability of the carbocation is influenced by the number of alpha hydrogens:
- For (A): The carbocation formed will be tertiary (3 alpha carbons), which is very stable.
- For (B): The carbocation formed will also be tertiary (3 alpha carbons), which is very stable.
- For (C): The carbocation formed will be primary (1 alpha carbon), which is less stable.

### Step 4: Count the Alpha Hydrogens
- (A) has 9 alpha hydrogens.
- (B) has 6 alpha hydrogens.
- (C) has 3 alpha hydrogens.

### Step 5: Rank the Stability of the Carbocations
The stability order based on the number of alpha hydrogens and the type of carbocation is:
1. (A) - Tertiary carbocation (most stable)
2. (B) - Tertiary carbocation (less stable than A)
3. (C) - Primary carbocation (least stable)

### Step 6: Match with Relative Rates
Now we can match these with their corresponding rates of SN1 reactions:
- (A) corresponds to R (highest rate).
- (B) corresponds to Q (middle rate).
- (C) corresponds to P (lowest rate).

### Final Matching:
- (A) - R
- (B) - Q
- (C) - P

### Conclusion:
The final matches are:
- A → R
- B → Q
- C → P