The decreasing order of reactivity of the following compounds in `S_(N)1` reactions is _________.
(i) `C_(6)H_(5)CH_(2)Br" " (ii) C_(6)H_(5)CH(C_(6)H_(5)Br`
`(iii) C_(6)H_(5)CH(CH_(3)Br" "(iv) C_(6)H_(5)C(CH_(3))(C_(6)H_(5))Br`

Options

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

Answer

To determine the decreasing order of reactivity of the given compounds in SN1 reactions, we need to analyze the stability of the carbocations formed during the reaction. The stability of carbocations is influenced by resonance and hyperconjugation effects. Here’s a step-by-step solution:

### Step 1: Identify the Compounds
The compounds given are:
1. \( C_6H_5CH_2Br \) (Compound 1)
2. \( C_6H_5CH(C_6H_5)Br \) (Compound 2)
3. \( C_6H_5CH(CH_3)Br \) (Compound 3)
4. \( C_6H_5C(CH_3)(C_6H_5)Br \) (Compound 4)

### Step 2: Understand the SN1 Mechanism
In an SN1 reaction, the rate-determining step is the formation of a carbocation after the leaving group (Br) departs. The more stable the carbocation, the faster the reaction will proceed.

### Step 3: Draw the Carbocations
1. **For Compound 1**:
- Upon breaking the C-Br bond, we get a benzyl carbocation:
\[ C_6H_5CH_2^+ \]
This carbocation is stabilized by resonance with the phenyl ring.

2. **For Compound 2**:
- Breaking the C-Br bond gives:
\[ C_6H_5C^+(C_6H_5) \]
This carbocation is stabilized by resonance from two phenyl groups.

3. **For Compound 3**:
- The carbocation formed is:
\[ C_6H_5C^+(CH_3) \]
This carbocation is stabilized by resonance from the phenyl group and hyperconjugation from the three hydrogens on the methyl group.

4. **For Compound 4**:
- The carbocation formed is:
\[ C^+(C_6H_5)(C_6H_5) \]
This carbocation is stabilized by resonance from two phenyl groups and hyperconjugation from the methyl groups.

### Step 4: Analyze Carbocation Stability
- **Compound 1**: Benzyl carbocation, moderate stability due to one phenyl group.
- **Compound 2**: More stable due to resonance from two phenyl groups.
- **Compound 3**: Stable due to one phenyl group and hyperconjugation from the methyl group.
- **Compound 4**: Most stable due to resonance from two phenyl groups and hyperconjugation from two methyl groups.

### Step 5: Order of Reactivity
Based on the stability of the carbocations:
1. **Compound 4**: Most stable → Most reactive
2. **Compound 2**: Second most stable → Second most reactive
3. **Compound 3**: Third most stable → Third most reactive
4. **Compound 1**: Least stable → Least reactive

### Final Answer
The decreasing order of reactivity in SN1 reactions is:
\[ \text{Compound 4} > \text{Compound 2} > \text{Compound 3} > \text{Compound 1} \]