Nucleophilic aliphatic substitution reaction is mainly of two types: `S_(N)1` and `S_(N)2`. The `S_(N)1` mechanism is a two step process. Reaction velocity of `S_(N)1` reaction depends only on the concentration of the substrate. Since product formation takes place by the formation of carbocation, optically active substrate gives (+) and (-) forms of the product. In most of the cases the product usually consits of `5-20%` inverted product and `80-95%` racemised species. The more stable the carbocation, the greater is the proportion of racemisation. In solvolysis reaction, the more nucleophilic the solvent, the greater is the proportion of inversion.
Which one of the following compound will give `S_(N)1` reaction predominantly?

Options

A.
B.
C. `H_(3)C-Br`
D. All of these

Answer

To determine which compound will predominantly undergo the \( S_N1 \) reaction, we need to analyze the stability of the carbocations formed from each compound. The \( S_N1 \) mechanism involves the formation of a carbocation intermediate, and the stability of this carbocation is crucial for the reaction to proceed efficiently.

### Step-by-Step Solution:

1. **Understanding the \( S_N1 \) Mechanism**:
- The \( S_N1 \) reaction is a two-step process where the first step involves the formation of a carbocation by the departure of the leaving group (in this case, a halide).
- The reaction rate depends only on the concentration of the substrate, not the nucleophile.

2. **Carbocation Stability**:
- The stability of carbocations follows this order:
- Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl.
- The more stable the carbocation, the more likely it is to form and thus favor the \( S_N1 \) pathway.

3. **Analyzing the Given Options**:
- Let's denote the compounds as A, B, C, and D.
- We will analyze the carbocation that would form from each compound upon losing the leaving group (e.g., bromide).

4. **Compound A**:
- Upon losing the bromine atom, the carbocation formed is a tertiary carbocation with additional resonance stabilization due to the adjacent aromatic ring.
- This makes it highly stable.

5. **Compound B**:
- The carbocation formed here is a primary carbocation, which is significantly less stable than a tertiary carbocation.

6. **Compound C**:
- This compound forms a methyl carbocation, which is even less stable than a primary carbocation.

7. **Compound D**:
- This option states "all of these," but since we are looking for the compound that predominantly undergoes \( S_N1 \), we need to choose the one that forms the most stable carbocation.

8. **Conclusion**:
- Among the options, Compound A forms the most stable carbocation (a tertiary carbocation with resonance), making it the compound that will predominantly undergo the \( S_N1 \) reaction.

### Final Answer:
The compound that will give \( S_N1 \) reaction predominantly is **Compound A**.