During `SN^1` reactions, the leaving group leaves the molecule before the incoming group is attached to the molecule.

Explanation

### Step-by-Step Solution:

1. **Understanding SN1 Reactions**:
- SN1 stands for Substitution Nucleophilic Unimolecular. This type of reaction involves two main steps: the formation of a carbocation and the nucleophilic attack.

2. **First Step - Leaving Group Departure**:
- In the first step of an SN1 reaction, the leaving group (in this case, bromine, Br) departs from the substrate (e.g., a haloalkane like CH3Br).
- This departure leads to the formation of a carbocation intermediate. The process can be represented as:
\[ \text{CH}_3\text{Br} \rightarrow \text{CH}_3^+ + \text{Br}^- \]
- Here, the bromine atom leaves, resulting in a positively charged carbocation (CH3^+).

3. **Second Step - Nucleophile Attack**:
- In the second step, the nucleophile (for example, OH^-) attacks the carbocation formed in the first step.
- This can be represented as:
\[ \text{CH}_3^+ + \text{OH}^- \rightarrow \text{CH}_3\text{OH} \]
- The nucleophile attaches to the carbocation, resulting in the final product, which is an alcohol (CH3OH).

4. **Conclusion**:
- The statement in the question is confirmed to be true: during SN1 reactions, the leaving group does indeed leave the molecule before the incoming nucleophile attaches to the molecule.