Answer
This compound is likely to participate both in Sn2 and Sn1 reactions.
Work Step by Step
1. Identify the leaving group:
- In this case, it is the "Br". (Normally they are electronegative groups).
2. Analyze the carbon that is connected to it, and count the number of other carbons connected to it.
- There are 2 other carbons, which means that this substrate is secondary. (2$^{\circ}$).
- Secondary substrates can participate in Sn2 or Sn1 reactions with the same effort
