Answer
This reaction will proceed via an $S_N2$
Work Step by Step
1. Substrate
- The carbon connected to the leaving group is connected to 1 other carbon, which makes the reaction only possible by $S_N2$
2. Nucleophile
- $Cl^-$ is a strong* nucleophile, which favors $S_N2$ reactions.
*(You can consult the table on page 216)
3. Leaving group
- "$Br^-$": Good leaving group, which favors $S_N2$.
4. Solvent
- DMSO: Polar aprotic, which makes it only $S_N2$
