Answer
This reaction will proceed via $S_N2$.
Work Step by Step
1. Substrate
- The carbon connected to the leaving group is connected to 2 other carbons.
This is not a factor to predict if the reaction is going to be $S_N1$ or $S_N2$
2. Nucleophile
- $OH^-$ is a strong* nucleophile, which fagvor $S_N2$ reactions.
*(You can consult the table on page 216)
3. Leaving group
- "$Cl^-$": Good leaving group, which favors $S_N2$.
4. Solvent
- DME: Polar aprotic, which makes it only $S_N2$.
