Answer
This reaction will proceed via $S_N1$.
Work Step by Step
1. Substrate
- The carbon connected to the leaving group is connected to 3 other carbons.
This favors $S_N1$ reactions.
2. Nucleophile
- $H_2O$ is a weak* nucleophile, which favors $S_N1$ reactions.
*(You can consult the table on page 216)
3. Leaving group
- "$Cl^-$": Good leaving group, which favors $S_N2$, but doesn't eliminate $S_N1$ reactions.
*(You can consulte the tables on pages 218 and 219)
4. Solvent
- Not mentioned: Not a factor.
