Answer:
2-iodo-2-methylpentane > 2-bromo-2-methylpentane > 2-chloro-2-methylpentane > 2-chloro-2-methylpentane
Explanation:
In this case, the Sn1 reaction would form a carbocation. So, the molecule that can generate a very stable carbocation will be more reactive. We have to remember that tertiary carbocations are the more stable ones. With this in mind, 3-chloropentane would be the least reactive molecules of all.
Then to decide which one is more reactive between the other ones, we have to check the leaving group. In this case, all the atoms are halogens, so if we have a larger atom the leaving group would leave more easily.
With his in mind the larger atom would be I, then Br and finally Cl, therefore the 2-iodo-2-methylpentane would be the more reactive one.
