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