Question

What are the products? CH3CH(OH)CH(CH3)CH2CH3 [3-methyl-2pentanol]+ HCl=?

Answer 1

This appears to be an SN1 reaction.

Given a strong acid in water, an oxonium ion will form, leave, the molecule will stabilize by undergoing a carbocation rearrangement, and the chloride ion will attack the electrophilic carbon.

If you want the mechanism, ask!

Answer 2

The product is 3-chloro-3-methylpentane.

Explanation:

The first steps are protonation of the `"OH"` group, followed by the loss of water.

Here's the critical point to remember: Whenever a 2° carbocation can become more stable by rearranging to a 3° carbocation, it will do so.

A hydride shift forms a 3° carbocation, which then adds the `"Cl"^"-"` ion as usual.