Organic Chemistry - Synthesis Problem

Looking at the reactions that esters normally undergo, we see that esters can undergo reduction to alcohols in the presence of Lithium Aluminum Hydride (in an acidic environment):
methyl-2-methylpentanoate + LiAlH4 -> 2-methyl-1-pentanol (image 1)

Next, an alcohol will undergo nucleophilic substitution with Thionyl Chloride (SOCl2) to form the corresponding alkyl halide:
2-methyl-1-pentanol + SOCl2 + heat -> 1-chloro-2-methylpentane (image 2)

I worked a bit on this question.  Don't know how to get the Cl on the 2nd carbon.  I might have made a mistake.  Just wanted to give you what I had done.  I hope this helps.

You definitely helped me Doctor-2-B although I'm still not 100% on understanding this completely...but thank you very much it definitely gave me an insight as to a possible mechanism!!!

I believe I might've missed a carbocation rearrangement or methyl shift but maybe not?

Hi,

I was thinking about the first step is the reduction of carboxylic acid by using 1)LiAlH4, then 2) water to get the primary alcohol product. Then on the 2nd step, i would add acid to promote elimination product, making 2-methyl-1-pentene. Then I would use HCl, so that the Cl would be added to the more substituted carbon (carbon #2).

Hope this helps,
Laser

That sounds completely reasonable.  I thought esters reduced to alcohols in Lithium Aluminum Hydride, though?

Hi Doctor-2-B,

You're right, it would; that's why we both used it in the first step. The water is just to protonate the alcohol. I did 1) and 2) on the first step, because the Lithium Aluminum Hydride violently reacts with water.

Best,
Laser