Question

What are the product(s) for the `"S"_N1` reaction of 2-iodo-3-methylbutane with ethanol?

I feel like I'm missing something, because I thought I got the major product right...

Answer 1

So, 2-iodo-3-methylbutane looks like this:

When it reacts with ethanol (the solvent)---indeed under an `S_N1` mechanism based on the steric hindrance---due to iodine's large radius, the first step is a slow, rate-determining step that promotes the creation of a carbocation intermediate.

When you see a tertiary alkyl chain right near a reactive/electrophilic site, it's a pretty good sign that there's probably a rearrangement that can happen to stabilize the intermediate the majority of the time.

In this case, you do get a rearrangement that promotes the more stable tertiary carbocation (unlike in an `S_N2` mechanism):

Also, remember, an INTRAmolecular 1,2-hydride shift is faster than the INTERmolecular nucleophilic attack by ethanol. If you're within a system, you act faster than if you were approaching the system from the outside, especially due to the steric hindrance at the electrophilic carbon.

Lastly, `HI` is too strong to form again in solution all that easily, (pKa `~~ -9`), so the solvent picks up the proton again, instead, and electrostatically interacts with `"I"^(-)`.

The final products end up being a mixture:

• 2-ethoxy-2-methylbutane (major)
• 2-ethoxy-3-methylbutane (minor, for no rearrangement)
• Protonated ethanol
• Iodide

I'm guessing you missed a methyl group? Also, just a sidenote, but even though your book seems to have specified that you focus on `S_N1`, it's not the only reaction that can occur; you would see some `E1` with this too, just not as much.