# SN1 reaction of ethanol at tertiary alkyl halide

I know this undergoes an $\ce{S_{N}1}$ reaction:

1. Ethanol is the solvent. Weak nucleophile. Neutral nucleophile. This points toward $\ce{S_{N}1/E1}$ (competing pathways).

2. The leaving group is on a tertiary carbon. This precludes backside attack, $\ce{S_{N}2}$.

3. The professor labeled this reaction as $\ce{S_{N}1}$.

My question: where's the substitution part of this reaction? I don't see any substitution. Is there supposed to be no reaction here? From what he wrote it looks like a bromine left as bromide ion and a hydrogen appeared out of nowhere.

My expected transition state is a tertiary carbocation, which will not rearrange, and then the nucleophilic oxygen part of the ethanol molecule attacks the carbocation. Bromide ion deprotonates the oxonium ion, and I get an ether.

Who's right?

• You're description is right.
– ron
Jul 30, 2014 at 23:52
• Sheesh, there was something so obviously wrong with this I was confused. Thanks for clarifying. Jul 30, 2014 at 23:59
• Yep, it's hard enough to learn this stuff without the teacher creating obstacles. You did well figuring it out!
– ron
Jul 31, 2014 at 0:02
• It's the reason I'm here on this site! Jul 31, 2014 at 0:38

Just to get your terminology right. The carbocation is a reactive intermediate, not a transition state. The $\ce{S_{N}1/E1}$ reaction have always (at least) two transition states. The first one corresponds to the carbon leaving group bond break and the second one corresponds to the nucleophile carbon bond formation, proton abstraction, respectively.
The product drawn is a reduction product. You can use $\ce{LiAlH4}$ or homolytic cleavage of the $\ce{C-Br}$ bond (think radicals) to prepare it (but good luck isolating this, isopentane's b.p. is 28 $^\circ$C ). But not $\ce{EtOH}$ hehe.