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

enter image description here

  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?

  • 2
    $\begingroup$ You're description is right. $\endgroup$
    – ron
    Jul 30 '14 at 23:52
  • $\begingroup$ Sheesh, there was something so obviously wrong with this I was confused. Thanks for clarifying. $\endgroup$
    – Dissenter
    Jul 30 '14 at 23:59
  • $\begingroup$ Yep, it's hard enough to learn this stuff without the teacher creating obstacles. You did well figuring it out! $\endgroup$
    – ron
    Jul 31 '14 at 0:02
  • $\begingroup$ It's the reason I'm here on this site! $\endgroup$
    – Dissenter
    Jul 31 '14 at 0:38

Just to leave this question not unanswered. As ron already pointed out, you reasoning is correct. Based on the information given, the ethanol may perform a nucleophilic attack at the carbocation. To a very minor extent you might find the elimination product, but I guess you will have to cook it quite well. The elimination pathway is more likely to occur, when there is a stronger base than ethanol.

reaction scheme

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.

reaction mechanisms


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.

To the point, I think a more appropriate example would be to use something like adamantyl bromide instead, as in this case you can not have competing elimination:

adamantyl bromide


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