this is an assignment question but I'm stumped.enter image description here

we have a polar protic solvent (also a weak base) interacting with the secondary carbon -- it looks like a good example of an sn1/sn2 to me. So why wouldn't the answer be "I and III"?

  • $\begingroup$ do you mean a tertiary carbon? In that case, Sn2 is out of context. $\endgroup$ Nov 10, 2015 at 4:55
  • $\begingroup$ Curious - What would the name of the starting compound be? 2-methyl-4-ethylether-hexane ?!? $\endgroup$
    – MaxW
    Nov 10, 2015 at 5:25
  • $\begingroup$ @Max (S)-4-bromo-2,4-dimethylhexane $\endgroup$ Nov 10, 2015 at 5:30
  • $\begingroup$ dah... I meant the name of the ether in answer (I) $\endgroup$
    – MaxW
    Nov 10, 2015 at 5:34
  • $\begingroup$ Well, all in all, it doesn't seem that a SN1 or a SN2 reaction can be happening at all. A pure SN2 reaction would give (III). A SN1 would give a racemic mixture of (I) and (III). Some combination of SN1 and SN2 would give a mixture of (I) and (III) with more (III). $\endgroup$
    – MaxW
    Nov 10, 2015 at 5:38

1 Answer 1


The major reasons I would give for going with the SN1 products are:

  • The amount of steric hindrance on tertiary alkyl halides in this context highly limits the amount of second-order reactions that can occur (which require participation from two molecules).
  • EtOH is not a strong nucleophile in this context. In contrast, EtO- would be, but it's not present. Had EtOH been a strong nucleophile, SN2 would be a tad more likely, but a primary alkyl halide situation, for instance, is much more likely to favor SN2. Therefore, an SN2 product cannot form as a major product. However, considering both I AND III, the product is not formed through SN2 (unless you pick only III, in which case you assume I is not formed).
  • III is one product in the racemic mixture that would result from forming the tertiary carbocation (which is nearly planar around carbon 3, allowing for both retention and inversion of the stereochemistry). Also, the 20 degrees C does not convince me that elimination is a major product since lower temperature makes elimination more difficult. It is harder for the molecules to orient themselves quickly enough before all reactant is used up if the environment is cold. Therefore, the answer is probably I AND III, assuming that somehow the answer key is incorrect.

I also agree with orthocresol that product II is strange. I would expect by Zaitsev's rule that the new pi bond would form across carbons 3 and 4 in the main chain since his rule says that typically, the elimination occurs by acquiring a proton from the carbon with fewer protons, forming typically the more substituted alkene.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.