1
$\begingroup$

Please look at the image on this website: http://www.physicsforums.com/showthread.php?t=765853

For (b) and (c), what the solution manual says makes no sense to me. You're reacting a primary alkyl halide with an unhindered base $\ce{EtONa}$. They say that the only product is the alkene. Why do they not consider the $\mathrm{S_N2}$ reaction that WILL take place in larger quantity? I'm uploading the question too on the other website

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
1
$\begingroup$

I agree with you, I'd expect some $\mathrm{S_N2}$ product too. I'm not sure about the relative quantities of the $\mathrm{E2}$ and $\mathrm{S_N2}$ products though.

I wouldn't flat out say that you'd expect more $\mathrm{S_N2}$ product than $\mathrm{E2}$ product, because although the substrate is unhindered, the solvent is ethanol, which is protic. The solvent interacts with the nucleophile through hydrogen bonding.

This hinders the nucleophile. This slows down the rate of the $\mathrm{S_N2}$ reaction because now the nucleophile is dragging along lots of solvent molecules, and it also increases the bulk of the nucleophile.

Improve this answer
$\endgroup$
8
  • $\begingroup$ How does the solvent being protic hinder the nucleophile? $\endgroup$
    – yolo123
    Commented Aug 13, 2014 at 17:14
  • $\begingroup$ The solvent can hydrogen bond with the nucleophile. $\endgroup$
    – Dissenter
    Commented Aug 13, 2014 at 17:15
  • $\begingroup$ Ok. Now, does that mean that E2 is favored in a strict sense? $\endgroup$
    – yolo123
    Commented Aug 13, 2014 at 17:16
  • $\begingroup$ Er that's a hard call to make without actually doing the experiment. $\endgroup$
    – Dissenter
    Commented Aug 13, 2014 at 17:16
  • 1
    $\begingroup$ Ok. I just consulted my textbook. It seems as though Sn1 and Sn2 are really impacted by solvent. E2 and E1 are indirectly concerned and as you say need the experiment. Great help! I'll post another question that was racking my brains. $\endgroup$
    – yolo123
    Commented Aug 13, 2014 at 17:18

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.