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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

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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.

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  • $\begingroup$ How does the solvent being protic hinder the nucleophile? $\endgroup$ – yolo123 Aug 13 '14 at 17:14
  • $\begingroup$ The solvent can hydrogen bond with the nucleophile. $\endgroup$ – Dissenter Aug 13 '14 at 17:15
  • $\begingroup$ Ok. Now, does that mean that E2 is favored in a strict sense? $\endgroup$ – yolo123 Aug 13 '14 at 17:16
  • $\begingroup$ Er that's a hard call to make without actually doing the experiment. $\endgroup$ – Dissenter Aug 13 '14 at 17:16
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    $\begingroup$ 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 Aug 13 '14 at 17:18

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