I uploaded pictures on http://www.physicsforums.com/showthread.php?t=765106

I was looking at the answers in my textbook. They seem to make incongruent choices of solvents.

In e) and j), why choose EtOH and H2O/CH3OH? I would rather choose an aprotic solvent like DMF to better allow Sn2 to occur. Especially with j), there is a lot of competition with E2 because OH- is a strong base and because we are attacking a secondary carbon.

I see this again in k). Secondary carbon attacked: why not add DMF to prevent E2 and get best yield?

enter image description here

  • $\begingroup$ I'm not seeing what you're seeing, OP. I don't see any methanol in e, for example. $\endgroup$
    – Dissenter
    Aug 13, 2014 at 17:49
  • $\begingroup$ sorry to poke but k) part is not there $\endgroup$
    – RE60K
    Aug 13, 2014 at 17:56
  • 2
    $\begingroup$ The competition from $\mathrm{E_2}$ in problem (j) is less of an issue than it may at first appear. The $\mathrm{E_2}$ elimination would require a ring flip that places the two substituents on the ring in axial positions (recall that $\mathrm{E_2}$ typically proceeds only when $\beta$-hydrogen and leaving group are anti-periplanar, and in a cyclohexane ring that can only occur if the two are axial). The conformer with both substituents axial is substantially less stable, so $\mathrm{E_2}$ is hindered (though not ruled out) by the energy barrier associated with the ring flip. $\endgroup$
    – Greg E.
    Aug 13, 2014 at 18:00
  • $\begingroup$ Sorry Aditya! I have no idea how to upload pictures. Can someone teach me? $\endgroup$
    – yolo123
    Aug 13, 2014 at 18:24
  • 2
    $\begingroup$ @yolo123, $\mathrm{E_2}$ is still possible for (j), it's just slower due to the pre-equilibrium for chair conformer interconversion. Maybe with careful temperature control, the ring flip is enough to favor the $\mathrm{SN_2}$ there, but I would expect a mixture even in the best case. The choice of protic solvents is theoretically sub-optimal for the reasons you've mentioned, though it's not really a problem in (e) because there are really no alternate pathways. Solubility is a possible practical reason, as jerepierre suggested in his answer. As for (k), I don't have an account on that site. $\endgroup$
    – Greg E.
    Aug 13, 2014 at 20:48

3 Answers 3


Your thinking is not wrong. For (e) DMF is a better choice of solvent - this doesn't mean that MeOH won't work just that DMF is better and is typically used in these reactions. For even better results you can use DMSO. If you dare, as DMSO/cyanide is a very toxic cocktail. (j) I don't think it is a practical synthetic reaction since it can possibly give you elimination and/or $S_N1$ (ie no inversion of stereochemistry) products.


For reaction (e), given the strength of the nucleophile, the primary benzylic leaving group, and the lack of any protons beta to the leaving group, SN2 is the only mechanism that can operate. The choice of ethanol as solvent likely has much more to do with getting everything into solution than influencing one pathway over another. According to Wikipedia, NaCN is soluble in EtOH but only slightly soluble in DMF.

For reaction (j), I would guess that the water/methanol mixture was chosen to get everything into solution. There might be some concern about hydroxide deprotonating methanol, and then methoxide carrying out the displacement. My answer would probably be to use a water/THF mixture, if you were really worried about that.


In both e and j, you would choose the protic solvent over aprotic solvent, because the protic solvent will stabilize the resulting bromide anion. Sometimes, the stability of the other product should also be concerned, not just the main product.


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