In my textbook (Organic Chemistry by Klein), it says that certain molecules can only act as nucleophiles and not as bases; thus they can only undergo substitution reactions, not elimination reactions. This group of molecules includes: $\ce{Cl- , Br- , I- , HS- , H2S , RS- , RSH}$. I know that while E2 reactions require strong bases, E1 reactions do not. Even a base like $\ce{H2O}$, which is pretty weak, can undergo E1 reactions. Then why can't something like $\ce{HS-}$ undergo a E1 reaction? For example, if we have a tertiary alkyl halide and $\ce{HS-}$, I would expect a mix of $S_N1$ and E1 products. However, my book claims there are only $S_N1$ products. Why is this the case?

  • $\begingroup$ I think "can not" is probably not the right description of its reactivity, and "does not" is more accurate. $\endgroup$
    – jerepierre
    Commented Mar 31, 2016 at 20:29
  • $\begingroup$ $\ce{H2S}$ is a much better nucleophile than $\ce{H2O}$. $\ce{HS^-}$ is even more so. The diffuse nature of sulfur's electronic radius 1. makes it more acidic (less attracted to $\ce{H^+}$) and 2. gives it a higher HOMO. This allows it to react with the low LUMO of a C-X bond $\sigma ^*_{\text{CX}}$ rather than the small, dense s-orbital of $\ce{H^+}$. $\endgroup$ Commented Mar 31, 2016 at 20:43
  • $\begingroup$ @DGS But isn't $\ce{HS-}$ still a stronger base than $\ce{H2O}$? So shouldn't it undergo E1 better than $\ce{H2O}$ can? I understand that for $\ce{HS-}$, substitution is the predominant reaction, but does this mean that elimination doesn't happen at all? $\endgroup$
    – carbenoid
    Commented Apr 1, 2016 at 16:56
  • $\begingroup$ Think about the competitive reactions. If it is a much better nucleophile, the nucleophilic on a carbocation attack would be a significantly faster reaction relative to deprotonation. $\endgroup$ Commented Apr 1, 2016 at 19:43

1 Answer 1


As has been pointed out in the comments, the use of the word only is a little misleading. What it should really say is that $E1$ is possible but $S_N1$ is so strongly favoured that to all intents and purposes the only product is the $S_N1$ product.

This can be explained in terms of the relative energies and sizes of the orbitals involved. $\ce{HS-}$ has a large, diffuse, high energy HOMO which can have much better overlap with the empty $2p$ orbital of the carbon than the small, low energy $\ce{C-H}~\sigma^*$. Therefore, the $S_N1$ reaction is strongly thermodynamically favoured and dominates.

If you compare this to something like $\ce{H2O}$, the HOMO is much more compact and lower in energy and so it has much better overlap with the $\ce{C-H}~\sigma^*$ and so $E1$ is comparatively much more favourable.


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