8
$\begingroup$

I went through the problems on the pictures:

enter image description hereenter image description here

The answer of the problem 119 is $\mathrm{S_N1'}$, and that of 120 is $\mathrm{S_N2'}$. I actually can't figure out what $\mathrm{S_N1'}$ ($\mathrm{S_N1}$ prime) and $\mathrm{S_N2'}$ ($\mathrm{S_N2}$ prime) are.

Also, how does the answer change by changing the nucleophile?

$\endgroup$
  • 5
    $\begingroup$ I would like to point out that the question is not about nucleophilic substitution in general, in the nutshell it's about the difference that "prime" ' sign adds (e.g. $\mathrm{S_N1}$ vs $\mathrm{S_N1'}$ and $\mathrm{S_N2}$ vs $\mathrm{S_N2'}$), from what I understand. IMO it's neither homework, nor too broad; I cannot find any duplicates, so I'm against closing it for now. Also, check out en.wikipedia.org/wiki/Allylic_rearrangement $\endgroup$ – andselisk Sep 19 '17 at 17:07
  • 1
    $\begingroup$ @andselisk Ah! I missed the prime. I've retracted my vote. O:) $\endgroup$ – paracetamol Sep 19 '17 at 17:13
  • 1
    $\begingroup$ @paracetamol I have a feeling that I saw something related on CSE, but searching for SN1' obviously returns tons of topics where SN1 has been discussed, same with Google. $\endgroup$ – andselisk Sep 19 '17 at 17:19
  • 4
    $\begingroup$ The ' indicates that the actual substitution site is not where the leaving group is attached. In the cases presented, the nucleophile would add at the upper end of the double bond. $\endgroup$ – Zhe Sep 19 '17 at 17:23
8
$\begingroup$

The primed version of $\mathrm{S_N1}$ and $\mathrm{S_N2}$ can only occur if there is a double bond in the vicinity of the leaving group as in your example. I have drawn both possibilities for the $\mathrm{S_N2}$ case in the scheme below.

SN2 and SN2' of methanolate with the starting material
Scheme 1: Comparison of the reaction products in an $\mathrm{S_N2}$ (top) and $\mathrm{S_N2'}$ (bottom) pathway.

The $\mathrm{S_N2}$ reaction is as you expect it to be. The $\mathrm{S_N2'}$ reaction uses the double bond as an electron relay system. Instead of the nucleophile (here: $\ce{OMe-}$) directly interacting with the $\unicode{x3c3}^*(\ce{C-Br})$ orbital, the π system interacts with the σ* orbital. The nucleophile then interacts with the π system’s LUMO (corresponding to the middle orbital of the allyl π system) to perform the attack. Since the nucleophilic attack and the leaving group are on different carbon atoms, relayed by the π system, this is not a direct $\mathrm{S_N2}$ but a derivative of it ($\mathrm{S_N2'}$).

The same logic can be applied to $\mathrm{S_N1'}$: here, the intermediate carbocation is not a localised one but an allyl cation.

The primed versions here are observed because not only the bimolecular attack on the highly substituted tertiary carbon ($\mathrm{S_N2}$) is unlikely — this carbon is tertiary which usually already implies no $\mathrm{S_N2}$ — but also the capturing of the carbocation under $\mathrm{S_N1}$ conditions is more likely to happen at the sterically much less hindered carbon atom.

As to why methanol predominantly attacks according to an $\mathrm{S_N1}$-type mechanism while methanolate predominantly follows an $\mathrm{S_N2}$-type mechanism I will refer you back to your textbook; the reasons are stated pretty often.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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