In the following substitution reaction, is compound I formed by $\mathrm{S_N}$2 and compound II formed by $\mathrm{S_N}$1 and carbocation rearrangement?

substitution reaction

Or is there some rearrangement I'm not aware of?

  • 1
    $\begingroup$ I would say yes. Your reasoning is OK. $\endgroup$
    – Maurice
    Commented May 17, 2021 at 10:04
  • 4
    $\begingroup$ $\mathrm{S_N2'}$ is also a possibility $\endgroup$
    – S R Maiti
    Commented May 17, 2021 at 11:16

3 Answers 3


Is compound I formed by $\mathrm{S_N}$2 and compound II formed via $\mathrm{S_N}$1 after a carbocation rearrangement? Or is there some rearrangement I'm not aware of?

For the first question, the answer is probably not because the mechanism of a given reaction is mainly based on the condition used in the reaction (at least at the undergraduate level of teaching). The the answer for the second question is yes, you are unaware of the mechanisms called $\mathrm{S_N}$1' and $\mathrm{S_N}$2' (as Shoubhik R Maiti mentioned in the comments section). This type of reactions are common when the allylic compounds are the substrate, which is appropriately called allylic rearrangement or allylic shift. Allylic shift is possible under either $\mathrm{S_N}$1 or $\mathrm{S_N}$2 conditions, and commonly known with $\mathrm{S_N}$1' and $\mathrm{S_N}$2' notations, respectively.

  • Strictly speaking, if you have used condition favoring $\mathrm{S_N}$1 reaction, the intermediate is a resonance stabilized allylic carbocation. Thus nucleophile would attack at two positions (either carbon 1 or 3) to give the resulting products, I and II. For instance, see your reaction in a protic solvent ($\mathrm{S_N}$1), which would given the same type of product mixture (for convenience, I have used the labelling to show the difference):

1-chloro-2-butene reaction with cyanide

Molecular modeling calculations would shows you the allylic secondary carbon position is more stable than the allylic primary carbon (recall your ab initio calculations for kinetic versus thermodynamic control in Org.Chem.II lab course; a simple and effective lecture demonstration of thermodynamic versus kinetic control, see: Ref.1) in intermediate carbocation. Accordingly, compound II would be the major product.

  • Similarly, if you are using the condition favoring $\mathrm{S_N}$2 reaction (strong nucleophile in a polar aprotic solvent) in this type of reaction, it is also possible to get two product due to $\mathrm{S_N}$2' mechanism. Usually, a bulky leaving group in $\mathrm{S_N}$2 conditions or bulky non-leaving substituent on the framework which give rise to significant steric hindrance would thereby increase the conjugate substitution ($\mathrm{S_N}$2'). Following scheme show the mechanism for two products:

3-chloropropene reaction with nucleophile

Late Edit:

Since it seems OP is still cannot see how compound II is formed, I included the arrow-pushing mechanism for $\mathrm{S_N}$2 versus $\mathrm{S_N}$2' products:

SN2 vs SN2'


  1. I. J. McNaught, "Thermodynamic versus kinetic control: A lecture demonstration," J. Chem. Educ. 1978, 55(11), 722 (https://doi.org/10.1021/ed055p722.1).
  • $\begingroup$ ahh I see. Yeah, I'm not aware of SN1' and SN2', I'm not from the US so maybe that wasn't in my course, please share some links ;-; ? I did think of allylic rearrangement, I went though the wiki page but I want to know more about it. I don't really know how compound II is formed. $\endgroup$
    – sam
    Commented May 18, 2021 at 13:13
  • $\begingroup$ I do not have access to the experiment at home, but it is based on molecular modeling using SPARTAN-8 program, which allows to do Ab initio calculations. Nice lab to show allyl shift in theoretical manner. $\endgroup$ Commented May 18, 2021 at 20:42
  • $\begingroup$ @sam: I did some editing to my answer to clear some issues you still have in this reaction. $\endgroup$ Commented May 18, 2021 at 21:22

Assuming an $\ce{S_N}1$ mechanism, there really is no rearrangement of atoms within the carbocation. When the chloride ion leaves, the resulting carbocation is allylic (which itself favors $\ce{S_N}1$), thus potentially electrophilic at both ends of a conjugated three-carbon chain. The mixture of products then corresponds to the cyanide ion adding at either end of the conjugated carbons.


You are right. You may also note that in polar protic solvent, such as water or alcohol, compound II would form the major product because of the stability of allylic carbocation, whereas in polar aprotic solvent compound I would form major product, because SN2 is preferred in such solvents due to its relatively poor ability to hydrate and stabilize ions.


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.