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Secondary allylic halides can be produced in high yield with the following Mitsunobu-type reaction:[1]

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The yield being $94~\%$. The last step in the mechanism is:

enter image description here

Why is the $\mathrm{S_N2}$ reaction favored so much over the $\mathrm{S_N2'}$ reaction?

Reference: Magid, R. M.; Fruchey, O. S.; Johnson, W. L.; Allen, T. G. Hexachloroacetone/triphenylphosphine: a mild reagent for the regioselective and stereospecific production of allylic chlorides from the alcohols. J. Org. Chem. 1979, 44 (3), 359–363. DOI: 10.1021/jo01317a011.

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  • $\begingroup$ By "Sn2-dash" do you mean conjugate addition ? $\endgroup$
    – J. LS
    Feb 16, 2015 at 17:47
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    $\begingroup$ Yes, it's the standard notation, Sn2-prime. $\endgroup$
    – EJC
    Feb 16, 2015 at 18:46
  • $\begingroup$ Presumably because $\ce{Cl-}$ is a hard nucleophile and is driven by electrostatic interaction to the $\ce{C-O}$ bond? $\endgroup$
    – Jori
    Feb 16, 2015 at 21:50
  • $\begingroup$ @Jori the chloride ion is a "hard" nucleophile? $\endgroup$
    – Dissenter
    Feb 16, 2015 at 23:20
  • $\begingroup$ But chloride doesn't seem so hard to give almost quantitative yield. $\endgroup$
    – EJC
    Feb 18, 2015 at 12:57

2 Answers 2

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While pi-based resonance effects are frequently important, you must always consider inductive effects in every reaction.

Electrostatically, there is a much higher positive charge on the ipso carbon than the beta carbon, because of inductive effects. The net effect is that there is a lower energy of activation for nucleophile attack at the electrophilic center because the coloumbic forces are more favorable for attack there. The softer the nucleophile and electrophile become, that less important this effect is, but it is still quite considerable.

Without doing the QM calcs, I would rate the phosphonium electrophile as quite hard.

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  • $\begingroup$ could you perhaps expand this answer a bit? can you give a more detailed explanation of why this is the case $\endgroup$
    – bon
    Feb 19, 2015 at 17:20
  • $\begingroup$ I no longer have a computational package to run a quick calculation on an analogous molecule to draw a nice picture. However some degree more elaboration has been added. $\endgroup$
    – Lighthart
    Feb 19, 2015 at 17:22
  • $\begingroup$ The ipso carbon will have a slight positive partial charge, while the beta carbon will most likely have a negative charge. However, the orbital coefficient of the LUMO is larger on the beta carbon, therefore soft and/or large nucleophiles will most likely attack there. After all, chlorine is quite small and probably also not well solvated in these reactions. (I might add some calculations supporting that on Monday, if I have the time.) $\endgroup$
    – Martin - マーチン ♦
    Feb 20, 2015 at 12:39
  • $\begingroup$ Could it be that Coulomb interactions between the chloride and phosphonium put the chloride near the ipso reactive center? $\endgroup$
    – EJC
    Feb 20, 2015 at 14:49
  • $\begingroup$ @Marko With molecules tumbling in solution, I think the use of coulombic forces to explain colloision frequency by location on a molecule is a much weaker argument. However, I can't dismiss the possibility. $\endgroup$
    – Lighthart
    Feb 20, 2015 at 15:04
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The $\mathrm{S_N2'}$ pathway has a strict requirement (similar to $\mathrm{E2}$) for the correct alignment of the orbitals involved in the reaction, in this case, the alkene pi-system and the C-X anti-bonding orbital (in this case C-O anti-bonding orbital). That means that only particular conformation(s) of the electrophile will be reactive.

In the direct $\mathrm{S_N2}$ attack of the electrophile, while certain conformations will be more reactive than others, the C-X is always oriented correctly for reaction to occur.

I suspect that the product distribution reflects the competition between the steric hindrance of the direct $\mathrm{S_N2}$ attack and the conformational requirements of the $\mathrm{S_N2'}$. The reaction given in the question uses a relatively unhindered electrophile and a small nucleophile, which likely explains the preference for the $\mathrm{S_N2}$ product.

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    $\begingroup$ It would be splendid if you post an image for the SN2' conformation requirements. $\endgroup$
    – EJC
    Jun 22, 2016 at 16:33

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