As explained by my instructor,

Allylic Halides form highly stable Transition States. They are stable because they are:

  • $sp^2$ hybridized
  • possess a conjugated system of pi bond and empty p orbital

However, if we consider the $S_N1$ mechanism, a carbocation intermediate is stabilized by resonance. Why is the formation of the transition state more favorable than the formation of the resonance stabilized intermediate?


1 Answer 1


You and your instructor are both correct, it is your premise that's wrong.

Your instructor's explanation about the allyl double bond stabilizing the $\mathrm{S_{N}2}$ transition state is correct. The following figure shows the stabilizing interaction.

enter image description here

image source

If we remove the attacking nucleophile and the leaving group and place a positive charge in the system, then we have the $\mathrm{S_{N}1}$ allyl intermediate. As you point out, this intermediate is resonance stabilized, and therefore, the transition state leading to it is also stabilized.

Your premise,

Why do Allylic Halides prefer SN2 reaction over Sn1?

is not true. There are many cases where allylic halides react preferentially by an $\mathrm{S_{N}1}$ process.

Since both the allylic $\mathrm{S_{N}1}$ and $\mathrm{S_{N}2}$ reactions are stabilized, there is a delicate balance between the two pathways. We can shift from one mechanism to the other by changing reaction conditions.

For example, allyl chloride reacts by an $\mathrm{S_{N}1}$ mechanism to produce allyl alcohol when we place it in a 50:50 mix of $\ce{H2O:EtOH}$ at 45°C. If instead we run the reaction in the presence of ethoxide ion in ethanol at 45°C, the corresponding ethyl ether is formed by an $\mathrm{S_{N}2}$ process.

  • $\begingroup$ Can you provide the condition when SN2 dominates SN1 among allyl halides? $\endgroup$
    – user14857
    Sep 18, 2016 at 6:55
  • $\begingroup$ @ZOZ See the last paragraph of the answer for an example. $\endgroup$
    – ron
    Sep 18, 2016 at 14:03
  • $\begingroup$ im asking for condition...not example $\endgroup$
    – user14857
    Sep 18, 2016 at 14:06
  • $\begingroup$ solvent and temperature (the conditions) are included. $\endgroup$
    – ron
    Sep 18, 2016 at 14:08
  • 4
    $\begingroup$ It's hard to say, but generally, lower temperatures, less polar solvent and stronger nucleophile will favor SN2 $\endgroup$
    – ron
    Sep 18, 2016 at 14:15

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