The SN1 solvolysis of one of the following alkenyl bromides is in MeOH approximately 100000 times faster than the other one. Which is the more reactive one? Draw the mechanism including the transition state, determine the rate determining step and draw the reaction product of each compound.

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Now what is apparently important here is the rotation of the substituent in each relative to the aromatic ring. In the second compound (with two methyl groups), the rotational profile shows destabilized molecule except when 90 degrees. That means the rotation in the second molecule helps with the leaving of Br and this (rate determining) step is much faster than when there are no methyl groups (first compound).

Is that a complete explanation or I am missing something important? Also is there anything special about the mechanism of the two reactions and the products in each case?


1 Answer 1


[...] SN1 solvolysis [...]

Let's recall what we know about these. We have a two-step reaction, consisting of

  1. heterolysis of the starting material
  2. association of a nucleophile $\ce{Nu}$

We remember that the heterolysis is not assisted by $\ce{Nu}$, there is no backside attack of $\ce{Nu}$, pushing out the leaving group. We also remember that the association step (addition of $\ce{Nu}$ to the carbocation intermediate) is usually pretty fast. Consequently, heterolysis, i.e. the formation of a carbocation is the rate-determining step.

Thank's to good old Arrhenius, we know that the rate is determined by the activation energy of this step.

The activation energy is lower when the resulting cation is more stable.

The cation is more stable when the delocalization of the charge is better.

Distribution of charge on the vinyl cation is better when it's actually a styryl cation.

styryl cation

What is the geometrical requirement for this interaction and what might prevent it?

  • $\begingroup$ The stabilisation is understood. But the geometry seems to be problematic as the cation is planar and the methyl groups are preventing the interaction. So what is the role of the rotation of this component in question that I referred to earlier? $\endgroup$
    – user40014
    Commented Feb 10, 2014 at 10:01
  • $\begingroup$ So, we agree that the stabilization is best when the double bond and the arene are coplanar? Now, what if this orientation is prevented by the methyl groups? Is the cation more or less stable then? What follows for the activation energy? Is full rotation really relevant? Or is it more important whether a full pi-overlap is possible? $\endgroup$ Commented Feb 10, 2014 at 10:26
  • 1
    $\begingroup$ If I got it right, a full pi-overlap is hindered by the methyl groups present in the second molecule. Then, stabilization most efficiently occurs for the first molecule and ${SN\_{1}}$ is then faster for the latter. Is that correct ? $\endgroup$
    – mannaia
    Commented Mar 1, 2014 at 10:10

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