My task is to explain, why molecule A reacts in a SN1-reaction faster than molecule B.enter image description here (I would do this by comparing their potential energy diagrams)

My idea is to have a closer look at the reaction mechanism at first:enter image description here

enter image description here

TS1 is here the intermediate with the carbo-cation. That's the step, where the reaction is slow, where one needs a lot of energy to remove the Bromine. Maybe I guess one does not need this lot of energy in molecule A compared to B because the Bromine is in the background (trans- isomer) there and not in the foreground? Furthermore, in SN1 reactions the reaction rate is determined by the substrate (the first step). So, the activation energy of molecule A is smaller then the activation energy needed in B and that's the reason, why the reaction ist faster?


An interesting question is how (or if) one really knows if both reactions proceed via bonafide SN1 type mechanism. Under the assumption that both reactions proceed via bonafide SN1 mechanism, we must infer that the carbocation (carbenium ion) intermediate 1 in both cases must be the same. While the previous answer correctly assumes that this intermediate may be described well as planar (at the carbon centre), the attack by water is the faster step and hence should not affect the overall rate of the reaction. Since both A and B must proceed through the same intermediate, (although not rate determining!), both must react at the same rate with the incoming nucleophile.

A better explanation can be deduced by looking at the most preferred conformations in the two starting materials. These are designated 2 and 3 below, for A and B respectively. In both cases, the bulky tertiary butyl group will likely prefer to be sat at an equatorial position to minimise 1,3 diaxial interactions. Once tertiary butyl group is "locked" at the equatorial position, we have to place the bromine axially (up) in A (2), and equatorially (down) in B (3).It is possible to argue (although still very qualitative and hand-wavy) that 3 is likely more stable than 2, since both groups are equatorial in 3.

Now, the rate of a reaction depends on the activation free energy barrier, which is the difference between the free energy of the reactant and that of the transition state. At this stage, without computation it is impossible to predict which transition state will be at a higher or lower energy. However, speculatively, to a first approximation the energies of the transition states would be very similar. Under this assumption, the activation energy barrier for A (higher energy conformation) should be less than that for B, which should explain the observed reaction rates.

Having said that, I would not necessarily think that energy diagrams are the best way to explain this difference in reactivity, or indeed if there is a single, simple explanation.

Proposed plausible potential energy diagram

  • $\begingroup$ Note: I have not shown the subsequent transition states and the intermediates in my energy diagrams, since these will not be rate determining, and hence irrelevant to the question (about rates). $\endgroup$ – Zachr May 25 at 23:55
  • $\begingroup$ At this level of abstraction it probably is fair to speak of a rate determining step, as it is (given that it actually is SN1) probably an elementary step. In more complex reactions this approximation will break down. A particular example for this is a Diels-Alder reaction with buta-1,3-diene, where the trans conformation is lower in energy. Only the cis conformation is viable for the TS though. The rate will therefore be determined through the energy span between trans and the TS. In any case, your analysis is good and should hold for this level of abstraction. ↑ $\endgroup$ – Martin - マーチン Jun 25 at 9:03

I would rather not solve this by comparing potential energy values.

Instead there is another theoretical way of doing so.

When SN₁ happens, either the attacking reagent( In this case water) can attack from above or below the planar carbocation.

In the first case, Br⁻ leaves from above, and now water can easily attack from below.

In the second case, Br⁻ leaves from below, and water has to attack from above(to avoid steric repulsions between the exiting Br⁻ and attacking water), but now there are steric repulsions between C(Me)₃ group and the attacking water, leading to a reduced SN₁ rate as compared to the first case.

  • $\begingroup$ But if you would solve it by comparing potential energy values how would it look like then? Like I did above? $\endgroup$ – Chemistryisfunny May 22 at 19:20
  • $\begingroup$ You are just speculating about the energies required. You don't know it for sure unless an experiment is conducted. Therefore, a potential energy diagram cannot be plotted. $\endgroup$ – user226375 May 23 at 4:17
  • $\begingroup$ @user226375: Your explanation that considers from which side the nucleophile (water) attacks, would be more suited if OP described the reaction as SN2. See answer below for details. $\endgroup$ – Zachr May 25 at 23:58

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