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I've been taught that a carbocation mainly rearranges because of:

  • increasing degree (1 to 2, 1 to 3, or 2 to 3)
  • +M stabilization
  • ring expansion (in an exceptional case, ring contraction as well).

However, I've never been taught whether the following carbocation A would rearrange:

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My professor would say that this carbocation would not rearrange, since at the alpha position we only have two hydrogen atoms, and either of the hydride shifts would still yield a two degree carbocation B:

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However I believe that B is stabilized by the strong inductive effect of three extra methyl groups (at the alpha position) that were absent in A. I believe this is a strong enough driving force for rearrangement.

So, can we say A will rearrange into B? Is there experimental evidence to support the fact that A rearranges/does not rearrange into B? Finally, based on these, can inductive effect be the sole driving force for carbocation rearrangement?

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  • $\begingroup$ What's the source $\endgroup$
    – Avyansh Katiyar
    Commented Mar 13, 2018 at 11:35
  • $\begingroup$ @AvnishKabaj Source for what? I never claimed A definitely rearranges into B? $\endgroup$
    – Gaurang Tandon
    Commented Mar 13, 2018 at 12:08

2 Answers 2

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The answer to this question is quite difficult to deduce logically. In the first place, the carbocation that is formed is secondary. Rearranging will only change the position of the carbocation, but it will be still secondary. But then there is also a stabilising +I effect.

Instead, if we take into account that that these carbocations are intermediates to form alkenes, a conclusion can be drawn. So, the best way is to consider the thermodynamic favour of the rearrangements.

Prior to rearrangement, the carbocation is secondary. If you consider that this carbocation is going to form an alkene, the major product from it (Saytzeff product) will be 4-methyl pent-2-ene, which is of the form $\ce{RCH=CHR}$. After rearrangement, this will be still a secondary carbocation, so there is a significant energy required for this hydride shift. This energy is not really compensated by the +I effect of the isopropyl group. But if we consider that this carbocation is an intermediate for the alkene formation, the major alkene (Saytzeff product) will be 2-methyl pent-2-ene, which is of the form $\ce{R2C=CHR}$, which is thermodynamically more favourable.

So it is seen that in the rearrangement, the energy of the intermediate is increased, but the overall energy of the product is decreased. So, I think that, at higher temperatures, this rearrangement can be possible, because the ultimate product (if we think this as an alkene) is thermodynamically favourable, while at normal temperature this rearrangement is energetically restricted a bit.

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  • $\begingroup$ "the energy of the intermediate is increased" but isn't the carbocation B more stable? So, its energy should be lower? $\endgroup$
    – Gaurang Tandon
    Commented Mar 13, 2018 at 13:48
  • $\begingroup$ The energy should be higher because the strength of $+I$ effect in this case is not so strong as you are thinking. The Hydride shift requires much more energy, than $+I$ can stabilise the system. $\endgroup$
    – Soumik Das
    Commented Mar 13, 2018 at 14:10
  • $\begingroup$ I'm not going in the direction of whether the hydride shift requires higher energy or lower energy to occur. Instead, please think of it in this way: whether a carbocation has higher or lower energy is solely decided by its stability. Do you agree? if yes, then why do you say that "the energy of the intermediate is increased"? Do you disagree that the carbocation B is more stable? $\endgroup$
    – Gaurang Tandon
    Commented Mar 13, 2018 at 14:17
  • $\begingroup$ How do you define stability ? Do you have any other way to define stability rather than thermodynamic stability ? $\endgroup$
    – Soumik Das
    Commented Mar 13, 2018 at 14:23
  • $\begingroup$ I am defining stability in carbocation to be due to aromaticity, resonance, or inductive effects. The first two are missing in both A and B, so I am forced to check inductive effects. $\endgroup$
    – Gaurang Tandon
    Commented Mar 13, 2018 at 14:26
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I think that without any experimental data, it is no possible to decide. Personally, I don't think the inductive effect differences between the two ($\ce{CH3}$ and $\text{iBu}$ vs. $\text{Et}$ and $\text{iPr}$) and are large enough so that the energy gain from the stabilization of the carbocation is enough to overcome the energy barrier of the hydride shift.

From Hammett parameters , the $\ce{CH3}$ would actually be more electron donating than the $\ce{iPr}$ group, and since $\text{Pr}$ and $\text{Et}$ have the same $\sigma_\text{p}$ value, I wouldn't expect a difference between $\text{iBu}$ and $\text{Et}$ either. Of course, Hammett parameters aren't necessary a direct measurement of inductive effect, and these are not aromatic rings, but I think it paints a picture of how the inductive effect difference isn't that large.

Overall, I wouldn't say that carbocation B will never form from carbocation A, but it certainly will depend on conditions. My answer in an academic situation would be that it will not form, since the stabilization effect, if any, is minimal.

Edit: Upon more thinking, the Taft polar parameter, $\sigma^*$, might be a better representation of this case. These actually show a higher electron donating ability for both $\text{iPr}$ vs. $\text{iBu}$, and $\text{Et}$ vs. $\text{Me}$, respectively. These are also empirical values, and not necessarily the same as we would observe, but again, they can give an idea. The differences are so small that I would stand on arguing towards the side that the energy barrier for the shift is large enough that it would be minimal for stabilization by shifting to be favored. Again, it would definitely depend on conditions.

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  • $\begingroup$ Thanks for your interesting answer! :) Actually, I was basing my assumption that longer alkyl chains have a larger +I effect based on this question. Could you review it and see if it's something what you agree/disagree with? $\endgroup$
    – Gaurang Tandon
    Commented Mar 14, 2018 at 4:10

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