Question: 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:

[![enter image description here](https://i.stack.imgur.com/QBYmn.png)](https://i.stack.imgur.com/QBYmn.png)

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**:

[![enter image description here](https://i.stack.imgur.com/m2jKT.png)](https://i.stack.imgur.com/m2jKT.png)

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?

Answer: 
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.
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](https://pubs.acs.org/doi/pdf/10.1021/jo01097a026) , 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](https://pubs.acs.org/doi/pdf/10.1021/ci060129d), $\sigma^\*$, 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.