# What happens to a molecule when it is unstable?

I'm just starting OChem and I see that for a SN1 reaction, the rate of the reaction changes with the number of alkyl groups attached to the carbocation that is formed. The textbook (Klein) explains that this is due to the electron-donating nature of alkyl groups which helps to stabilize the carbocation. I was wondering, why doesn't instability help to increase the rate of reaction? Wouldn't the instability of a primary carbocation give it more energy to attract a negative substituent given that it is now more positively charged than a tertiary carbocation?

Thanks,

Reference: Klein, David. Organic Chemistry as a Second Language, First Semester.

The carbocation is the intermediate $\mathbf{I}$. The relative energy for alkyl-substituted carbocations will be lower. This also practically means that $\mathrm{TS1}$ will have a lower relative energy as well. The fact that $\mathrm{TS1}$ and $\mathbf{I}$ are of lower energy for a substituted carbocation (vs. a primary or methyl carbocations) means that it will be formed more often under the quasi-equilibrium assumption of transition state theory. Basically, if the barrier is lower, then then the intermediate can form more often. And thus the reaction can happen more often.
I think you are correct as far as your thought goes. If you could somehow instantaneously create a beaker full of primary carbocations, they might well react more quickly than other carbocations. (And that's not only via the SN1 reaction we are considering but also via other pathways such as elimination as well.) But instantaneously creating a beaker full of primary carbocations is way more difficult than creating a corresponding beaker full of secondary or tertiary carbocations. So (referring to the diagram again) yes, iff you have lots of $\mathbf{I}$, for $\mathbf{I}=$ primary carbocations, reactions would probably be quicker than the case where $\mathbf{I}=$ higher carbocations. But in reality in the primary carbocation case essentially no $\mathbf{I}$ is ever generated. So the overall reaction rate is slower.