Why are tertiary carbocations the most reactive, if they're already stable?

This may seem silly, but doesn't it seem weird for a compound that's stable (in this context, the tertiary carbocation) to be the most reactive?

I mean, wouldn't it be the least, given that it's already stable and wouldn't want to leave that stable configuration. On the same lines, the methyl carbocation sounds like the most reactive due to its high instability, right? wouldn't it want to react in order to form a stable compound?

• A nice analogy I read once in a book." Dolomites (beautiful group in c the Alps) are very stable (indeed they are there since millions years) but very reactive. If you do not believe the latter go there with some HCl to see yourself. " – Alchimista Jan 7 '18 at 19:07
• Related (?): Longer the unsaturation, higher the stability and higher the reactivity. – Eashaan Godbole Sep 7 '19 at 12:40

This is an excellent question. Please correct me if I'm wrong, but I think this is what you are grasping at:

First, it is true that tertiary carbocations are generally more stable than primary carbocations (and secondary carbocations) due to having more inductively donating alkyl groups. The hyperconjugative effect can also be invoked to explain the relative stabilities of primary, secondary, and tertiary carbocations.

Second, transition states involving tertiary carbocations as opposed to primary carbocations are more favorable precisely because tertiary carbocations are more stable than primary carbocations.

Note that this does not mean that tertiary carbocations are more reactive. We don't generally say that something which exists in a transition state is reactive/unreactive. One reason is that the transition state complex cannot be directly captured or observed. Another reason is that it is the transition state ... of course it's going to be reactive - of course it'll change very rapidly!

We may however say that reactants and products are either reactive or unreactive. So if something goes through a transition state involving a tertiary carbocation, it might well be more reactive than something which goes through a transition state involving a primary carbocation.

Of course, before labeling anything as reactive or unreactive, you'd do well to note exactly what you mean by reactivity. To put it lightly, there are myriad reactions. An alkyne such as acetylene is reactive with regard to combustion; acetylene torches are commonplace. However, acetylene isn't going to be reactive with regard to, say, backside attack (for a variety of reasons).

Ultimately I think your issue lies with semantics rather than the chemistry ... still, a good question!

• Couldn't a certain carbocation exist on its own or will it only occur as a transition state? – awalllllll Jul 22 '20 at 21:28

Very interesting question. This also popped up in my head once. And this is how I convinced myself.

The confusing term here is stability. Tertiary carbocations are stable by inductive effect and hyper conjugation, and therefore have the tendency to sustain the positive charge on the carbon atom and stay like this for long. That is why we call it stable: it can stay that way longer.

Kinetic theory illustrates that the reactants present in the container randomly move in a container without any energy loss. Now imagine that there is a carbocation that is not so stable (take n-butane carbocation as an example) and now if a nucleophile would want to attack the cation, how could it attack it if it is not able to hold up by itself? The cation must hold on to itself in order to let the nucleophile come to it and then attack it. Tertiary carbocations have the stability to hold up like that and allow the nucleophile to attack it. Since it can hold up, that is why the reactions involving it favors $S_N1$ mechanisms, whereas a not so stable carbocation (like n-butane) follows $S_N2$, since there is "not enough time" to make a carbocation and proceed the reaction. $S_N2$ is a fast process compared to $S_N1$, since 1$^O$ have to react fast in order to form the product. This could also be supported by the fact that if we want to arrange carbocations in the order of reactivity by $S_N1$ and $S_N2$ mechanisms, they are usually the opposite.

To summarise, it was stable enough to hold on to its configuration and let the attack happen.

You should distinguish between stability — the thermodynamic concept with the adjectives stable and unstable — and reactivity — the kinetic concept with the adjectives reactive and inert.

Also, you should remember that thermodynamic stability is a meaningless concept unless you compare it to something else. Likewise, reactivity is only meaningful if it is compared. Unlike the concept of e.g. orbital energies which have a defined zero-point, there is no good zero-point for either reactivity or stability.

Now what does that mean for tertiary carbocations?

Any carbocation is a reactive species with respect to neutral carbon atoms. So the propan-2-yl cation is more reactive than propan-2-ol and the 2-methylpropan-2-yl cation (tert-butyl cation) is more reactive than 2-methylpropan-2-ol (tert-butanol). I would not want to classify the carbocations according to reactivity within their group. With the exception of very stable cations such as the trityl cation, most carbocations react with a diffusion-limited rate, so determining kinetic parameters is hard.

Within the carbocations, a tertiary carbocation is more stable than a secondary one which in turn is more stable than a primary one. So the tert-butyl cation is more stable than the propan-2-yl one — but remember that both are still very reactive.

However, never forget that with the exception of extreme cases (e.g. trityl), all carbocations are also less stable than any corresponding neutral compounds. In fact, the ‘stability’ of tertiary carbocations is nothing more than a very, very relativised measure of the low stability of different reactive intermediates.

• Why stability must be relative unlike orbital energies? We could say that compound A is more stable than compound B if Gibbs free energy of A is less than that of B. This makes perfect sense to me. What is wrong with that? – ado sar Dec 20 '20 at 13:46
• @adosar You just described a relative stability measure. Gibbs free energy relies on enthalpy and enthalpy has no defined zero-point, so Gibbs free energy also has no clear zero-point. (For the most stable form of an element, this is defined as having zero enthalpy but that leads to us working with negative and positive enthalpies of compounds with an arbitrarily chosen middle point.) – Jan Dec 24 '20 at 15:12

There is misconception related to stability. The actual fact should be whether the stability is before reaching the transition state or at the transition state or after reaching the transition state. If the stability occurs before reaching the transition state, then the reaction may not proceed at high rates. In the other two instances, the reaction may be facilitated by the stability of the intermediates of the reactions towards product formation. Imagine you are climbing a hill. If you have more load at the time of stating (due to stability you are pulled down), you may not reach the peak easily. on the other hand if some additional load is given to you after you reach the top, you are more likely to reach the other side quickly (due to stability you are pulled down). So stability disfavors reactivity, if it is before reaching the transition state and it favors reactivity if it is at or after the transition state.
Of course activation energy (rate/kinetics) and free energy (stability/thermodynamics) are two different terms. So comparing kinetics to thermodynamics is always not straight forward.

Good question.... Tertiary carbonation is more reactive than secondary carbonation which is in turn more reactive than primary carbocation due to stability... Primary carbocation is reactive and unstable so convert again back to a neutral reactant molecule instead of converting into products but tertiary carbocation is stable and there are more chances to form product by reacting with a nucleophile..

• Please elaborate. – A.K. Sep 2 '18 at 18:07