# Why are tertiary carbocations so stable?

Today our lecturer provided this data to illustrate the relative stability of carbocations.

\begin{array}{|c|c|c|c|c|} \hline \text{Substrate} & \ce{t-BuBr} & \ce{i-PrBr} & \ce{EtBr} & \text{MeBr}\\ \hline \mathrm{Relative~rate~of~S_N1~reaction} & 1.2 \times 10^6 & 12 & 1 & 1\\ \hline \end{array}

Additionally, in several places I have seen it said that tertiary carbocations are common and stable whereas secondary carbocations are only rarely stable and primary obviously not at all.

What is immediately obvious is that going from a primary to a secondary carbocation only produces a slight increase in rate, but the change from secondary to tertiary produces an enormous change. The number of bonds available to participate in hyperconjugation increases linearly along the series but the rate clearly doesn't so there must be some other factor at work. The only idea I had was that there might be some similar effect to Y-aromaticity taking place in the trigonal tertiary cation (there are six electrons involved in hyperconjugation after all) but I have no idea whether this is a plausible suggestion.

• This might help:-chemistry.stackexchange.com/questions/30893/… – Nilay Ghosh Jan 19 '16 at 12:11
• @NilayGhosh The way the passage was worded makes it sound like this is not a case of the alternative reaction ($\mathrm{S_N2}$ being more favourable. Also, I have heard before that secondary cations are not really stable. – bon Jan 19 '16 at 12:23
• @ShankRam I think part of your comment is missing. – bon Feb 3 '16 at 17:22
• This is just my theory. Please go through it completely and reply your view. Consider a typical $SN^1$ reaction. We have 3 intermediates: primary, sec. and tert. carbocation. Let us say the activation energy of the reactions are E1,E2 and E3. Now, we know that E3<E2<E1, which follows the order of stability of intermediates. From arrhenius equation, we know that rate constant is an exponential function of activation energy. So, even if the tertiary hydrocarbon we have taken has a few hundred kcal less activation energy than primary, the increase in rate constant would be huge. – ShankRam Feb 3 '16 at 17:23
• @bon, what do you think about the above comment? – ShankRam Feb 7 '16 at 7:25

What is immediately obvious is that going from a primary to a secondary cation only produces a slight increase in rate, but the change from secondary to tertiary produces an enormous change.

Uh-Oh!

When we see a table reporting that the solvolysis rate for an ethyl and methyl compound are both equal (both 1 in your table), well that's a tip-off that something is grossly wrong. No one would expect the methyl and ethyl carbocations to have anywhere near similar stabilities. Therefore we wouldn't expect the appropriate ethyl and methyl compounds to solvolyze (if they could) at anywhere near similar rates.

So if we accept that your table has bad data in it, then we really don't know that the change in going from primary to secondary is any different from the change on going from secondary to tertiary.

Carbocation Stability - Background

There are many factors that can affect carbocation stability (inductive effects, resonance effects, neighboring groups), but if we limit our discussion to the stability of saturated hydrocarbon carbocations then we need only consider two of these - inductive and resonance effects.

The carbocation center is roughly $sp^2$ hybridized with an empty p orbital; the alkyl substituents attached to this carbon are roughly $sp^3$ hybridized. There is a flow of electrons through the sigma bonds from the $sp^3$ carbon(s) to the more electronegative $sp^2$ carbon. We call this an inductive effect and it will stabilize the carbocation center.

Carbocation stabilization can also result from the flow of electrons through p orbitals. We call this a resonance effect, and in the particular case at hand it is a hyperconjugative resonance effect, as pictured in the diagram below.

(For more on hyperconjugation, particularly from an MO perspective, see the "Rotational barrier of ethane" section in this Wikipedia article.)

Prediction

The t-butyl carbocation has 3 methyl groups that can stabilize the carbocation center through inductive effects, there are also 9 beta hydrogens that can stabilize the carbocation center through hyperconjugation.

The 2-propyl carbocation has 2 methyl groups that can stabilize the carbocation center through inductive effects, there are also 6 beta hydrogens that can stabilize the carbocation center through hyperconjugation.

The ethyl carbocation has 1 methyl group that can stabilize the carbocation center through inductive effects, there are also 3 beta hydrogens that can stabilize the carbocation center through hyperconjugation.

The methyl carbocation has no inductive or hyperconjugative stabilization.

Consequently we might expect the relative carbocation stability to be

t-butyl > 2-propyl > ethyl > methyl

with the 2-propyl carbocation roughly midway between the t-butyl and ethyl carbocations.

Data

Chemistry is complex enough without adding extra unnecessary variables - like solvent. A relatively straight-forward method to compare carbocation stabilities is to measure the free energy ($\Delta G^o$) of the following equilibrium in the gas phase. The more negative the $\Delta G^o$ observed, the less stable the carbocation.

$$\ce{R^+ + H^- <=> R-H}$$

The following table shows the the results of this investigation and appears to confirm the above predictions on carbocation stability.

\begin{array}{|c|c|} \hline \text{carbocation} & \text{Delta} {~G^o} \\ & \text{(kcal/m)} \\ \hline \text{methyl} & -314 \\ \hline \text{ethyl} & -270 \\ \hline \text{2-propyl} & -252 \\ \hline \text{t-butyl} & -237 \\ \hline \end{array}

(data taken from Advanced Organic Chemistry, Part A: Structure and Mechanisms; fifth edition; Authors: Carey, Francis A., Sundberg, Richard J.; table 3.10, page 303)

The differences between the [t-butyl and 2-propyl] and the [2-propyl and ethyl] carbocations is 15 and 18 kcal/m respectively. That the 2-propyl cation is roughly in the middle of the t-butyl and ethyl carbocations is likely fortuitous since the ethyl carbocation is structurally quite different from the other two alkylated carbocations. The ethyl cation is non-classical in nature, involving a 3-center 2-electron bond with a bridging hydrogen as explained in this earlier answer.