# Why does carbocation stability increase in the order 1°, 2°, 3°?

I'm reading about $\:\mathrm{S_N1}$ and $\:\mathrm{S_N2}$ reaction mechanisms. 1° carbocations are unstable to the point of not having been observed in solution, ever. 2° are more stable, and 3° carbocations are the most stable.

I know that higher stability carbocations do also require less activation energy in their formation. Why is that? Why exactly are they more stable? And is this another way of saying they are also less reactive? Does this in turn mean they require a better nucleophile compared to less stable carbocations?

A small side question. How should I think about resonance structures? You don't view it as changing between the forms, spending more time being the more stable form, but rather as a seamless blend of the forms, with more characteristics of the more stable form, right?

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It may be better if you post your question about resonance in a separate post. Short answer: yep, it's a blend. More specifically a quantum superposition, but no need to get into that. Basically, the electrons of all the resonating bonds are "shared" over the given area. The probability of finding an electron is higher in the places where stability is more, but that doesn't mean that the electron is zipping back and forth in the area. You need to understand a bit of quantum mechanics to understand this. – ManishEarth Feb 14 '13 at 13:12

One word: hyperconjugation! The more carbons bound to the carbocation, the more bonds can take part in hyperconjugation. This effect happens when a bond between two atoms donates electron density to the electron-deficient carbocation, making it more stable.

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Why doesn't this work with hydrogen, considering hydrogen is less electronegative already to begin with? – Brian Feb 14 '13 at 3:03
@Brian: It is not the carbons attached to the carbocation center that donates electron density via hyperconjugation but the bonds attached to it (free electron pairs would also do). If you consider the $\cẹ{(CH_3)_3C^{+}}$ cation the carbocation center contains an empty p orbital. This can interact with vicinal $\ce{C-H}$ $\sigma$ bonds, that happen to be parallel to it, thus lowering its energy. – Philipp Feb 15 '13 at 0:32

From what I have read, the increasing stability of carbocations is attributed to the combined inductive and hyperconjugative effects, at least in case of alkanes. In other organic molecules, the mesomeric effect can also come into play. These effects are easy to understand, so i suggest you read up on them.

Subsequently, your next question deals with why more stable carbocations have a lower activation energy in their formation. In my opinion, you're getting the cause-effect mixed up. The carbocation is stable BECAUSE the energy required to form it is lower, not the other way around. And why is the energy lower? Because of the aforementioned stabilization effects!

This is from the point of view of a 17 year old kid who's been studying organic chemistry for about 3 years, and faced the same problems as you're facing. I won't be able to explain it as ManishEarth did though, using quantum mechanics.

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