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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
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up vote 3 down vote accepted

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
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@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
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