I recently asked a question about why the heat of formation of organic radicals and positive ions decreases with their size and degree of branching at the radical or ionic site. The user "Buttonwood" made the comment that hyperconjugation is one of the forms stabilising cations and radicals.

This begs two questions:

  1. What is the reasoning behind why hyperconjugation increases the stability of cations and radicals, and therefore decreases the heat of formation? Point 3. of this section of the Wikipedia article for hyperconjugation says that "the heat of formation of molecules with hyperconjugation are greater than sum of their bond energies and the heats of hydrogenation per double bond are less than the heat of hydrogenation of ethylene."; does this not imply that hyperconjugation increases the heat of formation of molecules?

  2. Why does an increase in stability imply a decrease in heat of formation? To a novice who is ignorant on this subject such as myself, it would intuitively seem that greater stability implies an increase in heat of formation, since the molecule is more "stable" in its current form and therefore it takes more energy to affect changes in its structure?

I would greatly appreciate it if people would please take the time to clarify these points.


1 Answer 1


Hyperconjugation is an electronic effect which is similar to resonance (one of the most fundamental concepts of organic chemistry). Here's how it works: $$\ce{H3C-CH2+ <-> H+H2C=CH2}$$

You can write the above structure three times for each hydrogen replaced. Note that only α-hydrogens (on the adjacent sp3 carbon atom) can be replaced.

  1. Now, answering your first question, hyperconjugation leads to stability because of the delocalisation of charge. The positive charge gets diffused away from the carbon atom, and starts to occupy a larger volume, decreasing the overall energy of the system caused by electronic repulsions. As a matter of fact, the greater the number of structures you can draw due to hyperconjugation, the greater the stabilisation. That explains why tertiary carbocations are much more stable than primary ones. $$\ce{(H3C)3C^+ <-> H+H2C=C(CH3)2}$$

    If you look closely there are nine replaceable hydrogens, so you can draw nine structures (explaining its stability).

    The Wikipedia page says:

    the heat of formation of molecules with hyperconjugation are greater than the sum of their bond energies

    Here, heat of formation refers to the heat released on the formation of these molecules. So it is negative.

    In the absence of hyperconjugation, you would expect that the heat released on formation is equal to the sum of all bond energies. But because of stabilisation due to hyperconjugation, the product molecule has lower energy than expected, so overall, more heat is released (higher magnitude enthalpy of formation).

  2. It's important to understand that enthalpy of formation (including its sign) decreases on increasing the stability of the molecule no matter what. But in the case that the enthalpy of formation is positive (energy absorbed), a decrease in the magnitude of heat of formation is observed, while if the enthalpy of formation is negative (energy released), an increase in magnitude is observed. In your other post, you include a list of carbocations, and their heats of formation are seen to be positive, while those of the non-ionic species are negative. So all the ionic species need to absorb heat in order to form (indicating a loss in stability), but that loss in stability (and hence the heat of formation) would be less if they were stabilised due to hyperconjugation.

  • $\begingroup$ [...] the greater the number of structures you can draw due to hyperconjugation, the greater the stabilisation. That is only true for very well-behaved molecules and as a zeroth order approximation. The contribution of different structures can be quite different and the overall superposition of all of them will only produce a good approximation. $\endgroup$ Apr 14, 2020 at 22:57

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