I am currently studying the textbook Mass Spectrometry, third edition, by Jürgen H. Gross. Chapter 2.4.3 Bond Dissociation Energies and Heats of Formation says the following:

The heat of formation of organic radicals and positive ions decreases with their size and even more importantly with their degree of branching at the radical or ionic site. A lower heat of formation is equivalent to a higher thermodynamic stability of the respective ion or radical. The corresponding trends are clearly expressed by the values given in Tables 2.2 and 2.3, This causes the fragmentation pathways of molecular ions proceeding by formation of secondary or tertiary radicals and/or ions to become dominant over those leading to smaller and/or primary radical and ionic fragments, respectively (Sect. 6.2).

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Why does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site?

I would greatly appreciate it if people would please take the time to explain this.

  • $\begingroup$ What about hyperconjugation comparing primary, secondary, tertiary radicals; or for the row of phenyl, phenoxyl, benzyl radicals? $\endgroup$
    – Buttonwood
    Commented Mar 14, 2020 at 21:49
  • $\begingroup$ @Buttonwood What are you referring to? I'm not sure how this relates to my question. $\endgroup$ Commented Mar 14, 2020 at 22:06
  • 1
    $\begingroup$ hyperconjugation is one of the forms stabilizing cations and radicals if a $\sigma$ orbital may interact with an adjacent empty or partially filled orbital, e.g., $\pi*$. (en.wikipedia.org/wiki/Hyperconjugation). One methyl group may not offer much stablization (two or three are increasingly better) which eventually lowers the enthalpy of formation for example of tertiary radicals over the one of primary ones. $\endgroup$
    – Buttonwood
    Commented Mar 14, 2020 at 22:17
  • $\begingroup$ @Buttonwood I have thought about your comments and posted a related question here chemistry.stackexchange.com/q/129076/75460 $\endgroup$ Commented Mar 15, 2020 at 14:39
  • $\begingroup$ There are essentially 2 ways to explain this, the modern and possibly more coherent way via „hyperconjugation“ as the other commentators mentioned, the other via a small electron donating effect of alkyl/aryle groups due to EN differences in C and H. QM I think it’s even wrong, but it still offers a nice model for many cases. $\endgroup$ Commented Jul 30, 2023 at 5:01

1 Answer 1


Here is a rather simple structure explanation presented on this site. To quote, where 'pieces' refer to an organic molecular with a non-symmetric branching structure:

You’ll notice something – the simpler the pieces are, the easier they are to stack together, which provides a tighter fit with fewer spaces. Here, by putting a kink in the block, we make them harder to stack.

What’s this got to do with chemistry?

When compounds freeze, the process is a lot like stacking bricks. The more symmetrical the molecules are, the easier it will be and the fewer spaces there will be between the molecules. Fewer spaces = better stacking. Hence, when you compare hexane to its structural isomer, 2-methylpentane, hexane has a much higher melting point due to the regular arrangement of its structure.

Better stacking, higher melting point. Case closed. Right? Not quite.

It’s also about surface area [...]

It looks like as we increase branching, we’re increasing melting point and decreasing boiling point. What’s going on? [...]

Going from “branched” to “highly branched” makes a molecule more compact and sphere-like. As the surface area of the molecule decreases (remember that spheres have the lowest surface area/volume ratio of any shape) they will become more compact and thus easier to pack. This explains the melting point phenomenon [...]

Boiling point is related to the forces between molecules, which in the case of hydrocarbons is Van Der Waals interactions. If you’ve ever seen microscope images of a gecko’s feet – which allow it to climb walls – you’ll see that there is no adhesive but the pads contain a tremendous amount of surface area. It’s all about the Van der Waals interactions.

As we decrease surface area, we are going to decrease intermolecular Van Der Waals interaction and therefore decrease boiling point.

The summary relationships are presented as follows:

linear versus branched —> higher melting/boiling points due to better stacking and surface area contact.

highly branched vs. branched —> more sphere-like –> better stacking –> higher melting point

highly branched vs. branched —>more sphere-like – -> lower surface area —> lower boiling point.

The above connects boiling point to branching structure, and this paper "Hypothetical Thermodynamic Properties: The Boiling and Critical Temperatures of Polyethylene and Polytetrafluoroethylene" found here links BP to thermodynamics. Per the opening reference statement: "A lower heat of formation is equivalent to a higher thermodynamic stability".

This completes the argument as to why the heat of formation decreases for organic radicals with decreased size and degrees of branching.


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