Which is the most stable free radical among the given species?
(1) $\ce{CH2=CH-CH2^.}$ (allyl radical)
(2) $\ce{C6H5-CH2^.}$ (benzyl radical)
(3) $\ce{(H3C)3C^.}$ (t-butyl radical)
(4) $\ce{C2H5^.}$ (ethyl radical)

The book claims the answer to be the allyl radical, option 1. But how can we definitely say that the allylic free radical is more stable, compared to the benzyl radical (option 2) which is also resonance-stabilised, or the tertiary radical (option 3) which has more hyperconjugative donation from α-hydrogen atoms?

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    $\begingroup$ I just want to say your book is correct. In simple words in this case benzyl free radical has resonance but the mesomeric effect is not transmitted to the free radical simply because it does not have a LONE PAIR of electrons. And besides Benzene has a -I effect thus further withdrawing the electrons and making the free radical unstable. (Coz obviously it needs electrons) $\endgroup$ Commented Jan 2, 2020 at 5:51

5 Answers 5


This is a tough question. I think it might even be unfair to ask such a question on a test in non-advanced classes. In advanced classes it could make an interesting topic of discussion, but I'm still not sure that the "real" answer is known.

What can be said is that due to resonance, both the allylic and benzylic radicals are more stable than the t-butyl or ethyl radicals which are not resonance stabilized. You can see from the following diagram that you can draw two resonance structures for the allyl radical.

Resonance structures of allyl radical

For the benzyl radical you can draw even more resonance structures.

Resonance structures of benzyl radicals

Superficially, this might suggest that the benzylic radical is more stable than the allyl radical. This was the "answer" and reasoning provided here.

However, if you do some thermochemical calculations (as done here), you arrive at the opposite conclusion that the allyl radical is roughly 2 kcal/mol more stable than the benzyl radical. The number of resonance structures alone is not a perfect indicator of stability.

For me, that is a pretty small difference in energy, and I think it would be fair to answer that the allyl and benzyl radicals have comparable stabilities.

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    $\begingroup$ Resonance makes a big difference in stability. $\endgroup$
    – ron
    Commented Feb 24, 2015 at 19:54
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    $\begingroup$ @Akashv No, you have read it incorrectly. The Link notes that "the allylic radical is stabilized by 12 kcal/mol relative to the nonallylic radical" and that "benzylic radicals are about 10 kcal/mol more stable than comparably substituted nonbenzylic radicals". Therefore, the allyl radical has approximately an extra 2 kcal/mol of stabilization compared to the benzylic radical. $\endgroup$
    – ron
    Commented Dec 19, 2017 at 18:32
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    $\begingroup$ @Shiva The order is allyl ~ benzyl > t-butyl. The question didn't address benzo or benzal (in any case, not sure what you mean by those terms). $\endgroup$
    – ron
    Commented Jul 5, 2022 at 15:16
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    $\begingroup$ @Shiva Confirmed, yes that order is correct. $\endgroup$
    – ron
    Commented Jul 7, 2022 at 13:04
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    $\begingroup$ @Shiva Your order looks correct except for the allyl~benzyl>*t*-butyl part. But remember that substituents can always cause exceptions. $\endgroup$
    – ron
    Commented Jul 10, 2022 at 18:05

Imagine your favorite activities are playing Video games (Most Favourite), and other ones are reading chemistry, spending time in a chemical lab. (these two are activities are those activities in between you can't choose one that you want to do, or say you like them equally.)

You dislike equally: reading history, reading civics, reading economics.

Let's assume two cases:

Case 1.

You have a big house. You have four rooms in your house containing these things:

  1. Video games
  2. History books
  3. Civics books
  4. Economics books

I don't think you are going to spend your so much time in rooms no. 2, 3 and 4. So The contribution of room 1 to your life is pretty high … you are going to live your life mostly living in room 1.

This the case of the benzylic radical. Room no. 1 is the canonical structure containing aromatic ring. In spite of having a big house you will not use it economically. Technically speaking, electrons have a big space to get delocalized but they are not delocalized efficiently.

Case No. 2

You have smaller house than in case 1, but you have two rooms containing these things:

  1. Chemistry books
  2. Chemical lab

So in this case you are going to spend your time in both of the rooms equally …

This is the case of the allylic radical. All rooms are equivalent, i.e. similar (or say identical), thus the resonance structures bring more stability to the system. You have a small house but you will use it economically. Technically speaking, electrons are delocalized efficiently.

That why both (benzylic and allylic) radicals have s similar stability order.

Note: The number of resonance structures alone doesn’t determine stability. Both quantity and quality are important.

In most case of chemistry, quantity wins.

Note: It's just an example for explaining the superposition principle in a simple manner, the analogy must not be taken too seriously or over-implicated.

Resonance is a static phenomenon, you can't say the electron is wandering from this carbon to that carbon (spending some time on that carbon or on this carbon). The only structure which exists is that of the resonance hybrid, which has a definite electronic distribution or precisely speaking has well defined time independent wavefunction.


Using the same concept as to distinguish whether the t-butyl carbocation is more stable than the benzyl carbocation, I calculated the isodesmic reactions of the form in $\eqref{isodesmic}$ at the DF-B97D3/def2-TZVPP level of theory. $$\ce{ R* + CH4 -> RH + *CH3 }\tag{1}\label{isodesmic}$$

I estimated the thermal correction at the standard conditions of $T=\pu{298.15 K}$ and $p=\pu{1 atm}$.

The results are within error compensation rate for the allyl and benzyl radicals. The level of theory might simply not be enough. If it is hard to distinguish the two with (more or less) elaborate calculations, then it clearly shows that this is a rather unfair question.

\begin{array}{llr} \ce{R*} & \ce{RH} & \Delta G / \pu{kJ mol-1}\\\hline \ce{*CH3} & \ce{CH4} & 0.0 \\ \ce{*CH2-CH3} & \ce{H3C-CH3} & -26.8 \\ \ce{*CH2-CH=CH2} & \ce{H3C-CH=CH2} & -75.7 \\ \ce{*CH2-C6H5} & \ce{H3C-C6H5} & -68.1 \\ \ce{*C(CH3)3} & \ce{HC(CH3)3} & -51.9 \\\hline \end{array}

Here are the resulting optimised geometries (click to enhance).

methyl methane
ethyl ethane
allyl propene
benzyl toluene
t-butyl isobutane

(I won't attach geometries or absolute energies this time, because that would exceed the character limit.)

  • $\begingroup$ Does "level of theory" mean the accuracy of the theory? Like higher level theory meaning more accurate calculations? $\endgroup$ Commented Feb 13, 2018 at 12:01
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    $\begingroup$ @ApoorvPotnis Not necessarily. It is more difficult, see here, in principle it means higher level more computational effort. Ideally this would mean less errors, but that's not always the case, since different approximations have different sources of errors. There are examples where a seemingly high level of theory fails completely: Transition metal chemistry is a graveyard for MP2 :D $\endgroup$ Commented Feb 13, 2018 at 12:12
  • $\begingroup$ @Martin-マーチン Please confirm that I have concluded the right order after reading your answer ( the first being most stable) : Benzo, Benzal, Allyl, Benzyl, t-Butyl. Please confirm. $\endgroup$ Commented Jul 4, 2022 at 13:06
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    $\begingroup$ @Shiva I cannot confirm or debunk, as the answer already clearly states, this is not possible. $\endgroup$ Commented Jul 5, 2022 at 20:09
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    $\begingroup$ @Shiva this answer already clearly states that with the data provided here this is not possible. I am also not intending to revisit this in a comment to this answer. $\endgroup$ Commented Jul 7, 2022 at 18:15

Indeed, this is a perplexing question. Stability compared to what? Are we considering their ease of formation from neutral molecule, i.e. the bond dissociation energy (BDE, or $DH^\circ$) of a carbon–hydrogen bond of propene, toluene and isobutane? [Ethane will be ignored.] Or are we comparing the three radicals from the same starting point, their constituent elements in the standard state, in which case we need to consider their heats of formation?

For the homolytic dissociation of a molecule $\ce{R–H}$ to two radicals $\ce{R^. + H^.}$, the sum of the heat of formation $\Delta_\mathrm f H^\circ$ of $\ce{RH}$ and the $\ce{R-H}$ bond dissociation energy (BDE) equals the sum of the heats of formation of the two radicals $\ce{R^.}$ and $\ce{H^.}$. This follows from Hess's law:

Reaction scheme

Gas-phase heats of formation are available from the NIST website, and the BDEs are available here. Comparison of $\ce{C-H}$ BDE's (in $\pu{kcal/mol}$) gives the order of stability as allyl ($+88.8$) < benzyl ($+89.7$) < t-butyl ($+96.5$).

If on the other hand the heat of formation of the three radicals is computed relative to the standard state, then the order is t-butyl ($+12.4$) < allyl ($+41.6$) < benzyl ($+49.7$). [The value of $\pu{+52 kcal/mol}$ for $\Delta_\mathrm f H^\circ(\ce{H^. (g)})$ is half the BDE of the hydrogen molecule.]

Energy diagrams for radical formation

$$\begin{align} & & \Delta_\mathrm f H^\circ (\ce{R^.}) &= \Delta_\mathrm f H^\circ (\ce{RH}) + \text{BDE}(\ce{R-H}) - \Delta_\mathrm f H^\circ (\ce{H^.}) \\ \ce{R} &= \text{allyl:} & \Delta_\mathrm f H^\circ (\ce{R^.}) &= \pu{4.8 kcal/mol} + \pu{88.8 kcal/mol} - \pu{52 kcal/mol} \\ & & &= \color{red}{\pu{+41.6 kcal/mol}} \\ \ce{R} &= \text{benzyl:} & \Delta_\mathrm f H^\circ (\ce{R^.}) &= \pu{12 kcal/mol} + \pu{89.7 kcal/mol} - \pu{52 kcal/mol} \\ & & &= \color{red}{\pu{+49.7 kcal/mol}} \\ \ce{R} &= \textit{t}\text{-butyl:} & \Delta_\mathrm f H^\circ (\ce{R^.}) &= \pu{-32.1 kcal/mol} + \pu{96.5 kcal/mol} - \pu{52 kcal/mol} \\ & & &= \color{red}{\pu{+12.4 kcal/mol}} \\ \end{align}$$

  • $\begingroup$ Stability, as defined by IUPAC Gold Book (goldbook.iupac.org/terms/view/S05900), should be measured in terms of the Gibbs free energies and not the enthalpy changes. The method which Martin used above is in agreement with the IUPAC definition while yours seems to be inconsistent. I am not trying to say that their method is flawed or incorrect. But perhaps, you may wish to explain how using absolute, quantitative measures of the heats of formation would serve as an acceptable and reasonable gauge of chemical stability? $\endgroup$ Commented Oct 25, 2019 at 15:55

The answer is benzylic radical.

  • It is the most stable because it resonates more compared to allylic and alkyl radicals.
  • Regardless of its radical class, be it primary, secondary or tertiary, benzenlic radicals will always be more stable than any other types of radical.
  • $\begingroup$ After all the evidence, which has already been presented here, this just restates a seemingly simple solution, which is likely to be wrong. And the choice of words is worse; it resonates more paints an absolutely wrong picture. The second point has been disproved on this very page. Apart from this, please support your claims with actual facts and evidence. $\endgroup$ Commented Feb 13, 2018 at 12:17

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