20
$\begingroup$

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?

$\endgroup$
15
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Anyways, but what about the 3rd option ? Its a tertiary free radical with lot of +I effecy from 3 methyl groups and also 9 hyper conjugations. Why is it not totally in the picture ? $\endgroup$ – Dhruba Banerjee Feb 24 '15 at 19:52
  • 3
    $\begingroup$ Resonance makes a big difference in stability. $\endgroup$ – ron Feb 24 '15 at 19:54
  • $\begingroup$ @ron as link you gave on your ans of pdf '.' As done here' its written below measurements results that, benzyl radical are 42 kJ ( 12k cal) more stable than comparably substituted NON BENZYLIC RADICALS. And you have said You arrive at the opposite conclusion that the allyl radical is roughly 2 kcal/m more stable than the benzyl radical. It was comparison between benzylic and non benzylic radical if iam not wrong. Please correct if I have wrongly interpreted. $\endgroup$ – Akashv Dec 19 '17 at 12:38
  • $\begingroup$ @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 Dec 19 '17 at 18:32
9
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ Does "level of theory" mean the accuracy of the theory? Like higher level theory meaning more accurate calculations? $\endgroup$ – Apoorv Potnis Feb 13 '18 at 12:01
  • 2
    $\begingroup$ @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 $\endgroup$ – Martin - マーチン Feb 13 '18 at 12:12
4
$\begingroup$

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

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

$\endgroup$
-5
$\begingroup$

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.
$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Martin - マーチン Feb 13 '18 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.