Free Radical Reactions

By James Ashenhurst

Bond Dissociation Energies = Homolytic Cleavage

Last updated: December 6th, 2022 |

Here’s a point which causes a lot of confusion.

Look at these two reactions.  What do you think is the stronger bond, O-H or C-H?

what-is-stronger-bond-o-h-or-c-h-o-h-easier-to-break-with-base-but-c-h-has-smaller-bond-dissociation-energy

According to this this table (PDF) the bond dissociation energy (BDE) of OH is 460 kJ/mol (110 kcal/mol) and the value for CH is 389 kJ/mol (93 kcal/mol). [For another table, see this page from Reusch]. So why is the stronger bond being broken here?

Another example:

alkyne-deprotonation-is-easier-than-alkane-deprotonation-even-though-alkyne-c-h-bond-dissociation-energy-is-higher

But the bond strengths here are alkyne C-H (523 kJ/mol or 125 kcal/mol) versus the tertiary C-H bond strength in this case (384 kJ/mol or 93 kcal/mol). So  why is the C-H bond with the lower bond dissociation energy formed and  the higher C-H bond is broken?

Here’s three clues about bond dissociation energies.

  • For C-H bonds, bond dissociation energies decrease as you add substitution to the carbon.
  • Water can interfere with acid-base reactions, but water tends not to interfere with free radical reactions. If you’ve done a Grignard reaction in the lab, you know how finicky they can be, because you need to remove all traces of water from the solvent in order for it to start. On the other hand, the same restraint doesn’t apply to free radical reactions! It’s possible to run free-radical reactions in the presence of water without any concern that the desired free-radical reaction will be trapped by the H-OH instead.
  • Another clue is in that it is much easier to form alkyl radicals than alkenyl and alkynyl radicals.

The answer is that bond dissociation energy = homolytic cleavage

The measured bond dissociation energies (BDE’s) in tables represent the breaking apart of the bond into two radicals. This is because of the way bond dissociation energies are measured – through calorimetry of radical reactions.

Therefore the bond dissociation energy reflects the stability of the radicals formed! R3C• is a more stable radical than HO• . R3C• is also a more stable radical than an alkynyl radical. It also helps to explain why the order of bond strengths goes  primary C-H > secondary C-H > tertiary C-H.

bond-dissociation-energies-represent-homolytic-bond-cleavage-so-they-are-good-proxies-for-radical-stabiity-tertiary-radicals-the-most-stable.


Notes

Note 1.  Why are homolytic bond strengths measured and not heterolytic? That’s a good question. It’s much easier to break C-H and C-C bonds in alkanes homolytically, for one. Secondly, radicals are neutral and don’t carry around a solvent shell with them, like anions. So they’re less sensitive to solvent effects. For a technical discussion, look here.

Note 2. Why might the OH radical be less stable than R radicals, and stability of alkyl radicals be greater than alkenyl and alkynyl radicals?


(Advanced) References and Further Reading

  1. Shortcomings of Basing Radical Stabilization Energies on Bond Dissociation Energies of Alkyl Groups to Hydrogen
    Andreas A. Zavitsas, Donald W. Rogers, and Nikita Matsunaga
    The Journal of Organic Chemistry 2010, 75 (16), 5697-5700
    DOI:
    1021/jo101127m
    Several textbooks, including some advanced ones, provide radical stabilization energies, and this paper discusses why that may not be the best way to quantify the stability of free radicals.
  2. On the Advantages of Hydrocarbon Radical Stabilization Energies Based on R−H Bond Dissociation Energies
    Matthew D. Wodrich, W. Chad McKee, and Paul von Ragué Schleyer
    The Journal of Organic Chemistry 2011, 76 (8), 2439-2447
    DOI:
    1021/jo101661c
    This paper addresses some of the shortcomings with the approach used in Ref #1 above. The late Prof. Schleyer was a very influential figure in organic chemistry, and was a pioneer in using computational methods to address interesting problems in organic chemistry.
  3. The Radical Stabilization Energy of a Substituted Carbon-Centered Free Radical Depends on Both the Functionality of the Substituent and the Ordinality of the Radical
    Marvin L. Poutsma
    The Journal of Organic Chemistry 2011, 76 (1), 270-276
    DOI:
    1021/jo102097n
  4. A Single Universal Scale of Radical Stabilization Energies Does Not Exist: Global Bond Dissociation Energies and Radical Thermochemistries Are Described by Combining Two Universal Scales
    Andreas A. Zavitsas
    The Journal of Organic Chemistry 2008, 73 (22), 9022-9026
    DOI:
    1021/jo8018768
  5. Bond Dissociation Energies by Kinetic Methods
    A. Kerr
    Chemical Reviews 1966, 66 (5), 465-500
    DOI: 10.1021/cr60243a001
    This paper describes experimental techniques for measuring homolytic BDEs.
  6. III – Bond energies
    Sidney W. Benson
    Journal of Chemical Education 1965, 42 (9), 502
    DOI: 10.1021/ed042p502
    This paper describes the empirical measurement of homolytic bond dissociation energies. This paper was written by Prof. Benson while at the Stanford Research Institute (now SRI International), a non-profit research center very close to Stanford University. In 1978, Prof. Benson joined Prof. George Olah at USC and helped established the Loker Hydrocarbon Research Institute there.
  7. From equilibrium acidities to radical stabilization energies
    Frederick G. Bordwell and Xian Man Zhang
    Accounts of Chemical Research 1993, 26 (9), 510-517
    DOI: 10.1021/ar00033a009
    This paper attempts to correlate the acidity of a proton with the BDE of the corresponding C-H or X-H bond.
  8. Ab Initio Calculations of the Relative Resonance Stabilization Energies of Allyl and Benzyl Radicals
    David A. Hrovat and Weston Thatcher Borden
    The Journal of Physical Chemistry 1994, 98 (41), 10460-10464
    DOI:
    1021/j100092a014
    The stabilization energy of a vinyl group (in the allyl radical) and a phenyl group (in the benzyl radical) has been calculated to be 15.7 kcal/mol and 12.5 kcal/mol, respectively.
  9. Effects of adjacent acceptors and donors on the stabilities of carbon-centered radicals
    G. Bordwell, Xianman Zhang, and Mikhail S. Alnajjar
    Journal of the American Chemical Society 1992, 114 (20), 7623-7629
    DOI:
    10.1021/ja00046a003
    Table I in this paper contains stabilization energies of methyl radicals with various substituents (e.g. ·CH2X).
  10. Bond Dissociation Energies of Organic Molecules
    Stephen J. Blanksby and G. Barney Ellison
    Accounts of Chemical Research 2003 36 (4), 255-263
    DOI: 10.1021/ar020230d

Comments

Comment section

23 thoughts on “Bond Dissociation Energies = Homolytic Cleavage

  1. Thanks for your response and input. (Also, MOC is an excellent website!)

    Using BDE for C-H activation prediction (or for example to identify the weakest bond likely to be targetted by CYP450-mediated metabolism) makes perfect sense.

    My question was already lengthy, but I was asking because in an enzyme-catalysed reaction, I (believe) I was seeing a thermodynamically uphill reaction, ie converting an alkyl ester of the ligand to the weaker phenol ester of a tyrosine residue on the enzyme, which by all intuition (pKa of phenol vs alkyl alcohol, delocalisation of charge by phenol, etc) should not occur.

    Thinking along the lines of Hess’s law, I was hoping by using the BDE of all formed/broken bonds to get an idea of how unfavorable the delta enthalpy is (as enthalpy is path invariant), which could provide insights into the catalytic driving force. This is of course, all heterolytic cleavage and in water (easier to separate charges), but the correction factors for the homolytic BDE (Acc Chem Res, 2003) should be roughly the same for each bond-breaking/forming step.

    A further wrinkle is that some of the products/reactants are cyclic, which has made predicting BDE’s unusual (usually you calculate the energy of the neutral radical fragments, but homolytic cleavage of a ring gives you one product, not two).

    I was hoping to estimate the enthalpy of the reaction in the usual way (bonds formed minus broken) using BDEs, which for the acrylic analogues of the ester to phenyl ester (tyrosine) transesterification is +5.7 kcal/mol. There’s an entropy component (a small group is cleaved, ie 1 reactant -> 2 products), so I’m trying to get a rough idea of whether this transesterification is possible, perhaps a different residue is involved (serine/threonine), or perhaps I’m way off.

    If using BDE values to validate/invalidate my hypothesis is a fool’s errand, please let me know! I’m in a synthetic/biology focused lab, not really physical organic, so I can’t quite tell if there is any utility to estimating the thermodynamics of the reaction, especially through BDE estimations.

  2. BDE is the value of the homolytic cleavage, and I get that heterolytic reactivity can be far different than neutral radical mechanism (eg, it’s easier to separate charges in polar solution), but I’m not quite sure if I get how to apply that. Does that mean BDE–when considering heterolytic reactions, acid/base mechanism, etc–is limited, or useless, or outright misleading?

    For example, if I was considering a transesterification reaction in water between para-cresyl acetate (CAS 140-39-6) and a benzyl acetate (CAS 140-11-4) and I know the BDE for the phenyl ester and alkyl ester (and the BDE of phenol/akyl deprotonation), would the BDE give me an idea of the favorability of the reaction? Is it possible for the reaction to reverse in this example where I tried to keep all else equal? Can it reverse in general? I get that reactions in solution may be different than breaking the lowest BDE-bond, but if we know the BDE of phenol vs alkyl O-H, can we assume this is the same? I’m also interested to know–in non/polar a/protic solvents, or for example, in the varied nonpolar and polar microenvironments of an enzyme–can we use BDE to compare a transesterification like the above or get any insight from the BDE?

    I’m asking many questions ultimately to try to clarify my main question: is BDE comparison to polar/heterolytic reactions acceptable (ceteris paribus), limited, useless, or outright misleading? I did read the article (Acc Chem Res, 2003, p255) and while I get that it’s problematic/more work needs to be done, I don’t think it addresses my main question.

    1. Hi Grant – sorry for late reply.

      BDE’s are extremely useful when planning free-radical reactions. One example would be C-H oxidations of alkanes. One time we were planning a late-stage C-H oxidation on a carbon adjacent to an amide and in order to choose a proper oxidant it was helpful to compare its BDE with those of C-H bonds that underwent oxidation with various oxidants.

      For a transesterification reaction, the mechanism will be heterolytic. Under basic conditions, your best proxy for deciding which ester is easiest to cleave will be the pKa of the corresponding alcohol. The more acidic the alcohol, the less basic its conjugate base will be, which will give you a fair idea of its ability to stabilize negative charge and thus act as a leaving group.

      To give a simple exanple, it is much easier to cleave phenolic esters than it is to cleave the corresponding aliphatic ester. Even more so if the phenolic ester has attached electron-withdrawing groups that can help to stabilize negative charge.

      So to answer your question, I would say, no, the BDE of the O-H bond is not going to tell you very much.

  3. Thank you for your reply. The only solvent used is hexane, and the leaving group is supposed to be an “OH” radical that would result from the homolytic cleavage of the hydroperoxide moiety of the oxidized molecule. I am using a wide-range WL LED as a source of photons (I checked the purity of the compound prior to the start of the experiment, and it was above 99%). Under these conditions, is there a possibility that heterolytic cleavage would be more favored? or cleavage of the hydroperoxide moiety other than of the O-O bond (?) … from the analyses, there was no sign of hydroxyl or alkoxy radicals …

  4. Thank you for the useful explanation. I have a question, I have been working on a photoinduced reaction, and during this reaction, I found that Heterolytic cleavage occurs rather than homolytic cleavage, the only energy source is photons, and there are no other reagents or catalysts in the solution. So, when is heterolytic cleavage more favored than homolytic cleavage? (given that the first requires more energy than the latter, I couldn’t find an explanation for this …)

    1. It’s hard to give a good answer without seeing a picture of the reaction. What solvent are you using? Do you have a good leaving group present that might ionize?

      I had a friend who was doing a reaction in a photobox with near-UV radiation and the solvent was CH2Cl2. Weird things happened. Turns out the light was generating HCl from CH2Cl2 and it was causing all kinds of polar reactions. Fun times

  5. When atoms combine to form molecules, energy is released as covalent bonds form. The molecules of the products have lower enthalpy than the separate atoms.

    1. Boo! :-) kcal/mol all the way IMO. A C-H eclipsing interaction is about 0.9 kcal/mol and a C-H bond strength is about 100 kcal/mol. So easy to remember, like the Celsius scale.

  6. Hello,

    I just want to make sure I am understanding this article right…So the tertiary bonds of a organic molecule will have the lowest bond dissociation energies because they produce the most stable radicals (tertiary radical)? Thanks!

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