Bond Dissociation Energies = Homolytic Cleavage

by James

in Chemical Bonds, Organic Chemistry 1, Understanding Electron Flow

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?

1-bde

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:

2-bde

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.

1) For C-H bonds, bond dissociation energies decrease as you add substitution to the carbon.

2) 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.

3) 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.

3-bde

PS 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.

PPS Why might the OH radical be less stable than R radicals, and stabiliity of alkyl radicals be greater than alkenyl and alkynyl radicals?

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{ 5 comments… read them below or add one }

Ayush Pateria

Just found a typo “stabiliity” in the PPS.

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Ashley

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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Chemboy

Really well presented!! (*Applause!)

Reply

JLS

You have a typo in the units of 3/4 of your C–H BDEs. They should all be kJ/mol. :-)

Reply

rahul

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.

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