Actually, where's the O-H bond in an aldehyde?

to ALL: sorry，I misstyped, it is the C-H bond

to Martin: I read just the opposite that the critical point is closer to  the less electronegative atom. hence, the O-H bond would be stronger? But never mind, I mistook the C-H bond in aldehyde with the O-H bond --|||

Consider the presence of electron withdrawing groups, EWG, (and their impact on pKa trends: methane, 55. alkene, 44, acetone, 20, acetic acid,5). An EWG induces charge delocalization. Try the particle in a box analogy, the box is widened-electrons delocalize in larger space, the frequency lowered, therefore it is a lower energy state relative to the system that does not have the EWG. In your case, you have H2C=CH2 versus O=CH2.  In the alkene, CH2 is inductively donating, the oxygen in formaldehyde is inductively withdrawing. The presence of the EWG (oxygen) stabilizes the corresponding anion/base. Therefore, it is enthalpically more favorable to remove the H from formaldehyde rather than ethene.

the radical R-C(•)=O: is stabilized by 3-electron interaction with the oxygen lone pair.

Sason

Thanks Dale for the very detailed illustrations^-^~

Thanks Sason, I have read your paper before, and it is very lucky to see you in the answer region !

You are welcome Rina.

Sason

It is simple.  The influence of  the electonegativity due to he CO bond that will stabilize any negative intermediate.

You can find a detailed answer here. The 2c,3e-bond in the radical product of C-H bond scission is quite strong:

Article Stereoelectronic power of oxygen in control of chemical reac...

Glad to read your answer, Igor - so we seem to agree on the stabilizing factor. I just asked for a copy of you paper.

Sason

interesting question, I would like to review the paper mentioned by Igor on the 2c,3e-bond on more details, although I think I agree it is the best explanation. Also, would it be consistent with the fact that acylium ion are particularly common and stable in MS (despite ion/radical differenc) ? Therefore, this is a simply a “fact”, clearly a favorable pathway in fragment ion.

Sure, one fact leads to an explanation but generate many more new puzzles. Chemistry advances...

Sason

Sason, yes, it is true that chemistry is based on facts and observations. Stable acylium ions and weak C-H bonds in aldehyde are those facts or observations, but what differentiate us from people of 100 years ago is our understanding (called chemistry), which we can rely on to reconcile different observations and even make predictions. sorry I didn’t read the whole threads carefully and I agree with you ;)

Thanks Yunlong,

Agreeing with one another is pleasing..

Sason

Interest Ask.

The short answer is this: the multiplicity of the C-H bond in aldehydes is less than one. To be precise, it is equal to 0.861. This can be shown rigorously and well explained.

There is an equation that uniquely describes the dependence of the energy and the multiplicity of the bond on the length of the chemical bond. Here's the equation:

E = a + b/r + c/r^2

Using the C-H bond data for ethylene, acetylene and ethane, the coefficients can be calculated. Thus, for the C-H bond, we get the equation:

E = a + b/r + c/r^2

a = -112104.88181312

b = 239481.07645903

c = -127352.04548799

If the bond lengths are taken in angstroms, then the bond energy is obtained in KJ/mol.

The values ​​are taken from the work: What are bond orders? http://dx.doi.org/10.13140/RG.2.2.28092.46726

Further, given that the energy of the C-H bond in the aldehyde group is 364.252 KJ/mol, that is, 87 Kcal/mol, from the corresponding equation we obtain the possible length of the C-H bond. Since this is a quadratic equation, we get two values:

L1 = 1.137787 Å.

L2 = 0.998437 Å.

Considering that the energy of the C-H bond in the aldehyde group is lower than in the C-H bond with a multiplicity of 1 (in ethane), we predict the length of the C-H bond in the aldehyde group to be 1.137787 Å.

CH3 – CH2 – H, Lc-н = 1.0914 Å, Ec-н = 405.848 KJ/mol.

H2C = CH − H, Lc-н = 1.085 Å, Ec-н = 435.136 KJ/mol.

HC≡C − H, Lc-н = 1.059 Å, Ec-н = 476.976 KJ/mol.

That is, the C-H bond in the aldehyde group should have an energy of 364.252 KJ/mol and a length of 1.138 Å.

It is easy to get a fold of 0.861 from here (see http://dx.doi.org/10.13140/RG.2.2.28092.46726).

The multiplicity of the C-H bond in ethane is taken to be 1.

Then for the aldehyde group:

F = E / L = 364.252 / 1.137787 = 320.14076 KJ/(mol * Å)

Taking into account the multiplicity of the C-H bond in ethane equal to 1 (F = 371.860 KJ / (mol * Å)), we obtain the multiplicity 0.861.

f = 320.14076 / 371.860 = 0.861

The decrease in the multiplicity of bond is easy to explain. Since in the aldehyde group, as in the amide group, there is a three-electron bond between oxygen and carbon (see the structural formula of urea), a part of the electron density of the C-H bond is drawn into the C-O bond. Hence, as a consequence, a decrease in the bond multiplicity, which means a decrease in the bond energy and an increase in the bond length. This is similar to how the basicity of the amine group decreases in amides.

This is indeed a beautiful proof.

Nevertheless, I prefer the simpler answer Igor and I gave, It is intuitively clear: you stabilize the radical and you lower the De.

The C-H bond in the alhedyde group is longer since the same interaction with the lone pair exists in the ground state but now applying Pauli repulsion between the lone-pair on O and the C-H bond.

I also like the proof through radical stabilization. This is a classic of chemistry, as it takes into account the stabilizing factors that are intuitive for chemists.

But, there is one caveat: theoretically, bond breaking can be heterolytic and homolytic, and this strictly depends on the conditions and mechanism of the reactions.

Therefore, the analysis by the structural formula, it seems to me, is more preferable, since we can immediately explain a lot by analyzing only the formula (in fact, X-ray structural analysis). For all mechanisms :).

I think that further development of the theory of chemical bond, for example, the development of electronegativity for organic chemistry similar to inorganics, would clarify a lot depending: properties - structure (biological activity, physical properties, etc.).

I was talking only about the C-H bond dissociation energy. This is by definition a homolyti bond breaking and not heterolytic.

Of course. You even have a radical depicted in your answer :).

However, the stability of radicals, carb-cations and carb-anions is explained on the basis of the same effects of electron density shift and conjugation (inductive and mesomeric effects, different types of superconjugation, etc.). Therefore, I summarized.