[macman104]

Going to continue a discussion here, so as not to derail a thread, thoughts appreciated.  To catch up:

Orgopete:

I don't agree with a) either. While fluorine has been reported as the more electronegative, it definitely is not the most electron withdrawing, see Hammett sigma values for an overall perspective, though you could infer this from the lower acidity of HF or the basicity of F-. If HBr gives a 2,2-dibromide upon addition to 2-butyne, then surely HBr will add to the intermediate fluoride in the same manner.

Macman104:

I think it would be incorrect to infer electron withdrawing ability based on acidity of HF, as it experiences the hydrogen bonding phenomenon that the other acid halides do not.  I can accept the precedent for 2-butyne, but how would you then explain this outcome then from the perspective of forming the most stable carbocation?  I don't think you could argue that inductively fluorine stabilizes a carbocation better than an alkyl group.

Orgopete:

@macman, granted that carbon is a better inductive electron donor than fluorine, but fluorine is a better resonance donor and resonance effects generally are stronger.

If fluoride can hydrogen bond, that simply indicates its electrons are available to hydrogen bond. That would be consistent with an increasing bond strength from oxygen to nitrogen and carbon (carbanion) with hydrogen bonds becoming covalent bonds. Hydrogen bonding effects are much harder to find in the third row because the greater nuclear charge increases the electron withdrawing character of those atoms and an increasing HX acidity. Therefore, while fluorine can hydrogen bond, hydrogen bonding virtually disappears for Cl, Br, etc.

I offered acidity, hydrogen bonding, and Hammett sigma values that show the other halogens are more electron withdrawing than fluorine. (I personally think acidity is perhaps the best method to determine heteronuclear ionization values. It is well studied, widely applied, water is virtually a universal solvent for a very large number of acids, quite consistent within the ionization range to which it can be applied.) Perhaps you could give me an example that shows fluorine more electron withdrawing than bromine (though perhaps in another thread or message as such a discussion would highjack the original posters question).

[macman104]

@macman, granted that carbon is a better inductive electron donor than fluorine, but fluorine is a better resonance donor and resonance effects generally are stronger.

I will agree here, and likewise, the resonance is the rationale for the dibromo compound.  However, bromine (being larger, more polarizable, and less electronegative), is much more likely to partake in resonance, than fluorine.  It may still be the predominant outcome, and I don't have a precedent to refute it.  But a reactivity difference that I think is important considering.

If fluoride can hydrogen bond, that simply indicates its electrons are available to hydrogen bond. That would be consistent with an increasing bond strength from oxygen to nitrogen and carbon (carbanion) with hydrogen bonds becoming covalent bonds. Hydrogen bonding effects are much harder to find in the third row because the greater nuclear charge increases the electron withdrawing character of those atoms and an increasing HX acidity. Therefore, while fluorine can hydrogen bond, hydrogen bonding virtually disappears for Cl, Br, etc.

Sorry, I was imprecise in my wording.  What I meant was that using acidity and the fact that in solution it goes from HF < HCl < HBr < HI is not a good indicator of electronegativity.  As dilute HF solutions, fluoride (F^-) strongly interacts with the hydronium species to reduce the effective amount of acid present, increasing the concentration however, greatly increases the acidity of the solution nearly beyond HI.  It's strong electronegativity and small atomic size make it an exception.

I offered acidity, hydrogen bonding, and Hammett sigma values that show the other halogens are more electron withdrawing than fluorine. (I personally think acidity is perhaps the best method to determine heteronuclear ionization values. It is well studied, widely applied, water is virtually a universal solvent for a very large number of acids, quite consistent within the ionization range to which it can be applied.) Perhaps you could give me an example that shows fluorine more electron withdrawing than bromine (though perhaps in another thread or message as such a discussion would highjack the original posters question).

Consider pKa of the trihaloacetic acids.  Trifluoroacetic acid is more acidic than trichloroacetic acid.

[orgopete]

Re: "Acidity … goes from HF < HCl < HBr < HI …" Agreed.
"…increasing the concentration…greatly increases the acidity…" While true, that does not prove the point. I concede that F-H-F^- H⁺ is a stronger acid than HI. Although that makes FHF^- a weaker base than F^-, it does not make F^- a weaker base than I^- however.

Re: Because electronegativity is not an analysis of heterolytic bond strength, hence "acidity … is not a good indicator of electronegativity." This is also correct. Electronegativity is derived from homolytic bond strengths and thus does not predict heterolytic bond strengths. Electron withdrawing properties are generally referred to in discussing heterolytic bond strengths, such as acidity. 

I am arguing that I>Br>Cl>>F as electron withdrawing properties are concerned. That does match the acidity, hydrogen bonding, and Hammett sigma values.

Re: trifluoroacetic acid, I agree that it is a stronger acid. However, I would like to compare monosubstituted acetic acids, hence we have the following pka values:


If I am to give credence to the pKa values reflecting the electronegativity, then glycolic acid (OH) should be the next most acidic compound. In actuality, even iodoacetic acid is a stronger acid, yet a much weaker electronegativity value. I am not disputing the pKa values. Fluoroacetic acid is a stronger acid than iodoacetic acid, but weaker than nitroacetic acid. My dispute is whether the acidity is corresponds with electronegativity. I argue that it does not. Since it also does not match the common electron withdrawing properties, acetic acid requires a different explanation for its acidity (which I cannot provide).

If we compare trichloro and trifluoroethanol, I found the measured values to be 12.02 and 12.40, respectively. If we compare trihalomethanes, the proton exchange rate is virtually the same as the HX acidities. CF₃H has the slowest exchange rate. Of the trihalomethanols, trifluoromethanol is the only one with any reasonable stability. If you did a solvolysis reaction of t-butyl halides, using identical conditions, I believe the fluoride would solvolyze the slowest (though I cannot find such data). If you did an SN2 substitution reaction, fluorides are poor leaving groups. If you wanted an alternate base, fluoride is a reasonably strong base.

I argue that if electronegativity were not written in any textbooks, then the order of electron withdrawing halogens would be I>Br>Cl>>F.  Pauling's 1932 paper on electronegativity is not good science. Those who have cited it over and over does not improve it. Fluorine bonds are indeed strong, but that does not make it a strong electron withdrawing group. Everyone virtually knows this intuitively if they pick up a bottle of iodine, it is heavy. If electron density per proton number were constant, then the density of all atoms in the periodic table would be similar (and would vary with neutron number) because electrons make up the majority of an atoms volume. Consequently, we know the electrons of iodine take up much less volume to account for its density. I argue this is a simple nuclear charge effect which is consistent with the many properties of periodic table (except electronegativity). 

[orgopete]

… Fluorine bonds are indeed strong, but that does not make it a strong electron withdrawing group. Everyone virtually knows this intuitively if they pick up a bottle of iodine, it is heavy. If electron density per proton number were constant, then the density of all atoms in the periodic table would be similar (and would vary with neutron number) because electrons make up the majority of an atoms volume. Consequently, we know the electrons of iodine take up much less volume to account for its density. I argue this is a simple nuclear charge effect which is consistent with the many properties of periodic table (except electronegativity).  

Much to my chagrin, as I have been researching this topic, this analogy is not correct. While it is true that iodine is more dense than fluorine, it doesn't necessarily correlate from a greater atomic electron density. I have to be ambiguous here because it is difficult to know or use very precise values. If halogen radii are estimated from the HX bond lengths (calculated as spheres), the volume per electron or electron pair is similar. This suggests two facts. One, as could have been predicted from Coulomb's formula, electrons cannot be concentrated. Two, the greater density of iodine is due to increased intermolecular forces.

[My apology for bumping this topic, may it now rest in peace.]

[opsomath]

I am a latecomer to this board and therefore to this thread, but I couldn't resist commenting on this one. First, density of a material is an extremely poor proxy for the properties y'all are interested in. For instance, the electronic structures of iodomethane, iodoform, and iodine are all very different, but these are all extremely dense compounds. In fact, density is more or less correlated with increasing atomic number in a very obvious way, as van der Waals radii scale with atomic number in a steadily decreasing fashion (maybe log?) while nuclear mass obviously scales more or less linearly. (There are exceptions to this, obviously, especially in crystalline materials, but the trend is clear.)

The apparent contradictions in the cases y'all cite can be rather straightforwardly resolved by the consideration of pi-donating ability vs. (inductive) sigma-withdrawing ability. All the halogens are pi-donors and sigma-acceptors. Fluorine happens to be the one in which the nonbonding lone-pair orbitals are energetically best suited to overlap with orbitals on neighboring carbon atoms, so that there is a substantial pi-donation effect destabilizing a carbanion alpha to a C-F bond. The same effect serves to stabilize a carbocation, resulting in the regioselectivity of HX addition to an alkyne.

Another good example is the Lewis acidities of the BX3 compounds. BBr3 is the most acidic of the four, with BF3 actually being relatively mild due to the excellent overlap between lone pair sp3 orbitals on F and the empty p orbital on B. OTOH, BI3 is mild because the inductive electron-withdrawing effect of I is less than that of the others, although there is relatively little overlap between the lone pairs on iodine and the electron-deficient centers. (A similar argument, going the other way, explains the trihalomethane H exchange rates.) 

[orgopete]

… In fact, density is more or less correlated with increasing atomic number in a very obvious way, as van der Waals radii scale with atomic number in a steadily decreasing fashion (maybe log?) while nuclear mass obviously scales more or less linearly. (There are exceptions to this, obviously, especially in crystalline materials, but the trend is clear.)

That is what I thought too. That was the point in the earlier post. The only way for density to increase was for volume to decrease. Since electrons take up the majority of atomic volume, it seemed as though the electrons are more tightly packed. Replacing a CH3 with I should result in a more dense substance. I am presuming the van der Waal's radii are an effect of the electrons. If you then discovered the volume per electron of a CH3 group were the same as those of I, then you need to use a different reason.

The apparent contradictions in the cases y'all cite can be rather straightforwardly resolved by the consideration of pi-donating ability vs. (inductive) sigma-withdrawing ability. All the halogens are pi-donors and sigma-acceptors. Fluorine happens to be the one in which the nonbonding lone-pair orbitals are energetically best suited to overlap with orbitals on neighboring carbon atoms, so that there is a substantial pi-donation effect destabilizing a carbanion alpha to a C-F bond. The same effect serves to stabilize a carbocation, resulting in the regioselectivity of HX addition to an alkyne.

Another good example is the Lewis acidities of the BX3 compounds. BBr3 is the most acidic of the four, with BF3 actually being relatively mild due to the excellent overlap between lone pair sp3 orbitals on F and the empty p orbital on B. OTOH, BI3 is mild because the inductive electron-withdrawing effect of I is less than that of the others, although there is relatively little overlap between the lone pairs on iodine and the electron-deficient centers. (A similar argument, going the other way, explains the trihalomethane H exchange rates.) 

The challenge is not to give new or different reasons why fluorine is not as electron withdrawing as the electronegativity values indicate, it was to find bona fide examples proving that it is the most electron withdrawing element. I do appreciate the BI3 example though. I've never used it and never investigated its properties. I can imagine other reasons than those you have suggested. None the less, a good example.

Oh, by the way, I think I have finally figured out the electronegativity issue yesterday. If you want to figure it out too, here is a clue. You need to wrestle with the Pauling premise:
Bond Energy = sum (covalent + ionic)

Pauling said in 1932, ionic contributions can only increase bond strength. The metal hydrides are weaker. That is, even though Na⁺H^- can be considered to have ionic properties, the bond energy is weaker than predicted. If you can explain that, you can explain the entire electronegativity conundrum.

[opsomath]

An example which just occurred to me:

Trifluoroethanol pKa 12.4
Trichloroethanol pKa 12.73
Tribromoethanol pKa 12.70
Triiodoethanol is not easy to find.

Seems to fit the classical pattern of fluorine being the most EWG, yes?

[orgopete]

Potentially, that could be another exception in which fluorine is actually the most electron withdrawing. That is a difficult example to discern what should be correct. In the pKa tables linked to Hans Reich's UW webpage, there are a number of values. I really don't know which should be the most acidic.

The values I had were
Trichloroethanol, pKa 12.02
Trifluoroethanol, pKa 12.4
http://openmopac.net/pKa_table.html (no original citation however)

However, my sense of this is that again, the trifluoro group is less acidic. Why might I think that? I had been interested in using a trifluoromethyl ketone as a masked carboxylic acid ala a trichloro-, tribromo- or triiodomethyl ketone. While trifluoromethyl ketone adducts were good mimics of tetrahedral intermediates, I could not find examples in which the trifluoromethyl group could be eliminated as readily as the others. Could it have been related to the electron withdrawing ability of the halogens?

I had also tried finding the monosubstituted ethanols. I don't recall what I had found. Epoxide formation potentially limit their assessment. However, if one wishes to use an electronegativity argument, it must hold for all members. That is, propanol, ethylene glycol, etc. I think this series fails at methanol, ethanol, propanol, isopropanol, and t-butanol.

I am interested in the reference that you are citing.

[opsomath]

I'll see if I can look up the references. It took some serious google-fu to find those three, and they are not all from the same source.

I am afraid I must disagree with your very bold assertion that an electronegativity series must work for all molecules; that is, that n-propanol and ethylene glycol (why not ethylene glycol monomethyl ether, or mono(tert-butyl) ether?) must follow the same trend based on electronegativity of the beta-substituent as the halogenated alcohol series for electronegativity to be useful as a parameter. There are simply too many structural differences in those systems for a direct comparison to be useful.

[orgopete]

I am afraid I must disagree with your very bold assertion that an electronegativity series must work for all molecules; that is, that n-propanol and ethylene glycol (why not ethylene glycol monomethyl ether, or mono(tert-butyl) ether?) must follow the same trend based on electronegativity of the beta-substituent as the halogenated alcohol series for electronegativity to be useful as a parameter.

I am fine with methoxyethanol or any of the others. The reason I wished to be more rigid in this is to avoid mixing another effect into the result. For example, if fluoroacetic acid is the strongest acetic acid due to its electronegativity, that statement implies acidity is due to electronegativity. If iodoacetic acid is more acidic than methoxyacetic or glycolic acid, then it implies the acidity is due to some other factor than electronegativity because oxygen is the second most electronegative atom, but not the second most acidic acid.

I understand that people like electronegativity, but liking it does not make it true. Why should you think that bonds that are homolytically strong are heterolytically strong (ionic)? For me, these are completely different reactions. Consequently, it is easy to find examples in which that rule is broken. Why don't organic chemistry textbooks discuss carbocation stability in the same chapter in which they discuss electronegativity?  Is carbon really a better electron donor and withdrawer than hydrogen? Well, I guess they have all possibilities covered. I may have well asked for examples of (an alkyl) carbon being a better electron withdrawer than hydrogen.

A theory of "ionic attraction" does not make sense. Perhaps it could in 1932, but we should abandon it today. It may work to explain NaCl being a neutral compound, but it fails to explain HCl or NaOH's properties. A theory of ionic attraction does not explain why a proton of HCl should be attracted to the electrons of a neutral water molecule. It suggests a chloride ion should be more basic than neutral ammonia.

For anyone who might be interested in a different approach to chemistry, I have discovered J. Michael McBride's lectures posted online, http://oyc.yale.edu/chemistry/freshman-organic-chemistry/content/class-sessions. I suggest anyone interested to listen to his first lecture. He talks about how we know things. He argues it is from experiments. I  would be flattering myself to think that I am making a similar argument, that the experiments are telling us one thing and the theory contradicts them. That is the challenge I am making, show me some experiments that show fluorine is more electron withdrawing then chlorine, bromine, or iodine. 

[opsomath]

I am fine with methoxyethanol or any of the others. The reason I wished to be more rigid in this is to avoid mixing another effect into the result.

But by using example compounds with such different structures, you inevitably do mix other effects into the results.

Furthermore,

For example, if fluoroacetic acid is the strongest acetic acid due to its electronegativity, that statement implies acidity is due to electronegativity.

Wait, what? That is a statement that assumes "Because A implies B, B implies A." This is a classical logical fallacy. Because Homer Simpson is yellow, all yellow persons are not Homer Simpson.

[orgopete]

An example which just occurred to me:

Trifluoroethanol pKa 12.4
Trichloroethanol pKa 12.73
Tribromoethanol pKa 12.70
Triiodoethanol is not easy to find.

Seems to fit the classical pattern of fluorine being the most EWG, yes?

Correction, I should have said, "If trifluoroethanol is the strongest acid due to EWG, that statement implies acidity is due to EWG."

…by using example compounds with such different structures, you inevitably do mix other effects into the results.

I agree, see my Nov 18 comment on the acetic acids. Since it does not follow the electronegativity trends, I cannot distinguish electron withdrawing properties from other effects on acidity. I would prefer to compare HF and HI, but then I would conclude that iodide is a stronger electron withdrawing group.