[dlf4196]

I'm sure I'm missing something pretty basic here but in my Ochem textbook regarding preparation of alkynes, they state that for geminal or vicinal dihalides, the first elimination reaction to form an alkene can be done with many different bases, while the second, to make the alkene an alkyne must be done with a strong one such as NH2-.  Wouldn't it take a stronger base to deprotonate the alkane than the alkene as ethane's pka is higher than ethlyene's?  Actually another thing that struck me as odd were that both pka's were higher than that of nh2-, so why would the nh2- be able to deprotonate either of them?  Thanks in advance for any insights!

[argulor]

I believe you are essentially trying to correlate bond strength with pka. This is awfully tempting to do and many people fall into this trap. The C-H bond of ethane is weaker (i.e. easier to break) than the C-H of ethylene which is weaker than the C-H bond of acetylene; this is the opposite trend of their respective pka values (ethane would be hardest to deprotonate and acetylene the easiest).

In the case of an E2 reactions it might appear that you are just "deprotonating" a substrate, but it's not really an acid-base reaction in the traditional sense because there is really much more going on than just shuttling a proton around. It's often necessary to take a look at the interacting orbitals involved in the chemical transformations.

http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf

[dlf4196]

Thanks for the response, I appreciate it!!  I've definitely found out there's a lot you just sort of have to accept in Ochem as a lot of the understanding is beyond the scope of a basic undergrad course.  Would I be correct in saying that the shorter sp orbital would be better able to hold onto the hydrogen by virtue of its shorter length than the sp2 or sp3 orbitals? 

My understanding of the acid base stuff here was that since the sp orbital was shorter, the nucleus exerted more influence on its electrons and therefore "didn't mind" as much if it became an ion (more stable ion) and therefore lower pka.  But I guess I extended that in my head to making for a weaker bond because if the ion was stable, it would lose the hydrogen more easily, which apparently isn't necessarily the case. 

So just to clarify, my previous understanding was wrong because I oversimplified what was going on, correct?

[argulor]

Would I be correct in saying that the shorter sp orbital would be better able to hold onto the hydrogen by virtue of its shorter length than the sp2 or sp3 orbitals? 

This is okay. I always think about the stability of the conjugate base when I rationalize relative acidity. There are many ways to stabilize a conjugate base, but they all focus around one thing: delocalizing electrons whether through resonance, the size of atom, induction, etc. I like to put the whole hybridization argument under induction (i.e. is an electron withdrawing group) and say that an alkyne is more acidic than an alkane because the lone pair of the acetylide anion (conj. base of alkyne) is stabilized via induction from the electron withdrawing sp bond. This is a similar argument of why triflouroacetic acid is a better acid that acetic acid. Honestly once you hear an organic chemist rationalize anything with having more "s-character" or more "p-character" it is extremely oversimplified and a handwaving argument.

A look at the atomic orbitals of hydrogen allows me to accept this handwaving argument:
http://staff.mbi-berlin.de/hertel/physik3/chapter8/8.3html/01__100.png
If you carefully compare the 3s and 3p (or 2s and 2p) you can see that the s orbital penetrates closer to the nucleus and this is why we say it has more s character and is therefore held "closer" to the nucleus.

So just to clarify, my previous understanding was wrong because I oversimplified what was going on, correct?

Essentially, yes.

[orgopete]

I believe you are essentially trying to correlate bond strength with pka. This is awfully tempting to do and many people fall into this trap. The C-H bond of ethane is weaker (i.e. easier to break) than the C-H of ethylene which is weaker than the C-H bond of acetylene; this is the opposite trend of their respective pka values (ethane would be hardest to deprotonate and acetylene the easiest).

In the case of an E2 reactions it might appear that you are just "deprotonating" a substrate, but it's not really an acid-base reaction in the traditional sense because there is really much more going on than just shuttling a proton around. It's often necessary to take a look at the interacting orbitals involved in the chemical transformations.

http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf

Re: estimating heterolytic pKa from homolytic bond strength
So true. These are completely different reactions. If acidity were correlated with bond dissociation energy, that would lead one to erroneously conclude NH₃ (386 kJ/mol) should be a stronger acid than HCl (432 kJ/mol). I have surmised that pKa values can be used in lieu of a heterolytic bond dissociation value. If so, then acetylene is simply more acid than an alkane.

Someone had posted the orbital argument and I had questioned this based upon a graph they had posted. The graph makes it appear that p-orbitals are actually closer to the nucleus, that is, the p-orbital electron density is closer. An s-orbital has a small blip of electron density that is closer, but if one must go by this blip, what is the value of the graph? I had invited comments on this, but I don't recall anyone took this up.

This what I had suggest in my class. The analogy for an electron pair’s ability to attack another nucleus is, ‘chemical reactivity is like a boxer. A boxer with a longer reach would have an easier time to hit the nucleus of a another atom’. If you look at the bond lengths of the CH, NH, OH, and HF, the electrons are held most tightly by fluorine, the bond length is shortest, and it is the most acidic. If you compare an alkane, alkene, and acetylene, the bond length is the shortest for an acetylene and it is the most acidic.

Re: dfl4196's quandary, alkane v alkene elimination
I agree with the premise, however, that may not tell us everything about the reaction. I know textbooks indicate reactions are concerted and that is interpreted as 'at the same time'. If we divide time into infinitely small divisions, then we must argue that a bond must begin to break virtually knowing an electron pair were attacking. It makes more sense to me that if a collision were to occur, we would find something like a chain reaction of autos in which the first collision caused the next etc. If we do that, then we can analyze how reactions might change. For an elimination reaction, we might ask whether a lengthening of a C-X bond precedes the rate determining deprotonation step. If that were the case, then a less electron withdrawing alkane may correspond with a greater lengthening of a C-X bond. The lengthening could correspond with a greater donation of neighboring electrons (see hyperconjugation). The protons removed would be the now more acidic neighboring protons.

If you read the reaction like this, then you can well rationalize that an alkene's C-X bond would be less lengthened and contribute less to an increase in the neighboring CH acidity. You may also rationalize that Zaitsev products are really 'SN1-like' and electron donation is greatest from the tertiary carbon, hence most substituted alkene.

[dlf4196]

Hey thank ya'll so much. Your explanations definitely helped me conceptualize this issue better.  I really appreciate it!

[dlf4196]

Oh crap, I forgot, one last question for orgopete. 

You said that HF was more acidic than CH etc. by virtue of its shorter bond length and tighter hold on its electrons. due to its electronegativity.  But isn't HI stronger than HF, while exerting a less strong hold on its electrons because of shielding?  My rationale for why HI was stronger was that it was able to spread the negative charge around enabled it to have a stronger hold but I understood that I was a larger atom than F.

[orgopete]

You said that HF was more acidic than CH etc. by virtue of its shorter bond length and tighter hold on its electrons. due to its electronegativity.  But isn't HI stronger than HF, while exerting a less strong hold on its electrons because of shielding?  My rationale for why HI was stronger was that it was able to spread the negative charge around enabled it to have a stronger hold but I understood that iodide was a larger atom than fluoride.

I suggest you count the number of electrons surrounding an iodine and fluorine atom and tell me which has a greater negative charge. I'm not counting protons, only electrons. I argue a proton elsewhere in an atom will not reverse the charge of an electron. Remember, ammonia is a stronger base than fluoride.

I argue acidity is an inverse square property. The further a proton is from the nearest electron pair, the more acidic it will be.

What has been confusing to some has been the bond length issue. As noted earlier, HF has the shortest bond and is the strongest acid. For the haloacids, we cannot use the same bond length argument because they have different numbers of electrons in their shells. Of course HCl has longer bond than HF because it has more electrons and they are in another shell. It isn't bond length that determines acidity or carbon should be the most acidic. It is the proton-electron pair distance that matters.

(Ugh, if I just completed my book, it is all explained therein with tables and graphs.)

[OrgXemProf]

Alternative explanation for the order of decreasing acidity: HI > HBr > HCl > HF:

Consider the equilibria for a series of hydrgen halides in solution, HX  H⁺ + X^-. Since HX is a neutral compound and H⁺ is a feature common to all equilibria in that series, the primary species that determines the position of equilibrium in each case is likely to be the (charged) anion, X^-.

Valence electrons in F- occupy 2p atomic orbitals, whereas valence electrons in I- (electronic configuration [Kr]5s²5p⁶) occupy much more diffuse orbitals whose probability density clouds remove them far from the nucleus.

Thus, the "volume" (probability density) that electrons occupy in F- is considerably smaller than that in I-. As a result, I- possesses a lower charge-per-unit-volume ("charge density") than F-.

Due to its lower charge density, I- in solution (where most organic reactions take place) demands less stabilization via solvation than does F-. We might consider this in terms of Debye-Hueckel theory, wherein solvation (in this case, interaction of solvent dipoles with a point charge) results in formation of a solvation shell or "ionic atmosphere". The solvation shell + anion continues to possess an overall negative charge, but the effect of solvent shell formation is to reduce the charge-per-unit-volume of the resulting assembly, which results in net stabilization of the anion in solution.

(Of course, there is also an entropy factor that must accompany reorganization of solvent molecules, but we could go on and on...)

If we agree that I- is more stable in solution than F-, then the equilbrium  HX  H⁺ + X^- must lie farther to the right for X = I- than for X = F-, consistent with the conclusion that HI is a stronger acid than HF.

A pragmatist might view the foregoing situation as follows: I may not be certain that the logic is 100% rigorous, but it leads to the correct conclusion, so maybe the model could be useful despite its obvious limitations.

[orgopete]

Thus, the "volume" (probability density) that electrons occupy in F- is considerably smaller than that in I-. As a result, I- possesses a lower charge-per-unit-volume ("charge density") than F-.

A pragmatist might view the foregoing situation as follows: I may not be certain that the logic is 100% rigorous, but it leads to the correct conclusion, so maybe the model could be useful despite its obvious limitations.

Let's check. The bond length of HF is 92 pm and HI, 160 pm. The volume of a sphere is 4/3*pi*r^3. HF has 10 electrons, HI 54. What is the volume per electron? I find them to be quite similar.

I never liked this kind of argument. If one were measuring electron distribution in this way, how does one explain ammonia is more basic than fluoride. It does not seem to be net charge. The local charge of a pair of electrons is minus two.

Now, calculate the repulsive force between a proton and its nucleus. The force is Q1*Q2/r^2. An iodine nucleus is nearly twice as repulsive compared to HF. I am more convinced this is an inverse square property. I also like the consistency with the bond lengths of CH4, NH3, H2O, and HF. HF has the greatest charge, shortest bond, and is most acidic.