Several students have asked for a clearer explanation of why one observes different products from 2-butyne isomerization with a very strong base (amide, NH₂^-) than with hydroxide. Here we lay out in gruesome detail the story of how to prepare the unstable terminal alkyne by equilibration.

In the following scheme one procedes from left to right by successive removal of a proton to form a resonance-stabilized anion and reprotonation to form an isomerized hydrocarbon.

First consider equilibration among the hydrocarbons in terms of their relative heats of formation (as given in Appendix I of the Text). As the diagram below shows, the internal alkyne, 2-butyne, is most stable, while the terminal alkyne, 1-butyne, is least stable. At room temperature the 2-butyne / 1,2-butadiene ratio would be about 10^(3/4 * 4.1) or about 1000, while the 1,2-butadiene / 1-butyne ratio should be about 10^(3/4 * 0.7) or about 3.

Catalysis by Hydroxide

Now consider how to achieve the equilibrium in the presence of a base that can remove protons and then give them back again. First we need the relative energies in the presence of hydroxide. For this purpose we lump each neutral hydrocarbon with hydroxide, and each carbanion with water, so as to compare the energies of comparable (isomeric) species. [One cannot determine an equilibrium constant between hydrocarbon and anion alone, because they are not isomers.] We also define the energy of 2-butyne plus base as 0 for purposes of considering relative energies.

In order to determine the energies of the anions (plus water), we need some pKa values. The pKa of water is 16. The text (p. 302) suggests that the pKa of ethylene is ~44. This means that the equilibrium constant for abstraction of a proton from ethylene by hydroxide is about 10^(16-44) = 10^(-28), or that the energy of water plus vinyl anion is about 4/3 * 28 = 37 kcal/mole higher than the energy of hydroxide plus ethylene.

[If you find the manipulation of pKa values confusing, translate them into two equilibrium constant expressions for dissociation of the two acids, rearrange the two equations to have the concentration of H⁺ on one side, and set the other sides equal to one another, because when they coexist there is a single H⁺ concentration.]

Because it is stabilized by resonance, the anion from 1,2-butadiene should be about 10 kcal/mole easier to form than that of ethylene. Let's assume that the energy of water plus an anion from 1,2-butadiene is about 27 kcal/mole higher than the energy of hydroxide plus 1,2-butadiene. Actually it is probably a little harder for hydroxide to remove the proton from C3 than from C1, but for our purposes the difference is negligible. The point is that the carbanions are WAY upstairs in energy.

Even the acetylenic proton on 1-butyne (pKa ~25) is hard for hydroxide to remove - equilibrium constant about 10^(16-25) = 10^(-9), i.e. energy increase about 4/3 * 9 = 12 kcal/mole.

This allows us to construct the following approximate energy diagram, where the wiggly line indicates a break in the vertical scale, above which the energies are approximate and not to scale.

Obviously the internal acetylene (2-butyne) will predominate at equilibrium, and the anions (plus water) that serve as intermediates between the isomeric hydrocarbons are so high in energy that their rate of formation is inconveniently slow at room temperature, at fastest about 10^13/sec * 10^(-3/4 * 26) = 10^(-7) /sec. Since 10^7 seconds is more than 3 months, the isomerization by hydroxide to prepare 2-butyne must be conducted at high temperature.

Amide (NH₂^-) is a much stronger base than hydroxide (pKa ~34 for NH₃). Abstraction of a proton by amide is easier than abstraction by hydroxide by a factor of 10^(34-16) = 10^18, which is to say by 4/3 * 18 = 24 kcal/mole. We can modify the energy scheme above to apply to amide by lowering each carbanion (plus ammonia) energy by 24 kcal/mole from the carbanion (plus water) analogue, as shown below:

Now equilibration among the hydrocarbons (plus amide) is much faster, since the intermediate carbanions are only 2 to 8 or so kcal/mole above the hydrocarbons and are quick to form. Even more important, the dominant species at equilibrium is none of the hydrocarbons (plus amide), but rather the anion of the terminal acetylene (plus ammonia). After the rapid equilibration one adds water which first converts any remaining NH₂^- to NH₃ plus OH^-. Next the water pronates the acetylide to give neutral 1-butyne. Since no really strong base remains (only OH^-), isomerization at normal temperature ceases and one has prepared the least stable of three hydrocarbon isomers.