2 problems with elimination-addition of halobenzene using NH2-

Question

When converting chlorobenzene to aniline, I drew the following mechanism and the book says it is correct, but there are 2 questions that I do not understand:

1. In the 2nd step, I am thinking that it is a nuclophilic attack, since $\ce{NH2^-}$ is a good nucleophile. However, I do not understand what makes the $\ce{C#C}$ bond in benzyne electrophilic. Doesn't a triple bond have high electron density and render it more nucleophilic instead? How can the reaction happen?

2. In the 3rd step, I don't know why proton transfer happens here. Isn't forming $\ce{NH2-}$ from $\ce{NH3}$ not thermodynamically viable as $\ce{NH2-}$ is a strong base?

My thought on (2): forming $\ce{NH2-}$ ion from $\ce{NH3}$ is more stable than leaving the carbanion (the aniline with an extra electron) there, but I am not sure about whether this is a correct explanation.

Answer 1

1. Electrophilicity of the triple bond

Hard-soft acid-base theory ties in with this question. The general trends in nucleophile-electrophile reactivity show that hard (highly charge dense) nucleophiles have a tendency to react well with hard electrophiles, and soft (low charge density) nucleophiles with soft electrophiles. This relationship is due to two competing factors; electrostatics and overlap of frontier orbitals. Hard-hard interactions lend their stability to electrostatic attraction, while soft-soft interactions rely on a good HOMO-LUMO overlap.

A π-bond is generally considered a soft species that doesn't reply on electrostatics very strongly. Good soft nucleophiles have relatively high HOMO's (highest occupied molecular orbitals) and good soft electrophiles have relatively low LUMO's (lowest unoccupied molecular orbitals). The third bond in benzyne is abnormal in that it doesn't involve overlap of 2 unhybridised p-orbitals, but 2 hybridised $sp^2$ orbitals that aren't completely parallel. From Clayden Organic Chemistry:

This will, as expected, result in a very weak orbital overlap. As a result, the split in energy between the resulting bonding and antibonding orbital is very small, giving a very low LUMO which can easily accept a nucleophile's electrons. From here:

This is the source of the triple bond's electrophilicity.

2. The final protonation step

(Image from Clayden Organic Chemistry)

Your assumption is correct that the anion is very unstable. According to this source, , benzene has a $\mathrm{p}K_\mathrm{a}$ of 43. This is due to the negative charge being situated in an $\pu{sp^2}$ orbital, which is highly charge dense, with no delocalisation into the ring (since it is not a part of the π-system). Ammonia has a $\mathrm{p}K_\mathrm{a}$ of 38, so its conjugate base, $\ce{NH2^-}$, is an extremely good base. This value is still lower than benzene's, so the benzene anion is still less stable. By a quick calculation, this gives the reaction above an equilibrium constant of roughly $\pu{10^5}$, heavily in favour of ammonia deprotonation.