Why Birch reduction of alkyne is E-selective?

Question

From Organic Chemistry(Clayden, Greeves and Warren, 2nd edition), pp. 682:

The sodium donates an electron to the LUMO of the triple bond (one of the two orthogonal $\pi^\ast$ orbitals). The resulting radical anion can pick up a proton from the ammonia solution to give a vinyl radical. A second electron, supplied again by the sodium, gives an anion that can adopt the more stable trans geometry. A final proton quench by a second molecule of ammonia or by an added proton source (t-butanol is often used, as in the Birch reduction) forms the E alkene.

I have some questions on this reaction:

(1) Double bond on the vinyl anion blocks rotation, but how can anion adopts trans geometry between cis and trans without rotation?

(2) Why anion with trans geometry is more stable? If it is because of steric repulsion between $R^1$ and $R^2$, why electronic repulsion between lone pair and $R^1$ can be ignored?

Answer 1

It's not rotation but inversion (remember amine lone pair inversion) from a linear $sp$ form to trigonal planar $sp^2$ that occurs. Lone pair and radical electron are held close to atom and repulse each other a lot so stable to keep them in anti dihedral positions. If the lone pair next to a radical electron was less repulsive, then it would have a lower energy transition state and be more readily formed according to Hammond's Postulate. But experimental data indicates that that is not the case.

Picture Citation: Kurti (2019)