Archive for April, 2022

Geometries of proton transfers: modelled using total energy or free energy?

Monday, April 18th, 2022

Proton transfers are amongst the most common of all chemical reactions. They are often thought of as “trivial” and even may not feature in many mechanistic schemes, other than perhaps the notation “PT”. The types with the lowest energy barriers for transfer often involve heteroatoms such as N and O, and the conventional transition state might be supposed to be when the proton is located at about the half way distance between the two heteroatoms. This should be the energy high point between the two positions for the proton. But what if a crystal structure is determined with the proton in exactly this position? Well, the first hypothesis is that using X-rays as the diffracting radiation is unreliable, because protons scatter x-rays very poorly. Then a more arduous neutron diffraction study is sometimes undertaken, which is generally assumed to be more reliable in determining the position of the proton. Just such a study was undertaken for the structure shown below (RAKQOJ)[cite]10/c3zxh2[/cite], dataDOI: 10.5517/cc57db3 for the 80K determination. The substituents had been selected to try to maximise the symmetry of the O…H…N motif via pKa tuning (for another tuning attempt, see this blog). The more general landscape this molecule fits into[cite]10.1039/C1RA00219H[/cite] is shown below:

(more…)

C2N2: a 10-electron four-atom molecule displaying both Hückel 4n+2 and Baird 4n selection rules for ring aromaticity.

Thursday, April 7th, 2022

The previous examples of four atom systems displaying two layers of aromaticity illustrated how 4 (B4), 8 (C4) and 12 (N4) valence electrons were partitioned into 4n+2 manifolds (respectively 2+2, 6+2 and 6+6). The triplet state molecule B2C2 with 6 electrons partitioned into 2π and 4σ electrons, with the latter following Baird’s aromaticity rule.[cite]10.1021/ja00769a025[/cite],[cite]10.1021/cr300471v[/cite]. Now for the final missing entry; as a triplet C2N2 has 10 electrons, which now partition into 4 + 6. But would that be 4π + 6σ or 4σ + 6π? Well, in a way neither! Read on.

(more…)