Figure 1

Equilibrated structures of five intermediates. Only the molecules in the doped chain are shown. I to VI are in the box; the others are in the periodic replicas. The $+$ indicates the charge. ($A$) The additional ${\mathrm{H}}^{+}$ is on III, but the charge is on V. ($B$) At 2 ps, the HB between III and IV is broken; III rotates and is almost orthogonal to the chain. The charge is on VI, while the N-H bonds of II and IV are aligned and point to the center of III. ($C$) At 3.5 ps, the HB between II and III is restored, and the $R$ defect is in the III/IV position (similar to $A$). ($D$) At 5 ps, the N-H bond of III is orthogonal to the chain, due to the in-plane rotation of III, while the N-H of II points towards the C-C bond of III. ($E$) At 8 ps, the $R$ defect is in the V/VI position and the charge is two molecules farther.Reuse & Permissions

Figure 2

Optimized geometries of four intermediates of doped Imi2eo (a), and corresponding schematic representations of the HBs’s pattern (b). Only the molecules in the chain and the first two segments of the spacers are shown. The relevant protons are labeled from 1 to 5 (H2: proton added in an intrapair site, $A$) and I to IV are the four molecules in the cell box (PBC are used). In (b) the arrow labels the molecule that is not aligned to the chain and forms the HB with one O (intra-$ABD$ or inter-$C$ pairs). ($A$) First minimum; H1 forms a HB with the opposite O atom, and H3 is shared. The $\mathrm{H}\cdots \mathrm{N}$ HB on the left side of I is elongated, due to the molecular bending. ($B$) By MD at 200 K, H3 moves from II to III and the Ch is on the III ring. ($C$) After 2 ps of MTD, II is orthogonal to the chain and H2 forms a weak HB with an O atom in a neighboring pair. The $\mathrm{N}\cdots \mathrm{H}3$ HB is elongated due to the rotation of II and Ch is on IV. ($D$) Intrapair $R$ defect; III is rotated and forms a HB with the opposite spacer. At about 11 ps, a new state ($E$) is found, by the rotation of IV. In $E$, IV is orthogonal to the chain and a $\mathrm{H}4\cdots \mathrm{O}$ is formed.Reuse & Permissions