Why ketonization is favored over enolization in 5-membered ring H-transfers by CH3C(OH+)CH2CH2 and in 6-membered ring H-transfers by CH3(COH+)CH2CH2CH2

The reactions leading to dissociation of the metastable enol isomers of 2-butanone have a puzzling feature: why does CH3C(OH+)CH2CH2 (2) isomerize to the less stable 2-butanone cation (3) and dissociate to the exclusion of conversion to CH2C(OH+)CH2CH3 (1) and CH3C(OH+)CHCH3 (4)? To answer this, the stationary points on the pathways established previously are located using ab initio and hybrid density functional theory and their geometries and energies determined. Rate curves for some of the reactions are also obtained by RRKM theory. It is found that 2 isomerizes to 3 rather than to its more stable enol isomers 1 and 4 because ketonization has the lowest barrier, despite its giving the highest energy product. For 2 → 1, this barrier arises from a strained, twisted transition state geometry being required in transfer of a hydrogen to the π-bonded methylene group in 1 whereas TS (2⇌3) is completely planar. The energy of the transition state for 4⇌2 is elevated by a different phenomenon, the need for an electron to go into an antibonding orbital in a 1,2-H-shift, raising its energy above that of TS (2⇌3). The critical energies for the 5-membered ring isomerizations of 1–2 and 2–3 are significantly higher than those of the 6-membered ring processes that ketonize and enolize CH3C(OH+)CH2CH2CH2 (6). However, similarly to 2, isomerization of 6 to CH3C(O+)CH2CH2CH3 has a lower energy transition state and is preferred to isomerization to the more stable CH2C(OH+)CH2CH2CH3. This is also attributable to twisting at the C1 methylene as hydrogen is transferred to and from it.