Match the column I and II.

Answer

`a -q; b-r; c-s; d-p`

Explanation

In some alkenes, steric crowding in the planar state may be so severe that the latter is no longer an energy minimum but becomes an energy barrier with the ground state becoming twisted. In this context the compounds in figure are of interest. The drop in rotational barrier from stilbene `(42.8 "kcal mol"^(-1),179 "kJ mol"^(-1))` to compound P `(21.1. "kcal mol"^(-1), 88.3 "kJ mol"^(-1))` to compound Q ("small barrier", cf.) may be explained by increasing stabilization of the triplet biradical transition state. However, the "negative" barrier in R, that is, the fact that the stable ground state of R is non-planar and that the planar conformation represents the transition state for rotation requires a different explanation, the difference between Q and R is presumably due to the steric interaction of the four chlorine substituents in the planar conformation of R. This situation resembles that in o, o.- tetrasubstituted biphenyls to alkenes with low or "negative” barriers.