Discuss the direct, concerted addition of \(\mathrm{H}_{2}\) to an alkene from the standpoint of orbital symmetry. (a) In the absence of catalyst, and (b) in the presence of photochemical stimulation.

In the absence of a catalyst, the hydrogen molecule is in its ground state with both hydrogen atoms having a pair of electrons in their 1s orbitals. An alkene molecule consists of sp2 hybridized carbon atoms, forming a pi bond by overlapping their unhybridized p orbitals. Now, let's consider the interaction between H2 and the alkene. A direct and concerted addition would require a simultaneous interaction between these orbitals. In the ground state, the 1s orbital of H2 is too low in energy and spherically symmetric, making it unfavorable for interaction with the pi bond of the alkene. Therefore, there is no suitable overlap of orbitals, and the reaction is symmetry-forbidden at the ground state of the hydrogen molecule.

In the presence of photochemical stimulation, the hydrogen molecule is excited, leading to the promotion of one of the electrons in the 1s orbital to a higher energy level. The resulting singly occupied orbitals become more asymmetric in shape and are suitable for interaction with the pi bond of the alkene. With this asymmetric orbital geometry, the overlap between the pi bond of the alkene and R-H sigma bond of the excited hydrogen molecule becomes possible, thus allowing the direct, concerted addition of H2 to the alkene. The reaction becomes symmetry-allowed due to the alignment of orbitals upon photochemical stimulation. In summary, the direct, concerted addition of H2 to an alkene is orbital symmetry-dependent, with the ground state of the hydrogen molecule making the reaction symmetry-forbidden in the absence of a catalyst. Photochemical stimulation allows for suitable orbital alignment, making the reaction symmetry-allowed.