(a) Partial dehydrohalogenation of either $(1 R, 2 R)-1,$ 2-dibromo-1,2-diphenylethane or $(1 S, 2 S)-1,$ 2-dibromo-1,2-diphenylethane enantiomers (or a racemate of the two) produces ( $Z$ )-1-bromo-1,2-diphenylethene as the product, whereas (b) partial dehydrohalogenation of $(1 R, 2 S)-1,2$ -dibromo-1, 2-diphenylethane (the meso compound) gives only $(E)$ -1-bromo-1,2-diphenylethene. (c) Treating $(1 R, 2 S)-1,2$ - dibromo-1,2-diphenylethane with sodium iodide in acetone produces only (E) -1,2-diphenylethene. Explain these results.