Question

Synthesis Based Experiment 1: Free Radical Bromination of 1,2-Diphenylethane 1. If the R,R and S,S enantiomers of 1,2-dibromo-1,2-diphenylethane have an observed melting point range of 113 -114'C and the meso 1,2-dibromo-1,2-diphenylethane has an observed melting point range of 236-237 C, which stereoisomer did you isolate? TLC 2. All three of these dibromo stereoisomers were produced in this reaction but you only isolated one of them. Explain this result taking into account the large differences in the three stereoisomers melting point range. Rto he developing chambe paper (aroundhaf of the wnes ethyl acetate) Swi nside of itwitht 3. Using any available source, draw the structure of your product showing the correct configuration of the two stereocenters. rom wn oo until Ihe soh too of the e from the nent with aconahe

Answer 1

(1) We can isolate meso 1 2-dibromo-1 2-diphenylethane from the mixture of three stereo isomer. (2) Because enantiomer have same physical properties but different biological reactivity. So, enantiomer can not isolate by distillation as it have same boiling & melting point. While diastereomer have different physical properties. So, it have different boiling point than other stereo isomer. (3) Br 1 2-dibromo-1 2-diphenylethane Br c (1R,2R)-1,2-dibromo-1,2-diphenylethane (15,25)-1,2-dibromo-1,2-diphenylethane Stereocenter in the compound have marked by"