Question

Bromination of Stilbene: Determining Stereospecificity in Organic Reactions In many cases, only particular stereoisomers are synthesized in an organic reaction, without the possibility of forming other stereoisomers. These reactions are called "stereospecific". One example of such reaction is the bromination of alkenes, which occurs through the formation of a cyclic "bromonium" ion and subsequent anti addition of a bromide ion to this intermediate, resulting in either threo or erythro enantiomers, depending on whether the starting alkene is cis or trans, respectively. Thus, depending on the desired stereoisomer of a particular compound, synthetic chemists must use their knowledge of mechanisms to design the correct synthesis using the appropriate starting reagents with the desired stereochemistry. In this experiment, you'll be carrying out a stereospecific organic synthesis using trans-stilbene and a source of bromine. The product will be recrystallized for purification, so that it can be used for identification of potential isomer(s) using melting point. In the first step, the addition of bromine to trans-stilbene (E-1,2-dipehnylethene) will be done by using a bromine "carrier" called pyridinium hydrobromide perbromide (a stable solid complex which liberates bromine in the presence of an alkene). The products could be either stilbene dibromide or meso-stilbene dibromide, or a mixture of the two BE oro. B12 tom.pl or Br trans-stilbene (mp 125 °C stilbene dibromide (mp 114 °C) meso dibromide (mp 238 °C)

1. Look up and draw the mechanism for the addition of Br2 to double bonds. Reference the source from which you acquired the mechanism.

2. Give the IUPAC name of the major product in the above-mentioned reaction.

3. Draw the major product of obtained in the addition of Br2 to stilbene.

4. Is this isomer chiral? What would be the optical rotation if this sample was placed in a polarimeter?

5. Would a student obtain the same product if you started with cis-stilbene instead of trans-stilbene? Explain your answer briefly

Answer 1

Br Brt Path-A S Br BrRY Ph Ph Path-B! Path-A Plane of symmetry Reference: Book: Carey, Francis A., Organic Chemistry, McGraw-Hill Companies, Inc., 3rd ed., 1996. Book: Mohrig, J.R.; Alberg, D.G.; Hofmeister, G.E.; Schatz, P.F.: Hammond, C.N.; Laboratory Techniques in -Organic Chemistry; Freeman, New York, 2014, 132-140; Meso compound Path-B 1,1'-[(1R,2S)-1,2-dibromoethane-1,2-diyl]dibenzene The trans stilibene reacts with Br2 to give electrophilic addition reaction via cyclic three membered bromonium ion ring. This ring can be opened via two pathways as shown where in each case the opening gives anti-addition product. Path-A gives 1R, 2S product. But the molecule contains plane of symmetry and hence it is meso compound. Path-B gives 19,2R product. The molecule also contains plane of symmetry and hence it Brs is also a meso compound. Thus both the products fromed from trans stilbene are identical and they are meso compounds. Hence they are achiral or the products will not show any optical activity. These products will have zero optical rotation if this sample was placed in a polarimeter. IUPAC names are given under each product. Meso compound 1,1'-[(1R, 2S)-1,2-dibromoethane-1,2-diyl]dibenzene Book: Schoffstall, A.M., Gaddis, G.A., Druelinger, M.L., Microscale and Miniscale Organic Chemistry Laboratory Experiments, 1st ed., Boston, McGraw-Hill Companies, Inc., 2000

If a student starts with cis-stilbene then they would get a racemic mixture of stilbene dibromide (1R,2R and 1S,2S) and hence it is not the "meso" isomer which can be obtained from trans stilbene.