Synthesis and Dehydrobromination of 1,2-Dibromo-1,2-diphenylethane

Background: McM sect 7.2, 9.6-9.8, 9.13 and 11.11. trans-1,2-Diphenylethene, more commonly named trans-stilbene is brominated with Br₂ to form 1,2-dibromo-1,2-diphenylethane:

The approaches of the two bromine atoms to the double bond lead to four possible stereoisomeric products:

Products I and II represent syn additions of the bromine atoms; if this were the mode of addition, the planar character of the alkene reactant requires that these two products would be formed in equal amounts. By making models one can see that these two products are actually mirror images of each other, and the mixture is known as a racemic mixture.

Products III and IV represent anti additions of bromine to the reactant; again they would be formed in equal amounts. Actually, when models are made for these drawings, one can see that by rotation about the central bond and reorientation, these two drawings depict the same product, called the meso product. The one meso compound is possible in this example because the carbon atoms of the central bond are attached to identical groups, H, Br and C₆H₅.

One of the goals of this laboratory is to determine which, anti or syn, the bromine atoms actually add to trans-stilbene. This question will be decided by comparing the melting point of your product with the known melting points of the two possible products. The melting point of the racemic mixture is 113^oC while the melting point of the meso product is 238^oC.

The other goal of this laboratory is to carry out an E-2 elimination on the 1,2-dibromo-1,2-diphenylethane to verify whether your dibromo product is the racemic mixture or the meso compound. The elimination in this case is a dehydrobromination, the loss of H and Br from 1,2-dibromo-1,2-diphenylethane:

In this reaction, NaOC₂H₅ serves as a strong base. It is known that in order for this elimination to proceed, the conformation of the molecule must be such that the H and the Br are oriented anti to each other and also the four H-C-C-Br atoms must lie on a single plane. In McM, p. 415, this arrangement is known as "anti periplanar" geometry (see Fig 11.17 and 11.18):

Depending on the 1,2-dibromo-1,2-diphenylethane compound(s) you synthesize, the product will be either (E)-1-bromo-1,2-diphenylethylene or (Z)-1-bromo-1,2-diphenylethylene. Which dibromo- compound forms which alkene can be worked out in advance. As an example, if the dehydrobromination is carried out on dibromo compound I, we first must orient the H-C-C-Br in the correct geometry. Subsequent loss of H and Br lead to the (Z)- alkene:

The above drawings depict the loss of the Br attached to the left carbon and the H attached to the right carbon. It is important that you draw out (Question 1, below) and recognize that the same (Z)- alkene will result from the loss of the left side H and the right side Br. Also you need to work with models to determine that the same (Z)- alkene would result from the dehydrobromination of II.

Now if the meso dibromo- compound is subjected to dehydrobromination, again with anti periplanar geometry, (E)-1-bromo-1,2-diphenylethylene results. This should be worked out with models (Question 2, below), but helpful drawings are found on p. 417 of McM.

¹H NMR spectrometry can clearly differentiate between (E)- and (Z)-1-bromo-1,2-diphenylethylene. A model of the (Z)- isomer shows that the bromine atom is in close proximity to the ortho hydrogens on the ring cis to it. This proximity results in the deshielding of these ortho hydrogens, whose signal at 7.7 ppm is clearly separated from signals for the regular aromatic hydrogens. In contrast, the bromine atom is relatively distant from benzene ring protons in the (E)- isomer, and as a result, the most deshielded signal for the (E)- isomer is centered at 7.4 ppm.

Procedure.

For the synthesis of 1,2-dibromo-1,2-diphenylethane: to a 25 mL Erlenmeyer flask, add 0.2g of trans-stilbene, 6 mL of ether and a stir vane. Swirl to dissolve the solid stilbene (warm it on the steam bath if necessary) and then add dropwise 2 mL of 1M Br₂ in CHCl₃. After 15 minutes, cool the flask in an ice bath for an additional 5 minutes. Using a shortened pipette, transfer the fine crystals to a prepared Hirsch funnel with suction. Transfer the remaining ether, then add 2 mL fresh cold ether to rinse the flask, and transfer the rinsings with any solids remaining to the Hirsch funnel. Repeat the rinsing process once more with one more mL of fresh, cold ether. Press the solid on the filter to remove any remaining ether, then transfer the solid to a small preweighed beaker, and determine the weight of the 1,2-dibromo-1,2-diphenylethane isolated. Determine the melting point of this product. Which product was formed?

For the dehydrobromination of your product, transfer 100 mg of the 1,2-dibromo-1,2-diphenylethane you formed to a large (25x150 mm) test tube, add a spin vane, 3 mL of ethanol and 0.3 mL of 2 M CH₃CH₂O^- Na⁺ in CH₃CH₂OH. Place the tube in a beaker containing 2-3 cm of water 80-90^o water on a stirring hot plate (this hot bath can be shared by several students). The stirring mixture should boil gently for 30 minutes. The test tube is then placed in a beaker of ice and 20 mL of water and 2 ml of CH₂Cl₂ are added. While it is still cold, use the spin vane to thoroughly mix the layers of the mixture. Then allow the layers to settle and transfer the lower layer with a pipette to a 16x125 mm test tube containing 1 mL of pH 7 buffer solution. After mixing, transfer the lower layer in the smaller test tube to a prepared filter-drying pipette clamped vertically above a clean, pre-weighed 25 mL Erlenmeyer flask. To the aqueous solution in the larger test tube is added another 2 ml of fresh CH₂Cl₂ and after stirring as before, the lower layer is transferred to the smaller test tube containing the buffer, and then as before, the lower layer layer is transferred to the filter-drying pipette. One more 2 ml portion of fresh CH₂Cl₂ is processed as before and added to the filter-drying pipette. After all the dried extracts are gathered in the flask, the CH₂Cl₂ is removed by evaporation on the steam bath (boiling stick) and the mass of the resulting oil is determined. Label this flask for analysis of the oil.

Questions

1. The stereochemical drawings, above, show only one example of the E-2 dehydrobromination for one component of the racemic mix. Make similar or sawhorse drawings for the dehydrobromination of the following, after first orienting the H and the Br into anti periplanar geometry:
(a) 1,2-dibromo-1,2-diphenylethane product I, but this time with the loss of the left side H and the right side Br.
(b) 1,2-dibromo-1,2-diphenylethane product II with the loss of left side H and the right side Br
(c) 1,2-dibromo-1,2-diphenylethane product II with the loss of left side Br and the right side H.

2. Set up sequences similar to those in question 1 for the dehydrobromination of meso-1,2-dibromo-1,2-diphenylethane

3. Set up a reaction table ("Common Practices…") to determine the theoretical yields of both your bromine addition reaction and your dehydrobromination reaction. Also calculate the percents of theoretical yields you obtained with your products

4. Using the sequence rules in McM, assign R or S configuration for the central carbon atoms in compounds I, II, III and IV.

5. Below are commonly encountered projections - called Fischer projections - of 1,2-dibromo-1,2-diphenylethane (McM sect 9.13). As suggested by the fonts, the atoms or groups attached to the central carbon atoms are either "behind" or "above" the page. Fischer projections, usually presented without the font size prompt, are convenient in displaying and organizing the stereochemical aspects of molecules. Using models, determine which of the two of the four drawings below are projections of the meso-1,2-dibromo-1,2-diphenylethane and the other two are Fischer projections of the racemic 1,2-dibromo-1,2-diphenylethane mixture:

6. The mode of addition you determined (syn or anti) is typical for all additions of Br₂ to alkenes. Make stereochemical drawings - as above or sawhorse - for the product(s) of bromination to cis-stilbene (cis-1,2-diphenylethene). Is this product a racemic mixture or a meso product?

7. Taking the product of the reaction in the previous question, draw the product of its E-2 dehydrobromination, remembering that the H and Br must leave anti to each other.

8. If the dibromo compound from trans-stilbene is subjected to NaOC₂H₅ for extended times and high temperatures, diphenylethyne is produced. On the other hand, the dibromo compound from cis-stilbene requires much less time and lower temperatures to produce the same diphenylethyne product. Using drawings of the 1-bromo-1,2-diphenylethylene compounds initially formed in each reaction, explain why diphenylethyne is formed easily in one case and with difficulty in the other.

Rev. Jan, 2000