Chemistry 350 Laboratory - Fall Semester 2010 – Professor T. Nalli, Winona State University

Expt #7. Dehydrobromination of 1,2-Dibromo-1,2-diphenylethane

Relevant textbook readings – Mohrig, Chapter 9. Smith, Chapter 8.8.

Overview – The stereoselectivity of alkyl halide dehydrohalogenation will be examined in this experiment.

The 1,2-dibromo-1,2-diphenylethane product of expt 6 will be dehydrohalogenated using sodium ethoxide (eq 2). The product, 1-bromo-1,2-diphenylethene, can exist as either the E or Z isomer. You will determine which stereoisomer(s) actually form. From this information you will determine if the dehydrohalogenation reaction has a stereochemical requirement. In other words, does the loss of HBr follow an anti or a syn pathway, or is it nonstereoselective?

Procedures

Running the reaction.

· Place 100 mg of 1,2-dibromo-1,2-diphenylethane to a 5-mL r. b. flask (Williamson kit), add a micro spin bar, 3 mL of ethanol and 0.3 mL of 2 M CH₃CH₂O^- Na⁺ in CH₃CH₂OH.

· Attach an air condenser and reflux the reaction mixture on a hot water bath (80-90^o) for 30 min.

Work-up procedures.

· Cool the reaction mixture and then transfer it to a large tube containing about 20 mL ice and water.

· After the ice has all melted, add 2 mL of CH₂Cl₂. Cap and shake the tube to extract the product, venting frequently.

· After the layers separate use a Pasteur pipet to transfer the lower organic layer with a pipette to a smaller test tube. Repeat steps 2 and 3 twice more. Each time transferring the lower layer from step 3 to the same test tube.

· Wash the contenst of the small test tube with 2 mL of pH 7 buffer solution. Transfer the lower layer to a small, dry Erlenmeyer flask. Dry the CH₂Cl₂ extracts over Na₂SO₄. Decant the liquid into a dry, pre-weighed vial.

· Evaporate off the solvent under N₂ stream with gentle warming. The final product is viscous oily liquid. Determine the mass and calculate the yield and percent yield.

Characterization of Product

· Obtain the IR spectrum. Make sure to annotate all key functional group peaks before printing.

· Obtain the ¹H NMR spectrum using CDCl₃ as the solvent. Literature data can be found in the following reference: Kropp, P. J.; Crawford, S. D. J. Org. Chem. 1994, 59, 3102-3112.

Report

Please try to integrate the answers to the assigned questions into your results and conclusions section for this report (rather than answering each sequentially).

Yield and Corrected Yield

Calculate the percent yield of your product based on the actual mass of 1,2-dibromo-1,2-diphenylethane used for the reaction. Also report a corrected yield which takes account the dichloromethane left in your product and visible in the NMR spectrum. To calculate the corrected yield note the following.

¹H NMR peak integrations are proportional to the moles of compound as well as to the number of Hs making up the peak. Therefore, to find the moles of compound you must correct for the number of hydrogens by dividing the integration by #Hs. Hence, the integration of the CH₂Cl₂ peak divided by two is proportional to the moles of CH₂Cl₂. Assign the peaks for the 1-bromo-1,2-diphenylethene (BDPE) and once you've done so, choose one of the peaks to represent the moles of BDPE. The integration of this peak divided by the number of Hs it gives the relative moles of BDPE product. Convert each of the mole values to mass using the MW of each compound. Convert the masses to mass percent purity by dividing the mass of product by the total mass of both compounds and multiply by 100. This gives you the correction factor which can just be multiplied by the % yield to give you the corrected percent yield.

Assigned Questions:

1. Which stereoisomer was formed in the reaction? Explain this result in terms of the accepted mechanism.

2. One of the main differences in the ¹H NMR spectrum of the (E) and (Z) isomers of 1-bromo-1,2-diphenylethene are the resonances at 7.72 ppm (2H, d) and 7.66 (2H, m) ppm in the spectrum of the (Z) isomer. These chemical shifts are 0.3 ppm higher than the highest frequency peaks in the (E) isomer. Use molecular models, peak integrations, and peak multiplicities to come up a proposal as to which hydrogens in the (Z) isomer cause the peaks at 7.72 and 7.66 ppm and explain why they experience greater deshielding than any of the hydrogens in the (E) isomer.