Question

Chem 2320 extra credit assignment-12 points total Suppose that you were asked to perform the following alkoxymercuration reaction with (S)-3- vinylcyclopentanone using the standard alkoxymercuration conditions (steps 1 and 2) shown below. 1. Hg(OAc)2 CH3OH 2. NaBH4 Actual product Expected product Note: (S)-3-vinylcyclopentanone has two strong IR peaks at 1748 cm and 1642 cm iH NMR peaks at δ 2.5-3 ppm (7H), 5-6 ppm (3H), and a 13C NMR peak at 210 ppm. a. (3 pts) In the first box provided above, draw the expected (stable) alkoxymercuration product if the complete alkoxymercuration reaction proceeds normally? The expected product would have a strong IR peak at 1748 cm and a peak in the C NMR spectrum at 210 ppm. No IR peaks are observed in the 1500-1700 cm range, nor are any peaks visible in the 'H NMR spectrum between 5-6 ppm. The expected product would also react with 2,4- dinitrophenylhydrazine to form a solid 2,4-dinitrophenylhydrazone derivative. If you're not 100% sure of your structure, in the space below, explain what led you to choose your structure. b. (3 pts) Instead of the expected alkoxymercuration product (first box), you actually isolated a different molecule at the end of this two-step, complete alkoxymercuration. The actual product shows a strong, broad IR peak at 3400 cml, but has no IR peak at 1749 cm1. It lacks downfield 1BC 100 ppm) and 'H NMR (5 ppm) peaks. The actual product does not react with 2,4-dinitrophenylhydrazine, but will react with NaH (sodium hydride) to form a gas. Draw the actual product of this reaction in the second box provided above. If you're not 100% sure of your structure, in the space below, explain what led you to choose your structure. c. (4 pts) Explain why the actual product was formed instead of the expected product. d. (2 pts) On the basis of your explanation for part c, suggest different reagents (you may have to invent a new reagent) or reaction conditions you might use to make the alkoxymercuration give the expected product instead of the actual product formed above.

Answer 1

НС £xpected ShnIR peak at 1748 i r3 due to C0 * Due to presence ot kuthu он No TR puuk at 1749 ci Since cavlenrt iu absent

Actual prodrut İstKmed due to presnu in <ie group to rNa8Hq a diol such we Cam wthicd will ask tr euuntl to OH HO H30h 2 NaBH H30