tbuo OtBu Br 1) B 2 H 6 /THF OH 2) OH - + H 2 O 2 NaNH 2 NH3 Na NH 3 /-35 C CH 3 CCH 2 CHCH 3

Transcription

1 EM 12A Name KEY EXAM II all (16 pts) Draw the structure of the major product expected from each of the following sets of reactants. Specify stereochemistry where appropriate. + 2 Pt enantiomer 1) g(oac) 2, 2 O, T O 3 2) NaB 4 + tbuo OtBu 1) B 2 6 /T O 2) O O NaN 2 N3 Na Na N 3 / (excess) 2

2 2.(8 pts) Is the compound below optically active? Explain your answer, being as concise and complete as possible, and include a definition of "optically active" in your explanation. O O An optically active compound will rotate the plane of plane-polarized light, due to a lack of symmetry. This compound has no elements (points, lines or planes) of symmetry, so it will rotate the plane of plane-polarized light. The absolute configuration of each chiral carbon is R, so a sample of this compound that is pure (that is, with none of its enantiomer present) will not have any mirror-image conformations present, which are necessary to cause a lack of optical activity. 3.(12 pts) Write the IUPA name for each of the following compounds. O 2 (S)-2-butanol 2 O (R)-2-bromo-2-chlorobutane (2S,3R)-3-fluoro-3-methyl-2-pentanol 3,7-dimethyl-4-nonyne

3 4.(8 pts) Show each step in the reaction mechanism for the reaction below. Include only reactants, intermediates and products - no transition states! O 2 SO 4 O O O OSO 3 + OSO 3 5.(12 pts) Addition of D 2 O to Z-3-methyl-2-pentene in the presence of D 2 SO 4 can occur by syn or by anti addition mechanisms. Draw the products of both types of addition, clearly indicating the stereochemistry around any chiral carbons that may result, and clearly labelling which is the syn product and which is the anti product. + D 2 O D 2 SO 4 + OD D OD D syn product anti product (there are more than one of each of these possible)

4 6.(12 pts) A high yield of 2-bromo-3-methylbutane is desired. Explain, using both words and a mechanism, why the reaction below would not be successful in forming a high yield of 2-bromo-3-methylbutane, and draw the structure of the main side product Once the initial carbocation (2 ) is formed, bromide ion could immediately bond to form the desired product. owever, a more stable 3 carbocation can be formed by a 1,2-hydride shift. This leads to substantial amounts of 2-bromo-2-methylbutane being formed. 7.(12 pts) or each of the following pairs of structural formulas, tell whether they represent identical species, conformers of the same species, constitutional isomers, enantiomers or diastereomers. 2 2 O 2 2 3O identical enantiomers diastereomers diastereomers

5 8.(8 pts) The enzyme aconitase catalyzes the hydration of aconitic acid to two products, citric acid and isocitric acid. Isocitric acid is optically active; citric acid is not. What are the respective structures (use perspective formulas or ischer projections) of citric acid and isocitric acid? O 2 2 O 2 O 2 O 2 aconitic acid 2 O 2 O O 2 citric acid O 2 2 O 2 O 2 O isocitric acid 9.(12 pts) The alkane formed by total hydrogenation of (S)-4-methyl-1-hexyne is optically active, while the one formed by total hydrogenation of (S)-3-methyl-1-pentyne is not. a) Show these products and explain this difference. Structures of the alkynes obviously (I hope!) do not have their stereochemistry specified. (S)-4-methyl-1-hexyne (S)-3-methyl-1-pentyne /Pd 2 2 /Pd The product of hydrogenation of (S)-4-methyl-1-hexyne is chiral: (S)-3-methylhexane. The product of hydrogenation of (S)-3-methyl-1-pentyne has two ethyl groups bonded to the "chiral" carbon, rendering it achiral, and the molecule achiral. b) Would each of the products of hydrogenation of these two compounds by a poisoned catalyst (e.g., Lindlar catalyst) be expected to be optically active? YES!