14C_w11_finalB_key

14C_w11_finalB_key - Chemistry 14C Winter 2011 Final Exam...

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Chemistry 14C Winter 2011 Final Exam Part B Solutions Page 1 Statistics : High score, average and low score will be posted on the course web site after exam grading is complete. The exam is ready to be picked up when these numbers are posted. A note about exam keys : The answers presented here may be significantly longer than expected from a student taking the exam. An exam key serves not only to reveal what was expected, but to instruct you as well. To see the final course grade cutoffs, consult the grading scale on the Chem 14C course web page. 1. Benzene ring, alcohol, and ketone. 2. Either of the first two can be labeled as most significant. The third resonance contributor can be based on either of the Kekulé structures and either oxygen. Any other possibilities are less significant. O O HN CH 3 O O HN CH 3 O O HN CH 3 3. Twelve atoms are conjugated. 4. Ninhydrin and MDMA are colorless whereas the test reaction product is red because neither MDMA nor ninhydrin have enough conjugated p orbitals to provide a HOMO-LUMO gap that is small enough for the molecules to absorb visible light, whereas the product does. 5. Both rings or Their aromaticity cannot be considered separately. 6. Ninhydrin has zero stereocenters. Including the structure shown at the top of the page, there are two stereoisomers, two MDMA enantiomers, and zero MDMA diastereomers. Enantiomers come in pairs and diastereomers groups of two or more . 7. Ninhydrin is achiral (it has no stereocenters). MDMA is chiral (it has one stereocenter). 8. The only other stereoisomer that can be drawn is the enantiomer. O O H N H CH 3 9. Biological effects cannot be predicted. 10. O O HN CH 3 H O H H O O H 2 N CH 3 + H 2 O 11. (a) MDMA accepts a proton on nitrogen instead of oxygen because nitrogen is less electronegative than oxygen. In the base, nitrogen is more willing to shares it electron density. In the conjugate acid, nitrogen can accommodate the positive formal charge more effectively than oxygen. In addition, the oxygen lone pairs enjoy conjugation and resonance delocalization with the benzene
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This note was uploaded on 09/06/2011 for the course CHEM 140A taught by Professor Whiteshell during the Fall '04 term at UCSD.

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14C_w11_finalB_key - Chemistry 14C Winter 2011 Final Exam...

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