14C_s11_lec01_finalB_key

14C_s11_lec01_finalB - Chem 14C Lecture 1 Spring 2011 Final Exam Part B Solutions Page 1 Statistics High score average and low score will be posted

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Unformatted text preview: Chem 14C Lecture 1 Spring 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 (or phenyl), amine, and ester. 2. Molecule A has five lone pairs and three atoms that are more electronegative than carbon. 3. C–O This bond is composed of the atoms with the largest electronegativity difference. Note that not all the answer choices are present in molecule A . 4. H N O O CH 3 H N O O CH 3 All other choices violate at least one resonance contributor preference rule, and are therefore less significant . 5. (c) Both structures are of approximately equal significance. In fact, they are degenerate . 6. (a) Conjugation : Stabilization due to delocalization of pi electrons in three or more parallel, adjacent, overlapping p orbitals. (b) Aromaticity : Stabilization due to delocalization of 4n+2 (n = any integer) pi electrons in a closed loop of three or more parallel, adjacent, overlapping p orbitals. (c) Anomeric carbon : In a cyclic carbohydrate, the carbon atom that becomes the carbonyl carbon in the carbohydrate's acyclic form. 7. Many answers are possible. For example: O OH HO OH 8. Any molecule adopts a shape, geometry, or conformation that allows conjugation because conjugation increases molecular stability. 9. (a) I and III; (b) I; and (c) I and III. 10. H N H CO 2 CH 3 H Chem 14C Lecture 1 Spring 2011 Final Exam Part B Solutions Page 2 11. In the set of molecules that includes Concerta and all its stereoisomers, there are four molecules that are enantiomers and four molecules that are diastereomers. Each stereoisomer in the set is an enantiomer as well as a diastereomer of some other stereoisomer in the set . 12. H N OCH 3 O H Cl N OCH 3 O H H Cl 13. Acid and electrophile 14. p K a = 10 Concerta's positive formal charge is neutralized upon deprotonation. Molecule A gains a negative formal charge upon deprotonation. Therefore Concerta is more acidic (lower pK a ) than molecule A ....
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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_s11_lec01_finalB - Chem 14C Lecture 1 Spring 2011 Final Exam Part B Solutions Page 1 Statistics High score average and low score will be posted

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