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Unformatted text preview: Chemistry 14C 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. Molecule F has five sp 3 carbons . 2. Molecule E has seven lone pairs. Two on each oxygen atom and one on the nitrogen atom . 3. Ketone, alkene, nitrile, and ester. 4. C O O N C O O N This is the only other contributor in which the negative formal charge is on an atom more electronegative than carbon . Contributor I Contributor J 5. Between contributors I and J , contributor J is the less significant contributor because its negative formal charge is on a less electronegative atom (nitrogen in contributor J versus oxygen in contributor I ). 6. Molecule E has eleven conjugated atoms. In molecule E the greatest number of atoms that always lie in the same plane is twelve . C H 3 C O OCH 3 O N C H 3 C O OCH 3 O N H H H Conjugated atoms circled. Always coplanar atoms circled. 7. Molecules G and H . Both have twelve conjugated p orbitals. All other choices have fewer conjugated p orbitals . 8. Molecule F . Molecules D and E each have one sp 3 stereocenter; the remaining choices have none . 9. Constitutional isomers. They differ in the sequence of atom attachment . 10. Molecules D and E . Molecules without sp 3 stereocenters cannot have enantiomers. The presence of more than one stereocenter allows for diastereomers . Chemistry 14C Spring 2011 Final Exam Part B Solutions Page 2 11. C H 3 C O OCH 3 N 12. Molecules A , G , and H . The molecules with OH groups are hydrogen bond donors. No NH bonds are present . 13. Molecule B is more soluble in CH 3 OH than molecule A is soluble in CH 3 OH because molecule B 's negative formal charge makes for stronger hydrogen bonds and dipole-dipole interactions with the polar solvent (CH 3 OH). Stronger attractions result in higher solubility....
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