chem236-2009fa-hex4-key - Fall 2009 Chemistry 236 Hour Exam...

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Chemistry 236 Hour Exam 4 Fall 2009 Last Name _______________________________ First Name _______________________________ 1. _______ 20 pts 2. _______ 12 pts 3. _______ 36 pts 4. _______ 10 pts 5. _______ 10 pts Total: __________ 100 pts 6. _______ 15 pts KEY ANSWER 20 12 36 10 10 15 103
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1 H NMR spectrum is shown below has the molecular formula C 5 H 10 O 2 a. (5 points) Calculate the degrees of unsaturation - show your work. b. (5 points) Given the NMR above, the degrees of unsaturation that you calculated, and that the 13 C NMR shows peaks at δ 146.1, 110.8, 71.1, and 29.4 ppm, propose a structure. degrees of unsaturation = 2 x 5 - 10 + 2 2 = 1 1 degree of unsat means C=O or C=C or ring. C13 does not indicate carbonyl but does show alkene. Only four peaks means two carbons are probably magnetically equivalent, which is consistent with 1 H NMR. Looks like three alkene protons in NMR spectrum and broad peak at 3.6 ppm could be -OH. Really only one structure fits this:
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This note was uploaded on 11/07/2011 for the course CHEM 236 taught by Professor Silverman during the Fall '08 term at University of Illinois, Urbana Champaign.

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chem236-2009fa-hex4-key - Fall 2009 Chemistry 236 Hour Exam...

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