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CHEM14C_f08_e02_key

CHEM14C_f08_e02_key - Chemistry 14C Fall 2008 Exam 2...

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Chemistry 14C Fall 2008 Exam 2 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. Some questions have more than one answer, even though only one answer is listed here. To see the projected course grade cutoffs, consult the grading scale on the Chemistry 14C course web page. 1. M+1 : (17 x 1.107%) + (21 x 0.015%) + (1 x 0.366%) + (4 x 0.037%) = 19.5-19.7% depending upon rounding. M+2 : 0%, 1%, and 2% all acceptable. No sulfur, chlorine or bromine present . 2. (a) M, (d) A molecular ion, and (g) An ion that does not contain 13 C. We cannot determine which peak is the base peak without seeing the entire spectrum, and examining the intensity of every ion in the spectrum . 3. 339 amu (R 3 NH + X - ) - 303 amu (R 3 N) = 36 amu for HX = HCl 4. Peaks in zone 1 are due to alcohol O–H, amine/amide N–H, or terminal alkyne C–H. These are present in Δ 9- THC (OH), tramadol (OH), and lidocaine (NH). The O–H stretch peaks for aspirin, ibuprofen, and naproxen appear in zone 2 because these are carboxylic acids OHs. 5. Most deshielded proton = any of the benzene ring protons. The principle cause of this deshielding is magnetic induction by the benzene ring pi electrons (with a contribution by the aryl ester carbonyl pi electrons).
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