14C_s10_lec01_e02_key

14C_s10_lec01_e02_key - Chemistry 14C Lecture 1 Spring 2010...

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Chemistry 14C Lecture 1 Spring 2010 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. 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 projected course grade cutoffs, consult the grading scale on the Chem 14C course web page. 1. All but TNT. Nitrogen rule: When a molecule has an even number of nitrogen atoms in its formula, m/z for M is an even number . 2. Plasticizer 3. PETN. Among carbon, hydrogen, nitrogen, and oxygen, only oxygen makes an appreciable contributor to M+2. (The natural abundance of 18 O is 0.2%). PETN has the most oxygen atoms, and therefore (relative to the intensity of its M peak) the most intense M+2 peak . 4. All of these. By definition, the base peak is the most intense peak in the spectrum, regardless of the molecule . 5. TNT 6. TNT and plasticizer. 3050 cm -1 is due to an aryl/vinyl sp 2 C–H stretch. 2950 cm -1 is due to an sp 3 C–H stretch. TNT and plasticizer are the only aromatic molecule choices . 7. (a) 3410 cm -1 (alcohol O–H stretch) and 1680 cm -1 (too low for an ester, even considering conjugation; no alkene is present). 3069 cm -1 is due to an aryl/vinyl sp 2 C–H bond. 2932 cm -1 is due to an sp 3 C– H bond (not necessarily an aldehyde C–H). 1729 cm -1 is too low for an ester, but not an ester conjugated with a benzene ring . (b) sp 3 C–H (because there are so many of them) and C=O. 8. Alcohol : Absent - no broad peak in zone 1. Aldehyde
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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_s10_lec01_e02_key - Chemistry 14C Lecture 1 Spring 2010...

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