CHEM14C_w09_e02_key - Chemistry 14C Winter 2009 Exam 2...

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Chemistry 14C Winter 2009 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 may be listed here. To see the projected course grade cutoffs, consult the grading scale on the Chemistry 14C course web page. 1. sp 3 C–H (2960–2850 cm -1 ) and C=O (1750–1735 cm -1 ) 2. 3.8 ppm singlet (OCH 3 ), 2.2–3.0 ppm quartet (CH 2 ) and 0.9 ppm triplet (CH 3 ) 3. (a) The role of TMS in this sample is a reference point. (b) 0.00 ppm (c) More deshielded (d) The molecular phenomenon that has the greatest influence on the chemical shift of benzene’s hydrogen atoms is magnetic induction. 4. The 1 H-NMR spectrum of theobromine has four signals. When the smallest integral is 1.0, this spectrum has two signals of integral = 1.0, zero signals of integral = 1.5, and two signals with an integral of more than 1.5. 5. (a) It is a base peak. By definition, the peak with the highest relative ion abundance is the base peak. It is not necessarily a molecular ion or a fragment. It might have small or large m/z. Therefore (a) is the only correct answer . 6. ~99:1 Theobromine lacks sulfur, chlorine, and bromine, so when M = 100%, M+2 < 4%. 7. Theobromine and theophylline. For xanthine (C 5 H 4 N 4 O 2 ), M has m/z = 152. Theobromine and theophylline are isomers (both are C 7 H 8 N 4 O 2 ), with M having m/z = 180. For caffeine (C 8 H 10 N 4 O 2 ) m/z for M = 194 . 8. (a) 42 amu Three carbons and six hydrogens . (b) 3500–3300 cm -1 An N–H stretch .
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CHEM14C_w09_e02_key - Chemistry 14C Winter 2009 Exam 2...

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