CHEM14C_f07_e02_key

CHEM14C_f07_e02_key - Chemistry 14C Fall 2007 Exam 2...

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Chemistry 14C Fall 2007 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. H 2 O and Br 2 2. Formula C 10 H 13 BrO so M has m/z = (10 x 12) + (13 x 1) + 79 + 16 = 228 3. (a) (10 x 1.107%) + (13 x 0.015%) + (1 x 0.037%) = 11.3% (b) 100% or more precisely, (49.31% / 50.69%) = 97%. 100% worth one point; 97% worth two points . 4. H O H (from removal of lone pair electron) or H O H (from removal of O–H bond electron). 5. Other answers are possible. (a) One significant similarity between the mass spectra of molecules Y and Z is the m/z and intensities for the molecular ions. (b) One significant difference between the mass spectra of molecules Y and Z is their fragmentation patterns. 6. (a) Peak ~1600 cm -1 due to alkene C=C stretch; also changes in fingerprint region (b) Peak ~3400 cm -1 due to O–H stretch; also changes in fingerprint region 7. Your IR spectrum must include a broad OH stretch ~3400 cm -1 , aryl sp 2 C–H stretches ~3050 cm -1 , alkyl sp 3 C–H stretches ~2950 cm -1 , as well as peaks ~1600 cm -1 and ~1450 cm -1 for the benzene ring. 8. Zone 1 through zone 5: nearly identical. The molecules have the same functional groups . Fingerprint region: significantly different. Molecules with similar structure and functional groups can have widely different fingerprint regions .
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CHEM14C_f07_e02_key - Chemistry 14C Fall 2007 Exam 2...

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