Download Exam Solutions for Chemistry 30B Winter 2004, Exam 2 and more Exams Organic Chemistry in PDF only on Docsity! Chemistry 30B Winter 2004 Exam 2 Solutions Page 1 Statistics: High score = 95 Average = 65.1 Standard Deviation = irrelevant as it does not control grade distribution in this class. A note about exam keys: The answers presented here are usually 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. Exam key posted 9 AM, Friday March 5, 2004. To see the estimated course grade cutoffs, consult the grading scale on the Chem 30B course web page. 1. (a) M (b) M+1 (c) More than one peak needed (d) More than one peak needed (e) M+2 (f) M 2. Spectroscopy Aldehyde Ketone IR C-H stretch at 2700 and 2900 cm-1 No C-H stretch IR C=O stretch 1720 โ 1740 cm-1 C=O stretch 1705 โ 1750 cm-1 1H-NMR C-H at 9.5 โ 11 ppm No C-H 13C-NMR CH in DEPT C (no hydrogens) in DEPT 3. DBE = C - (H/2) + (N/2) + 1 = 7 โ [(6 + 1)/2] + (1/2) + 1 = 5 The molecule has some combination of five ฯ bonds and/or rings. A benzene ring is a good possibility, but not definite. 4. (a) Region of IR spectrum 1450 โ 450 cm-1. Characteristic for a given compound. (b) Distortionless Enhancement by Polarization Transfer. A NMR method that reveals the number of hydrogens bonded to individual carbon atoms. Not an actor in a recent pirate movie. (c) Tetramethylsilane, (CH3)4Si. NMR reference compound. By definition, its 1H and 13C chemical shifts are both 0.00 ppm. Chemistry 30B Winter 2004 Exam 2 Solutions Page 2 (d) When comparing chemical shifts, the lower chemical shift (closer to TMS) is said to be upfield. (e) 6.5 โ 8.0 ppm. 5. (a) A carboxylic acid absorption is broad due to hydrogen bonding. (b) A carbonyl stretch is strong because a C=O bond is highly polar. (c) The fundamental molecular phenomenon that IR spectroscopy depends upon is absorption of a photon causes excitation to a higher vibrational quantum state. 6. OCH3 OH OCH3 H3C CH3H3C Furthest downfield 7. (a) 8 signals. (b) 4 singlets. (c) No triplets. 8. Mass spectrum M: Molecular weight = 192 Even number of nitrogens. M+1: 13.51 % / 1.1 % ~ 12 carbons M+2: No sulfur, chlorine or bromine Formula 192 โ 144 (12 carbons) = 48 amu left over for oxygen, nitrogen and hydrogen. Oxygens Nitrogens 48 โ O โ N = H Formula Notes None None 48 โ 0 โ 0 = 48 C12H48 Violates H-rule One None 48 โ 16 โ 0 = 32 C12H32O Violates H-rule Two None 48 โ 32 โ 0 = 16 C12H16O2 Reasonable Three None 48 โ 48 โ 0 = 0 C12O3 Rejected: Hโs show in NMR, IR None Two 48 โ 0 โ 28 = 20 C12H20N2 Reasonable One Two 48 โ 16 โ 28 = 4 C12H4N2O Rejected: More than 4 Hโs in NMR IR shows C=O so C12H20N2 is rejected because it has no oxygens.
