CHEM 14C Winter 12 Final Exam Part B Solutions,100% CORRECT

CHEM 14C Winter 12 Final Exam Part B Solutions 1. HO O O 2. Influence #1: Resonance causes some or the entire molecule to be planar. Influence #2: Resonance increases stability. Influence #3: Resonance causes electron delocalization. 3. Coplanar The carboxylic acid and ring B have this geometry because it allows for conjugation between the C=O and the B ring aromatic electron cloud. H 4. 5. Ten 6. A, B, and C 7. Of the carbon stereocenters in penicillin G, two have the R configuration, one has the S configuration, and zero are not R or S. Not including the structure shown above, there are 23 - 1 = 7 penicllin G stereoisomers. 8. Several answers are possible. Invert at least one, but not all, of the stereocenters of penicillin G. For example invert only the H N S CH3 stereocenter bearing the carboxylic acid: N O COOH CH3 9. (d) No conclusions can be drawn concerning the drug properties of the enantiomer. 10. The carboxylic acid of penicillin G is a stronger acid than acetic acid because the electron-withdrawing inductive effect of the nearby nitrogen atom stabilizes the penicillin carboxylate (the conjugate base of the penicillin carboxylic acid). 11. H N S CH3 N CH3 O COOH H3C 12. My modified structure for penicillin G is less acidic than acetic acid because the carboxylate moeity of its conjugate base has greater electron density due the methyl group’s electron-donating inductive effect. Alkyl groups (i.e., CH3) are weak electron donors. 13. (a) H HO HO (b)  COOH O OH anomeric carbon (c) D-glucuronic acid 14. D-glucopyranose 15. Nicotine 16. The nicotine-glucuronic acid conjugate is more soluble in water than nicotine itself because the conjugate has more areas for attractive interactions with water through hydrogen bonding, dipole-dipole, and cation-dipole forces than nicotine itself does. 17. Amide 18. A significant reason why this oxidation occurs at the five-membered ring of nicotine instead of the six- membered ring is oxidation of the six-membered ring disrupts aromaticity (more energetically expensive) whereas oxidation of the five-membered ring does not (less energetically expensive). 19. Waxes, triacylglycerols (or triglycerides), and phospholipids 20. Steroids and prostaglandins 21. Phospholipids 22. Numerous answers are possible. Feature #1: All amino acids are chiral, except glycine, which is achiral. Feature #2: All amino acids are primary amines (RNH2), except proline, which is a secondary amine (R2NH). 23. Mass spectrum: m/z = 188 (M): zero or even number of nitrogens; MW (lowest mass isotopes) = 188 m/z = 189 (M+1): 11.33% / 1.1% = 10.3 = 10 carbons m/z = 190 (M+2): < 4% so no S, Cl, or Br Formula: 188 - 120 (C10) = 68 amu for oxygen, nitrogen, and hydrogen. IR shows a distinct C=O at 1711 cm-1 so the formula must have at least one oxygen. Oxygens Nitrogens 68 - O - N = H Formula Comments 1 0 68 - 16 - 0 = 52 C10H52O Violates H-rule 2 0 68 - 32 - 0 = 36 C10H36O2 Violates H-rule 3 0 68 - 48 - 0 = 20 C10H20O3 Acceptable 4 0 68 - 64 - 0 = 4 C10H4O4 Rejected: more than four signals in 1H-NMR 1 2 68 - 16 - 28 = 24 C10H24N2O Rejected: does not fit 1H-NMR integration 2 2 68 - 32 - 28 = 8 C10H8N2O2 Rejected: does not fit 1H-NMR integration 0 4 68 - 0 - 56 = 12 C10H12N4 Rejected: does not fit 1H-NMR integration A very thorough student might also include C11 candidates, in which case C11H24O2 and C11H8O3 are also valid formulas. However these are both rejected as inconsistent with the 1H-NMR integration. DBE: 10 - (20/2) + (0/2) + 1 = 1 One ring or pi bond IR: Zone 1 Alcohol O–H: Uncertain - obscured by carboxylic acid O–H Amide or amine N–H: Absent - no nitrogen in formula C–H: Absent - not enough DBE; no CC stretch in zone 3 Zone 2 Aryl or vinyl C–H: Absent - not enough DBE for C=C plus C=O Alkyl C–H: Uncertain - obscured by carboxylic acid O–H Carboxylic acid O–H: Present - very broad; zone 4 C=O consistent Aldehyde C–H: Absent - zone 4 C=O inconsistent; not enough DBE for 2 C=O Zone 3 CC or CN: Absent - no peaks; not enough DBE Zone 4 C=O: Present. Not enough DBE to be conjugated. 1711 cm-1 consistent with carboxylic acid (seen in zone 2 and 13C-NMR) or ketone. Not enough DBE for two C=O (13C-NMR also confirms only one C=O present), and carboxylic acid is a better fit with the IR than ketone. Zone 5 Alkene or benzene ring: Absent - no peaks ~1600 cm-1; not enough DBE 1H-NMR: Shift Splitting Integral # H Implications 11.5 ppm singlet 1 1H COOH 3.83 ppm doublet 2 2 H CH2 in CH2CH 2 x CH in CHCH 3.62 ppm triplet 2 2 H CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH 2.58 ppm six lines or triplet of quartets 1 1 H CH in CH3CHCH2 CH in (CH2)2CHCH CH in CH3CH(CH)2 1.56 ppm triplet 2 2 H CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH 1.20 ppm doublet 3 3 H CH3 in CH3CH 3 x CH in CHCH 1.00 ppm singlet 9 9 H 3 x CH3 9 x CH Totals 20 20 H COOH + CH2 + CH2 (3 x CH3) = C9H20O2 + CH + CH2 + CH3 + 13C-NMR: Only one C=O present (181 ppm singlet). This could be a ketone or carboxylic acid. IR supports carboxylic acid. There are two less 13C-NMR signals than carbons in the formula, so some symmetry is present. Atom check: C10H20O3 (formula from mass spectrum) - C9H20O2 (1H-NMR) = CO. This carbon and oxygen are not bonded to any hydrogen atoms, nor are they part of any functional group that appears in the IR. DBE check: One DBE calculated for C10H20O3 used by C=O. Pieces: COOH CH in CH3CHCH2 3 x CH3 CH2 in CH2CH CH2 in CH2CH2 C CH2 in CH2CH2 CH3 in CH3CH O Assembly: The 1H-NMR splitting patterns suggest CH in CH3CHCH2, CH2 in CH2CH, and CH3 in CH3CH combine to form CH3CHCH2. The splitting patterns also suggest that CH2 in CH2CH2 and CH2 in CH2CH2 combine to make CH2CH2. COOH CH3CHCH2 3 x CH3 C CH2CH2 O The only way to have 3 x CH3 be equivalent (as required by the 1H-NMR) is to attach them to the C, making a tert-butyl group. COOH CH3CHCH2 (CH3)3C CH2CH2 O These pieces can be assembled in several ways that are consistent with the 1H-NMR splitting patterns. Deciding between them requires a detailed analysis of 1H-NMR and 13C-NMR chemical shifts, which is beyond the scope of Chem 14C. Any of these answers earned full credit. COOH O O HOOC O COOH COOH O