hem 14 Lecture 2 Spring 2017 Final Part B Solutions Page 1 Statistics: igh score, average, and low score will be posted on the course web site after exam grading is complete. 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 hemistry 14 course web page. 1. 18 13 3 6. E hydrogen (2.1) < E carbon (2.5) < E nitrogen (3.0) < E oxygen (3.5). 2. Molecule B contains five lone pairs. ne pair on each nitrogen atom and two pairs on the oxygen atom. 3. 4. When in its most stable (lowest energy) conformation, twelve (all) atoms of molecule B lie in the same plane. All conjugated atoms, plus all atoms directly bonded to these conjugated atoms, lie in the same plane. Verify with a model. 5. (a) Among molecules A D, molecule D has the largest M-LUM gap. Molecule D has the smallest number of conjugated p orbitals. (b) Among molecules A D, molecules A and are most likely to be colored. These molecules are tie for the largest number of conjugated p orbitals. 6. 2 2 2 2 Any other contributors are less significant and therefore earned less credit. 7. 8. Total number of pi electrons in molecule : Twelve. Two each in four pi bonds, and two each in two lone pairs. nly two of the molecule's five lone pairs are in p orbitals. umber of p orbitals in molecule 's closed loop: ine. Does molecule obey ückel's rule? Yes. The closed loop hosts ten pi electrons: eight from pi bonds and two from a nitrogen atom lone pair.
hem 14 Lecture 2 Spring 2017 Final Part B Solutions Page 2 Is molecule aromatic? Yes. Molecule has a closed loop of parallel, adjacent, overlapping p orbitals hosted by a planar atom loop, and ückel's rule is obeyed (ten pi electrons; n = 2). 9. β, galactose, pyranose, and hexose. 10. (c) ptically active but the direction of rotation cannot be ascertained. Molecule D is chiral and therefore optically active. owever one cannot determine if the molecule is dextrorotatory or levorotatory from just the molecular structure. 11. (h) Diastereomers. Molecule D and allose are stereoisomers differing in the configuration of at least one, but not all, stereocenters. 12. (f) o predictions can be made. We cannot make any predictions about the biological properties of one molecule based on the biological properties of its stereoisomer. 13. (a) False. Resolution is separation of enantiomers. Molecule A and molecule E are diastereomers. (b) False. A meso compound is achiral, whereas molecule D is chiral. (c) True. Molecule D has five stereocenters, and therefore 2 5 = 32 total stereoisomers (including molecule D itself). 14. Is molecule F a lipid? Yes. Molecule F is an example of a steroid. Is molecule G a lipid? Yes. Molecule G is an example of a hydrophobic vitamin. "Vitamin" is not sufficiently specific, because not all vitamins are lipids. 15. Best fit = molecule K. Molecules, I, and J are eliminated because these molecules have less than five signals in their 13 -MR spectra. Molecule L is eliminated because this molecule has two quartets (due to two nonequivalent methyl groups) in its 13 -MR spectrum. 16. (c) Electron density map. (b) Diffraction pattern and (e) Photo 51. 17. Diagram B is the best representation of a micelle. 18. (a) Molecule A ( 3 ). This is the only choice with hydrogen bonding. (b) Molecule D. This is the most polar molecule. (c) Molecule D, ( 3 2 ) 2 S. All of the molecules are nonpolar. This choice has the largest surface area, and therefore the strongest London dispersion forces.
hem 14 Lecture 2 Spring 2017 Final Part B Solutions Page 3 (d) Molecule D. This is the most polar molecule. 19. (a) Molecule D This molecule has the most resonance stabilization for the oxygen negative charge. (b) Molecule B ( 3 ). itrogen is less electronegative than oxygen. A molecule with a positive formal charge is unwilling to share electrons, and thus gain a +2 charge. (c) Molecule D (F 3 2 ). This molecule has the strongest electron-withdrawing inductive effect because it has the greatest number of fluorine atoms. (d) Molecule ( 3 + ). This molecule has a positive formal charge, and oxygen is more electronegative than nitrogen. 20. onacidic (the side chain is not more acidic than water) and hydrophilic. 21. A peptide or protein disulfide bridge is formed by the amino acid called cysteine, and is part of the tertiary structure of that peptide or protein. 22. 2 P itrogen atom missing 3 P itrogen atom missing DA is 2'-deoxy DA is 2'-deoxy P 23. Mass spectrum: m/z = 270 (M): Molecular mass (lowest mass isotopes) = 270 amu. Even number of nitrogen atoms. m/z = 271 (M+1): 13.73% / 1.107% = 12.40. Twelve or thirteen carbon atoms. m/z = 272 (M+2): omine present. o sulfur or chlorine. Formula ( 12 ): 270-144 ( 12 ) - 79 ( 79 ) = 47 amu for oxygens, nitrogens, and hydrogens.
hem 14 Lecture 2 Spring 2017 Final Part B Solutions Page 4 xygens itrogens 47 - - = Formula omments 0 0 47-0 - 0 = 47 12 47 Violates -rule. 1 0 47-16 - 0 = 31 12 31 Violates -rule. 2 0 47-32 - 0 = 15 12 15 2 Reasonable. 0 2 47-0 - 28 = 19 12 19 2 o oxygen for = in IR. 1 2 47-16 - 28 = 3 12 3 2 More than three signals in 1 -MR. Formula ( 13 ): 270-156 ( 13 ) -79 ( 79 ) = 35 amu for oxygens, nitrogens, and hydrogens. xygens itrogens 35 - - = Formula omments 0 0 35-0 - 0 = 35 13 35 Violates -rule. 1 0 35-16 - 0 = 19 13 19 Does not fit 1 -MR integrals. 2 0 35-32 - 0 = 3 13 3 2 More than three signals in 1 -MR. 0 2 35-0 - 28 = 7 13 7 2 o oxygen for = in IR. DBE: 12 - [(15+1)/2] + 0/2 + 1 = five DBE. Five pi bonds and/or rings. Possible benzene ring. IR: Zone 1: Alcohol : Absent - no strong peak. Amine/amide : Absent - no nitrogen in formula. Terminal alkyne : Absent - no strong peak; no peak in zone 3. Zone 2: Aryl/vinyl : Present - peaks > 3000 cm -1. Alkyl : Present - peaks < 3000 cm -1. Aldehyde : Absent - no peak ~ 2700 cm -1. 13 -MR confirms (no = doublet 180 220 ppm). arboxylic acid : Absent - not broad enough. 13 -MR confirms (no = 175 185 ppm) Zone 3: Alkyne and nitrile : Absent - no peak. Zone 4: =: Present at 1737 cm -1. Might be conjugated with carbon-carbon pi bond. 1737 cm -1 suggests only ester, aldehyde, or ketone. 13 -MR chemical shift confirms ester. Zone 5: Benzene ring: Present - peaks at ~1600 cm -1 and ~1500 cm -1. 13 -MR confirms with four signals 139 126 ppm. Alkene =: Absent - not enough DBE for = plus benzene ring plus alkene. 1 -MR:
hem 14 Lecture 2 Spring 2017 Final Part B Solutions Page 5 hemical shift Splitting Integral # Implications 7.27 7.20 ppm multiplet 2.5 5 Phenyl group ( 6 5 ) 4.21 ppm quartet 1.0 2 2 in 2 3 2 in 2 2 2.47 ppm quartet 1.0 2 2 in 2 3 2 in 2 2 2 x in 3 2 x in 2 2 x in () 3 2 x in 3 2 x in 2 2 x in () 3 1.28 ppm triplet 1.5 3 3 in 3 2 3 x in 2 3 x in 0.96 ppm triplet 1.5 3 3 in 3 2 3 x in 2 3 x in Totals 7.5 15 6 5 + 2 + 2 + 3 + 3 = 10 15 13 -MR: 173 ppm singlet is ester carbonyl carbon. 139 126 ppm is phenyl group. There are ten signals in the 13 -MR spectrum versus twelve carbons in the formula, so two carbons are equivalent to other carbons (i.e., there is some symmetry). Atom check: 12 15 2 (from mass spectrum) - 10 15 (from 1 -MR) - 2 (ester from IR) = and. This carbon is not attached to a hydrogen atom (otherwise it would have been accounted for in the 1 -MR), nor is it part of a functional group that appears in the IR. DBE check: Five DBE (calculated from 12 15 2 ) - benzene ring - = = all DBE used. Pieces: Phenyl group ( 6 5 ) 2 in 2 3 2 in 2 3 3 in 3 2 3 in 3 2 2 (ester) Assembly: We begin as we have every other time, with the 1 -MR splitting. 3 in 3 2 and 2 in 2 3 combine to form an ethyl group, 2 3. A second ethyl group, not equivalent to the first, is also formed. Phenyl group ( 6 5 ) 2 3 2 3 2 (ester) The ethyl groups cannot be attached to the phenyl group or the bromine atom because either of these would violate the one molecule rule. Both ethyl groups cannot be bonded to the ester because this would also violate the one molecule rule. Both ethyl groups cannot be attached to the because this would make the ethyl groups equivalent. So by elimination, one ethyl is attached to the and the other to the ester group. The 4.21 ppm 1 -MR chemical shift of the ester ethyl group is more consistent with = 2 3 than with the ethyl group bonded to the ester carbonyl carbon. Phenyl group ( 6 5 ) = 2 3 2 3 The phenyl group cannot be bonded to the bromine atom or to the = 2 3 because either of these would violate the one molecule rule. Thus the phenyl group must be bonded to the 2 3. = 2 3 Ph 2 3
hem 14 Lecture 2 Spring 2017 Final Part B Solutions Page 6 The remaining pieces can only be assembled in one way: 2 3 2 3 This was the only structure accepted for full credit.