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14C_s11_lec01_e02_key

# 14C_s11_lec01_e02_key - Chemistry 14C Lecture 1 Spring 2011...

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Unformatted text preview: Chemistry 14C Lecture 1 Spring 2011 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. (13 x 12) + (9 x 1) + (3 x 35) + (2 x 14) + (1 x 16) = 314 2. (13 x 1.107%) + (9 x 0.015%) + (2 x 0.366%) + (1 x 0.037%) = 15.295% when M = 100%. 7.6% is half of 15.295%, so the intensity of M is scaled by the same factor. Intensity of M = 50%. 3. (a) Molecular ion: No. In question 1, we determined M has m/z = 314. Any ion with m/z < 314 is a fragment. (b) Base peak: Cannot determine. The base peak is the most abundant ion in the spectrum, and not necessarily a molecular ion. We do not know the relative ion abundances, so we cannot determine which peak in the spectrum is the base peak. (c) Fragment ion: Yes. In question 1, we determined M has m/z = 314. Any ion with m/z < 314 is a fragment. 4. Amide N-H ~3350 cm-1 Benzene ring C-H ~3050 cm-1 Benzene ring C=C ~1600 cm-1 and ~1500 cm-1 C=O ~1670 cm-1 5. Triclosan: Absent - No alcohol O–H 3650–3200 cm-1; no benzene C–H 3100–3000 cm-1. Triclocarban: Absent - No amide N–H 3500–3200 cm-1; no benzene C–H 3100–3000 cm-1; no C=O 1690–1650 c m -1 . Metacide 38: Present - C≡N at ~2200 cm-1. EDTA: Absent - No carboxylic acid O–H 3000–2500 cm-1; no C=O 1725–1700 cm-1. 6. (a) The 1H-NMR spectrum of EDTA includes three signals. Of these signals, three are singlets and none are triplets. When the smallest integral = 1, the total number of signals with integral = 1 is two. (b) The 13C-NMR spectrum of EDTA has three signals. Of these signals one is a singlet. Chemistry 14C Lecture 1 7. Exam 2 Solutions Page 2 COOH Most deshielded H N HOOC Spring 2011 COOH N HOOC Closest to TMS Most deshielded carbon 8. (a) Number of signals equal. Both have three signals in their 1H-NMR spectra. (b) Metacide 38. None of the carbons in metacide 38 are equivalent to other carbons in the molecule, resulting in six signals in its 13C-NMR spectrum. EDTA has some equivalent carbons, resulting in just three signals in its 13 C-NMR spectrum. 9. X-ray diffraction can determine the position of atoms in space and therefore any aspect of molecular structure than can be described by atomic positions. The technique cannot count electrons, determine formal charges, or measure energy. (a) That all carbon-chlorine bonds lengths are not equal can be determined (b) That the rings are aromatic cannot be determined. X-ray diffraction cannot ascertain if Hückel's rule is obeyed. (c) The number of atoms lying in the same plane can be determined. (d) The magnitude (in kcal mol-1) of the C–O barrier to rotation cannot be determined. 10. (a) m/z: Mass-to-charge ratio. (b) M: Molecular ion. 11. Silicon. This element causes shielding because of its low electronegativity (EN = 1.8). 12. (a) Splitting is caused by spin-spin coupling between neighboring nuclei within a molecule. (b) For a proton, its proton neighbor can have I = +1/2 or I = -1/2. (c) When a proton Ha has only one neighbor Hb, Ha has two spin flip energies. Hb's spin axis can be parallel or antiparallel to Ha's spin axis. 13. (a) Mass spectrometry does depend on a magnetic field to sort (or focus) ions according to their m/z. (b) Infrared spectroscopy does not depend on a magnetic field. (c) NMR does depend on a magnetic field to influence (or amplify) the spin flip energy. (d) X-ray crystallography does not depend on a magnetic field. 14. Mass spectrum: m/z = 236 (M): Zero or even number of nitrogens. m/z = 237 (M+1): 15.7%/1.1% = 14.3 C14 or C15. m/z = 238 (M+2): < 4% so no sulfur, chlorine, or bromine. Formula (C14): 236 - 168 (C14) = 68 amu for oxygen, nitrogen, and hydrogen. The 1749 cm-1 peak in the IR spectrum is due to a C=O stretch, so the formula must contain at least one oxygen atom. Chemistry 14C Lecture 1 Spring 2011 Exam 2 Solutions Page 3 Oxygens Nitrogens 68 - O - N = H Formula Comments 1 0 68 - 16 - 0 = 52 C14H52O Rejected; violates H-rule 2 0 68 - 32 - 0 = 36 C14H36O2 Rejected; violates H-rule 3 0 68 - 48 - 0 = 20 C14H20O3 Reasonable 4 0 68 - 64 - 0 = 4 C14H4O4 Rejected; more than four signals in 1H-NMR 1 2 68 - 16 - 28 = 24 C14H24N2O Rejected; poor fit for 1H-NMR integration 2 2 68 - 32 - 28 = 8 C14H8N2O2 Rejected; poor fit for 1H-NMR integration Formula (C15): 236 - 180 (C14) = 56 amu for oxygen, nitrogen, and hydrogen. The 1749 cm-1 peak in the IR spectrum is due to a C=O stretch, so the formula must contain at least one oxygen atom. Oxygens Nitrogens 56 - O - N = H Formula Comments 1 0 56 - 16 - 0 = 40 C15H40O Rejected; violates H-rule 2 0 56 - 32 - 0 = 24 C15H24O2 Rejected; poor fit for 1H-NMR integration 3 0 56 - 48 - 0 = 8 C15H8O3 Rejected; poor fit for 1H-NMR integration 1 2 56 - 16 - 28 = 12 C15H12N2O Rejected; poor fit for 1H-NMR integration DBE: 14 - (20/2) + (0/2) + 1 = 5 Five rings and/or pi bonds. Benzene ring possible. IR: Zone 1: Alcohol O–H: Present - strong, broad peak ~3450 cm-1. Amine/amide N–H: Absent - no peak in zone 1. Terminal alkyne ≡C–H: Absent - no peak in zone 1. Zone 2: Aryl/vinyl sp2 C–H: Present - peaks above 3000 cm-1. Alkyl sp3 C–H: Present - peaks below 3000 cm-1. Aldehyde C–H: Absent - no peak ~2700 cm-1. Carboxylic acid O–H: Absent - zone 2 peaks not broad enough. Zone 3: Alkyne C≡C: Absent - no peak ~ 2200 cm-1; not enough DBE for C=O plus C≡C plus benzene ring. Nitrile C≡N: Absent - no peak ~ 2200 cm-1; not enough DBE for C=O plus C≡N plus benzene ring. Zone 4: C=O: Present @ 1749 cm-1. Could be ester or ketone; 13C-NMR confirms ester. Zone 5: Benzene ring: Present - peaks at ~1600 cm-1 and ~ 1500 cm-1. Alkene C=C: Absent - peak at ~1600 cm-1 but not enough DBE for C=C plus C=O plus benzene ring. Chemistry 14C Lecture 1 1 Spring 2011 Exam 2 Solutions Page 4 H-NMR: Chemical shift Splitting Integral #H Implications 7.42–7.26 ppm multiplet 5 5H Phenyl group C6H5 3.82 ppm doublet 1 1H CH in CHCH 3.58 ppm singlet 1 1H CH or OH 2.57 ppm pentet 1 1H CH in CH3CHCH* or CH in CH2CHCH2 *This implication is selected because elsewhere in the 1HNMR data there is a CH3CH implication (1.14 ppm) but not a CH2CH2 implication. 1.14 ppm doublet 3 3H CH3 in CH3CH or 3 x CH in CHCH 0.94 ppm singlet 9 9H 3 x C H 3 or 9 x C H Totals 20 20 H C6H5 + CH + OH + CH + CH3 + (3 x CH3) = C12H20O 13 C-NMR: 175.4 ppm (singlet) is an ester C=O. The 151.3 ppm through 121.6 ppm signals are consistent with a phenyl group. The molecule has 14 carbons whereas the 13C-NMR has 10 signals, indicating the molecule has some equivalent carbons (some symmetry). Atom check: C14H20O3 (from mass spectrum) - C12H20O (from 1H-NMR) - CO2 (ester from IR or 13C-NMR) = C. That this carbon is not accounted for in the 1H-NMR means it is not bonded to a hydrogen atom. That this carbon is not accounted for in the IR means it is not part of a functional group that appears in the 4000–1450 cm-1 range of the IR. DBE check: Five (calculated for C14H20O3) - 1 (C=O) - 4 (benzene ring) = 0. All DBE used. Pieces: Ph CH in CHCH OH CH in CH3CHCH CH3 in CH3CH 3 x CH3 O=C–O (ester) C Assembly: The 1H-NMR splitting suggests that the CH3, CH, and CH pieces join to form CH3CHCH. Ph OH CH3CHCH O=C–O (ester) C 3 x CH3 The methyl groups of 3 x CH3 are equivalent, and not coupled. This can happen only if they are part of a tert-butyl group. Ph OH CH3CHCH O=C–O (ester) (CH3)C There are several ways to assemble these pieces into a structure that is consistent with the given data, any of which was accepted for full credit. (Note the C=O stretching frequency is not consistent with the carbonyl directly bonded to the benzene ring.) The actual structure is: OH O C(CH3)3 O CH3 ...
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