Chemistry 14C Winter 2017 Exam 2 Solutions Page 1

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1 Chemistry 14C Winter 2017 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. 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 Chemistry 14C course web page. 1. (a) m/z = 30. Formula CH 2. (1 x 12) + (2 x 1) + (1 x 16) = 30. (b) (1 x 1.107%) + (2 x 0.015%) + (1 x 0.037%) = 1.174% rounds to 1.17%. (c) 0. Not enough oxygen, sulfur, chlorine, or bromine to give an intensity of more than 0.5%, so rounds to zero. 2. Base peak Relative ion abundance (%) Fragment ions M M+1 Mass-to-charge ratio (m/z) The x-axis scale does not go far enough to include an M+2 peak. 3. Answers other than those given here may be possible. (a) ne important difference between the IR spectra of formaldehyde and 1,3,5-trioxolane is that the formaldehyde spectrum has aldehyde C H stretch peaks at ~2700 cm -1 and 2900 cm -1 whereas the trioxane spectrum lacks these peaks. (b) ne important difference between the IR spectra of formic acid and formaldehyde is that the formic acid spectrum has a very broad carboxylic acid H stretch in zone 2 whereas the formaldehyde spectrum lacks this peak. 4. The 1 H-NMR spectrum of a mixture containing equal amounts of 1,3,5-trioxane, formaldehyde, and formic acid contains four signals total. (ne each for 1,3,5-trioxane and formaldehyde, and two for formic acid.) f these signals, four occur at a chemical shift greater than 3.5 ppm, and four are singlets. Check the 1 H-NMR chemical shifts table. Note that CH 3 typically occurs around at 3.8 ppm, so CH 2 must be more deshielded than 3.8 ppm. 5. ther explanations may be possible in each case. (a) This mass spectrum is not conclusive for alizapride because the alizapride M has m/z = 315. (b) This mass spectrum is not conclusive for tramadol because it might be a constitutional isomer of tramadol (same molecular formula, different connectivity of the atoms).

2 Chemistry 14C Winter 2017 Exam 2 Solutions Page 2 6. Alcohol -H ~3400 cm -1 Benzene ring ~1600 cm -1 Aryl C-H ~3050 cm -1 sp 3 C-H ~2900 cm -1 Benzene ring ~1500 cm cm cm cm cm cm cm When tramadol contains one atom of carbon-14 the m/z for M will be 265. 'Normal' tramadol having 12 C, 1 H, 14 N, and 16 has m/z = 263 for M. So when one carbon-12 is replaced with one carbon-14 the m/z is increased by two mass units. 8. All possible present except molecule B. The IR spectrum lacks a broad carboxylic acid H stretch in zone 2, and the observed C= stretching frequency of 1687 cm -1 is too low for a carboxylic acid, which is typically in the cm -1 range. 9. (a) Has the most signals total: Molecule F (nine signals). Has the least number of signals total: Molecule A (one signal). (b) Has the most signals with chemical shift less than 6.0 ppm: Molecule F (six signals). For molecules A F only the benzene ring and carboxylic acid proton signals have a chemical shift of 6.0 ppm or more. (c) Has the most deshielded proton: Molecule B (the carboxylic acid H proton). Has the proton with the highest chemical shift: Molecule B. Highest chemical shift = most deshielded. (d) Has the most singlets: Molecules A, B, or E. Each has just one singlet. Has the most total singlets with chemical shift less than 6.0 ppm: Molecule E. Benzene ring proton and carboxylic acid H protons both have chemical shifts > 6.0 ppm. 10. In the 1 H-NMR spectrum of molecule C, when the smallest integral = 1.0, the integral for the CH 3 group is three. The smallest integral belongs to a single proton (the benzene ring proton furthest from the carbonyl), so a threeproton group has an integral of three. 11. (a) None of these. (b) Molecule C. 12. (a) The IR data indicates the oxygen atom is not an H or a carbonyl group, so it must be an ether. ne DBE calculated from the formula. 1 H-NMR data indicates two sets of coupled CH 2 groups. (b) CH 2 CH 3 The IR indicates a carbonyl, which uses the only DBE. The 1 H-NMR indicates a CH 3 and CH 2 CH 3.

3 Chemistry 14C Winter 2017 Exam 2 Solutions Page Mass spectrum: m/z = 159 (M): Molecular weight (lowest mass isotopes) = 159 amu. dd number of nitrogen atoms. m/z = 160 (M+1): 9.55% / 1.107% = 8.62 Eight or nine carbons. m/z = 161 (M+2): < 4% so no sulfur, chlorine, or bromine. Formula (C 8 ): (C 8 ) = 63 amu for oxygens, nitrogens, and hydrogens. xygens Nitrogens N = H Formula Notes = 49 C 8 H 49 N Violates H-rule = 33 C 8 H 33 N Violates H-rule = 17 C 8 H 17 N 2 Reasonable = 1 C 8 HN 3 Rejected: more than one 1 H-NMR signal = 21 C 8 H 21 N 3 Violates H-rule = 5 C 8 H 5 N 3 Rejected: more than five 1 H-NMR signals. Formula (C 9 ): (C 9 ) = 51 amu for oxygens, nitrogens, and hydrogens. xygens Nitrogens N = H Formula Notes = 37 C 9 H 37 N Violates H-rule = 21 C 9 H 21 N Rejected: Does not fit 1 H-NMR integrals = 5 C 9 H 5 N 2 Rejected: more than five 1 H-NMR signals = 9 C 9 H 9 N 3 Rejected: Does not fit 1 H-NMR integrals. DBE: 8 - (17/2) + (1/2) + 1 = one DBE. ne ring or one pi bond. No benzene ring. IR: Zone 1: Alcohol H: Absent - no peak. Amine/amide N H: Absent - no peak. Terminal alkyne C H: Absent - no peak; not enough DBE; no C C in zone 3. Zone 2: Aryl/vinyl C H: Absent - no peaks > 3000 cm -1. Alkyl C H: Present. Aldehyde C H: Absent - no peak ~ 2700 cm -1. Carboxylic acid H: Absent - not broad enough. Zone 4 inconsistent with carboxylic acid. Zone 3: Alkyne C C: Absent - no peak; not enough DBE. Nitrile C N: Absent - no peak; not enough DBE. Zone 4: C=: 1661 cm -1. Not enough DBE for conjugation with a carbon-carbon pi bond. nly tertiary amide is possible (no amide N H in zone 1). Peak is too strong for a typical alkene, which would need to be R 2 C=CR 2 due to lack of vinyl =C H stretch in zone 2, and lack of vinyl hydrogen signals in H-NMR. Zone 5: Benzene ring: Absent - no peak ~ 1600 cm -1 ; not enough DBE. Alkene C=C: Absent - no peak ~ 1600 cm -1 ; not enough DBE.

4 1 H-NMR: Chemistry 14C Winter 2017 Exam 2 Solutions Page 4 Chemical shift Splitting Integral # H Implications 3.48 ppm quartet H CH 2 in CH 2 CH 3 CH 2 in CH 2 CH 2 CH 3.35 ppm triplet H CH 2 in CH 2 CH 2 CH 2 in CHCH 2 CH 2 x CH in CHCH 3 2 x CH in CH 2 CHCH 2 x CH in CH(CH) 3 2 x CH in CHCH 2 2 x CH in CHCHCH 2.98 ppm singlet H 2 x CH 3 or 3 x CH 2 or 6 x CH 2.34 ppm triplet H CH 2 in CH 2 CH 2 CH 2 in CHCH 2 CH 1.84 ppm pentet H CH 2 in CH 2 CH 2 CH 2 * CH 2 in CHCH 2 CH 3 2 x CH in CHCH 2 2 x CH in CHCHCH 2 x CH in CH 2 CHCH 2 2 x CH in CHCHCH 3 2 x CH in CH 2 CH(CH) ppm triplet H CH 3 in CH 2 3 x CH in CHCH 2 3 x CH in CHCHCH Totals H CH 2 + CH 2 + (2 x CH 3 ) + CH 2 + CH 2 + CH 3 = C 7 H 17 * Selected as most likely implication for the 1.84 ppm pentet signal because no CHCH 2 is present in any other selected implication. Atom check: C 8 H 17 N 2 (from mass spectrum) - C 7 H 17 (from 1 H-NMR) - N C= (tertiary amide from IR) = one oxygen atom left over. There is no H and only one C= in IR (due to DBE count) so this oxygen must be an ether. DBE check: ne (calculated for C 8 H 17 N 2 ) - one (C= in IR) = all DBE used. Pieces: CH 2 in CH 2 CH 2 in CH 2 CH 2 N C= (tertiary amide CH 2 in CH 2 CH 2 CH 2 in CH 2 CH 2 CH 2 (ether) 2 x CH 3 in CH 2 Assembly: As always, we begin with the 1 H-NMR splitting. CH 2 in CH 2 CH 3 and CH 3 in CH 2 join to form one ethyl group. CH 2 CH 2 in CH 2 CH 2 N C= (tertiary amide CH 2 in CH 2 CH 2 CH 2 in CH 2 CH 2 CH 2 (ether) 2 x CH 3 CH 2 in CH 2 CH 2 and CH 2 in CH 2 CH 2 join to form one CH 2 CH 2. This CH 2 CH 2 joins with CH 2 in CH 2 CH 2 CH 2 to form CH 2 CH 2 CH 2. CH 2 CH 3 N C= (tertiary amide CH 2 CH 2 CH 2 CH 2 CH 2 (ether) 2 x CH 3 The 2 x CH 3 signal indicates these methyl groups are equivalent and have no neighbors. This can only be accommodated by attaching them to the tertiary amide. CH 2 CH 3 (CH 3 ) 2 N C= (tertiary amide CH 2 CH 2 CH 2 CH 2 CH 2 (ether)

5 Chemistry 14C Winter 2017 Exam 2 Solutions Page 5 The (CH 3 ) 2 N C= group cannot be bonded to CH 2 CH 3, because this creates one complete molecule, and thus violates the one molecule rule. This leaves just two possible final structures, both of which were accepted for full credit: N or N