Chem 14C Lecture 1 Spring 2016 Exam 2 Solutions Page 1

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1 Chem 14C Lecture 1 Spring 2016 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. Infrared spectroscopy, 1 H-NMR, and 13 C-NMR. Mass spectrometry does not involve photon absorption. X-ray crystallography depends on photon diffraction, not photon absorption. 2. (a) Molecule A (sucrose) and molecule B (lactose). A quick check of the formulas answers the question, and calculation verifies if necessary. (b) Molecule A (sucrose) and molecule B (lactose). These have the most carbons; carbon is the greatest contributor to M+1. For M+1 intensities, four nitrogen atoms ~ one carbon atom. No calculations are needed. (c) Molecule A (sucrose) and molecule B (lactose). Molecular ion portion of the mass spectrum (M, M+1, and M+2) is identical when formula is identical, regardless of the exact molecules involved. (d) Molecule E (phenethylamine). An ion with m/z = 131 could be a molecular ion or a fragment. Thus this ion is possible in the mass spectrum of any molecule with a molecular weight (lowest mass isotopes) of 129 or more. m/z = 131 would be M+2 when M has m/z =129. Phenethylamine is the only choice with M having m/z of less than 129. (e) Molecule C (caffeine) and molecule D (theophylline). Both have six DBE. 3. If theobromine has a bromine atom in its structure, this will cause the mass spectrum to have an M+2 peak that is about the same intensity as the M peak. 4. (a) N (none of these). The cm -1 range hosts peaks due to C= stretch. The cm -1 range contains peaks due to sp 3 C H, aldehyde C H, and carboxylic acid H. All of the answer choices have sp 3 C H. (b) Molecule F (calone), molecule H (raspberry ketone), molecule I (nootkatone), and molecule K (carvone). Zone 4 hosts peaks due to C= stretch, while a peak in zone 5 indicates benzene ring or alkene. 2- Methylundecanal has a C= but no benzene ring or alkene. Limonene has no C=. (c) Molecules F K. Any proton near a pi bond is influenced by magnetic induction. Each of the choices has one or more pi bonds. 5. Aryl C-H cm-1 Alkyl sp 3 C-H cm-1 Ketone C= cm -1 ~1600 cm -1 plus cm

2 Chem 14C Lecture 1 Spring 2016 Exam 2 Solutions Page 2 6. Molecule H (raspberry ketone). The zone 1 peak is due to alcohol H or amine/amide N H. The 1717 cm -1 is due a C=. f the given choices, only raspberry ketone fits. 7. Calone has three hydrogen atoms that are not equivalent to any other hydrogen atoms in the molecule, and ten carbon atoms that are not equivalent to any other carbon atoms in the molecule. The three methyl group hydrogen atoms are equivalent to each other. In each CH 2 group the two hydrogen atoms are equivalent to each other. 8. (a) Molecule F (calone; has three) or molecule H (raspberry ketone; has two or three, depending upon exact chemical shift of the phenol H proton). Signals of 6.5 ppm or higher are due to benzene ring protons, aldehyde protons, or carboxylic acid protons (b) Molecules I (nootkatone), J (limonene), and K (carvone). None of these have 1 H-NMR signals > 6.5 ppm. Calone has three signals > 6.5 ppm, 2-methyundecanal has one, and raspberry ketone has two (benzene ring protons) or three (if the phenol H proton signal is > 6.5 ppm). (c) Molecule I (nootkatone). Nootkatone has 16 signals < 6.5 ppm. Calone has 3, 2-methylundecanal has 11, raspberry ketone has 4, and limonene has 12, and carvone has 10. (d) Molecule F (calone). A 13 C-NMR signal of 100 ppm or more is due to sp 2 carbons. Calone has seven 13 C- NMR signals of at least 100 ppm. 2-Methylundecanal has one, raspberry ketone, nootkatone, and carvone each have five, and limonene has four. (e) Molecule G (2-methyundecanal). 13 C-NMR triplets are due to CH 2 groups. Calone, raspberry ketone, and carvone have two CH 2 groups, 2-methylundecanal has eight, nootkatone has four, and limonene has three. (f) Molecule G (2-methylundecanal). Most deshielded proton = highest chemical shift. (g) Molecule G (2-methylundecanal). f the given choices, the aldehyde proton is more deshielded. 9. Numerous answers are possible by changing a hydrogen atom into something more electronegative. The change needs to be close to the aldehyde proton, but not so close as to change the splitting pattern. (Changing the aldehyde into a carboxylic acid changes a doublet into a singlet.) ne acceptable answer is: CH 3 (CH 2 ) 8 H 10. (a) The 1 H-NMR spectrum of 2-methylundecanal has zero singlets, two doublets, one triplet, and nine signals whose splitting is more complex than a triplet. (b) When the smallest integral is set to 1, the 1 H-NMR spectrum of 2-methylundecanal has two signals with integral = 1, eight signals with integral = 2, and two signals whose integral is neither 1 nor 2. The smallest integral for this molecule is one hydrogen. (c) Assuming all signals are resolved, the 1 H-NMR spectrum of raspberry ketone has six signals. 11. (a) =C C=C bond lengths, (b) =C C=C planarity, and (e) configuration of the stereocenters. X-ray crystallography measures the relative position of atoms in space, so any molecule feature that is defined solely by atomic positions can be measured. ther aspects might be confirmed or refuted based on bond angles, planarity, etc. but these other features cannot be directly measured. CH Mass spectrum: m/z = 194 (M): Molecular mass (lowest mass isotopes) = 194. Even number of nitrogen atoms.

3 Chem 14C Lecture 1 Spring 2016 Exam 2 Solutions Page 3 m/z = 195 (M+1): 9.2% / 1.1% = 8.4 Eight or nine carbons. m/z = 196 (M+2): Chlorine present. No bromine (intensity too small) and no sulfur (because Dr H promised M+2 would only be due to one atom). Formula (C 8 ): (C 8 ) - 35 ( 35 ) = 63 amu for oxygen, nitrogen, and hydrogen. xygens Nitrogens N = H Formula Notes = 63 C 8 H 63 Violates H-rule = 47 C 8 H 47 Violates H-rule = 31 C 8 H 31 2 Violates H rule = 15 C 8 H 15 3 Reasonable = 35 C 8 H 35 N 2 Violates H-rule = 19 C 8 H 19 N 2 Does not fit 1 H-NMR integrals = 3 C 8 H 3 N 2 2 Rejected - more than three signals in 1 H-NMR = 7 C 8 H 7 N 4 Rejected - no oxygen for C= in IR Formula (C 9 ): (C 9 ) - 35 ( 35 ) = 51 amu for oxygen, nitrogen, and hydrogen. xygens Nitrogens N = H Formula Notes = 51 C 9 H 51 Violates H-rule = 35 C 9 H 35 Violates H-rule = 19 C 9 H 19 2 Does not fit 1 H-NMR integrals = 3 C 9 H 3 3 Rejected - more than three signals in 1 H-NMR = 23 C 9 H 23 N 2 Violates H-rule = 7 C 9 H 7 N 2 Does not fit 1 H-NMR integrals DBE: 8 - [(15+1)/2] + (0/2) + 1 = 1 ne ring or pi bond. No benzene ring. IR: Zone 1 Alcohol H: Present - strong, broad peak. Amine/amide N H: Absent - no nitrogen. Terminal alkyne C H: Absent - not enough DBE; no C C at ~2200 cm -1. Zone 2 Aryl/vinyl C H: Absent - no peaks > 3000 cm -1 ; not enough DBE. Alkyl sp 3 C H: Present. Aldehyde C H: Absent - no peak ~2700 cm -1. Carboxylic acid H: Absent - not broad enough; C= stretching frequency too high. Zone 3 Alkyne C C and nitrile C N: Absent - no peak; not enough DBE. Zone 4 C=: present. Not enough DBE for conjugation with a carbon-carbon pi bond functional group cm -1 could be ester or ketone. 13 C-NMR ppm confirms ester. Zone 5 Benzene ring: No peak ~1600 cm -1 ; not enough DBE. Alkene C=C: No peak ~1600 cm -1 ; not enough DBE for C= and C=C.

4 Chem 14C Lecture 1 Spring 2016 Exam 2 Solutions Page 4 1 H-NMR: Chemical shift Splitting Integral # H Implications 4.13 ppm triplet 2 2 H CH 2 in CH 2 CH 2 CH 2 in CHCH 2 CH 2 x CH in CHCH 2 2 x CH in CHCHCH 3.65 ppm singlet 1 1 H CH or H 3.50 ppm triplet 2 2 H CH 2 in CH 2 CH 2 CH 2 in CHCH 2 CH 2 x CH in CHCH 2 2 x CH in CHCHCH 1.82 ppm singlet 6 6 H 2 x CH 3 or 3 x CH 2 or 6 x CH 1.62 ppm pentet 2 2 H CH 2 in CH 2 CH 2 CH 2 CH 2 in CHCH 2 CH ppm pentet 2 2 H CH 2 in CH 2 CH 2 CH 2 CH 2 in CHCH 2 CH 3 2 x CH in CH 2 CHCH 2 2 x CH in CHCHCH 3 2 x CH in CH 2 CH(CH) 2 2 x CH in CH 2 CHCH 2 2 x CH in CHCHCH 3 2 x CH in CH 2 CH(CH) 2 Totals H CH 2 + H + CH 2 + (2 x CH 3 ) + CH 2 + CH 2 = C 6 H C-NMR: ppm singlet is C= of ester. There are seven 13 C-NMR signals and eight carbons in the formula, so two carbons are equivalent. These equivalent carbons are the two methyl groups whose 1 H-NMR signal is at 1.82 ppm. Atom check: C 8 H 15 3 (from mass spectrum) - C 6 H 15 (from 1 H-NMR) - C 2 (ester from IR or 13 C-NMR) = C and. There are no hydrogen atoms bonded to this carbon atom, otherwise it would have been accounted for in the 1 H-NMR. This carbon atom is not part of a functional group that appears in the IR spectrum. DBE check: ne DBE (calculated from C 8 H 15 3 ) - one (C=) = all DBE used Pieces: CH 2 in CH 2 CH 2 2 x CH 3 C 2 (ester) H CH 2 in CH 2 CH 2 CH 2 C CH 2 in CH 2 CH 2 CH 2 in CH 2 CH 2 CH 2 Assembly: We begin as usual with the 1 H-NMR splitting. The CH 2 in CH 2 CH 2, CH 2 in CH 2 CH 2 CH 2, CH 2 in CH 2 CH 2, and CH 2 in CH 2 CH 2 CH 2 pieces combine to form CH 2 CH 2 CH 2 CH 2. CH 2 CH 2 CH 2 CH 2 2 x CH 3 C 2 (ester) H C The 2 x CH 3 cannot be attached to CH 2 CH 2 CH 2 CH 2 because this would violate the observed splitting patterns. The 2 x CH 3 cannot be bonded to the H or because this would violate the one molecule rule. The 2 x CH 3 are equivalent, and the ester group does not have two points to attach equivalent CH 3 groups. By elimination, the 2 x CH 3 must be bonded to the C. CH 2 CH 2 CH 2 CH 2 H 3 C C CH 3 C 2 (ester) H The 1 H-NMR chemical shifts for the end CH 2 groups of the CH 2 CH 2 CH 2 CH 2 too high for these to bonded to the (CH 3 ) 2 C unit or the C= side of the ester. They must be bonded to the, or the H, or the oxygen side of the ester. This leaves just two ways (that do not violate the one molecule rule) to assemble the remaining pieces:

5 Chem 14C Lecture 1 Spring 2016 Exam 2 Solutions Page 5 H H Choosing between these two with the given data is not possible, so either answer is acceptable.