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, 1H-NMR, and 13C-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 1800–1650 cm-1 range hosts peaks due to C=O stretch. The 3000-2700 cm-1 range
contains peaks due to sp3 C–H, aldehyde C–H, and carboxylic acid O–H. All of the answer choices have sp3
C–H.
(b) Molecule F (calone), molecule H (raspberry ketone), molecule I (nootkatone), and molecule K (carvone).
Zone 4 hosts peaks due to C=O stretch, while a peak in zone 5 indicates benzene ring or alkene. 2Methylundecanal has a C=O but no benzene ring or alkene. Limonene has no C=O.
(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.
Alkyl sp3 C-H
2960-2850 cm-1
Aryl C-H
3100-3000 cm-1
Ketone C=O
1750-1705 cm-1
~1600 cm-1 plus
1500-1450 cm-1
4000
3500
3000
2500
2000
1500
Chem 14C Lecture 1
Spring 2016
Exam 2 Solutions
Page 2
6.
Molecule H (raspberry ketone). The zone 1 peak is due to alcohol O–H or amine/amide N–H. The 1717 cm-1 is
due a C=O. Of 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 CH2 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 OH 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 1H-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 OH 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 13C-NMR signal of 100 ppm or more is due to sp2 carbons. Calone has seven 13CNMR 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). 13C-NMR triplets are due to CH2 groups. Calone, raspberry ketone, and
carvone have two CH2 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). Of 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.) One acceptable answer is:
O
CH3(CH2)8
H
CH2Cl
10. (a) The 1H-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 1H-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 1H-NMR spectrum of raspberry ketone has six signals.
11. (a) O=C–C=C bond lengths, (b) O=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. Other aspects might be confirmed or refuted based on bond angles,
planarity, etc. but these other features cannot be directly measured.
12. Mass spectrum: m/z = 194 (M): Molecular mass (lowest mass isotopes) = 194.
Even number of nitrogen atoms.
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 (C8): 194 - 96 (C8) - 35 (35Cl) = 63 amu for oxygen, nitrogen, and hydrogen.
Oxygens
Nitrogens
63 - O - N = H
Formula
Notes
0
0
63 - 0 - 0 = 63
C8H63Cl
Violates H-rule
1
0
63 - 16 - 0 = 47
C8H47ClO
Violates H-rule
2
0
63 - 32 - 0 = 31
C8H31ClO2
Violates H rule
3
0
63 - 48 - 0 = 15
C8H15ClO3
Reasonable
0
2
63 - 0 - 28 = 35
C8H35ClN2
Violates H-rule
1
2
63 - 16 - 28 = 19
C8H19ClN2O
Does not fit 1H-NMR integrals
2
2
63 - 32 - 28 = 3
C8H3ClN2O2
Rejected - more than three signals in 1H-NMR
0
4
63 - 0 - 56 = 7
C8H7ClN4
Rejected - no oxygen for C=O in IR
Formula (C9): 194 - 108 (C9) - 35 (35Cl) = 51 amu for oxygen, nitrogen, and hydrogen.
Oxygens
Nitrogens
51 - O - N = H
Formula
Notes
0
0
51 - 0 - 0 = 51
C9H51Cl
Violates H-rule
1
0
51 - 16 - 0 = 35
C9H35ClO
Violates H-rule
2
0
51 - 32 - 0 = 19
C9H19ClO2
Does not fit 1H-NMR integrals
3
0
51 - 48 - 0 = 3
C9H3ClO3
Rejected - more than three signals in 1H-NMR
0
2
51 - 0 - 28 = 23
C9H23ClN2
Violates H-rule
1
2
51 - 16 - 28 = 7
C9H7ClN2O
Does not fit 1H-NMR integrals
DBE: 8 - [(15+1)/2] + (0/2) + 1 = 1 One ring or pi bond. No benzene ring.
IR: Zone 1 Alcohol O–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 sp3 C–H: Present.
Aldehyde C–H: Absent - no peak ~2700 cm-1.
Carboxylic acid O–H: Absent - not broad enough; C=O stretching frequency too high.
Zone 3 Alkyne C≡C and nitrile C≡N: Absent - no peak; not enough DBE.
Zone 4 C=O: present. Not enough DBE for conjugation with a carbon-carbon pi bond functional group.
1739 cm-1 could be ester or ketone. 13C-NMR 171.0 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=O and C=C.
Chem 14C Lecture 1
1
Spring 2016
Exam 2 Solutions
Page 4
H-NMR:
Chemical shift
Splitting
Integral
#H
Implications
4.13 ppm
triplet
2
2H
CH2 in CH2CH2
CH2 in CHCH2CH
3.65 ppm
singlet
1
1H
CH or OH
3.50 ppm
triplet
2
2H
CH2 in CH2CH2
CH2 in CHCH2CH
1.82 ppm
singlet
6
6H
2 x CH3 or 3 x CH2 or 6 x CH
1.62 ppm
pentet
2
2H
CH2 in CH2CH2CH2
CH2 in CHCH2CH3
2 x CH in CH2CHCH2
2 x CH in CHCHCH3
2 x CH in CH2CH(CH)2
1.53 ppm
pentet
2
2H
CH2 in CH2CH2CH2
CH2 in CHCH2CH3
2 x CH in CH2CHCH2
2 x CH in CHCHCH3
2 x CH in CH2CH(CH)2
Totals
15
15 H
CH2 + OH + CH2 + (2 x CH3) + CH2 + CH2 = C6H15O
2 x CH in CHCH2
2 x CH in CHCHCH
2 x CH in CHCH2
2 x CH in CHCHCH
13
C-NMR: 171.0 ppm singlet is C=O of ester. There are seven 13C-NMR signals and eight carbons in the
formula, so two carbons are equivalent. These equivalent carbons are the two methyl groups whose 1H-NMR
signal is at 1.82 ppm.
Atom check: C8H15ClO3 (from mass spectrum) - C6H15O (from 1H-NMR) - CO2 (ester from IR or 13C-NMR) =
C and Cl. There are no hydrogen atoms bonded to this carbon atom, otherwise it would have been accounted for
in the 1H-NMR. This carbon atom is not part of a functional group that appears in the IR spectrum.
DBE check: One DBE (calculated from C8H15ClO3) - one (C=O) = all DBE used
Pieces:
CH2 in CH2CH2
OH
CH2 in CH2CH2
2 x CH3
CH2 in CH2CH2CH2
CH2 in CH2CH2CH2
CO2 (ester)
C
Cl
Assembly: We begin as usual with the 1H-NMR splitting. The CH2 in CH2CH2, CH2 in CH2CH2CH2, CH2 in
CH2CH2, and CH2 in CH2CH2CH2 pieces combine to form CH2CH2CH2CH2.
CH2CH2CH2CH2
OH
2 x CH3
CO2 (ester)
C
Cl
The 2 x CH3 cannot be attached to CH2CH2CH2CH2 because this would violate the observed splitting patterns.
The 2 x CH3 cannot be bonded to the OH or Cl because this would violate the one molecule rule. The 2 x CH3
are equivalent, and the ester group does not have two points to attach equivalent CH3 groups. By elimination,
the 2 x CH3 must be bonded to the C.
CH2CH2CH2CH2
OH
H3C–C–CH3
CO2 (ester)
Cl
The 1H-NMR chemical shifts for the end CH2 groups of the CH2CH2CH2CH2 too high for these to bonded to the
(CH3)2C unit or the C=O side of the ester. They must be bonded to the Cl, or the OH, 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:
Chem 14C Lecture 1
Spring 2016
Exam 2 Solutions
O
O
HO
Cl
O
Cl
OH
O
Choosing between these two with the given data is not possible, so either answer is acceptable.
Page 5