Answer Key to Assignment #2

Chem 2600
Assignment 2 Answer Key
1. Deduce the structure of the molecule that gives the following NMR spectra. The formula is C 11H12O3 and peaks were
observed at 1748 and 1817 cm -1 in the IR spectrum. Assign all 1 H and 13 C NMR signals. Indicate how the spectra
identified key structural features of the unknown.
1 H NMR Spectrum:
7.11
3.70
7.11
7.11 7.11
2.34
O
The singlets that integrate for 3H are indicate of methyl groups. The singlet that integrates for 2H is a methylene which is
not adjacent to other protons. The signal at 7.11 ppm is suggestive of a doubly (para) substituted benzene ring.
13 C NMR Spectrum:
129.6
129.4
40.2
131.7
137.2
21.3
O
129.6
129.4
O
O
163.1 167.0
The resonances at 163.1 and 167.0 ppm are indicative of carbonyl groups. Combined with the IR data, it can be
concluded that an anhydride group must be present. The 4 signals between 129.4 and 137.2 ppm indicate a para
substituted benzene ring (two different groups).
O
O
2.24
20.2

2. Match the 1 H NMR spectra (1-4) with the correct chemical structure (A-D).
Spectrum 1: A. The aromatic region is indicative of a meta substituted benzene ring (two same groups). The presence of
only one septet and one doublet indicated these groups must be the same as well.
Spectrum 2: C. The aromatic region is indicative of a para substituted benzene ring (two same groups). The septet
appears as 2.6 ppm, which agrees with the following environment: H – Csp 3 – Csp 2 .
Spectrum 3: B. The aromatic region contains several overlapping signals. Clearly this must be one of the two meta
substituted benzene rings (A or B). The two different septets and doublets indicates that the two groups bound to the
benzene ring are different.
Spectrum 4: D. The aromatic region is indicative of a para substituted benzene ring (two same groups). The septet
appears as 5.3 ppm, which agrees with the following environment: H – Csp 3 – O – Csp 2 .
3. In the 1 H NMR spectra of the following two compounds, HA appears as a 1:2:1 triplet, but HB appears as a double of
doublets. Briefly explain why this is the case.
HA has the same magnitude of J coupling to each adjacent proton. Thus, it appears as a 1:2:1 triplet (n+1 rule applies).
HB has a different sized J coupling ( 3 J and 4 J) to each of the other two protons on the benzene ring. This causes a doublet
of doublets as explained by the following tree diagram.

4. In the 60 MHz spectrum of a certain compound there are two signals separated by 2 ppm.
i) What is this separation in Hz?
2 ppm (Hz/MHz) x 60 MHz = 120 Hz
ii) In a 250 MHz instrument, what is the separation between these signals in ppm?
2 ppm
iii) In a 250 MHz instrument, what is the separation between these signals in Hz?
500 Hz
5. For each of the following spectra, determine the structure.
i)
7
C14H14O. Relative integration is 5:2.
6
5
4
PPM
3
O
2
1
0

ii)
C5H12O. Relative integration is 2:1:1:2:6.
iii)
10
C9H10O3. Relative integration 1:2:2:2:3.
3
8
2
PPM
6
PPM
6. How many proton signals, and how many carbon signals would you expect from each of the following molecules?
A) 3C and 4H
B) 4C and 3H
C) 5C and 6H
OH
O O H
O CH 2CH 3
4
1
2
0
0

7. You wish to carry out the following reactions, and you have the 1 H NMR spectrum of each starting material. In each
case explain how the NMR spectrum of the product and starting material would be expected to differ.
a) The spectrum would simplify from two signals to one.
b) There would be a slight upfield shift of the lone signal.
8. How would you distinguish between the compounds within each set below using their 1 H NMR spectra? Explain
carefully, briefly and explicitly what features of the NMR spectra you would use.
a) Due to its symmetry, the first compound has only 3 signals while the second has 6 signals.
b) The first has two singlets while the second will have a singlet, a doublet and a multiplet.
c) All three will exhibit two doublets in the downfield vinyl region. The coupling constant, however, will be
approximately 10 Hz in the first case, 16 Hz in the second and 2 Hz in the last case.
9. There are four isomers with the formula C 4 H 9 Cl. The proton-decoupled 13 C NMR spectrum of one of the isomers is
shown below.
i) Of the four possible isomers, which might give this spectrum?
The four isomers are:
Cl
Cl
CH2Cl
Of these, only the first and last will exhibit four carbon signals. The second compound will have only two; the
third three.
ii) How would the proton-coupled 13 C NMR spectrum help you decide which isomer it is?
The proton coupled spectrum will give quartets for methyls, triplets for methylenes and doublets for methine
protons. Thus, the proton coupled spectrum for the first compound will have one quartet and three triplets while
the last compound will have two quartets, a doublet and a triplet.
Cl