Using H NMR Spectroscopy for Structure Determination

Example #1
Using 1 H NMR Spectroscopy for Structure Determination
Deduce the structure of a compound with the formula C3H7Br that has the following 1 H NMR
spectrum: 3.1 ppm (triplet; integral = 1), 1.7 ppm (sextet; integral = 1), and 1.0
ppm (triplet; integral = 1.5).
To Start: calculate the number of double bond equivalents.
DBE = C - (H/2) + (N/2) + 1 = 3 - (7 + 1)/2 + 0/2 + 1 = 0. We conclude that the molecule has no
rings or pi bonds.
1 H-NMR Analysis
Step 1: summarize the 1 H-NMR data in a table.
Chemical Shift Splitting Integral # of H Implications
3.1 ppm triplet 1
1.7 ppm sextet 1
1.0 ppm triplet 1.5
Totals 3.5
Step 2: The total of the integrals is 1 + 1 + 1.5 = 3.5. Because there are seven hydrogens, we
conclude that each unit of the integral corresponds to two hydrogens.
Chemical Shift Splitting Integral # of H Implications
3.1 ppm triplet 1 2 H
1.7 ppm sextet 1 2 H
1.0 ppm triplet 1.5 3 H
Totals 3.5 7 H
Step 3: For each 1 H-NMR signal, use the number of hydrogens to decide if each signal
corresponds to a CH3, CH2, CH, OH, NH, etc. For example, a three hydrogen signal might arise
from a CH3, 3 x CH, or 3 x OH. OH is possible only when there is enough oxygen in the
formula, if the IR does not rule it out, and if the 1 H-NMR signal is not split.
Chemical Shift Splitting Integral # of H Implications
3.1 ppm triplet 1 2 H CH2
1.7 ppm sextet 1 2 H CH2
1.0 ppm triplet 1.5 3 H CH3
Totals 3.5 7 H
1

Step 4: Use the n+1 rule and the observed splitting patterns to determine all the possibilities that
would give the observed 1 H-NMR signals. Example: The 2H triplet at 3.1 ppm could be due to a
CH2, or 2 CH groups. The CH2 or 2 x CH groups must have two neighbors, as the observed
splitting pattern is a triplet. Keeping things simple, it would appear that the CH2 group has
another CH2 group next to it; this accounts for the triplet. The sextet at 1.7 ppm must arise from
a total of five neighboring hydrogens. Since a single carbon can’t have five hydrogens, there
must be two different sets of hydrogens that sum to five; let’s use a CH3 and a CH2 group. We
make record of this in the table. (In this table the underlined hydrogens correspond to the given
chemical shift.) The triplet at 1.0 ppm is determined accordingly.
.
Chemical Shift Splitting Integral # of H Implication
3.1 ppm triplet 1 2 H CH2 in CH2CH2
1.7 ppm sextet 1 2 H CH2 in CH2CH2CH3
1.0 ppm triplet 1.5 3 H CH3 in CH3CH2
Totals 3.5 7H
Step 5: We now sum up the atoms that are underlined. We do this as a check to make sure we
have used all the hydrogens and to look for atoms that do not show up in the 1 H-NMR spectrum,
such a COO portion of an ester.
Chemical Shift Splitting Integral # of H Implication
3.1 ppm triplet 1 2 H CH2 in CH2CH2
1.7 ppm sextet 1 2 H CH2 in CH2CH2CH3
1.0 ppm triplet 1.5 3 H CH3 in CH3CH2
Totals 3.5 7H CH2 + CH2 + CH3 = C3H7
Step 6: Atom Check. Look for unused atoms. Given formula - atoms from 1 H-NMR or other
sources such as IR = unused atoms. For this example: C3H7Br – C3H7 = Br. Therefore 1 H-NMR
data accounts for all of the atoms in the formula except the bromine. This bromine atom must be
included when assembling the pieces.
Step 7: DBE Check. Look for unused DBEs. DBEs calculated from formula – DBE accounted
for (by 1 H-NMR, etc.) = unused DBEs. In this example the formula has no DBEs and the 1 H-
NMR data does not require any DBEs, so there are no left over DBEs to consider when
assembling the pieces.
Step 8: List and assemble the pieces. At this point you’ve done 90% of the work. Now we
assemble the pieces to come up with the final structure. The pieces suggested by the 1 H-NMR
data are:
CH2 in CH2CH2
CH2 in CH2 CH2CH3
CH3 in CH3CH2
Br
2

Also include pieces suggested by other means such as IR data.
There is often many ways to go about assembling the pieces in solving NMR spectroscopy
problems. Use the clues provided by the spectra, such as coupling patterns. Sometimes trial and
error is all that works. In this case, the middle fragment uses all three of the available carbons
and fits with the other fragments as shown below.
CH2 in CH2CH2
CH2 in CH2 CH2CH3
CH3 in CH2CH3
The fragments have been reduced to CH3CH2CH2 and Br, which can only be assembled one way:
CH3CH2CH2Br
Step 9: Check to make sure the proposed structure is consistent with the original data, i.e.,
chemical shift (use table), splitting, and integration. In this case, examine CH2 next Br: is the
chemical shift consistent with what we expect?
This process might seem long and unnecessarily complicated for such a simple molecule, but can
be very useful in organizing your thoughts for more complex examples. The process becomes
much faster with practice and a little bit of experience.
Example #2:
Deduce the structure of a compound with the formula C9H12 that has the following 1 H NMR
spectrum: 7.40 – 7.02 ppm (multiplet, integral = 2.5); 2.57 ppm (triplet, integral = 1); 1.64
ppm (sextet, integral = 1) and 0.94 ppm (triplet, integral = 1.5).
To Start: calculate the number of double bond equivalents.
DBE = C - (H/2) + (N/2) + 1 = 9 - (12)/2 + 0/2 + 1 = 4. We conclude that the molecule has
either four double bonds or four rings, or some combination of the two. Given the number of
carbon atoms, we suspect a benzene ring.
1 H-NMR Analysis
Step 1: summarize the 1 H-NMR data in a table.
Chemical Shift
7.40 to 7.02 ppm
2.57 ppm
1.64 ppm
0.94 ppm
Splitting
multiplet
triplet
sextet
triplet
Totals
Integral
2.5
1
1
1.5
6
# of H
Implications
3

Step 2: The total of the integrals is 1 + 1 + 1.5 + 2.5 = 6. Because there are 12 hydrogens, we
conclude that each unit of the integral corresponds to two hydrogens.
Chemical Shift
7.40 to 7.02 ppm
2.57 ppm
1.64 ppm
0.94 ppm
Splitting
multiplet
triplet
sextet
triplet
Totals
Integral
2.5
1
1
1.5
6
# of H
5
2
2
3
12
Implications
Step 3: Again, for each 1 H-NMR signal, use the number of hydrogens to decide if each signal
corresponds to a CH3, CH2, CH, OH, NH, etc.
Chemical Shift
7.40 to 7.02 ppm
2.57 ppm
1.64 ppm
0.94 ppm
Splitting
multiplet
triplet
sextet
triplet
Totals
Integral
2.5
1
1
1.5
6
# of H
5
2
2
3
12
Implications
Step 4: As before, use the n+1 rule and the observed splitting patterns to determine all the
possibilities that would give the observed 1 H-NMR signals. Example: The 2H triplet at 2.57 ppm
is most likely due to a CH2 bonded to another CH2 group, as the observed splitting pattern is a
triplet. Determine the rest of table accordingly.
Chemical Shift
7.40 to 7.02 ppm
2.57 ppm
1.64 ppm
0.94 ppm
Splitting
multiplet
triplet
sextet
triplet
Totals
Integral
2.5
1
1
1.5
6
# of H
5
2
2
3
12
CH2
CH2
CH3
C 6 H 5
Implications
C 6 H 5
CH2 in CH2CH2
CH2 in CH2CH2CH3
CH3 in CH3CH2
4

Step 5: We now sum up the atoms that are underlined. Check to make sure we have used all the
hydrogens and to look for atoms that do not show up in the 1 H-NMR spectrum.
Chemical Shift
7.40 to 7.02 ppm
2.57 ppm
1.64 ppm
0.94 ppm
Splitting
multiplet
triplet
sextet
triplet
Totals
Integral
2.5
1
1
1.5
6
# of H
5
2
2
3
12
Implications
CH2 in CH2CH2
CH2 in CH2CH2CH3
CH3 in CH3CH2
C6H5 + CH2 + CH2 + CH3 =
C9H12
Step 6: Atom Check. Given formula - atoms from 1 H-NMR or other sources such as IR =
unused atoms. For this example: C9H12 – C9H12 = nothing. Therefore 1 H-NMR data accounts for
all of the atoms in the formula.
Step 7: DBE Check. For this example the formula has 4 DBEs and the 1 H-NMR data accounts
for 4 DBE. No additional DBE.
Step 8: List and assemble the pieces. The pieces suggested by the 1 H-NMR data are:
CH2 in CH2CH2
CH2 in CH2 CH2CH3
CH3 in CH3CH2
The fragments have been reduced to CH3CH2CH2 and mono-substituted benzene ring with a total
of two open valences (places to make a bond). There is only one way to assemble pieces. Final
structure is propylbenzene.
CH 2 CH 2 CH 3
C 6 H 5
Step 9: Check to make sure the proposed structure is consistent with the original data, i.e.,
chemical shift (use table), splitting, and integration, and IR data (if given).
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Example #3:
Deduce the structure of a compound with the formula C4H8O2 that has the following 1 H NMR
spectrum: 3.67 ppm (singlet, integral = 1.5); 2.32 ppm (quartet, integral = 1); 1.15 ppm (triplet,
integral = 1.5).
To Start: calculate the number of double bond equivalents.
DBE = C - (H/2) + (N/2) + 1 = 4 - (8)/2 + 0/2 + 1 = 1. The molecule has either one double bond
or one ring.
1 H-NMR Analysis
Chemical Shift
3.67 ppm
2.32 ppm
1.15 ppm
Splitting
singlet
quartet
triplet
Totals
Integral
1.5
1
1.5
4
# of H
Atom Check: C4H8O2 (given formula) – C3H8 (from 1 H-NMR) = CO2.
3
2
3
8
Implications
CH3
CH2 in CH2CH3
CH3 in CH3CH2
CH3 + CH2 + CH3 = C3H8
DBE Check: The 1 H-NMR data does not account for the one DBE. Now suspect an ester
functional group, since have CO2 remaining and one DBE. Being able to make inferences like
this is very important and requires practice.
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List and Assemble the Pieces:
CH 3
CH 2 in CH 2 CH 3
O
CH 3 in CH 3 CH 2 ester
We have an ethyl group and a methyl group, each with an empty valence. Since there are two
empty valences on the ester functional group, the ethyl and methyl groups can go on either end
of the ester; thus, there are two possible ways to assemble the fragments:
O
H 3 C C
O CH 2 CH 3
C
CH 3 CH 2
A B
O
O
C
O CH 3
Use chemical shift and splitting patterns to distinguish between these. Compound A would have
a singlet integrating for three at ~2.2 ppm for the CH3 group bonded to the carbonyl carbon.
Compound A would also have a quartet at ~3.8 ppm for the CH2 group bonded to the ether
oxygen. Our NMR data is not consistent with these predicted chemical shifts and splitting
patterns. We conclude, therefore, that compound B is the correct answer.
H 3 C C
Predict:
O
singlet @ 2.2 ppm
Predict:
quartet @ 3.8 ppm
Final Check for Consistency:
O CH 2 CH 3
CH 3 CH 2
A B
O
C
Predict: triplet @ ~0.9 ppm
Observe: triplet @ 1.15 ppm
Observe quartet @ 2.32 ppm
Predict: quartet @ 2.2 ppm
O CH 3
Predict: singlet @ 3.8 ppm
Observe: singlet @ 3.67 ppm
Compound B should have a singlet integrating for 3 protons at about 3.8 ppm; we have a singlet
at 3.67 ppm. We should have a quartet integrating for two protons at ~2.2 ppm; we have such a
signal at 2.32. We predict a triplet integrating for three protons at 0.9; we observe a triplet at
1.15 ppm. The data is consistent with the proposed structure; it is not consistent with compound
A.
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