1H
NMR Aid
This is an aid to 1H NMR analysis. Use it to ensure that you cover each step in your
analysis of a 1H NMR spectrum.
Given a 1H NMR and molecular formula, find the structure of the compound.
3H (s)
C5H10O2
6H (d)
1H (septet)
Step 1:  + r
Calculate how many pi () bonds plus rings (r) are in the compound.
3H (s)
C5H10O2
6H (d)
+r= 1
Step 1
1H (septet)
1
The compound must contain one  bond or one ring.
Step 2: Formulate Partial Structures for Each Signal
3H (s)
H
H C
H
C5H10O2
+r= 1
6H (d)
Step 2b
Step 2a
H
H
H C
H
H
H C
H
Step 2c
1H (septet)
The compound contains C, H and O atoms. From your knowledge that carbon has four
bonds, hydrogen one bond and oxygen two bonds, draw partial structures for each signal.
Step 2a: The hydrogen must be bonded to O or C. Step 2b: The three equivalent H atoms
must be bonded to carbon. Step 2c: The six equivalent H atoms must be bonded to two
carbon atoms (i.e., two methyl groups).
.
Step 3: Join Partial Structures That Have Neighbors
The number of H neighbors is determined by the number of lines in a signal. Subtract one
from the number of lines to get the number of neighboring H atoms. For a singlet (s),
which is one line, the number of neighbors is 1 – 1 or 0 (no neighbors). For a doublet (d),
the number of neighbors is 2 – 1 or 1 H neighbor, etc.
signal
singlet
doublet
triplet
quartet
septet
multiplet
symbol
s
d
t
q
septet
m
lines
one
two
three
four
seven
many
neighbors
zero
one
two
three
six
many
2
H
H C
H
C5H10O2
+r= 1
H
H C H
H
H C
H
3H (s)
6H (d)
Step 3a
C
H
Step 3b
1H (septet)
Step 3a: The 3H (s) has zero neighbors. Step 3b: The 1H septet has six equivalent
hydrogen neighbors, and the 6H (d) has one hydrogen neighbor, giving us an isopropyl
group. Note: The 1H cannot be bonded to O, or it would be a singlet. We have accounted
for 4 carbon and all 10 hydrogen atoms.
3
Step 4: Determine the remaining partial structures.
The remaining pieces are one carbon atom, two oxygen atoms and a ring or  bond,
which we cannot “see” in the NMR spectrum.
3H (s)
C5H10O2
+r= 1
6H (d)
O Step 4
C
O
H
H C
H
CH3
C H
CH3
1H (septet)
Note that the methyl and isopropyl groups identified in Step 3 define the ends of the
molecule. The remaining pieces must be interior pieces inside of these ends. One of the
oxygen atoms is probably part of a double bond to account for the  bond. The other
oxygen must have two single bonds. The partial structures account for all of the atoms
and the  bond.
4
Step 5: Combine the Partial Structures According To Their Chemical Shifts
Put the individual pieces together in the correct order to fit all the data—this gives the
structural formula of a compound.
3H (s)
C5H10O2
6H (d)
O
CH3
C
O
CH(CH3)2
1H (septet)
5.0
downfield
2.0 1.2
upfield
The three signals are centered at  = 5.0 (1H);  = 2.0 (3H) and  = 1.2 (6 H). The
downfield H must be bonded to a carbon bonded directly to oxygen. Then the four partial
structures can only go together as shown above.
Step 6: Ensure The Chemical Shifts Of All H Atoms Are Consistent With The
Proposed Structure.
After you have assigned chemical shifts to each signal, correlate the chemical shift with
known structural features. Those that you should memorize are given here. Others are
given in Sec. 13.9 of McMurry.
 0.9 to 1.9
 2.1 to 2.6
 3.4 to 5.5
 4.6 to 5.9
 7 to 8
 9 to 10.5
 10 to 13
Alkyl (or alkane-like) hydrogens (C-C-CH3)
Hydrogens bonded to a carbon that is bonded to a carbonyl (H-C-C=O)
Hydrogens bonded to a carbon bonded to a hetero atom (H-C-O) (H-C-Br)
Vinylic hydrogens (H-C=C-C) (i.e., H’s bonded to a double bond C)
Aromatic hydrogens (i.e., H’s bonded to a benzene or phenyl ring)
Aldehyde hydrogens (H-C=O) hydrogen bonded directly to a carbonyl C
Carboxylic acid hydrogens (C-COOH)
Do the correlation after finding as many partial structures as possible. The proposed
structure is shown below with the chemical shifts highlighted. They are consistent with
the values shown above.
5
H O
CH3
H C C O C CH3
H
s
= 2.0
6H (d)
= 1.2
H
1H (septet)
= 5.0
6