# PMR Lesson 6 Coupling Diagrams

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```					PMR Lesson 7: Coupling Diagrams
A visual image of that explains how a multiplet is generated is called a coupling diagram.
The steps behind construction of a coupling diagram are really pretty simple. With it you
can explain why the n+1 rule works and you can explain more complex coupling patterns
where more than one J value comes in to play. The steps used to generate a coupling
diagram are described below.

Step 1) Draw a hash mark at the top of the
diagram to represent the original NMR
signal for an uncoupled proton. This hash
mark represents the chemical shift of the
hydrogen atom and will be the midpoint of
any multiplet pattern generated. See Figure
PMR7-1.

down field     ppm scale     up field

Figure PMR7-1. The uncoupled signal.

Step 2) Using the largest J value, split the
original line into two lines equally apart
from the center line to represent the positive
and negative interaction of the interacting
(coupling) nucleus with the hydrogen being
described. If you have a J value of 7 Hz,
then the two new lines should be 3.5 Hz on                                J = 7 Hz
either side of the original hash mark. (Pick a
convenient scale for drawing coupling
diagrams, e.g., 1 Hz = 1 cm or 1 Hz = 1
mm.) If the hydrogen experiences no other
coupling interactions the resulting pattern
would be the doublet shown in step 2. The                 down field                      up field
ppm scale
size of each line in the doublet pattern
(excluding any perturbation caused by the                     Figure PMR7-2 Evolution of a
leaning effect) is a direct reflection of the                 doublet by coupling with a single
number of pathways leading to each line. In                   neighbor where J = 7 Hz.
this example, the lines in the doublet are
equal (1:1), as predicted using Pascal’s
triangle. See Figure PMR7-2.
Step 3) If the hydrogen experiences another
coupling interaction then it must be
described separately. The previous step
describes the interaction of the two                                  J = 7 Hz
methylene protons in BrCH2-CHCl2 with the
adjacent methine. But the coupling of the
methine proton in BrCH2-CHCl2 with the
two methylene protons requires another step,
one step for each neighbor. This is done by
independently treating each of the lines
J = 7 Hz                J = 7 Hz
formed in step 2 as you treated the original
line in step 1. The new lines are centered
about the line from which they evolve by the
appropriate     distance,    indicating     the
magnitude (J value) of the interaction.              down field       ppm scale        up field
Since the second nucleus couples with the
proton where the same J value (7 Hz) center        Figure PMR7-3. Evolution of a triplet
lines overlap and the result is a 1:2:1 triplet,   pattern where each line derived from the
as predicted by use of the n+1 rule.               doublet of the first interaction (J = 7Hz) is
further split into a second doublet by
coupling to a second nuclei, with J = 7 Hz. A
triplet is really a doublet of doublets.

Alternate Step 3) If the hydrogen does
experience another coupling interaction but
the J value of the second interaction is not
the same as in the first, you should generate
J = 7 Hz
two new lines as before, but the result will
be different. This can happen for an
aromatic proton that experiences both an
ortho and a meta coupling interaction. If the                 J = 3 Hz                  J = 3 Hz
second nucleus couples with the proton
where J = 3 Hz, then the diagram shown
here would be generated. This coupling                   down field        ppm scale         up field
diagram represents the pattern expected
when proton Ha couples with proton Hb and                Figure PMR7-4. Evolution of a dd
Hc, where Jab = 7 Hz and Jac = 3 Hz. The                 pattern where each line derived from
pattern generated is called a doublet of                 the doublet of the first interaction (J =
doublets and is abbreviated as d,d. See                  7Hz) is further split into a second
doublet by coupling to a second nuclei,
Figure PMR7-4.
with J = 3 Hz.
The pattern shown in Figure PMR7-4 could NOT be predicted by use of Pascal’s triangle.
Pascal’s triangle only applies to interaction of nuclei where all the J values are the same.
The n+1 rule (from 2nI + 1, where I is spin, for 1H and 13C, I = ½) is also limited to
describing coupling patterns between nuclei with one J value.

One limitation of predicting coupling patterns (i.e., Pascal’s triangle, n+1 rule, coupling
diagrams) is that differences in chemical shift (as measured in Hz) between coupling
nuclei must be much larger than the J values. J values are constant and independent of
field strength. Chemical shifts, in terms of ppm, are also independent of field strength.
However, the corresponding Hz value of a particular ppm chemical shift is not
independent of field strength. The separation (∆υ) in Hz of two signals in a NMR
spectrum is dependent upon field strength. As a general rule of thumb, a spectrum will
be predictably recognizable (known as first order) if the value of J/∆υ is less than 0.1. J
is the coupling constant and ∆υ is the difference in Hz of the chemical shifts of the two
coupling nuclei.

For a 300 MHz NMR, ortho (J = 7 Hz) and meta (J = 3 Hz) coupling interactions of
aromatic protons should appear as first order patterns provided that the difference in
chemical shift of the coupling aromatic protons is greater than 0.23 ppm (69 Hz). If a
larger field NMR was used, first order spectra could be obtained for aromatic protons
whose chemical shifts are closer than 0.23 ppm.

For a review of coupling diagrams, consider the analysis of one of the signals in the 1H
NMR spectrum of methyl 3-nitro-4-methyl benzoate. (Figure PMR7-5, next page) The
aromatic proton labeled Hb experiences two different types of coupling. The two
couplings are independent of each other. The larger, 7 Hz coupling is an ortho coupling
to Hc, and the smaller 3 Hz coupling is a meta coupling to Ha. The coupling diagram for
Hb is shown in Figure PMR7-5.

An Aside: You have acquired three tools to deal with coupling interactions. They each have their
own advantages and special uses. Here is a quick review of how each tool is used.

Pascal’s triangle: Used to determine the ideal relative intensity of peaks within a multiplet in case
your memory fails you at a critical moment. Greatest value when looking at spectral data.
Coupling diagrams: Used to predict complex, first order coupling patterns or to interpret complex
first order coupling patterns. Greatest value when trying to match spectral data to a structure.
Multiplicity rule (n+1): Used to predict or interpret simple first order spectra. Greatest value is
ease of use, especially when applied to aliphatic signals in NMR.
O–
Ha       O
N+
O
OCH3

Hb
H3C
Hc
Chemical shift is measured at
the center of the multiplet pattern.

Figure PMR7-5. Hb split into a doublet of doublets (d,d) by ortho
coupling to Hc and meta coupling to Ha.

QPMR7-1.       Describe the coupling pattern
expected for Ha in Figure PMR7-5.

QPMR7-2.       Describe the coupling pattern APMR7-1. Ha couples only with Hb. Jab = 3 Hz.
expected for Hc in Figure PMR7-5.            Ha will appear as a doublet.

J = 3 Hz

QPMR7-3. If the ppm values of Ha, Hb and Hc APMR7-2. Hc couples only with Hb. Jcb = 7 Hz.
were predicted to be 8.78, 8.24 and 7.43 ppm Hc will appear as a doublet.
respectively, what would the NMR spectrum of
the aromatic portion of the compound in Figure
19 look like?

J = 7 Hz
QPMR7-4. Would the following protons exhibit APMR7-3.
a first order spectrum at 60 MHz?

7.3 ppm (2H, d, Jab = 7 Hz, Ha),
7.2 ppm (2H, d, Jba = 7 Hz, Hb)
Ha        Hb           Hc

QPMR7-5. Would the system described in Q53 APMR7-4. No. These two signals would appear
be first order at 300 MHz?                 as an unresolved bump (aromatic bundle).

Why?
(7.3 – 7.2)(60) = 6 Hz
J = 7 Hz
J / ∆υ = 7/6 = 1.16 >> 0.1

QPMR7-6. Would two nuclei that have a J of 7 APMR7-5. No. Even at 300 MHz, this would be
Hz show first order coupling at 300 MHz if they unresolved.
had chemical shifts of 7.15 and 7.39 ppm?
Why?
(7.2 ppm)(300) = 2160 Hz
(7.3 ppm)(300) = 2190 Hz
∆υ =  (2160 Hz) – (2190 Hz) = 30 Hz
J = 7 Hz
J / ∆υ = 7/30 = 0.23 >> 0.1

APMR7-6. YES. At 300 MHz, a difference of
only 0.23 ppm in chemical shift is required if J
values are 3 – 7 Hz.

Following the analysis of the previous problems:
(7.15 ppm)(300) = 2145 Hz
(7.39 ppm)(300) = 2217 Hz

∆υ =  (2145 Hz) – (2217 Hz) = 72 Hz
J = 7 Hz

J / ∆υ = 7/72 = 0.09 < 0.1

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