Instrumental Analysis Sheet#13 Dr.Emad Hamdan Done by:Noor Aswad
Last time we talked about NMR spectrum, and we talked about the modification on the x-axis,
the delta scale (δ) or chemical shift scale, and it is scaled from 0 to 17 or most commonly from 0
to 12.
And we said that we’ll get signals proportional to the number of
chemically different protons (e.g. 5 chemically different protons then
we’ll get 5 signals and so on).
Also we said that methyl (CH3) must appear between 0 and 1, unless
it is affected by other adjacent electronegative groups (like O,
halogens, double bonds, elkenes, alkynes, carbonyl, or aromatic ring)
that will cause shift to the downfield.
If we want to compare between Cl and F, F is more electronegative,
so it’ll cause larger shift to the downfield.
Conclusion: as the electronegativity increases, then the shift to the downfield is larger.
Example:
b
In this molecule, Ha will be more downfield shifted than Hb, because it
is attached to a carbon with 2 Cl instead of one Cl.
a
Examples in the handout:
1- Acetone has one type of protons for the six
protons, so we expect for it only one signal and it
will be slightly shifted downfield because of the
carbonyl.
2- Chloroethane has two types of protons, so we
expect to see two signals on the spectrum. This
happened because one carbon is attached to Cl.
The downfield shifted signal will be for the CH2
attached to the Cl, and the signal near the zero is
for CH3. (The signal for CH3 has larger than that of
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Instrumental Analysis Sheet#13 Dr.Emad Hamdan Done by:Noor Aswad
CH2 because the ratio between electrons is 3/2).
Conclusion: the peak areas are proportional to the number of protons causing these
signals.
3- Ethyl methyl ether
Here there are 3 different types of protons.
In NMR spectrum we’ll get 3 different signals, containing a 3-proton
singlet, a 2-proton quartet, and a 3-proton triplet.
The first signal (most upfield) is for CH3 that is not attached to the
oxygen.
The second signal is for the CH3 that is attached to the oxygen.
The third signal (most downfield) is for the CH2 .
Now, why the CH2 that is attached to the oxygen is more downfield
shifted than CH3?
This is a nature of CH2, normally it is shifted to the downfield more than
CH3, even though there was no electronegative near group.
For more examples refer to the handout (NMR page2).
Integrations of the spectrum
In old NMR spectrophotometers:
An NMR spectrophotometer automatically integrates the
area under the peaks, and prints out a stepped curve
(integral) on the spectrum.
The height of each step is proportional to the area under
the peak, which in turn proportional to the number of
absorbing protons.
So, we can get the ration between lengths of steps (e.g. 3 to
2) and so we know the ratio between the number of protons for each
signal
In new NMR spectrophotometers:
Integration is better, but we have the same problem of not having the exact (integer)
ratio between the steps (e.g. 3.01 to 1.95 so we’ll understand that the ratio is 3 to 2).
Example (from the handout – NMR page3)
Here in this structure, we have 3 different types of protons (CH3,
CH2, aromatic protons).
We’re not very accurate when we say that all aromatic protons are
equivalent, but we’ll assume them equivalent.
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Instrumental Analysis Sheet#13 Dr.Emad Hamdan Done by:Noor Aswad
Always, aromatic protons will appear in the area between 6.5 and 8.
Notes for the spectrum of previous structure:
-
Note that CH3, CH2 signals, both are shifted downfield (CH3→ because it is attached
to the carbonyl, CH2→ because it is attached to the oxygen and aromatic ring).
Remember: the real life use (actual use) of NMR is the structure elucidation. I.e. we
have the spectrum, and we want to know the structure.
Spin-spin coupling
Usually, there are splits in the peaks seen in the NMR, and this happens because of the
“spin-spin coupling” that is the magnetic interaction between nonequivalent protons on
adjacent carbons.
Take the next structure as example:
Ha and Hb are not equivalent. Signals of both are splitted. But why?
Because Ha will be affected by 2 states of Hb (states are with or against
H°) and the same thing for Hb. so, the signal will split (appear on several
frequencies). For each, the signal will appear as a doublet (split to 2
signals).
Doublet for Hb
Doublet for Ha
When we want to read the frequency of the signals then we take the average of the two
splitted signals (the doublet) (e.g. the first split of the doublet at 3.02 and the second split of the
doublet at 3.09, then we’ll take the average of both frequencies as the correct reading).
What if the structure was like the following?
We have 3 protons, but with 2 different chemically types (Hb and Hc are
chemically equivalence).
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Instrumental Analysis Sheet#13 Dr.Emad Hamdan Done by:Noor Aswad
Signals of Hb and Hc will appear as doublet (because they are affected by the two states of Ha).
Signal of Ha : Ha will be affected by the state of both Hb and Hc
Use the possibilities to understand such state, here we have 4 possibilities:
1234-
Both Ha and Hc with H° (lowest energy state)
Both Ha and Hc against H° (highest energy state)
Ha with H° and Hc against H° (intermediate energy state)
Hc with H° and Ha against H°. (intermediate energy state)
Possibilities 3+4 are the same (because they have the same energy content), so we have 3 states
(different energy states).
So, Ha signal will split into 3 splits (depending on 3 states of both Hb and Hc) and this signal is
called “triplet”.
In the triplet signal, why the second split is longer than the two other signals?
Because it is for the 3rd and 4th possibilities combined (more electrons are in this intermediate
state of energy). The triplet is a characteristic to know that this proton has 2 adjacent protons.
So, as we have singlet, doublet,
triplet, we also have quartet, pentet,
etc. (depending on the number of
adjacent protons).
Rule: number of splittings = n+1 ,
where n= number of adjacent
protons.
Example of pentet:
Note: to assign the proton as adjacent
proton, then it must be only 3 bonds far from the signaling proton.
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Instrumental Analysis Sheet#13 Dr.Emad Hamdan Done by:Noor Aswad
If the adjacent proton is 2 bonds far from the signaling proton, then in MOST CASES it is a
chemically equivalent proton attached on the same carbon, but we have exceptions in which
the two protons attached on the same carbons are not chemically equivalent, study the next
example to understand:
In this structure, Ha and Hb are attached on the same carbon, but
they are not chemically equivalent (Ha is trans to aromatic ring, and
Hb is cis to the aromatic ring).this happened because of the double
bond (no free rotation). So, here there is a spin-spin coupling
between Ha and Hb, but more details are not included.
Review the example in the handout (NMR- page3).
Degree of unsaturation
This helps us in structure elucidation
Rings + π bonds = 𝒄 −
𝑯
𝟐
−
𝑿
𝟐
+
𝑵
+
𝟐
𝟏
Where c= number of carbons, H= number of hydrogens, X=number of halogens, N= number of
nitrogens.
(Rings count as one degree of unsaturation, double bonds count as one degree of unsaturation,
triple bonds count as two degrees of unsaturation).
At least here, we need the molecular formula.
Example:
C6H12O6
This formula has many isomers, but to know if there is a double bond or ring structure in this
compound, we calculate the degree of unsaturation
Rings + π bonds = 6 −
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12
2
−
0
2
+
0
2
+ 1= 1 , so in this structure there is a π bond or one ring