Spring 2014
CHEM 212 – NMR Spectroscopy
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Spectral Analysis – 1H NMR
NMR Spectroscopy
NMR Spectral Analysis – Introductory 1H NMR
1. NMR is rarely used in a vacuum to do a “forensic” analysis of an unknown
2. NMR (all nuclei) is usually used:
–
To analyze the product of a chemical reaction along with IR
–
To elucidate the structures of natural products (like the spice lab
compounds in CHEM 213) in conjunction with mass spectrometry
(which gives molecular weights and formulas), IR and UV
3. In this course, you will be given one of three pieces of data with an 1H
NMR for consideration:
–
A molecular formula
–
An IR spectrum
–
The first part of a chemical reaction – for example:
O
NaBH4
EtOH
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Spectral Analysis – 1H NMR
NMR Spectroscopy
NMR Spectral Analysis – Introductory 1H NMR
Step 1: Do a quick assessment of the information you are given
Molecular formula – one of the most important pieces of information
Use the index of hydrogen deficiency (HDI) to determine the possible
number of rings, double and triple bonds in the molecule:
For a chemical formula:
CxHyNzO (halogens count as Hs)
HDI = x – ½ Y + ½ Z + 1
Example: C4H8O --
HDI = 4 – ½ (8) + ½ (0) + 1
HDI = 1 This compound contains either one
double bond or one ring
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Spectral Analysis – 1H NMR
NMR Spectroscopy
NMR Spectral Analysis – Introductory 1H NMR
Step 2: Do a quick assessment of the 1H NMR you are given
–
Is the molecule simple or complex?
–
Is the molecule aromatic, aliphatic or both?
–
What are the total number of resonances that you observe
Be careful with overly simple spectra – remember a large molecule may
appear to be small and simple if it is highly symmetrical
Consider: Durene, C10H14
1H
4
NMR spectrum consists of two singlets!
Spectral Analysis – 1H NMR
NMR Spectral Analysis – Introductory 1H NMR
Example: 1H NMR for C4H8 (which has an HDI of 1)
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NMR Spectroscopy
Spectral Analysis – 1H NMR
NMR Spectroscopy
NMR Spectral Analysis – Introductory 1H NMR
Step 3: Use the integration along with the molecular formula to make sure
you can find all of the 1Hs (this is 1H NMR after all!)
If you do not have a molecular formula, use the integration to attempt to
tabulate the number of 1Hs in the formula (does it make sense?)
If one hydrogen appears to be “missing”, you may suspect it is acidic or
exchangeable with the dueterated NMR solvent
Remember, integration gives you the least common denominator of the
total number of protons of each type
Keep in mind organic molecules contain –C-H’s, -CH2- ’s, -CH3 ’s and
multiples of chemically equivalent ones!
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Spectral Analysis – 1H NMR
NMR Spectroscopy
NMR Spectral Analysis
Continue our example:
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For C4H8O we need to find 8
protons
The integral for the quartet
at d 2.5 measures 30 mm,
the singlet at d 2.2 measures
44 mm, the triplet at d 1.1
measures 43 mm
The ratio:
30:43:42 is roughly 2:3:3
2+3+3=8
We found all 8 protons
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44
30
Spectral Analysis – 1H NMR
NMR Spectroscopy
NMR Spectral Analysis – Introductory 1H NMR
Step 4: Classify each of the proton resonances by using the general
correlation table
Reconsider the number of rings, double or triple bonds that are possible
given the HDI, and reconcile this data with what the 1H chemical shifts are
telling you
Some hints:
If you calculate an HDI > 4, you probably have an aromatic ring, and this
should show on the spectrum
If you calculate an HDI of 1 or 2 and see no protons that are part of an
alkene and alkyne – suspect rings if no oxygen's are present, carbonyls if
(C=O) they are
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Spectral Analysis – 1H NMR
NMR Spectroscopy
NMR Spectral Analysis
Continue our example:
For C4H8O we have 3 families
of chemically equivalent
protons in a ratio of 2:3:3 or
–CH2-, -CH3, -CH3
EWG CH3
Both the d 2.2 and 2.5
resonance correspond to
protons on carbons next to
electron withdrawing groups
EWG
The d 1.1 resonance
corresponds to protons on
carbons bound to other
aliphatic carbons
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R CH2
R CH3
Spectral Analysis – 1H NMR
NMR Spectroscopy
NMR Spectral Analysis
Continue our example:
We found –CH2-, -CH3 and
–CH3, subtract these from
our original formula:
C4H8O - CH2 = C3H6O
C3H6O – CH3 = C2H3O
C2H3O – CH3 = CO
We needed an HDI of 1 and
there is no evidence for
1H-C=C on the 1H NMR, our
missing HDI, and EWG is a
C=O!
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EWG CH3
EWG
R CH2
R CH3
Spectral Analysis – 1H NMR
NMR Spectroscopy
NMR Spectral Analysis – Introductory 1H NMR
Step 5: Analyze the spin-spin coupling multiplets to elucidate the carbon
chains of the molecule
Hints:
•
Singlets indicate you have protons on carbons that have no
chemically non-equivalent protons on any adjoining atom
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•
Multiplets mean you have chemically non-equivalent protons on
adjoining carbons (or atoms), use the n+1 rule in reverse to find
out how many
•
Spin-spin coupling or splitting is MUTUAL, if you observe a multiplet
there must be another multiplet it is related to (split by)
Spectral Analysis – 1H NMR
NMR Spectroscopy
NMR Spectral Analysis
For our example, this is
trivial; we have concluded
the C=O is the electron
withdrawing group
The –CH3 singlet at d 2.2
obviously has no 1Hs on
adjoining carbons, as it is
next to the carbonyl
The d 2.5 –CH2- quartet is
next to a –CH3 (n+1 = 4, so
n = 3, it is next to a –CH3)
The d 1.2 –CH3 is a triplet, so
it is next to a –CH2
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EWG CH3
EWG
R CH2
R CH3
Spectral Analysis – 1H NMR
NMR Spectroscopy
NMR Spectral Analysis – Introductory 1H NMR
Step 6: Construct the molecule and double-check consistency
•
Does the HDI match? Have you accounted for all atoms in the
formula?
•
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From your constructed molecule, pretend you are trying to verify if
that spectrum matches, and quickly re-do the problem
Spectral Analysis – 1H NMR
NMR Spectral Analysis
We concluded we have:
O H
2
H3C C C CH3
or 2-butanone
C4H8O, HDI = 1
3 proton resonances
2 mutually coupled (split)
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 2: C9H9BrO
HDI =
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 2: C9H9BrO
HDI = 9 – ½ (10) + 0 + 1 = 5
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 2: C9H9BrO
HDI = 9 – ½ (10) + 0 + 1 = 5
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 2: C9H9BrO
HDI = 9 – ½ (10) + 0 + 1 = 5
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NMR Spectroscopy
O
Br
Spectral Analysis – 1H NMR
Example 3: C4H10O
HDI =
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 3: C4H10O
HDI = 4 – ½ (10) + 0 + 1 = 0
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 3: C9H9BrO
HDI = 0
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 3: C4H10O
HDI = 0
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NMR Spectroscopy
HO
Spectral Analysis – 1H NMR
Example 4: C5H10O2
HDI =
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 4: C5H10O2
HDI = 5 – ½ (10) + 0 + 1 = 1
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 4: C5H10O2
HDI = 1
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NMR Spectroscopy
Spectral Analysis – 1H NMR
NMR Spectroscopy
Example 4: C5H10O2
O
HDI = 1
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HO
Spectral Analysis – 1H NMR
Example 5: C10H12O2
HDI =
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 5: C10H12O2
HDI = 10 – ½ (12) + 0 + 1 = 5
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 5: C10H12O2
HDI = 5
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 5: C10H12O2
HDI = 5
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NMR Spectroscopy
O
OH
Spectral Analysis – 1H NMR
Example 6: C6H4ClNO2
HDI =
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 6: C6H4ClNO2
HDI = 6 – ½ (5) + ½ (1) + 1 = 5
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 6: C6H4ClNO2
HDI = 5
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NMR Spectroscopy
Spectral Analysis – 1H NMR
Example 6: C6H4ClNO2
NMR Spectroscopy
O
O
N
HDI = 5
Cl
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