Post on 23-Dec-2015
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Identification of Unknown Identification of Unknown CompoundsCompounds
1
Steps in the Identification of UnknownSteps in the Identification of Unknown• Identify Molecular ion MM..++• Determine Molecular FormulaMolecular Formula (oddodd / eveneven mass)• Analyze heteroatomheteroatom (M+1M+1 and M+2M+2 …)
– S, Si, Cl, Br, ….• Use rule of 13 to determine # Carbons (M+1M+1 and M+2M+2 …) • Compare with 1313C-NMRC-NMR (# carbons(# carbons) with APTAPT experiment (J-MOD)
( # protons# protons)• Compare with proton NMRproton NMR ( # protons# protons)• Identify base peakbase peak (note if eveneven / oddodd)
– One or two bond fragmentation• Test your conclusions: in lab make derivatives (TMS … or Na or K
complexes mass shift)
2
Solving problems in MSSolving problems in MS
• Try to identify the Molecular Ion or decide if it is present (most critical step in solving a structure)– Check if [M+1]+ ion is too large to accommodate reasonable number of
carbons. (the [M+1]+ ion might be the very small M+ instead!)– Determine the first loss from proposed molecular ion. Some loss are
impossible (e.g 12, 14, 23 daltons)– Does the spectrum appear dirty? (lots of small peaks even at high mass)– If GC of the comopund Is available, compare retention time
• Is the molecular weight even or odd? – An odd mass can be associated with an odd number of Nitrogen– An even mass means no Nitrogen or an even number of Nitrogen– This Rule is applicable only to Molecular ion and to odd-electron ions
• Examine ion cluster for isotopic natural abundance (look for special heteroatom pattern). Try to calculate number of carbons
3
Solving problems in MSSolving problems in MS
• From the overall appearance: is it a fragile compound? Is it likely to be aromatic or aliphatic?
• Look in the low mass ions. Do you see any clues of the family of compounds that you might be dealing with?
• Make a list of suggested losses from the molecular ion and try to make a pattellsrn from them.
• Look for intense odd-electron ions in the spectrum: this is almost impossible in compounds containing Nitrogen! These provides clues for rearrangements (retro Diels Alder, McLafferty…)
• Speculate on the structure using all that information
Index of Hydrogen deficiency/Degree of Unsaturation (DU)Index of Hydrogen deficiency/Degree of Unsaturation (DU)
CCxxHHyyNNzzOOnn IndexIndex = x – = x – ½ y½ y + + ½ z½ z +1+1
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Index of Hydrogen deficiencyIndex of Hydrogen deficiency
Nitrogen ruleNitrogen rule
MM+ + even even even # of N 0, 2, 4, … even # of N 0, 2, 4, …
MM+ + odd odd odd # of N 1, 3, 5, … odd # of N 1, 3, 5, …
Index = C – H/2 –X/2 + N/2 + 1Index = C – H/2 –X/2 + N/2 + 1
e.g. Ce.g. C77HH77NO NO Index = 7 - 3.5 + 0.5 + 1 = 5 Index = 7 - 3.5 + 0.5 + 1 = 5
Hydrogen deficiency can be unsaturation (multiple bonds)Hydrogen deficiency can be unsaturation (multiple bonds) or cyclic structure or cyclic structure
I=4 (3 DB + cycle)I=4 (3 DB + cycle)
R-CN
I=2I=2I=2 (1 DB + cycle)I=2 (1 DB + cycle)
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Neutral losses and Ion series Neutral losses and Ion series
M-1M-1 HH
M-15M-15 CHCH33
M-16M-16 O O (rare) , NH(rare) , NH2 2
M-17M-17 OH OH , NH, NH33 (rare) (rare)
M-18M-18 HH22O O
M-19M-19 F F
M-20M-20 HF (very rare)HF (very rare)
M-26M-26 HCCH , CNHCCH , CN
M-27M-27 HCNHCN
M-28M-28 HH22C=CHC=CH22 , CO , CO
M-29M-29 CHCH33CHCH2 2 , , HCO HCO
M-30M-30 NO NO (Nitro compounds), (Nitro compounds), HH22 CO CO (anisoles)(anisoles)
M-31M-31 CHCH33O O
M-32M-32 CHCH33OH OH
M-35M-35 Cl Cl
M-36M-36 HCl HCl
M-42M-42 CHCH22=C=O, CH=C=O, CH22=CH-CH=CH-CH33
M-43M-43 CHCH33CO CO , C, C33HH77
M-44M-44 COCO22
M-45M-45 CHCH33CHCH22O O , CO, CO22H H
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Neutral losses and Ion series Neutral losses and Ion series Figuring out which peak is molecular ionmolecular ion can be supported by identifying what fragment is lost.fragment is lost.
There can be sometimes 2 consecutive loss:
In steroid, M-33M-33 is often observed: comes from the loss of Meloss of Me and HH22OO
The ions lossions loss are only useful from molecular ionmolecular ion
There is no fragment in organic compounds between M-1M-1 and M-15M-15
Loss of M-14M-14 is never observed!
Other gaps in mass loss are: between 21-2521-25, 33-3433-34, 37-4137-41
IonsIons in these areasin these areas should be viewed suspiciouslyviewed suspiciously: either compound is not pure or postulated molecular ion is wrongmolecular ion is wrong
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Neutral losses and Ion series Neutral losses and Ion series
Among the losses: most common are
Loss of HH, , CHCH33 , , HH22OO (from some oxygenated compounds), (from some oxygenated compounds),
HCHCCHCH (from aromatic compounds), (from aromatic compounds),
HCHCN N (from aromatic compounds containing Nitrogen), (from aromatic compounds containing Nitrogen),
CCOO and and CHCH22=CH=CH22 (both at 28! Difficult to tell which one is lost) (both at 28! Difficult to tell which one is lost)
EthylEthyl radical ( radical (2929))
MethoxyMethoxy radical ( radical (3131))
ClCl and and HClHCl ( (3535, , 3636))
AcetylAcetyl ( (4343) accompanied by ) accompanied by m/z 43m/z 43 prominent and prominent and propylpropyl ( (4343) radical) radical
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Neutral losses and Ion series Neutral losses and Ion series
Among the losses: most common are
Loss of HH, , CHCH33 , , HH22OO (from some oxygenated compounds), (from some oxygenated compounds),
HCHCCHCH (from aromatic compounds), (from aromatic compounds),
HCHCN N (from aromatic compounds containing Nitrogen), (from aromatic compounds containing Nitrogen),
CCOO and and CHCH22=CH=CH22 (both at 28! Difficult to tell which one is lost) (both at 28! Difficult to tell which one is lost)
EthylEthyl radical ( radical (2929))
MethoxyMethoxy radical ( radical (3131))
ClCl and and HClHCl ( (3535, , 3636))
AcetylAcetyl ( (4343) accompanied by ) accompanied by m/z 43m/z 43 prominent and prominent and propylpropyl ( (4343) radical) radical
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Generally, with only three pieces of data
1) empirical formula (or % composition)
2) infrared spectrum
3) NMR spectrum
a chemist can often figure out the completestructure of an unknown molecule.
SPECTROSCOPY IS A POWERFUL TOOLSPECTROSCOPY IS A POWERFUL TOOL
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FORMULA
Gives the relative numbers of C and H and other atoms
INFRARED SPECTRUM
Reveals the types of bonds that are present.
NMR SPECTRUM
Reveals the enviroment of each hydrogenand the relative numbers of each type.
EACH TECHNIQUE YIELDS VALUABLE DATAEACH TECHNIQUE YIELDS VALUABLE DATA
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Typical Infrared AbsorptionTypical Infrared AbsorptionRegionsRegions
C-Cl2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 1800 1650 1550 650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O C=NVeryfewbands
C=C
C-ClC-O
C-NC-CX=C=Y
(C,O,N,S)
C N
C C
N=O N=O*
12
How to Use an Infrared SpectrumHow to Use an Infrared Spectrum
Molecular formula:
Check for carbonyl:
Check for O-H, N-H
Check for triple bonds
Check for C=C, benzene rings
calculate index of hydrogen deficiency
note any shift from 1715 cm-1
1)
2)
3)
4)
5)
13
How to Use an Infrared SpectrumHow to Use an Infrared Spectrum
Look below 1550 cm-1;
Go back over spectrum for refinements; check the C-H region for aldehydes and for peaks above 3000 cm-1
CONTINUED
check for C-O and
(alkenes and terminal alkynes)
6)
nitro
7)
14
acid
THE MINIMUM YOU NEED TO KNOWTHE MINIMUM YOU NEED TO KNOW
OH 3600NH 3400
CH 3000
C N 2250C C 2150
C=O 1715
C=C 1650
C-O 1100
3300 3100 2900 2850 2750
3000
1800 1735 1725 1715 1710 1690
=C-H -C-H-CHO
C-H
ketoneesteracidchloride
aldehydeamide
anhydride : 1810 and 1760
CH2 and CH3 bend : 1465 and 1365
BASE VALUES
Know also the effects of H-bonding, conjugation and ring size.
benzene C=C : between 1400 and 1600
EXPANDED CH
EXPANDED C=O
15
C=O present ?
2 C=O Peaks OH present ?
OH present ?
NH present ?
NH present ?
C-O present ?
C-O present ?
CHO present ?
C=N present ?
C=C present ?
C=C present ?
anhydride
acid
amide
ester
aldehyde
ketone
alcohol
amine
ether
nitrile
alkyne
alkene
aromatic
NO2 present ? nitro cpds
C-X present ? halides
(benzene ?)
YES
YES
NO
YES
NO
=
=
16
17
NMR Correlation ChartNMR Correlation Chart
12 11 10 9 8 7 6 5 4 3 2 1 0
-OH -NH
CH2FCH2ClCH2BrCH2ICH2OCH2NO2
CH2ArCH2NR2
CH2SC C-HC=C-CH2
CH2-C-O
C-CH-C
C
C-CH2-CC-CH3
RCOOH RCHO C=C
H
TMS
HCHCl3 ,
(ppm)
DOWNFIELD UPFIELD
DESHIELDED SHIELDED
Ranges can be defined for different general types of protons.This chart is general, the next slide is more definite. 18
19
AldehydesKetones
Acids AmidesEsters Anhydrides
Aromatic ringcarbons
Unsaturated
carbon - sp2
Alkyne carbons - sp
Saturated carbon - sp3
electronegativity effects
Saturated carbon - sp3
no electronegativity effects
C=O
C=O
C=C
C C
200 150 100 50 0
200 150 100 50 0
8 - 30
15 - 55
20 - 60
40 - 80
35 - 80
25 - 65
65 - 90
100 - 150
110 - 175
155 - 185
185 - 220
Correlation chart for 13C Chemical Shifts (ppm)
C-O
C-Cl
C-Br
R3CH R4C
R-CH2-R
R-CH3
RANGE
/
20
Problem 1: C3H5BrO2, MW = 152
CCxxHHyyNNzzOOnn
Index (DU)Index (DU) = x – = x – ½ y½ y + + ½ z½ z +1+1 = = 3 – 3 – (5+1)/2 (5+1)/2 + 1+ 1 = 1 = 1
21
Typical Infrared AbsorptionTypical Infrared AbsorptionRegionsRegions
C-Cl2.5 4 5 5.5 6.1 6.5 15.4
4000 2500 2000 1800 1650 1550 650
FREQUENCY (cm-1)
WAVELENGTH (m)
O-H C-H
N-H
C=O C=NVeryfewbands
C=C
C-ClC-O
C-NC-CX=C=Y
(C,O,N,S)
C N
C C
N=O N=O*
22
23
Problem 1: C3H5BrO2, MW = 152
NMR Correlation ChartNMR Correlation Chart
12 11 10 9 8 7 6 5 4 3 2 1 0
-OH -NH
CH2FCH2ClCH2BrCH2ICH2OCH2NO2
CH2ArCH2NR2
CH2SC C-HC=C-CH2
CH2-C-O
C-CH-C
C
C-CH2-CC-CH3
RCOOH RCHO C=C
H
TMS
HCHCl3 ,
(ppm)
DOWNFIELD UPFIELD
DESHIELDED SHIELDED
Ranges can be defined for different general types of protons.This chart is general, the next slide is more definite. 24
AldehydesKetones
Acids AmidesEsters Anhydrides
Aromatic ringcarbons
Unsaturated
carbon - sp2
Alkyne carbons - sp
Saturated carbon - sp3
electronegativity effects
Saturated carbon - sp3
no electronegativity effects
C=O
C=O
C=C
C C
200 150 100 50 0
200 150 100 50 0
8 - 30
15 - 55
20 - 60
40 - 80
35 - 80
25 - 65
65 - 90
100 - 150
110 - 175
155 - 185
185 - 220
Correlation chart for 13C Chemical Shifts (ppm)
C-O
C-Cl
C-Br
R3CH R4C
R-CH2-R
R-CH3
RANGE
/
25
ChemNMR 1H Estimation
3.57
2.73 11.0Br
O
OH
Estimation quality is indicated by color: good, medium, rough
024681012PPM
26
27
28
Problem 2: C8H14O4, MW = 174
CCxxHHyyNNzzOOnn
Index (DU)Index (DU) = x – = x – ½ y½ y + + ½ z½ z +1+1 = = 8 – 8 – 14/2 14/2 + 1+ 1 = 2 = 2
29
acid
THE MINIMUM YOU NEED TO KNOWTHE MINIMUM YOU NEED TO KNOW
OH 3600NH 3400
CH 3000
C N 2250C C 2150
C=O 1715
C=C 1650
C-O 1100
3300 3100 2900 2850 2750
3000
1800 1735 1725 1715 1710 1690
=C-H -C-H-CHO
C-H
ketoneesteracidchloride
aldehydeamide
anhydride : 1810 and 1760
CH2 and CH3 bend : 1465 and 1365
BASE VALUES
Know also the effects of H-bonding, conjugation and ring size.
benzene C=C : between 1400 and 1600
EXPANDED CH
EXPANDED C=O
30
31
Problem 2: C8H14O4, MW = 174
ChemNMR 1H Estimation
0.82 3.39
2.47
2.47
1.06
1.06
O
O
Estimation quality is indicated by color: good, medium, rough
01234PPM
32
Problem 3: C9H12, MW = 120
CCxxHHyyNNzzOOnn
Index (DU)Index (DU) = x – = x – ½ y½ y + + ½ z½ z +1+1 = = 9 – 9 – 12/2 12/2 + 1+ 1 = 4 = 4
33
acid
THE MINIMUM YOU NEED TO KNOWTHE MINIMUM YOU NEED TO KNOW
OH 3600NH 3400
CH 3000
C N 2250C C 2150
C=O 1715
C=C 1650
C-O 1100
3300 3100 2900 2850 2750
3000
1800 1735 1725 1715 1710 1690
=C-H -C-H-CHO
C-H
ketoneesteracidchloride
aldehydeamide
anhydride : 1810 and 1760
CH2 and CH3 bend : 1465 and 1365
BASE VALUES
Know also the effects of H-bonding, conjugation and ring size.
benzene C=C : between 1400 and 1600
EXPANDED CH
EXPANDED C=O
34
35
Problem 3: C9H12, MW = 120
NMR Correlation ChartNMR Correlation Chart
12 11 10 9 8 7 6 5 4 3 2 1 0
-OH -NH
CH2FCH2ClCH2BrCH2ICH2OCH2NO2
CH2ArCH2NR2
CH2SC C-HC=C-CH2
CH2-C-O
C-CH-C
C
C-CH2-CC-CH3
RCOOH RCHO C=C
H
TMS
HCHCl3 ,
(ppm)
DOWNFIELD UPFIELD
DESHIELDED SHIELDED
Ranges can be defined for different general types of protons.This chart is general, the next slide is more definite. 36
AldehydesKetones
Acids AmidesEsters Anhydrides
Aromatic ringcarbons
Unsaturated
carbon - sp2
Alkyne carbons - sp
Saturated carbon - sp3
electronegativity effects
Saturated carbon - sp3
no electronegativity effects
C=O
C=O
C=C
C C
200 150 100 50 0
200 150 100 50 0
8 - 30
15 - 55
20 - 60
40 - 80
35 - 80
25 - 65
65 - 90
100 - 150
110 - 175
155 - 185
185 - 220
Correlation chart for 13C Chemical Shifts (ppm)
C-O
C-Cl
C-Br
R3CH R4C
R-CH2-R
R-CH3
RANGE
/
37
ChemNMR 1H Estimation
7.40
7.27
7.40
7.29
7.29 2.62
1.65
0.90
Estimation quality is indicated by color: good, medium, rough
012345678PPM
38
Problem 4: C7H13Br, MW = 176
CCxxHHyyNNzzOOnn
Index (DU)Index (DU) = x – = x – ½ y½ y + + ½ z½ z +1+1 = 7= 7 – – (13+1)/2 (13+1)/2 + 1+ 1 = 1 = 1
39
acid
THE MINIMUM YOU NEED TO KNOWTHE MINIMUM YOU NEED TO KNOW
OH 3600NH 3400
CH 3000
C N 2250C C 2150
C=O 1715
C=C 1650
C-O 1100
3300 3100 2900 2850 2750
3000
1800 1735 1725 1715 1710 1690
=C-H -C-H-CHO
C-H
ketoneesteracidchloride
aldehydeamide
anhydride : 1810 and 1760
CH2 and CH3 bend : 1465 and 1365
BASE VALUES
Know also the effects of H-bonding, conjugation and ring size.
benzene C=C : between 1400 and 1600
EXPANDED CH
EXPANDED C=O
40
41
Problem 4: C7H13Br, MW = 176
ChemNMR 1H Estimation
1.53;1.43
1.49;1.47
1.53;1.43
1.52;1.27
1.85
1.52;1.27 3.26
Br
Estimation quality is indicated by color: good, medium, rough
01234PPM
42
Problem 5: C5H8O, MW = 84
CCxxHHyyNNzzOOnn
Index (DU)Index (DU) = x – = x – ½ y½ y + + ½ z½ z +1+1 = 5= 5 – – 8/2 8/2 + 1+ 1 = 2 = 2
43
acid
THE MINIMUM YOU NEED TO KNOWTHE MINIMUM YOU NEED TO KNOW
OH 3600NH 3400
CH 3000
C N 2250C C 2150
C=O 1715
C=C 1650
C-O 1100
3300 3100 2900 2850 2750
3000
1800 1735 1725 1715 1710 1690
=C-H -C-H-CHO
C-H
ketoneesteracidchloride
aldehydeamide
anhydride : 1810 and 1760
CH2 and CH3 bend : 1465 and 1365
BASE VALUES
Know also the effects of H-bonding, conjugation and ring size.
benzene C=C : between 1400 and 1600
EXPANDED CH
EXPANDED C=O
44
45
Problem 5: C5H8O, MW = 84
ChemNMR 1H Estimation
4.00
1.90
1.90
4.60
6.40O
Estimation quality is indicated by color: good, medium, rough
01234567PPM
46
Problem 6: C9H13NO, MW = 151
CCxxHHyyNNzzOOnn
Index (DU)Index (DU) = x – = x – ½ y½ y + + ½ z½ z +1+1 = 9= 9 – – 13/2 13/2 + + 1/21/2 + 1+ 1 = 4 = 4
47
acid
THE MINIMUM YOU NEED TO KNOWTHE MINIMUM YOU NEED TO KNOW
OH 3600NH 3400
CH 3000
C N 2250C C 2150
C=O 1715
C=C 1650
C-O 1100
3300 3100 2900 2850 2750
3000
1800 1735 1725 1715 1710 1690
=C-H -C-H-CHO
C-H
ketoneesteracidchloride
aldehydeamide
anhydride : 1810 and 1760
CH2 and CH3 bend : 1465 and 1365
BASE VALUES
Know also the effects of H-bonding, conjugation and ring size.
benzene C=C : between 1400 and 1600
EXPANDED CH
EXPANDED C=O
48
49
Problem 6: C9H13NO, MW = 151
ChemNMR 1H Estimation
7.37
7.27
7.37
7.30
7.30 3.09
3.94;3.69
3.06;2.81
3.65
5.11
OH
NH2
Estimation quality is indicated by color: good, medium, rough
012345678PPM
50