identification of unknown compounds

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

4

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)

5

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

6

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

7


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

8


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

9

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

11

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

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

AldehydesKetones

Acids AmidesEsters Anhydrides

Aromatic ringcarbons

Unsaturated

carbon - sp2

Alkyne carbons - sp

Saturated carbon - sp3

electronegativity effects


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


12 11 10 9 8 7 6 5 4 3 2 1 0

-OH -NH


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


AldehydesKetones



Unsaturated

carbon - sp2

Alkyne carbons - sp





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


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

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


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


aldehydeamide



BASE VALUES



EXPANDED CH

EXPANDED C=O

30

31

Problem 2: C8H14O4, MW = 174


0.82 3.39

2.47

2.47

1.06

1.06

O

O


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


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


aldehydeamide



BASE VALUES



EXPANDED CH

EXPANDED C=O

34

35

Problem 3: C9H12, MW = 120


12 11 10 9 8 7 6 5 4 3 2 1 0

-OH -NH


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


AldehydesKetones



Unsaturated

carbon - sp2

Alkyne carbons - sp





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


C-O

C-Cl

C-Br

R3CH R4C

R-CH2-R

R-CH3

RANGE

/

37


7.40

7.27

7.40

7.29

7.29 2.62

1.65

0.90


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


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


aldehydeamide



BASE VALUES



EXPANDED CH

EXPANDED C=O

40

41

Problem 4: C7H13Br, MW = 176


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


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


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


aldehydeamide



BASE VALUES



EXPANDED CH

EXPANDED C=O

44

45

Problem 5: C5H8O, MW = 84


4.00

1.90

1.90

4.60

6.40O


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


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


aldehydeamide



BASE VALUES



EXPANDED CH

EXPANDED C=O

48

49

Problem 6: C9H13NO, MW = 151


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


012345678PPM

50

identification of unknown compounds

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