identification of unknown compounds [compatibility mode]

8/11/2019 Identification of Unknown Compounds [Compatibility Mode]

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Identification of Unknown

Compounds

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Steps in the Identification of Unknown

Identify Molecular ion M.+

Determine Molecular Formula (odd / even mass)

Analyze heteroatom (M+1 and M+2 )

S, Si, Cl, Br, .

Use rule of 13 to determine # Carbons (M+1 and M+2 )

Compare with13

C-NMR (# carbons) with APT experiment (J-MOD)( # protons)

Compare with proton NMR ( # protons)

Identify base peak (note if even / odd)

One or two bond fragmentation

Test your conclusions: in lab make derivatives (TMS or Na or K

complexesmass shift)

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Solving 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

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Solving 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)

CxHyNzOn Index = x y + z +1

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Index of Hydrogen deficiency

Nitrogen rule

M+ even even # of N 0, 2, 4,

M+ odd odd # of N 1, 3, 5,

Index = C H/2 X/2 + N/2 + 1

e.g. C7H7NO Index = 7 - 3.5 + 0.5 + 1 = 5

Hydrogen deficiency can be unsaturation (multiple bonds)

or cyclic structure

I=4 (3 DB + cycle)

R-CN

I=2

I=2 (1 DB + cycle)5


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Neutral losses and Ion series

M-1 H

M-15 CH3

M-16 O (rare) , NH2

M-17 OH , NH3 (rare)

M-18 H2O

M-19 F

M-20 HF (very rare)

M-26 HCCH , CN

M-27 HCN

M-28 H2C=CH2 , CO

M-29 CH3CH2 ,HCO

M-30 NO (Nitro compounds),

H2 CO (anisoles)

M-31 CH3O

M-32 CH3OH

M-35 Cl

M-36 HCl

M-42 CH2=C=O, CH2=CH-CH3

M-43 CH3CO , C3H7

M-44 CO2

M-45 CH3CH2O , CO2H

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Figuring out which peak is molecular ion can be supported by identifying what fragment is lost.

There can be sometimes 2 consecutive loss:

In steroid, M-33 is often observed: comes from the loss of Me and H2O

The ions loss are only useful from molecular ion

There is no fragment in organic compounds between M-1 and M-15

Loss of M-14 is never observed!

Other gaps in mass loss are: between 21-25, 33-34, 37-41

Ions in these areas should be viewed suspiciously: either compound is not pure or postulated

molecular ion is wrong

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Among the losses: most common are

Loss of H

, CH3

, H2O (from some oxygenated compounds),

HCCH (from aromatic compounds),

HCN (from aromatic compounds containing Nitrogen),

CO and CH2=CH2 (both at 28! Difficult to tell which one is lost)

Ethyl radical (29)

Methoxy radical (31)

Cl and HCl (35, 36)

Acetyl (43) accompanied by m/z 43 prominent and propyl (43) radical

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Among the losses: most common are

Loss of H

, CH3

, H2O (from some oxygenated compounds),

HCCH (from aromatic compounds),

HCN (from aromatic compounds containing Nitrogen),

CO and CH2=CH2 (both at 28! Difficult to tell which one is lost)

Ethyl radical (29)

Methoxy radical (31)

Cl and HCl (35, 36)

Acetyl (43) accompanied by m/z 43 prominent and propyl (43) 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 complete

structure of an unknown molecule.

SPECTROSCOPY 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 DATA

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Typical Infrared Absorption

RegionsC-Cl

2.5 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=NVery

few

bands

C=C

C-ClC-O

C-N

C-CX=C=Y

(C,O,N,S)

C N

C C

N=O N=O*

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How to Use an Infrared Spectrum

Molecular formula:

Check for carbonyl:

Check for O- , N-

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)

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How to Use an Infrared Spectrum

Look below 1550 cm

-1

;

Go back over spectrum for refinements;

check the C- region for aldehydes

and for peaks above 3000 cm

-1

CONTINUED

check for C-O and

(alkenes and terminal alkynes)

6)

nitro

7)

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acid

THE MINIMUM YOU NEED TO KNOW

OH 3600

NH 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

ketoneesteracid

chloride

aldehyde

amide

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

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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

=

=

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NMR Correlation Chart

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

-OH -NH

CH2F

CH2Cl

CH2Br

CH2I

CH2OCH2NO2

CH2Ar

CH2NR2CH2S

C C-H

C=C-CH2CH2-C-

O

C-CH-C

C

C-CH2-C

C-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


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Aldehydes

Ketones

Acids Amides

Esters Anhydrides

Aromatic ring

carbons

Unsaturated

carbon - sp2

Alkynecarbons - 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

/

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Problem 1: C3H5BrO2, MW = 152

CxHyNzOn

Index (DU) = x y + z +1= 3 (5+1)/2 + 1

= 1

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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

ketoneesteracid

chloride

aldehyde

amide



BASE VALUES



EXPANDED CH

EXPANDED C=O

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Problem 1: C3H5BrO2, MW = 152


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

-OH -NH

CH2F

CH2Cl

CH2Br

CH2I

CH2OCH2NO2

CH2Ar

CH2NR2CH2S

C C-H

C=C-CH2CH2-C-

O

C-CH-C

C

C-CH2-C

C-CH3

RCOOH RCHO C=C

H

TMS

HCHCl3 ,

(ppm)

DOWNFIELD UPFIELD

DESHIELDED SHIELDED




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Aldehydes

Ketones

Acids Amides

Esters Anhydrides

Aromatic ring

carbons

Unsaturated

carbon - sp2

Alkynecarbons - 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

/

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ChemNMR 1H Estimation

3.57

2.73 11.0Br

O

OH

Estimation quality is indicated by color: good, medium, rough

024681012PPM

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Problem 2: C8H14O4, MW = 174

CxHyNzOn

Index (DU) = x y + z +1= 8 14/2 + 1

= 2

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acid


OH 3600NH 3400

CH 3000

C N 2250C C 2150

C=O 1715

C=C 1650

C-O 1100

3300 3100 2900 28502750

3000

1800 1735 1725 1715 1710 1690

=C-H -C-H

-CHO

C-H

ketoneesteracid

chloride

aldehyde

amide



BASE VALUES



EXPANDED CH

EXPANDED C=O

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31

Problem 2: C8H14O4, MW = 174


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0.82 3.39

2.47

2.47

1.06

1.06

O

O


01234PPM

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Problem 3: C9H12, MW = 120

CxHyNzOn

Index (DU) = x y + z +1= 9 12/2 + 1

= 4

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acid


OH 3600NH 3400

CH 3000

C N 2250C C 2150

C=O 1715

C=C 1650

C-O 1100

3300 3100 2900 28502750

3000

1800 1735 1725 1715 1710 1690

=C-H -C-H

-CHO

C-H

ketoneesteracid

chloride

aldehyde

amide



BASE VALUES



EXPANDED CH

EXPANDED C=O

34


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35

Problem 3: C9H12, MW = 120


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

-OH -NH

CH2F

CH2Cl

CH2Br

CH2I

CH2OCH2NO2

CH2Ar

CH2NR2CH2S

C C-H

C=C-CH2CH2-C-

O

C-CH-C

C

C-CH2-C

C-CH3

RCOOH RCHO C=C

H

TMS

HCHCl3 ,

(ppm)

DOWNFIELD UPFIELD

DESHIELDED SHIELDED




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AldehydesKetones

Acids Amides

Esters Anhydrides

Aromatic ring

carbons

Unsaturated

carbon - sp2

Alkynecarbons - 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

/

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Problem 4: C7H13Br, MW = 176

CxHyNzOn

Index (DU) = x y + z +1= 7 (13+1)/2 + 1

= 1

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acid


OH 3600NH 3400

CH 3000

C N 2250C C 2150

C=O 1715

C=C 1650

C-O 1100

3300 3100 2900 28502750

3000

1800 1735 1725 1715 1710 1690

=C-H -C-H

-CHO

C-H

ketoneesteracid

chloride

aldehyde

amide



BASE VALUES



EXPANDED CH

EXPANDED C=O

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41

Problem 4: C7H13Br, MW = 176


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Problem 5: C5H8O, MW = 84

CxHyNzOn

Index (DU) = x y + z +1= 5 8/2 + 1

= 2

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acid


OH 3600NH 3400

CH 3000

C N 2250C C 2150

C=O 1715

C=C 1650

C-O 1100

3300 3100 2900 28502750

3000

1800 1735 1725 1715 1710 1690

=C-H -C-H

-CHO

C-H

ketoneesteracid

chloride

aldehyde

amide



BASE VALUES



EXPANDED CH

EXPANDED C=O

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45

Problem 5: C5H8O, MW = 84


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4.00

1.90

1.90

4.60

6.40O


01234567PPM

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Problem 6: C9H13NO, MW = 151

CxHyNzOn

Index (DU) = x y + z +1= 9 13/2 + 1/2 + 1

= 4

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acid


OH 3600NH 3400

CH 3000

C N 2250C C 2150

C=O 1715

C=C 1650

C-O 1100

3300 3100 2900 28502750

3000

1800 1735 1725 1715 1710 1690

=C-H -C-H

-CHO

C-H

ketoneesteracid

chloride

aldehyde

amide



BASE VALUES



EXPANDED CH

EXPANDED C=O

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49

Problem 6: C9H13NO, MW = 151


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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

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identification of unknown compounds [compatibility mode]

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