Instrumental Analysis Lecture 36 - Chemistry

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Instrumental Analysis

Lecture 36

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Mass Spectroscopy (MS)

Mass Spectrum

The pattern of ion intensities is characteristic

(fingerprint) of the original molecule.

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MS Interpretation

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MS Interpretation

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MS Interpretation

Atomic Masses and Isotopes

Atomic masses on the periodic table represent an

average for all the naturally occurring isotopes.

For example chlorine is a mixture of:35Cl - 75.77% and 37Cl - 24.23%

For MS the most abundant isotope is set to 100%

and all other isotopes are normalized relative to

the most abundant.

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MS Interpretation

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MS Interpretation

Isotope Classes

The A, A+1, and A+2 represent the types of

isotopes present.

A - only a single isotope, i.e. F

A+1 - two isotopes, i.e. C and N

A+2 - at least two isotopes with the highest mass

isotope being +2 from the lowest mass

isotope, i.e. O, S, Cl, Br.

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MS Interpretation

Looking at other A+1 abundances, notice that

nitrogen has a very small number, that is why we

use the nitrogen rule.

For some A+2 elements, O also has a very small

abundance, but sulfur is easier to see.

Cl and Br are A+2 elements and have relatively high

abundances for the isotopes. So it is important to

look at the patterns to identify Cl and Br

containing compounds.

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MS Interpretation.

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MS Interpretation

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MS InterpretationLogical Losses - radical losses

Table of common neutral fragments

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1 H 19 F31

OCH3/CH2OH

15 CH3 26 CN 35 Cl

16 NH2 27 C2H3

43

OC2H5/COOH

17 OH 29 C2H5/CHO 46 NO2


MS InterpretationLogical Losses - radical losses

Table of neutral losses – not as common

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2 H2 27 HCN 36 HCl

17 NH2

28

CO/C2H2

44 CO2

18 H2O 30 CH2O 74 C3H6O2

20 HF 34 H2S 80 HBr


Structural Interpretation

Rings Plus Double Bonds Rule

Knowing the number of rings (R) and double bonds (DB) in an ion can help to determine the structure.

The R+DB value can be calculated as follows:

R + DB = C – (H + X)/2 + N/2 + 1 (round down)

Where C, H, X, and N stand for the number of carbons, hydrogen, halogen, and nitrogen atoms in the ion.

Si is treated as a carbon and P as nitrogen. Neither oxygen or sulfur are considered in the

calculation.

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Rings Plus Double Bonds Rule

Examples:

Ion Formula Calculation

CH3COCH2+ R+DB = 3 – (5+0)/2 + 0/2 + 1 = 1.5

= 1

C5H4NCl+ R+DB = 5 – (4-1)/2 + ½ + 1 = 4

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Nitrogen Rule

Molecules that contain an odd number of nitrogen

atoms have an odd nominal mass.

Molecules with no nitrogen atoms or an even

number of nitrogen have an even nominal mass.

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MS Interpretation

1. Assume that we are working with good spectrum.

2. Evaluate the General appearance of spectrum Consider:

-degree of fragmentation

-presence of clusters

-general shape

-odd or even major lines

3. Find the molecular ion

The molecular ion must be:

-the highest mass ion - exclusive of isotope related ions

-an odd electron ion

-consistent with the rest of the spectrum.

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MS Interpretation

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For the following spectral data:

m/e Abundance m/e abundance

26 3472 41 14044

27 20593 42 7575

28 48287 43 27536

29 79823 44 31440

39 14675 45 1026

Which line would be the molecular ion?


MS Interpretation

The molecular ion would be at 44.

4. Determine the elemental composition

Next step is to normalize the data. The standard spectra are normalized to the largest (base peak) which may not be the molecular ion.

Steps to normalize data:-select the potential molecular ion-normalize the lines setting the molecular ion to 100.-construct a normalization table

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MS Interpretation

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100/31440 = .00318


MS Interpretation

-put the masses and abundances in the first two

columns

-calculate the normalization factor (NF), where

NF = 100/rel. abund. of A

-calculate the normalized abundances for the

isotope lines

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MS Interpretation

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31440/1026 = 100/x x=3.26


MS Interpretation

If have 2 carbons, assume for each carbon A+1 = 2x1.1 = 2.2 and so on. (from previous table)

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MS Interpretation

The mass of the parent ion 44 - 3 carbons

= 36 leaves a remaining mass of 8. Since 8

is small, the potential formula may be C3H8

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MS Interpretation

Example:

m/z Rel. Abundance57 100.0084 0.1085 0.4086 15.5187 1.00

86 appears to be the parent ion.

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MS Interpretation

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MS Interpretation

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MS Interpretation

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MS Interpretation

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Matches our proposed structure


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4x1.1

+0.8 =

5.2


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R + DB = C – (H + X)/2 + N/2 + 1 (round down)

Where C, H, X, and N stand for the number of carbons, hydrogen, halogen, and nitrogen atoms in the ion. Si is treated as a carbon and P as nitrogen. Neither oxygen or sulfur are considered in the calculation.


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Instrumental Analysis Lecture 36 - Chemistry

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