Chapter 14

Mass Spectroscopy

Mass Spectrometer

Positive ions are detected. Neutral species are undetected.

p.545

Figure 14.2, p.546

Parent peak due to molecular radical cation.

Detecting nitrogen, N

Consider some simple molecules and their nominal mass.

CH4 16

CH3NH2 29

CH3OH 32

CH3F 34

CH3Cl 50, 52

CH3SH 48

What is unusual about the N compound?

The parent peak should have an odd mass for an odd number of nitrogens.

p.546

p.547

One way to distinguish between molecules having the same about the same mass is to measure their mass very precisely.

About the same mass for these species formed from the most common isotopes. In these cases we can actually determine the molecular formula from high resolution mass spectroscopy.

Table 14.1, p.548

Recall that the atomic weight is the average mass for all isotopes found in nature.For example chlorine….

35.453 = (100 * 34.9689 + 31.98 * 36.9659) / 131.98

Figure 14.4, p.550

Bromine has two isotopes 79Br and 81Br in about equal amounts.

Obtain two peaks at 122 and 124.

Low resolution mass spec does not involve itself with precise mass measurements. Low resolution examines the various peaks produced.

First consider the parent radical cation: if an element has two naturally occurring isotopes then two peaks will be produced.

Further comments on presence of chlorine and bromine.

Both Cl and Br have two common isotopes separated by two mass units.

Given the natural abundances we may calculate the ratio of the M and M+2 peaks for various combinations of Cl and Br being present.

The presence of peaks at X, X+2… for the molecular ion or fragment hopefully with close to the expected ratio is taken as indication of Cl or Br.

Ratio of peaks calculated as35Cl2

35Cl37Cl & 37Cl35Cl 37Cl2

1.00*1.00 1.00*.324+.324*1.00 .324*.324

Ratio of peaks calculated as35Cl79Br 37Cl79Br & 35Cl81Br 37Cl81Br

1.00 *1.00 .324 *1.00+1.00 *.979 .324*.979

.767 1.00 .243

Molecular Peaks, M+1

Have seen that for Cl and Br, having two common isotopes, two radical cation peaks produced. What about other elements having more than one isotope?

We know what the isotopes are and their natural occurrence.

For the M+1 peak, one atom must be using an isotope heavier by one.

Here is the data. We will use isotopic occurrence data for H, C, O for the M + 1 peak.

The M+2 peakRecap: The M+1 peak has contributions from one atom being a heavier isotope by 1.

The M+2 peak can have contributions from•One atom being a heavier isotope by 2.•Two atoms being heavier by 1 each.

M+2 peak, contributions from one atom and two atoms.

Recap: The M+1 peak has contributions from one atom being a heavier isotope by 1.

(M+1)/M = ca. 1.1% * no. of C atoms + 0.36% * no. of N atomsThe M+2 peak can have contributions from two sources

•One atom being a heavier isotope by 2. Mainly O (excluding S, Cl and Br)•Two atoms being heavier by 1 each. Mainly C atoms.

(M+2)/M = ca. (0.20% * no. of O atoms) + (1.1 * no. of C atoms)2/200%

Example 1: C5H5N[(A + 1)+]/[A+] = 5 x 1.1% + 1 x 0.36% = 5.9%[(A + 2)+]/[A+] = 5.52/200 % = 0.15%

Example 2: C7H5O[(A + 1)+]/[A+] = 7 x 1.1% = 7.7%[(A + 2)+]/[A+] = 7.72/200 % + 0.20% = 0.50%

Technique to obtain molecular formula using intensities of M, M+1, M+2 peaks.

Consider the M+1 peak, nominal mass + 1. If we know the formula we should be able to calculate the relative intensity of that peak due to the contributions from each of the atoms present. Here are the major contributors to M+1.

Here are major contributors to M+2.

Example. Given the data.

Peak Intensity

150 (M) 100

151 (M+1) 10.2

152 (M+2) 0.88

Looking at M+2 there is no Br, Cl or S. There could be oxygen.Even mass for M means there could only be even number of Nitrogen

Technique to obtain molecular formula using intensities of M, M+1, M+2 peaks.

Example. Given the data.

Peak Intensity

150 (M) 100

151 (M+1) 10.2

152 (M+2) 0.88

EquationsM+1: (1.11% x # of C) + (0.38 x # of N+ small contributions from OM+2: (0.20 x # of O) + (1.1 x # of C)2/200

We can have 0 or 2 nitrogens. Even number.

We can have 0,1,2,3,4 oxygens. 0.88/0.2 < 5Can have 0,1,2,3,4,5,6,7,8,9 carbons. 10.2/1.11 <10

Find molecular formulas having reasonable M+1 peaks

M+1 M+2

C7H10N4 9.25 0.38

C8H10N2O 9.61 0.61

C9H10O2 9.96 0.84

C9H14N2 10.71 0.52

Examine reasonable formulae. Calculate M+1, M+2 peaks

First distinguish between some species

•Radical/cation produced by ejection of electron from structure. It contains all the

atoms of the original molecule just minus one electron. Example C4H10+.

•Carbocation which is not a radical, is electron deficient and is a fragment of a stable molecule. Example C4H9

+

•Radical is not charged and is a fragment of a stable molecule. Example C4H9.

The highly energetic radical/cation can undergo fragmentation.

Return to Fragmentation of Molecular Radical Cation, M+

Example. Identify this molecule

m/e Abundance

1 <0.1

16 1.0

17 21

18 100

19 0.15

20 0.22Due to heavier isotopes

Molecular radical ionEjection of an H

H2O

Example 2

m/e Abundance

12 3.3

13 4.3

14 4.4

15 0.07

16 1.7

28 31

29303132

100891.30.21

Heavier isotopes

parent

H ejection

Oxygen

carbon

CH2O

p.550

Fragmentation of the radical/cation can lead to

2. Rearrangements can occur including elimination of neutral molecules to produce a different radical/cation.

1. A radical and carbocation as separate species. Usually a bond is split. Choice of bond to split is frequently controlled by the carbocation stability.

OH

H2O +

H

.

Carbocation, CH3CH2

+

Radical/ cation,

C4H10+.

Radical,

CH3CH2.

Radical/ cation,

C5H12O+.

Radical/ cation,

C5H10+.

Remember only the positive species are detected. Neutral species are invisible.

How to think about it…

.For this

fragmentation

Think of it this way…

One electron in this bond. When it splits we get a carbocation (observable in MS) and a radical (not observable in MS).

The fragmentation is written this way.

This C becomes +

This C bears the .

p.551

In fragmentation to produce a carbocation stability of the carbocation is an important factor in determining where the fragmention will occur.

Figure 14.5, p.552

For simple linear alkanes fragmentation will occur towards the middle of the chain.

Figure 14.6, p.552

Caution. Sometimes a peak will occur which cannot be explained such as the ethyl peak below.

Figure 14.7, p.553

Technique: recall that the neutral molecules split out do not produce a peak.

The mass of the neutral particle (invisible) may sometimes be obtained by subtracting the mass of the newly formed positive ion (detected) from the mass of the original radical carbocation.

In this example the parent molecule has mass of 84. The mass of a positive fragment is 56.

The peak at 56 is subtracted from the mass of the original ion, 84, yielding 28, the mass of ethylene which is taken as the invisible neutral molecule.

p.553

.Ionization followed by fragmentation

Splitting out of ethylene.

Ejection of electron

Radical/cation

fragmentation

Alkenes can yield allylic stabilized carbocations by fragmentation, splitting out a radical.

+ CH3

.

Difference is 15, the methyl radical

p.555

1. An alcohol radical/cation can undergo α fragmentation to produce a radical and a resonance stabilized carbocation.

C

R

R'

R"

O H

+.

R.

CR'

R"

O H CR'

R"

O H

The one “electron bond”

Alcohols have several characteristic fragmentation patterns.

Figure 14.10, p.555

2. Alcohol radical/cations can split out water to produce a new alkene radical/cation which may be detected.

Elimination of water.

74 – 56 = 18 (water).Elimination of propyl radical.

74 – 31 = 43 (C3H7)

Here is an example demonstrating both processes.

Several characteristics reactions the radical/cations

1. α cleavage: break bond to carbonyl group

Note that an α cleavage of an aldehyde could produce a peak at M – 1 by eliminating H atom.

This is useful in distinguishing between aldehydes and ketones.

Aldehydes and Ketones

p.557

2. McLafferty Rearrangement: splitting out an alkene (neutral molecule) and producing a new radical/cation.

Note that the process involves a six membered ring for a transition state.

Mass Spec of 2-octanone displays both α cleavage and McLafferty

CH3CO+

resulting from α cleavage at this bond.

CH3CH2CH2CH2CH2CH2CO+ resulting from α cleavage here.

The “invisible” radical C6H13

CH3 radical“Invisible” pent-1-ene

Carboxylic Acids can also undergo α cleavage and McLafferty rearrangement.

The peak at 60 is usually prominent for a carboxylic acid.

Likewise for esters, 2 α cleavages (around C=O) and McLafferty

C

p.559

An unexpected observation.

Figure 14.14, p.559

Some general principles

The relative height of the parent radical cation peak is greatest for straight chain compounds and decreases for branched structures. Easier cleavage.

Cleavage is favored at branched carbons due to increased stability of substituted carbocations.

Double bonds or aromatic rings stabilize the parent radical cation increasing the size of its peak.

Double bonds (aromatic rings) favor cleavage yielding allylic (benzylic or tropylium) carbocations.

CH2

R

RH H

Bonds beta to a hetero atom having lone pairs are frequently cleaved. N

HH

HElimination of small, stable molecules (water, alkenes, etc.) can occur to yield a new radical cation.

Example

Given spectra for an unknown compound: IR, MS, and NMR. Identify the compound.

First, let’s do the MS

Mass Spectrum

Base peak, the largest, others are relative to it.

Molecular peak, parent peak.

Want M+1 and M+2 relative to M not base peak.

Observations:1. Parent mass is even.

Even number of nitrogens: 0, 2, 4

2. No Br or Cl since M+2 is much too small.

3. At most about 9 carbons since 9.9/1.1 =9.

What molecular formulas could we have with good values for M+1 between 9 and 11?

Best fit for both M+1 and M+2.Maybe aromatic!

Now the IR….

Infra Red. Assumed formula C9H10O2

Carbonyl, no OH C-O stretch

Looks as if it may be an ester with an aromatic group.

Next the NMR….

NMR . Assumed formula C9H10O2

First the hydrogen counts.

Chemical Arithmetic. Assumed formula C9H10O2

C9H10O2

- C6H5

- CH2

- CH3

Subtract to get CO2

O

O

O

O

Consistent with the lack of splitting could be either.

Tentatively identified parts,