+ 2 e- radical cation

Basics of Mass Spectroscopy

The roots of mass spectroscopy (MS) trace back to the early part of the 20th century. In 1911

J.J. Thomson used a primitive form of MS to prove the existence of isotopes with 20 and

neon-22. Current, easy-to-use, table-top instruments of today are a very recent luxury. In less than a day,

you could be running samples on a mass spectrometer. However, it would take you longer to learn

the many intricacies of MS, something we cannot pursue in a book such as this. We will mainly look

at electron impact mass spectrometry (EI) and briefly mention chemical ionization (CI) as they

pertain to determining an organic structure.

The technique of MS only requires very small amounts of sample (



g-ng) for high quality data.

For that reason, it is the preferred method to evaluate product structures in combinatorial chemistry,

forensic laboratories and with complicated biological samples. Generally, in these situations, you

have some indication of the structure(s) possible. MS can be coupled to separation techniques such

as gas chromatography (GC) and high pressure liquid chromatography (HPLC) to make a

combination technique (GC-MS and LC-MS). GC can separate components in relatively volatile

mixtures and HPLC can separate components in relatively less volatile mixtures. There are also

options for direct inlet of solid samples and sampling methods for high molecular weight

biomolecules and polymers. But, these are beyond the scope of this book.

MS is different from the other spectroscopies (UV-Vis, IR, NMR) in that absorption or emission

of electromagnetic radiation is not used. Rather, the sample (molecule) is ionized by some method

(often a high energy electron beam = electron impact = EI). An electron is knocked out of a bonding

molecular orbital (MO), forming a radical cation. Dications and anions can also be formed, but we

will not consider these possibilities.

R

H

e-high

energy

R

H

radical cation

+

EI mass spec

+

2

e-The cations formed are accelerated in a high voltage field, focused and separated by mass to

charge ratio (m/z or m/e) using a magnetic and/or electric fields. A detector indicates the intensity of

each mass signal and the mass data (x axis) are plotted against this intensity (y axis) to produce a

spectrum similar to that shown below. It is also possible that this same data can be printed in a

tabulated, numerical form (shown in the side box).

The most useful information from the MS is the molecular weight (the M+ peak), which can

indicate what the formula is. The formula provides the degree of unsaturation, which gives important

clues to the possible structures (rings and pi bonds). Fragment peaks that are detected provide hints

as to the nature of the carbon skeleton, heteroatoms and functional groups present.

The most abundant peak (largest) in the mass spectrum is called the

base peak

. It is assigned a

value of 100% and all other detectable masses are indicated as a percent of the base peak. The

molecular weight peak is called the

mass peak or molecular ion peak

or parent peak

and

symbolized with an M. Since this peak is a radical cation, it often also has a + or +

.

(plus sign and a

dot) superscript as well. We will use M+. There is often ambiguity in the other fragment peaks

because of high energy rearrangements that are possible. It is usually very difficult to assign a

structure to a completely unknown molecule based solely on mass spectroscopy. But a mass

spectrum can help provide a very important piece of the puzzle, the molecular weight.

base peak = largest peak in MS spectrum = 100% peak, other peaks are reported as a

percent of this peak

molecular ion = M = M+ = M+



= parent peak

Only specific isotopic masses are found in the molecular formula. We do

not see “average” masses that are listed in the periodic table. Also present

will be M+1, M+2, etc. peaks due to other isotopes. On low resolution MS

these peaks can help decide what the molecular formula is.

In the MS example below, some of the peaks are very ‘logical’ (57, 85 and 91 are logical) and

some are less so (39, 41, 42, 51 and 55). It is also true that peaks that are ‘logical’ are sometimes

small or completely missing (119). Many of the other peaks will be explainable with certain

assumptions about fragmentations discussed later in this chapter. .

Mass percent Tabulated Data 0 25 50 75 100 125 150 175 200 0 25 50 75 100 percent relative intensity mass charge = me 85 = base peak 57 29 41 65 91 Many smaller peaks are not shown, but listed in data table to the left. 176 M+ peak O 1-phenyl-2-hexanone C12H16O , MW = 176 91 85 57 27 _{58 86 92} 119 39 27 6 28 2 29 24 39 7 41 26 42 1 43 1 50 1 51 3 55 3 57 99 58 5 60 1 63 3 65 11 77 2 85 100 86 6 89 2 90 2 91 36 92 6 176 7 177 1 (base) = M+

Typical MS Instrument Features.

The moving charged cations (R-H

+

) can be made to curve in their direction of flight in a

magnetic or electric field. The amount of curvature is determined by the mass (m) of the ions as

shown in the following equations (assuming the charge, e, is constant = +1). The magnetic field (B)

and/or accelerator plate voltage (V) can be altered to cause each possible mass to impact the detector.

The charged masses must survive about 10

-6

to 10

-5

seconds to make this journey to the detector.

Often there is some rational feature to explain each peak’s special stability that allows it to last long

enough to reach the detector, where it becomes part of the data we examine. We will look at some of

these features later in this discussion. We will not discuss other possibilities, such as metastable ions

or +2 and negatively charged ions. Our main goal in this book is interpretation.

m e B 2_r2 2V

=

r

=

mV e 1 B



m = mass e = charge (usually +1) B = size of magnetic field r = radius of curvature

V = voltage on accelerator plate

Besides just seeing a positively charged mass at the detector, we must resolve it from nearby

mass values. MS instruments can be either low resolution (LRMS) or high resolution (HRMS). Low

resolution MS instruments can generally resolve single amu values as high as about 2000 amu’s (e.g.

they can distinguish 300 amu from 301 amu). An atomic mass unit is defined as 1/12 the mass of a

neutral carbon-12 atom (

12

C = 12.0000, by definition). High resolution MS instruments can resolve

masses as close as the fourth decimal place (XXX.XXXX). With such accuracy, an exact molecular

formula can be determined by a computer. A molecular formula can also be obtained from LRMS,

through a slightly more involved procedure. HRMS instruments tend to be more expensive and less

common.

Exact Masses

We need to be precise in our calculation of possible masses for each collection of atoms because

the atoms in any cation hitting the detector are specific isotopes. The atomic weights listed in the

periodic table are average weights based on the abundance and mass of all of the naturally occurring

isotopes of each element. For example, the atomic weight of bromine in the periodic table is 79.9,

even though there is no bromine isotope with a mass of 80. The 79.9 atomic weight is a result of an

approximate 50/50 mixture of two stable isotopes of mass 78.9 and 80.9. Because of this

complication, we will require data on the exact masses and the relative abundance of the common

isotopes that we expect to encounter. Those most useful to us in organic chemistry and biochemistry

are listed below.

Average

Element

Atomic

Weight Nuclides

Exact Mass Relative

Abundance*

hydrogen

1.00797

1

H

1.00783

100.0

2

H (D)

2.01410

0.015

carbon

12.01115

12

C

12.00000 100.0

13

C

13.00336

1.11

nitrogen

14.0067

14

N

14.00307 100.0

15

N

15.00011

0.37

oxygen

15.9994

16

O

15.9949

100.0

17

O

16.9991

0.04

18

O

17.9992

0.20

fluorine

18.9984

19

F

18.9984

100.0

silicon

28.086

28

Si

27.9769

100.0

29

Si

28.9765

5.06

30

Si

29.9738

3.36

phosphorous

30.974

31

P

30.9738

100.0

sulfur

32.064

32

S

31.9721

100.0

33

S

32.9715

0.79

34

S

33.9679

4.43

chlorine

35,453

35

Cl

34.9689

100.0

37

Cl

36.9659

31.98

bromine

79.909

79

Br

78.9183

100.0

81

Br

80.9163

97.3

iodine 126.904

127

I

126.9045

100.0

*The most abundant nuclide is assigned 100% and the others assigned a fractional percent of that value.

Coincidently, in the examples listed in the table above with more than one isotope, the lowest mass isotope is

the 100% isotope.

Obtaining a molecular formula from a HRMS is relatively straight forward Each possible

molecular mass is unique when calculated to 3-4 decimal places and computers can do the

calculations for us. Try the problems below. Unfortunately, here you have to do the calculations

yourself.

Problem 1 - A low-resolution mass spectrum of 1,10-phenanthroline showed the molecular weight to

be 180. This molecular weight is correct for the molecular formulas C

14

H

12

, C

13

H

8

O and C

12

H

8

N

2

. A

high-resolution mass spectrum provided a molecular weight of 180.0688. Which of the possible

molecular formulas is the correct one? What is the degree of unsaturation in 1,10-phenanthroline?

Problem 2 – Isopalhinine A, a natural product was found by low-resolution mass spectrometry to

have a molecular weight of 291. Possible molecular formulas include C

15

H

17

NO

5

, C

16

H

21

NNO

4

, and

C

17

H

25

NO

3

. High-resolution mass spectrometry indicated that the precise molecular weight was

291.1472. What is the correct molecular formula of isopalhinine? What is the degree of

unsaturation?

To obtain a molecular formula from a LRMS requires more sophistication. Various possible

formulas can be generated using the molecular ion peak and the

rule of 13

. The first possible formula

assumes that only carbon and hydrogen are present. The molecular mass (M+) is divided by 13

generating an integer (n) and a remainder (r). The number 13 represents the mass of one carbon atom

and one hydrogen atom. The CH formula becomes C

n

H

n+r

. All molecular hydrocarbons have even

mass molecular weights.

C

H

H

H

C

H

H

C

H

H

C

H

H

C

H

H

C

H

H

H

Each of these masses = 13 amu = C + H

(We assume there are "n" of them if

the unknown was a hydrocarbon.

This is our starting point formula.)

These are left over hydrogen atoms = r

M

13

=

n +

M = molecular weight

n = number of CH units = quotient

r = left over hydrogens = remainder

Possible hydrocarbon molecular formula = C

_n

H

_n+r

(as a hydrocarbon always an even mass)

r

The degree of unsaturation can be calculated for this formula and possible rings and/or pi bonds

can be considered (discussed in the introduction, p 10). If oxygen and/or nitrogen (and other

elements) are present, the C/H numbers in the molecular formula must be changed by an amount

equal to the new element’s isotopic mass. It is assumed, when substituting atoms, that the major

isotope is used in all cases (always the lowest mass isotope, for us), H=1, C=12, N=14, O=16, S=32,

Cl=35, Br=79. Since oxygen weighs 16, we can subtract CH

4

(= 16) from the formula and substitute

in the oxygen atom. If two oxygen atoms were present, we would subtract 2x(CH

4

) = C

2

H

8

and so

forth. Nitrogen-14 would substitute for CH

2

and n nitrogen atoms would substitute for (CH

2

)x(n). If

we did not have enough hydrogen atoms for some reason (it happens), we could take away one

carbon atom and add in 12 hydrogen atoms, or if there were too many hydrogens, you could do it the

other way around and add one carbon and take away 12 hydrogen atoms.

Information concerning the possible number of nitrogen atoms in the molecular formula is also

available in the molecular mass. If the molecular mass is an even number, then the number of

nitrogen atoms has to be zero or an even number (= 0, 2, 4...). If the molecular mass is an odd

number, then the number of nitrogen atoms has to be odd (= 1, 3, 5...). Remember, each nitrogen

atom in the formula adds an extra bonding position.

C C C C O C C C C N N C C C C N CnH2n+2Ox CnH2n+3N1 CnH2n+4N2 (N is odd) (N is even) C = even mass H = even mass O = even mass MW = even mass C = even mass H = odd mass O = even mass MW = odd mass C = even mass H = even mass O = even mass MW = even mass

Problem 3 - An unknown compound produces a molecular weight of 108. What are all possible

formulas having only carbon and hydrogen or having carbon, hydrogen and an oxygen atom (…two

oxygen atoms) or having carbon hydrogen and nitrogen (what is the minimum of nitrogen atoms that

would have to be present)? What is the degree of unsaturation for each of these possibilities? Is it

possible that the formula has only a single nitrogen? If so what would the formula be? If not, why

not? What if the molecular weight was 107? (Same questions.)

To choose among the various formulas generated from the rule of 13, we can consider the other

possible isotopes present and their relative abundances to calculate the size of the peaks just one mass

unit (M+1) and two mass units (M+2) larger than the molecular ion peak (M+). For each possible

formula, percents of the M+1 and M+2 peaks versus the M+ peak are calculated. In this calculation

the M+ peak is assumed to be 100% for comparisons with M+1 and M+2, regardless of the base

peak. These calculated values are compared to the experimental values to determine the most likely

formula. The reason for this is that the relative sizes of the M+1 and M+2 peaks are determined by

the number and isotopic abundance of the elements present. The presence of either chlorine, bromine

or sulfur significantly changes the M+2 peak. If there are multiple halogens (Cl and Br), the M+2,

M+4, M+6 and beyond can be calculated and compared to the experimental mass spectrum. This

approach only works if the M+ peak is large enough so that M+1 and M+2 are significant. If the M+

peak is too small, we can’t tell what the relative fractions of M+1 and M+2 are. Let’s take a look at

how one could calculate the relative size of these peaks (M+1 and M+2).

Sample calculation using M+, M+1, M+2 peaks to identify the molecular formula by LRMS

We will assume an actual formula that is C

4

H

10

O. However, we will pretend we don’t know

this. How could the M+1 and M+2 lead us to the correct formula? The molecular mass of C

4

H

10

O is

74 and that would produce our molecular ion peak, M+. We would have an extra amu in the mass if

we had a different isotope one amu higher. We could do this 4 ways with carbon (because there are

four

13

C atoms) 10 ways with hydrogen (

2

H = D) and 1 way with oxygen (

17

O). The probabilities for

these possibilities are shown below for the M+1 peak. If we add all of these together we can see the

total probability for getting an M+1 peak relative to 1.0000 for getting the M+ peak. Using a similar

strategy we can estimate the probability for getting an M+2 peak, which will be considerably lower

since we have to get two

13

C or two

2

H or one

13

C and one

2

H. The main contribution to the M+2

peak is the

18

O isotope. Taken together, these three peaks would predict the indicated distribution for

M+, M+1 and M+2 for this collection of atoms (C

4

H

10

O).

molecular ion peak = M+ = 4x(12C) + 10x(1H) + 1x(16O) = 74 amu

as a fraction = 1.000 as a percent = 100%

Whatever the size of this peak, it is assumed to be 100% for comparison with the M+1 and M+2 peaks.

M+1 peak - arises from different possibilities of one additional amu = 75 amu

one 13C = _101.111.11 (4 ways) = 0.0439 one 2_{H =} 0.015 100.015 (10 ways) = 0.0015 one 16_{O =} 0.04 101.24 (1 ways) = 0.0004 13_C 12_{C +} 13_C 2_D 1_{H +} 2_D 17_O 16_{O +} 17_O+ 18_O sum of possibilities = (0.0439) + (0.0015) + (0.0004) = 0.0458

M+1 peak as a percent of M+ peak = (0.0458)x(100%) = 4.58%

"mini" probability theory There are 4 ways of picking the first carbon and 3 ways of picking the second carbon (=4x3) and since all carbon is the same, we can't tell what carbon was picked first and second, so we divide by two facorial (2x1).

M+2 peak - arises from different possibilities of two additional amu = 76 amu

two 13C = _101.111.11 4 x 3_{2 x 1} two 2_{H =} 0.015 100.015 one 18_{O =} 0.20 101.24 (1 ways) = 0.0020 sum of possibilities = (0.0007) + (0.0020) + (0.0001) = 0.0028

M+2 peak as a percent of M+ peak = (0.0028)x(100%) = 0.28% 2

= (0.0439)2_{(6 ways) = 0.0007} 2 _{10 x 9}

2 x 1 = (2.25x10-8)(45 ways) = 1 x 10-6 _{= 0.000001 = too small to consider}

one 13_{C and one} 2_{H =} 1.11

101.11 (4 ways) x 100.0150.015 (10 ways) = 1 x 10-6 _{= 0.000065 = 0.0001} M+ M+1 M+2 100% 4.58% 0.28%

M+ = molecular ion peak Exact Mass (M+1) (M+2) (formulas) 74 CH₂H₂O₂ 74.0117 1.95 0.41 CH4N3O 74.0355 2.33 0.22 CH₆N₄ 74.0594 2.70 0.03 C2H2O3 74.0004 2.31 0.62 C2H4NO2 74.0242 2.69 0.42 C₂H₆NO 74.0480 3.06 0.23 C3H6O2 74.0368 3.42 0.44 C₃H₁₀N₂ 74.0845 4.17 0.07

C4H10O 74.0003 4.52 0.28 Here is our compound.

M+ = molecular ion peak Exact Mass (M+1) (M+2) (formulas) 75 CH₂H₂O₂ 74.9956 1.60 0.61 CH4N3O 75.0320 2.70 0.43 CH₆N₄ 75.0798 3.45 0.05 C2H2O3 75.0684 3.81 0.25

etc. Data tables exist with many values _{already calculated for comparisons.}

Since the molecular weight is even, the number of nitrogens atoms must be even (0,2,4...). Any formulas with an odd number of nitrogen atoms must be part of a fragment.

To find a possible molecular formula using the M+1 and M+2 peaks, we first find the correct molecular weight for our molecule (in this case mass = 74). Then we look through the M+1 and M+2 values for two values that match our mass spec data. In this case we see that C4H10O is a very close match and it becomes our best guess.

Problem 4 –

a.

Calculate the relative intensities (as a percent) of M

+

, M+1 and M+2 for propene

(CH

3

-CH=CH

2

) and diazomethane (CH

2

=N=N). Can these two formulas (C

3

H

6

vs CH

2

N

2

) be

distinguished on the basis of their M+1 and M+2 peaks? Calculate the exact mass (four decimal

places) for both of these formulas. Can they be distinguished on the basis of exact mass? Helpful

data are on page 4.

b.

Both CHO

+

and C

2

H

5+

have fragment masses of approximately 29, yet CHO

+

has a M+1 peak of

1.13% and M+2 peak of 0.20%, whereas C

2

H

5+

has a M+1 peak of 2.24% and M+2 peak of

0.01%. High resolution mass spec shows CHO

+

to have a different fragment mass than C

2

H

5+

.

Explain these observations and show all of your work. Helpful data are on page 4.

Chlorine, bromine and sulfur, when present, have very characteristic M+2 peaks (32.6% for Cl,

96.9% for Br and 4.4% for S). If multiple Cl’s and/or Br’s are present M+2, M+4 and beyond are

indicative of the number and type of halogen(s) present. The various patterns are available in many

references. However, you can calculate these values yourself, as was done above for the M+1 and

M+2 peaks above.

one Cl – comparison of M+ peak (

35

Cl) to M+2 peak (

37

Cl)

M+ peak

relative size

probability of

35

Cl =

100

100 + 32

(1 way) = 0.758

(assigned a referenced value of 100%)

M+2 peak

relative size

probability of

37

Cl =

32

100 + 32

(1 way) = 0.242

percent of M+ peak =

0.242

0.758

(100%) = 32%

M+

M+1

M+2

100%

32%

one Br – comparison of M+ peak (

79

Br) to M+2 peak (

81

Br)

M+ peak relative size

probability of 79_{Br =} 100

100 + 97 (1 way) = 0.508

(assigned a referenced value of 100%)

M+2 peak relative size

probability of 81Br = 97

100 + 97 (1 way) = 0.492

percent of M+ peak = 0.492

0.508 (100%) = 97% M+ M+1 M+2

one S – comparison of M+ peak to M+1 to M+2 peak

M+ peak relative size

probability of 32_{S = 100}

100 + 0.79 + 4.43 (1 way) = 0.950 (assigned a referenced value of 100%)

M+1 peak relative size

probability of 33S =

percent of M+ peak = 0.008_0.950 (100%) = 0.8%

M+ M+1 M+2

100%

4.4%

M+2 peak relative size

probability of 34_{S =} percent of M+ peak = 0.042_0.950 (100%) = 4.4% 0.79 100 + 0.79 + 4.43 (1 way) = 0.008 4.43 100 + 0.79 + 4.43 (1 way) = 0.042 0.8%

one Br and one Cl – comparison of M+ peak to M+2 and M+4 peaks

M+ peak relative size

probability of 79_{Br = 0.508 (from above) probability of} 35_{Cl = 0.758 (from above)}

(probability of 79_{Br)(probability of} 35_{Cl) = (0.508) (0.758)(1 way) = 0.385}

(assigned a referenced value of 100%)

M+2 peak relative size

M+ M+2 M+4 100%

31% 129%

probability of 81_{Br = 0.492 (from above) probability of} 37_{Cl = 0.242 (from above)}

(probability of 79_{Br)(probability of} 37_{Cl)(1 way) = (0.508) (0.242)(1) = 0.123}

(probability of 81_{Br)(probability of} 35_{Cl)(1 way) = (0.492) (0.758)(1) = 0.373}

total = 0.496 percent of M+ peak = (0.496/0.373)x100% = 129%

M+4 peak relative size

(probability of 81_{Br)(probability of} 37_{Cl)(1 way) = (0.492) (0.242)(1) = 0.119}

percent of M+ peak = (0.119/0.373)x100% = 31%

two Cl – comparison of M+ peak to M+2 peak to M+4 peaks

M+ peak relative size

probability of two 35_{Cl = (0.758)}2 _{(1 way) = 0.602}

(assigned a referenced value of 100%)

M+2 peak relative size

M+ M+2 M+4 100%

10% 61%

probability of 37Cl = 0.242 (from above)

(probability of 35_{Cl)(probability of} 37_{Cl)(2 ways) = (0.758) (0.242)(2) = 0.367}

percent of M+ peak = (0.367/0.602)x100% = 61%

M+4 peak relative size

(probability of 37_{Cl)(probability of} 37_{Cl)(1 way) = (0.242)}2_{(1) = 0.059}

Problem 5 - Calculate the relative intensities (as a percent) of M+, M+2 and M+4 for Br

2

. Use the

probabilities from above.

Problem 6 - Calculate the relative intensities (as a percent) of M

+

, M+2, M+4 and M+6 for BrCl

2

and

Br

2

Cl. Hint: All of the data you need to perform these calculations are in the examples above. Use

the probabilities from above.

Energetics of Fragmentation of simple hydrocarbon patterns

Bonds are broken in fragmentations, forming radicals and/or cations. The energy costs for

radicals and cations of common hydrocarbon patterns are worked out in the tables that follow. We

first assume a C-H bond is homolytically broken (each atom gets one electron, no charge is formed).

Next, we take away the cost of making the hydrogen atom (the same for every C-H bond) to find out

what the cost is for forming only the carbon free radical. Lower energy possibilities are favored over

higher energy possibilities. A few problems are provided just below the following tables to illustrate

these points.

A similar diagram is constructed to estimate the energy costs of forming carbocations. We start out the

same, but in this diagram we include the ionization potential of the carbon free radical, a value that can be

measured experimentally. We again take away the energy to make the hydrogen free radical and also take

away the energy change when the hydrogen atom attracts the extra electron (electron affinity) to become a

hydride. What remains is an estimate of the energy to make only the carbocation. This is a considerably larger

amount of energy than to make the carbon free radical (because we are stealing away an electron).

General Energy Cycle for Carbocations- relative energy to form carbocations (all energy values in kcal/mole) C H heterolytic bond energy C H R-H R H homolytic bond energy

H_fo_{(H ) = -52 heat of formatio of hydrogen}

atom, common to all cycles

H-H H3C-H CH₃CH₂-H (CH3)2CH-H (CH3)3C-H CH₂=CHCH₂-H C6H5CH2-H Compound Radical H (hydrogen carbocation) H3C (methyl carbocation) CH₃CH₂ (primary carbocationl) (CH3)2CH (secondary carbocation) (CH₃)₃C (tertiary carbocation) CH₂=CHCH₂ (allylcarbocation) C6H5CH2 (benzyl carbocation) (104) + (313) - (17) - (52) = +348 Hfo(R ) = [BE+IP-EA- Hfo(H )] = energy to make R 104 105 98 95 92 86 ionization potential of R R e- H H Hf o_{(H electron affinity) = -17} H_fo_{(R ) = + value} (see table) Energy to form carbocation 88 313 227 193 169 154 186 165 -17 -17 -52-52 I.P. E.A.(H) H_fo_{(H )} (BE) -17 -17 -17 -17 -17 -52 -52 -52 -52 -52 (105) + (227) - (17) - (52) = +263 (98) + (193) - (17) - (52) = +222 (95) + (169) - (17) - (52) = +195 (92) + (154) - (17) - (52) = +177 (86) + (186) - (17) - (52) = +203 (88) + (165) - (17) - (52) = +184

Common arguments for relative stabilities of free radicals and carbocations are inductive

effects/hyperconjugation and resonance. Inductive effects and hyperconjugation argue that switching

out a hydrogen for a carbon group allows greater electron donation to the electron deficient carbon

atom (free radical or carbocation) because of increased pairs of electrons polarized towards the

electron deficient centers. Carbocations are much more electron deficient than free radicals and

benefit much more from this effect. The resonance argument states that an adjacent pi bond or lone

pair can spread electron density through parallel p orbitals, thus reducing the energy to form a cation

or free radical.

The differences in relative carbocation stabilities parallel the trend seen in free radicals, but are greatly

enhanced versus the free radical stabilities.

One could also make a steric argument for tertiary being the most stable free radical or

carbocation. The geometry changes from 109

o

(sp

3

) bond angles to 120

o

bond angles (sp

2

). The

ground state of a tertiary C-H bond would start at higher potential energy from crowding, which

would be relieved somewhat when the fourth group is removed, providing, perhaps, part of the

advantage in the tertiary reaction over secondary over primary over methyl when forming tertiary free

radicals and carbocations.

C R

R R R

more crowded as sp3 _{center =higher}

potential energy starting point with 3-4 larger groups around

tetrahedral carbon

less crowded as sp2_{with 3 groups}

around trigonal planar carbon is slightly more stable than it would be if groups were smaller

C R R

R

R Breaking a bond is a large

uphill energy transformation, but less so with a sterically crowded starting point, so E_a is a little smaller than expected.

Problem 7 – Consider the possible fragmentation of 2-methylbutane (isopentane). There are 3 types

of C-C bonds that could break (b,d,f) and 4 types of C-H bonds that could break (a,c,e,g). Only

consider breaking the C-C bonds (b,d,f) and the tertiary C-H bond (c). Each bond could break in two

ways: either atom could be a cation and either atom could be a free radical. Calculate the energy cost

for each possibility (each bonded atom as a radical and each atom as a cation). For each possibility

what are the masses that would be observed at the detector (we only see cations)? This problem will

require eight calculations for the four bonds considered.

CH₃ H2C H H C CH2 H H H a b c d e f g high energy electron beam 2-methylbutane (isopentane) CH₃ H₂C H H C CH₂ H H H a b c d e f g radical cation Possible fragmentations? Energy to rupture bonds (eight calculations).

b _b c c d d f f

Actual Mass Spectrum – tabulated and graphical.

15 2 26 4 27 43 28 6 29 60 30 1 37 1 38 3 39 30 40 5 41 88 42 95 43 100 44 7 50 2 51 3 53 4 55 10 56 40 57 95 58 6 71 5 72 16 mass percent = base = M+

Peaks 15, 29, 43, 57 and 72 are logical. In our discussions of

fragmentation we will see how many of the other peaks are explainable.

0 25 50 75 100 0 25 50 75 100 percent relative intensity mass charge = me 29 41,42 43 = base peak 57 _{Many smaller} peaks not shown.

72 M+ peak isopentane C5H12 75 eV CH CH3 H₃C C H2 CH3 57 15 43 29 39 MW = 72

Problem 8 – Consider the possible fragmentation of 2,2,4-trimethylpentane. There are four types of

C-C bonds that could break (a, b, d, f) and 4 types of C-H bonds that could break (a, c, e, g). Only

consider breaking the C-C bonds (a, b, c, d). Each bond could break in two ways: either atom could

be a cation and either atom could be a free radical. Calculate the energy cost for each possibility

(each bonded atom as a radical and each atom as a cation). For each possibility what are the masses

that would be observed at the detector? This problem will require eight calculations for the four

bonds considered (we only see cations).

b b c c d d C CH₃ H₃C CH₃ H₂ C HC CH₃ a _b c _d high energy electron beam

2,2,4-trimethylpentane radical cation

Possible fragmentations? Energy to rupture bonds (eight calculations). CH_{3 C} CH₃ H₃C CH₃ H₂ C HC CH₃ a _b c _d CH3 a a

Actual Mass Spectrum tabulated and graphical.

mass percent = base = M+ (missing) 27 5 29 8 39 5 40 1 41 21 42 1 43 18 53 1 55 3 56 33 57 100 58 4 99 6 114 0

Problem 9 - Predict reasonable fragmentation patterns for n-octane and where the major ion peaks

should appear. Rationalize your predictions on the basis of energetics. The mass spectrum is

provided for comparison. Some of the less logical peaks will become explainable after our

discussions on fragmentation. Is there a ‘logical’ peak that is missing?

Actual Mass Spectrum tabulated and graphical.

mass percent = base = M+ 27 20 28 4 29 27 39 12 40 2 41 44 42 15 43 100 44 3 53 2 55 11 56 18 57 34 69 2 70 12 71 20 84 7 85 26 86 2 114 6 0 25 50 75 100 0 25 50 75 100 percent relative intensity mass charge = mZ base peak 29 71 43 57 Many smaller peaks not shown.

M+ peak octane C₈H₁₈ 75 eV 114 H3C H2 C C H2 H2 C C H2 H2 C C H2 CH3 120 85 41

Special patterns of fragmentation from organic functional groups

Alkanes - Key Points (see examples above)

1.

Lower mass alkyl branch fragments (2-6 C’s, masses = 29, 43, 57, 71, 85) are more intense than

higher mass fragments (



6). The loss of the smaller branch as the cation more commonly reaches

the detector.

2.

The major carbocations that form follow carbocation stabilities (R

+

= 3

o

> 2

o

> 1

o

> Me). It is

also quite possible that less stable carbocations rearrange to more stable carbocations before they

reach the detector. We can’t tell by only observing the mass since they have the same number.

R proposed fragmentation R C4H9 probable rearrangement C4H9 less stable

primary carbocation tertiary carbocationmore stable

can't tell which

mass = 57 mass = 57

3.

Linear alkanes more often have observable molecular ion peaks, while increased branching

weakens the molecular ion peak. Fragmentation is more common at branch points. Loss of a

methyl from a straight chain is considerably weaker than loss of a methyl at a branch point.

M+ = 114 (6%) base peak = 43 (M - 15) = 99 peak (0%) M+ = 114 (3%) base peak = 43 (M - 15) = 99 peak (1%) M+ = 114 (0%) base peak = 57 (M - 15) = 99 peak99 peak (6%)

4.

Linear fragments often differ by 14 amu (different size branches split off between carbons in different

molecules, CH

2

= 14). Take another look at problem 9, just above.

5.

There are often clusters of peaks around main peaks. Very large fragment peaks will have a

trailing M+1 peak due to

13

C isotopes (about 1% for every carbon present). A rough guide for

any large peak is that it will have “M+1” peak that is about 1% its size for every carbon in the

fragment due to 1%

13

C isotopes at each carbon. For example, if a fragment mass had an 80%

value in a five carbon fragment, the next mass peak would be expected to be 0.05x80%



4% size

based on

13

C isotopes. If there were 10 carbons, the next mass peak would be expected to have

0.10x80%



8% size just based on the

13

C isotopes (in addition to any real fragments that might

come at that value.

6.

Cycloalkanes tend to have stronger molecular ion peaks (two bonds have to break) and their

fragment patterns are more complicated to interpret (and we won’t try to interpret every

possibility). Alkene fragmentation peaks are often subfeatures of the fragmentation pattern. Loss

of “CH

2

CH

2

“ (= 28) is common, if present.

M+ = 112 (59%)

M-28 = 84 (39%) M+ = 114 (6%)M-28 = 86 (2%)

7.

Two masses that seem to show up in nearly every mass spectrum are 39 and 41. These may arise

from resonance stabilized carbocations formed by rearrangements in the high energy electron

beam. Look for peaks that extend those patterns by units of 14 (insertion of a CH

2

),. which are

also commonly observed masses.

8.

Even masses of 30, 44, 58, 72, etc. on occasion can be due to “radical-cation alkanes” that form

from high energy rearrangements. Some of these masses form from other fragmentations too.

But if there is no other logical reason to see one of these masses, this could be a possible

explanation.

Common alkane fragmentations occur at branch points; more branches lead to more stable

carbocations. However, skeletons can rearrange in almost any conceivable way possible to form

more stable carbocations (e.g. 3

o

R+ > 2

o

R+ > 1

o

R+ > H

3

C+). Also, alkanes can lose H

2

or R-H to

form alkenes, so we have to consider possible alkene rearrangements for alkanes too (see our next

functional group). Smaller masses tend to be more prominent than larger masses in the mass

spectrum. Perhaps they don’t have as many options for falling apart as the larger fragments do.

Also, when larger fragments fall apart, they make smaller fragments.

mass % 15.0 1 26.0 1 27.0 20 28.0 4 29.0 27 39.0 12 40.0 2 41.0 44 42.0 15 43.0 100 44.0 3 51.0 1 53.0 2 54.0 1 55.0 11 56.0 18 57.0 34 58.0 2 69.0 2 70.0 12 71.0 20 72.0 1 84.0 7 85.0 26 86.0 2 99.0 none 114.0 6 115.0 1 M+ = 114 C8H18 8529 71 43 57 C8H18 C6H13 C5H11 C4H9 C3H7 C2H5 C1H3 M+ = 114 85 71 57 29 15 99 15 57 C₆H₁₃ 99 20 30 40 50 60 70 80 90 100 110 120 27

Mostly peaks greater than 4% of the base peak are shown. 43 = base 71 39 41 M+ = 114 85 57 42 29 28 55 56 70

Only cations reach detector, so only the part with positive charge is observed at the detector. A positive charge is written on all fragments to indicate that either part could retain the positive charge (in a rearranged stable form). Often you can see the mass of both cations of a possible fragmentation. It is useful to look for both fragment masses in the mass spectrum. Peaks related to alkene fragmentations are discussed in the next functional group.

The typical appearance of a mass spectrum is shown below. Data is also often presented as shown to the right. The intensity of the peaks tends to decrease as the fragment masses get larger. Larger fragments are less likely to survive the 10-5second trip to the detector.

not observed in octane octane - all alkane fragments

are observed, except 99.

MW = 114 actual peaks in octane M+ base alkane peaks 43

(-H₂)

Loss of hydrogen (H-H) or an alkane (R-H) fragment generates alkenes so alkene fragmentation patterns are also observed from alkane structures (see on next page). 3,4-dimethylhexane - has branches

15 / 99 29 / 85 43 / 71 57 / 57 MW = 114 (-RH) possible alkyl fragments H _{58 (4%)} 56 (100%) elimination reaction similar to

-H₂O in alcohols to form alkene

The base peak (56) is likely from an alkene, C₄H₈.

mass % 27.0 10 28.0 1 29.0 26 39.0 7 40.0 1 41.0 43 42.0 2 43.0 58 44.0 2 51.0 1 53.0 2 55.0 8 56.0 100 57.0 81 58.0 4 69.0 3 70.0 1 71.0 1 84.0 7 85.0 41 86.0 3 99.0 none 114.0 2 115.0 0.2 actual peaks in 3,4-dimethylhexane alkenes

(see the next functional group)

M+

20 30 40 50 60 70 80 90 100 110 120

27

Mostly peaks greater than 4% of the base peak are shown. 56 = base 39 41 M+ = 114 130 84 57 43 29 55 MW = 114 85 alkane peaks the base peak is not expected

Remarkably, it is the major peak in the spectrum!

It is very common to see alkene fragments in the mass spectra of alkanes, though it is very

surprising to see one as the base peak, as is the case here. In the next functional group, we will

compare fragmentations of alkenes and alkanes.

Alkenes - Key Points

1.

A pi electron is likely to be ionized first from the HOMO of the alkene as the least tightly held electrons.

Alkenes often produce stronger molecular ion peaks than alkanes because of this.

R R

+

e- Remaining sigma bond_{holds skeleton together.}

octane, MW =114 (M+ = 6%) oct-1-ene, MW =112 (M+ = 20%)

2.

The double bond can migrate through the skeleton (this makes it difficult to distinguish among positional

isomers sharing a common skeleton).

3.

Allylic cleavage is common due to resonance stabilization of cation fragment. The mass can vary

depending on the groups attached to the allylic part.

Look for peaks that extend this pattern by units

of 14 (insertion of CH

2

x1, x2, …).

resonance stabilized carbocation R' ionization R' R' fragmentation free radical is sucked away R _R R R mass = 41 (R = H) 55 (R = CH3) 69 (R= CH2CH3) 83 (R = C₃H₇) etc.

4.

McLafferty-like rearrangements are possible (similar to carbonyl pi bonds). Again, bond migration is

possible. Also look for some of these fragment peaks in alkane mass spectra that have lost H

2

.

C R H2C H CH₂ R C H2 C C H CH₂ mass = 42 (R = H) 56 (R = CH₃) 70 (R= CH2CH3) 84 (R = C3H7) C C fragmentation McLafferty-like rearrangement R R R R 28 (R = H) 42 (1 extra C) 56 (2 extra C) 70 (3 extra C) It is possible to see the cation charge on either fragment. Both fragments will be even unless an odd number of nitrogen atoms is present.

even mass

5.

Cyclohexenes often undergo retro Diels-Alder reactions.

R₁ R₂ fragmentation is a retro-Diels-Alder reaction R1 R2 diene dienophile

Only cations reach the detector. Either fragment could be positive, but usually the diene would be the more stable cation. Both fragments will be even unless an odd number of nitrogen atoms is present.

Alkenes Fragmentation Patterns

(Many of those below can also be found in octane, an alkane.)

Only cations reach detector, so only the part with positive charge is seen at the detector. A positive charge is

written on both fragments to indicate that either could retain the positive charge (in a rearranged stable form).

Often you can see both as cations from different fragmentations. The following peaks are explained by

common alkene fragmentations (data on the right). Many of them are found in fragment peaks of octane, an

alkane (see data on the following page). A pi bond can migrate through the skeleton to almost any conceivable

position, leading to almost any variation conceivable.

H 70 (12%) 42 (15%) H 56 (18%) H 42 (15%) 69 (2%) 43 (100%) 57 (34%) 71 (20%) 41 (44%) 55 (11%) McLafferty rearrangements _{allylic fragmentations}

OR OR H 84 (7%) 28 (4%) OR OR 112 (0%) OR OR OR OR OR CH₃ 83 (0%) 97 (0%) 29 (27%) 15 (0%) actual peaks from octane mass % 26.0 1 27.0 20 28.0 4 29.0 27 39.0 12 40.0 2 41.0 44 42.0 15 43.0 100 44.0 3 53.0 2 55.0 11 56.0 18 57.0 34 58.0 2 69.0 2 70.0 12 71.0 20 72.0 1 84.0 7 85.0 26 86.0 2 114.0 6 112 (0%) 112 (0%) 112 (0%) 112 (0%) 112 (0%) 112 (0%) 112 (0%) 112 (0%) 70 (12%) 56 (18%)

Similar fragmentation patterns for C₈H₁₆ alkenes. Notice that octane (an alkane) has many of these same fragments. 15.0 1 26.0 1 27.0 25 28.0 5 29.0 35 30.0 32.0 1 38.0 1 39.0 28 40.0 5 41.0 82 42.0 66 43.0 100 44.0 3 50.0 51.0 2 52.0 1 53.0 8 54.0 9 55.0 99 56.0 87 57.0 19 58.0 59.0 63.0 65.0 1 66.0 67.0 6 68.0 7 69.0 44 70.0 86 71.0 12 72.0 77.0 79.0 81.0 1 82.0 6 83.0 34 84.0 22 85.0 2 86.0 97.0 4 112.0 20 113.0 2 15.0 1 26.0 1 27.0 18 28.0 4 29.0 33 30.0 1 32.0 38.0 1 39.0 19 40.0 3 41.0 64 42.0 34 43.0 11 44.0 50.0 1 51.0 2 52.0 1 53.0 8 54.0 9 55.0 100 56.0 52 57.0 21 58.0 1 59.0 63.0 65.0 1 66.0 67.0 5 68.0 4 69.0 29 70.0 43 71.0 4 72.0 77.0 79.0 81.0 1 82.0 2 83.0 16 84.0 7 85.0 86.0 97.0 2 112.0 28 113.0 3 15.0 1 26.0 1 27.0 25 28.0 4 29.0 45 30.0 1 32.0 38.0 1 39.0 22 40.0 4 41.0 81 42.0 44 43.0 15 44.0 1 50.0 1 51.0 2 52.0 1 53.0 8 54.0 8 55.0 100 56.0 63 57.0 25 58.0 1 59.0 63.0 65.0 1 66.0 67.0 6 68.0 5 69.0 34 70.0 56 71.0 6 72.0 77.0 79.0 81.0 1 82.0 3 83.0 22 84.0 10 85.0 1 86.0 97.0 2 112.0 36 113.0 3 15.0 2 26.0 2 27.0 25 28.0 4 29.0 19 30.0 32.0 38.0 2 39.0 26 40.0 5 41.0 100 42.0 38 43.0 18 44.0 1 50.0 2 51.0 4 52.0 2 53.0 11 54.0 8 55.0 95 56.0 54 57.0 16 58.0 1 59.0 1 63.0 1 65.0 2 66.0 1 67.0 10 68.0 7 69.0 47 70.0 48 71.0 6 72.0 77.0 2 79.0 2 81.0 3 82.0 2 83.0 24 84.0 7 85.0 2 86.0 97.0 2 112.0 36 113.0 3 15.0 1 26.0 2 27.0 23 28.0 3 29.0 17 30.0 32.0 38.0 2 39.0 24 40.0 4 41.0 93 42.0 29 43.0 15 44.0 50.0 2 51.0 3 52.0 2 53.0 9 54.0 9.2 55.0 100 56.0 46 57.0 14 58.0 59.0 63.0 1 65.0 2 66.0 1 67.0 9 68.0 5 69.0 36 70.0 44 71.0 5 72.0 77.0 1 79.0 2 81.0 2 82.0 2 83.0 24 84.0 7 85.0 1 86.0 97.0 2 112.0 36 113.0 3 15.0 1 26.0 1 27.0 16 28.0 2 29.0 14 30.0 32.0 38.0 1 39.0 16 40.0 2 41.0 78 42.0 25 43.0 12 44.0 50.0 1 51.0 2 52.0 1 53.0 6 54.0 7 55.0 100 56.0 43 57.0 12 58.0 59.0 63.0 65.0 2 66.0 67.0 8 68.0 4 69.0 32 70.0 42 71.0 4 72.0 77.0 1 79.0 1 81.0 2 82.0 2 83.0 29 84.0 7 85.0 86.0 97.0 1 112.0 33 113.0 3

1-octene trans-2-octene cis-2-octene trans-3-octene cis-3-octene _{cis-4-octene trans-4-octene}

octane 15.0 1 26.0 1 27.0 20 28.0 4 29.0 27 30.0 32.0 - 38.0 - 39.0 12 40.0 2 41.0 44 42.0 15 43.0 100 44.0 3 50.0 - 51.0 1 52.0 - 53.0 2 54.0 1 55.0 11 56.0 18 57.0 34 58.0 2 59.0 - 63.0 - 65.0 - 66.0 - 67.0 - 68.0 - 69.0 2 70.0 12 71.0 20 72.0 1 77.0 - 79.0 - 81.0 - 82.0 - 83.0 - 84.0 7 85.0 26 86.0 2 97.0 - 112.0 - 113.0 - 114.0 - 115.0 1 not available A C E G B D F H H G A B C D E F

alkyl branch fragments = 15, 29, 43, 57, 71, 85, 99 allylic fragments = 27, 41, 55, 69, 83, 97

Another Alkene Example (C7 alkene)

H

H

H

McLafferty rearrangements A pi bond can migrate _{through the skeleton to allylic fragmentations} almost any conceivable

position. 15 (1%) 29 (56%) 43 (16%) 57 (31%) 71 (3%) 85 (0%)

alkenes (1-heptene, 2-heptene, 3-heptene, all of them look similar because the pi bond can migrate through the skeleton)

C₇H₁₄ = 98 (14%) _{42 (55%) 56 (100%)} C7H14 = 98 (14%) 41 (97%) 57 (31%) C7H14 = 98 (14%) C7H14 = 98 (14%) 56 (100%) 42 (55%) _{C7H14 = 98 (14%)} C7H14 = 98 (14%) 70 (44%) _{28 (5%)} 55 (68%) 43 (16%) 69 (31%) _{29 (56%)} alkyl branches C7H14 = 98 (14%) CH3 = 15 (1%) 83 (31%) This example starts with hept-1-ene

15.0 1 18.0 1 26.0 2 27.0 26 28.0 5 29.0 56 30.0 1 38.0 2 39.0 30 40.0 5 41.0 97 42.0 55 43.0 16 50.0 2 51.0 2 52.0 1 53.0 6 54.0 8 55.0 68 56.0 100 57.0 31 58.0 1 67.0 2 68.0 4 69.0 31 70.0 44 71.0 2 83.0 4 98.0 14 allylic

fragments McLafferty_fragments 27 (26%) 41 (97%) 55 (68%) 69 (31%) 83 (4%) 28 (5%) 42 (55%) 56 (100%) 70 (44%) all peaks > 1%

Alkynes - Key Points

1.

Terminal alkynes have weak or missing M+ peaks (they often lose radical hydrogen), though M-1

can be very strong.

R H H R H H M+ (M-1)+

2.

The triple bond can migrate through the skeleton (this makes it difficult to distinguish among positional

isomers sharing a common skeleton).

These alkynes all look similar.

3.

All alkynes give a reasonably strong m/e = 39 peak from propargylic cleavage (resonance is OK,

but more electronegative sp carbocation resonance form reduces contribution). This mass can

also be explained by rearrangement to from a very stable aromatic cyclypropenyl carbocation. If

you look at a lot of mass spectra, this mass always shows up, even if no alkyne is present. Look

for peaks that extend this pattern by units of 14 (insertion of CH

2

x1, x2, …).

mass = 39 (R = H) 53 (R = CH₃) 67 (R= CH2CH3) R also works for Only cations reach the detector. Mass 39 is in every EI mass spectrum.

This could be because the cation is really an aromatic carbocation. R CH C C H radical cation fragmentation R' C H C C H C H C C H resonance R _R R'

4.

Small peaks at M=26 are probably ethyne (acetylene).

H

C C

H

M = 26

5.

McLafferty-like rearrangements are possible (similar to the alkene above and a carbonyl pi bond)

R H R radical cation fragmentation C C C H H R H

Either fragment can be observed and both show an even mass. R C C H H H on one or the other.

even mass 40 (R = H) 54 (R = CH3) 68 (R= CH2CH3) 82 (R= C3H7) 28 (R = H) 42 (R = CH3) 56 (R= CH2CH3) 70 (R= C3H7)

Example peaks from hept-1-yne:

alkynes (1-heptyne, 2-heptyne, 3-heptyne, all of them look similar because the pi bonds can migrate through the skeleton)

H

H

H

McLafferty rearrangements

allylic fragmentations A pi bond can migrate

through the skeleton to almost any conceivable position. 15 (0.5%) 29 (46%) 43 (4%) 57 (28%) 71 (0.2%) 85 (0%) C₇H₁₂ = 96 (1%) _{40 (12%) 56 (26%) 39 (30%)} 57 (28%) 54 (35%) 42 (8%) 68 (30%) _{28 (4%)} 53 (18%) 43 (4%) 67 (44%) _{29 (46%)} C7H12 = 96 (1%) C7H12 = 96 (1%) C7H12 = 96 (1%) C7H12 = 96 (1%) C7H12 = 96 (1%) 81 (100%) 15 (0.5%) C7H12 = 96 (1%) CH₃ C7H12 M+ = 96

mass % mass % mass % mass % mass %

1-heptyne 26.0 1 27.0 18 28.0 4 29.0 46 30.0 1 37.0 1 38.0 3 39.0 30 40.0 12 41.0 71 42.0 8 43.0 4 45.0 1 50.0 3 51.0 6 52.0 3 53.0 18 54.0 35 55.0 51 56.0 26 57.0 28 58.0 1 63.0 2 65.0 7 66.0 3 67.0 44 68.0 30 69.0 2 70.0 2 77.0 3 79.0 11 80.0 1 81.0 100 82.0 7 95.0 9 96.0 1 C7H12 M+ = 96

mass % mass % mass % mass % mass %

2-heptyne 15.0 1 18.0 2 26.0 3 27.0 40 28.0 7 29.0 9 37.0 2 38.0 4 39.0 51 40.0 8 41.0 68 42.0 7 43.0 26 50.0 6 51.0 12 52.0 9 53.0 47 54.0 82 55.0 22 56.0 8 57.0 1 62.0 2 63.0 3 65.0 10 66.0 6 67.0 43 68.0 42 69.0 4 77.0 5 78.0 1 79.0 14 80.0 3 81.0 100 82.0 8 91.0 1 95.0 5 96.0 18 97.0 2 C7H12 M+ = 96

mass % mass % mass % mass % mass %

3-heptyne 39.0 43 40.0 12 41.0 84 42.0 10 43.0 3 50.0 6 51.0 12 52.0 7 15.0 2 18.0 1 26.0 3 27.0 23 28.0 1 29.0 14 37.0 2 38.0 4 53.0 49 54.0 25 55.0 26 56.0 5 61.0 1 62.0 3 63.0 5 64.0 1 65.0 21 66.0 11 67.0 100 68.0 29 69.0 2 74.0 1 77.0 9 78.0 2 79.0 32 80.0 4 81.0 93 82.0 6 91.0 2 93.0 1 95.0 7 96.0 70 97.0 6

Benzenoid Structures - Key Points

1.

Generally, aromatics compounds show a strong M+ peak.

2.

A side chain alkyl branch (RCH

2

-) can fragment at the benzylic position, which is proposed to

rearrange to the tropylium ion showing a m/e = 91 peak. Analogous rearrangements are possible

in more substituted benzenoid compounds producing different, but predictable, masses.

CH2 R radical cation fragmentation CH2 R lots of resonance rearrangement tropylium ion, an aromatic cation (lots of resonance) Only cations reach the detector. This mass is 91 (if

R = H) and even though it is a very stable cation, it rearranges to a more stable 'tropylium' carbocation. Any branches or heteroatoms would change the '91' mass. R' _R' R' R' = mass H 91 CH3 105 C2H5 119 HO 107 H2N 106

3.

Isomeric benzenes are difficult to distinguish among, as a group. Even though the structures are

different, the mass spectra of the compounds are pretty much alike due to high energy

rearrangements.

These isomers have similar looking mass spectra.

4.

McLafferty-like rearrangements are possible, if a simple alkyl chain of three more carbons is

present (oxygen can also be in the branch) and a hydrogen atom is on the gama atom. This

fragmentation produces an even mass of m/e = 92 for an unsubstituted carbon chain. Substituted

rings will have different masses depending on the additional atoms. Remember that part of the 92

peak is C-13 isotopes in the 91 peak (about 7x0.01 = 0.07).

C C_ C_ H H _H H H H R C C_ C_ H _H H H H R H H or

can be on either fragment

R = mass H 92 CH3 106 C2H5 120 HO 108 H2N 107 R = mass H 28 CH3 42 C2H5 56 C3H7 70 Even mass, if there is not an

odd number of nitrogen atoms.

Both have even masses, if there is not an odd number of nitrogen atoms.

O C_ C_ H H H H H O C_ C_ H H H H H H or

can be on either fragment

M+ = 122 (35%) 94.0 = 100% 28.0 = 1%

Examples:

H C11H16 = 148 (27%) H H

McLafferty rearrangements benzylic fragmentations

92 (74%) 56 (0.4%) 91 (100%) 57 (4%) 15 (0.2%) 29 (6%) 43 (1%) 57 (4%) 71 (0%) 85 (0.2%) 105 (11%) 105 (11%)

bridging phenyl group

43 (1%)

65 (9%) 77 (4%)

C12H18

M+ = 162

mass % mass % mass % mass % mass %

hexylbenzene 27.0 5 29.0 6 39.0 6 41.0 8 42.0 1 43.0 17 50.0 1 51.0 3 52.0 1 55.0 4 56.0 1 63.0 2 65.0 9 71.0 2 77.0 5 78.0 6 79.0 4 82.0 1 83.0 2 89.0 1 91.0 100 92.0 95 93.0 8 103.0 2 104.0 3 105.0 11 106.0 2 115.0 1 117.0 1 119.0 3 133.0 5 162.0 33 163.0 5 *

* Only about 7% is due to 13C isotopes. 162.0 21 163.0 3 27.0 1 29.0 2 39.0 2 41.0 6 51.0 2 53.0 1 57.0 1 63.0 1 64.0 2 65.0 4 66.0 1 77.0 4 78.0 2 79.0 6 89.0 1 91.0 14 92.0 1 103.0 2 104.0 1 105.0 4 107.0 2 115.0 5 116.0 2 117.0 4 119.0 21 120.0 2 128.0 2 129.0 1 131.0 3 133.0 1 147.0 100 148.0 12 C12H18 M+ = 162

mass % mass % mass % mass % mass %

1-t-butyl-3-ethylbenzene

Notice that "91" is not logical, but it shows up. 27.0 2 39.0 3 41.0 2 51.0 2 53.0 1 63.0 1 65.0 2 77.0 6 C10H14 M+ = 135

mass % mass % mass % mass % mass %

p-propyltoluene 78.0 2 79.0 5 91.0 6 92.0 2 103.0 3 104.0 2 105.0 100 106.0 9 115.0 1 117.0 1 119.0 1 134.0 23 135.0 2.6

Halogenated Compounds - Key Points

1.

Fluorine (mass = 19) and iodine (mass = 127) have only one naturally occurring isotope, loss of

either of these masses is informative (M-19, M-127). Fluorine compounds tend to show weak

M+ peaks (or none at all). When iodine is lost, there can be a big hole (= 127) in the middle of

the mass spectrum.

2.

Chlorine has two isotopes (35 and 37) which occur in a 3:1 ratio; this is easily observed when

there is a molecular ion and in any fragments that retain the chlorine. An M-35 peak is

informative, and M-36 corresponds to loss of HCl.

3.

Bromine has two isotopes (79 and 81) which occur in a 1:1 ratio; this is easily observed when

there is a molecular ion and in any fragments that retain the bromine. An M-79 peak is

informative, and M-80 corresponds to loss of HBr.

4.

Loss of “X” is common (see above) and loss of HX can occur with fluorine (M-20), chlorine

(M-36), bromine (M-80).

5.

Loss of an alkyl radical and formation of a five atom ring or three atom ring is possible with

chains of C5 and longer with bridging chlorine, bromine or iodine (also true for sulfur).

X _X R _        fragmentation R

Free radicals are sucked away by the vacuum pump.

X = mass Cl 91 Br 135 I 183 X R fragmentation X R

Cations reach the detector, will see this mass.

Free radicals are sucked away by the vacuum pump. Cations reach the detector,

will see this mass.

X = mass Cl 63 Br 107 I 155

Examples

15.0 1 26.0 2 27.0 27 28.0 5 29.0 32 39.0 17 40.0 3 41.0 59 42.0 45 43.0 72 44.0 2 49.0 3 53.0 4 54.0 4 Cl alkyl branches 15 (1%) 29 (32%) 43 (72%) 57 (15%) 71 (3%) 85 (0.7%) Cl 91 (100%) Cl 63 (5%) 84 (4%), minus HCl other alkene fragments McLafferty allylic 27 (27%) 41 (59%) 63 (81%) 28 (5%) 42 (45%) 56 (56%) 70 (3%) 84 (1%) C₆H₁₃Cl = 120 1-chlorhexane mass % 55.0 81 56.0 56 57.0 15 63.0 5 65.0 2 67.0 3 69.0 22 70.0 2 71.0 3 84.0 4 91.0 100 92.0 4 93.0 32 94.0 1 mass % 91.0 100 93.0 32 35_{Cl and} 37_Cl 15.0 1 26.0 1 27.0 16 28.0 3 29.0 21 39.0 11 40.0 2 41.0 42 42.0 10 43.0 66 44.0 2 53.0 2 54.0 1 55.0 6 56.0 5 57.0 100 58.0 4.9 69.0 .5 70.0 3 71.0 3 81.0 1 83.0 1.5 84.0 1 85.0 18 86.0 1 99.0 14 100.0 1 107.0 1 109.0 1 135.0 8 137.0 8 mass % mass % Br alkyl branches 15 (1%) 29 (21%) 43 (66%) 57 (100%) 71 (3%) 85 (18%) Br 135 (8%) Br 107 (1%) 84 (4%), minus HBr other alkene fragments McLafferty allylic 27 (16%) 41 (42%) 63 (0%) 28 (3%) 42 (10%) 56 (5%) 70 (3%) 84 (1%) C₆H₁₃Br = 165 1-bromorhexane 91.0 100 93.0 32 35Cl and 37_Cl 91.0 100 93.0 32 35_{Cl and} 37_Cl mass % mass % I alkyl branches 15 (1%) 29 (15%) 43 (100%) 57 (11%) 71 (0%) 85 (50%) I 183 (0%) I 107 (2%) 84 (4%), minus HI other alkene fragments McLafferty allylic 27 (14%) 41 (25%) 63 (0%) 28 (3%) 42 (3%) 56 (2%) 70 (0%) 84 (0%) C6H13I = 212 1-iodohexane 27.0 14 28.0 3 29.0 15 39.0 7 40.0 1 41.0 25 42.0 3 43.0 100 44.0 3 53.0 1 55.0 6 56.0 2 57.0 11 85.0 50 86.0 3 155.0 2 212.0 4 mass % mass % I 15 (2%) (1%) 29 (0%) (0%) 43 (100%) (100%) 57 (0%) (0%) 71 (0%) (0%) 85 (0%) (0%) I not possible I 155 (0%) 42, minus HI other alkene fragments McLafferty allylic 27 (32%) (28%) 41 (37%) (36%) 63 (0%) (0%) 28 (3%) (2%) 42 (3%) (4%) 56 (0%) (0%) 70 (0%) (0%) 84 (0%) (0%) C3H7I = 170 1-iodopropane 15.0 2 26.0 2 27.0 32 28.0 2 38.0 2 39.0 11 40.0 2 41.0 37 42.0 3 43.0 100 44.0 3 127.0 5 128.0 1 170.0 24 15.0 1 26.0 1 27.0 28 28.0 2 38.0 2 39.0 12 40.0 2 41.0 36 42.0 4 43.0 100 44.0 3 127.0 6 128.0 2 170.0 24 I 2-iodopropane C₃H₇I = 170 1-iodopropane 2-iodopropane alkyl branches I = 127.0 (5%) (6%) HI = 128.0 (1%) (2%) Almost identical mass spectra.

Alcohols - Key Points

1.

Alcohols generally have weak M+ peaks. Tertiary alcohols often do not have an M+ peak.

However, if you had an IR, you would know an alcohol was present from the OH and CO bands.

Additional evidence would be present in the proton and carbon 13 NMR spectra, if available.

2.

Loss of water (M-18) is common; more so with straight chains and less so with branched

alcohols.

R' H O H fragmentation R' O H H M-18

This can lead to alkene fragmentations.

OH OH OH OH 15.0 3 26.0 3 27.0 33 28.0 12 29.0 16 31.0 83 39.0 11 40.0 4 41.0 66 42.0 32 43.0 59 45.0 7 53.0 1 55.0 14 56.0 100 57.0 6 59.0 0.3 74.0 0.6 15.0 2 26.0 2 27.0 10 28.0 52 29.0 6 31.0 17 39.0 3 40.0 1 41.0 12 42.0 1 43.0 9 45.0 100 53.0 1 55.0 2 56.0 2 57.0 2 59.0 20 74.0 0.2 15.0 3 26.0 1 27.0 4 28.0 1 29.0 6 31.0 27 39.0 6 40.0 1 41.0 21 42.0 1 43.0 9 45.0 1 53.0 1 55.0 2 56.0 3 57.0 8 59.0 100 74.0 0 15.0 2 26.0 2 27.0 23 28.0 12 29.0 8 31.0 40 39.0 14 40.0 3 41.0 57 42.0 59 43.0 100 45.0 4 53.0 1 55.0 6 56.0 5 57.0 3 59.0 6 74.0 13 linear has branch has branch has branch

(M-18) = H2O

M+ peak (M-15) = CH3

(M-29) = C2H5

The base peak is bolded in each example.

H2C=OH

3.

“Alpha” cleavage is common because a resonance stabilized carbocation can form three possible

ways in tertiary alcohols where R

1

≠

R

2

≠

R

3

. (two ways with 2

o

alcohols). Often all are

observed, when present.

O H C R1 R2 R3 fragmentation _{H O C} R₁ R2 R3 O H C R2 R3

"X" lone pair electrons fill in loss of electrons at carbocation site. This is a common fragmentation pattern for any atom that has a lone pair of electrons (oxygen = alcohol, ether, ester; nitrogen = amine, amide; sulfur and halogens). Loss of R₁, R₂ or R₃ is possible.

4.

Cyclic alcohols tend to show stronger M

+

peaks than linear chains.

OH OH OH M+ = 100 (3%) M-18 = 82 (46%) M-28 = 72 (7%) base peak = 57 (100%) M+ = 86 (9%) M-18 = 68 (7%) M-28 = 58 (14%) base peak = 57 (100%) M+ = 72 (1%) M-18 = 54 (1%) M-28 = 44 (100%) base peak = 44 (100%) OH M+ = 114 (2%) M-18 = 96 (23%) M-28 = 86 (4%) base peak = 57 (100%) OH _OH OH _OH M+ = 74 (0.6%) M-18 = 56 (100%) base peak = 56 (100%) M+ = 88 (0%) M-18 = 70 (51%) base peak = 42 (100%) M+ = 102 (0%) M-18 = 84 (9%) base peak = 56 (100%) M+ = 116 (0%) M-18 = 98 (6%) base peak = 70 (100%)

5.

When oxygen is present in any molecule, it is likely that mass 31 will be present.

O H C H H O H C H H

Mass = 31 is almost always present when oxygen is present, especially in alcohols.

Example

not logical, but observed (-H2O)

See alkene fragmentations earlier. The pi bond can move around the carbon skeleton, which can also rearrange.

CH2 O H M+ = 116 C7H16O 98 31 71 C5H11 HC O H 45 CH O H

Many types of skeletal rearrangements are possible using a such high energy electron beam. The "31" fragment does not make sense at a 2o _{or 3}o _{ROH, but is often observed (in ethers too).}

OH CH3 101 M+ = 116 C₇H₁₆O OH a b CH3 15 a b H loss of water from either side

mass % 27.0 5 29.0 5 31.0 2 39.0 3 41.0 10 42.0 4 43.0 8 44.0 7 45.0 100 46.0 2 55.0 15 56.0 7 57.0 4 69.0 3 70.0 5 83.0 9 98.0 4 101.0 4 actual peaks 98 (4%), minus H₂O

other alkene fragments McLafferty allylic 27 (5%) 41 (10%) 55 (15%) 69 (3%) 28 (0%) 42 (4%) 56 (7%) 70 (5%) 84 (0%) alkyl branches 15 (1%) 29 (5%) 43 (8%) 57 (4%) 71 (1%) 85 (0%)

Ethers - Key Points

1.

Ethers tend to have stronger M+ peaks than alcohols, but still can lose ROH the way that alcohols

lose H

2

O.

R' H O R fragmentation R' O R H R = mass H 18 CH3 32 C₂H₅ 46 C3H7 60

from either side

2.

Alpha cleavage is common from either side and further loss of the carbonyl fragment is possible.

O R' C R₁ R2 R3 fragmentation _{R' O C} R1 R2 R3 O R' C R2 R3

"X" lone pair electrons fill in loss of electrons at carbocation site. This is a common fragmentation pattern for any atom that has a lone pair of electrons (oxygen = alcohol, ether, ester; nitrogen = amine, amide; sulfur and halogens). Loss of R1, R2 or R3 is

possible. radical cation

3.

Loss of an oxygen carbon branch is also possible (from either side).

O R' C R1 R2 R3 fragmentation

We only see the cations. The fragmentation could potentially occur from either side. radical cation O R' C R1 R2 R3 loss of alcohol from either side

15 (1%) CH3 O (-ROH) _OH 2 O O C6H14O 46 (0%) 56 (24%) 59 (100%) 87 (2%) H2C H2 C CH3 O c d c 43 (6%) d O C6H14O e f O O e f 29 (27%) 73 (8%) 45 (10%) 57 (31%) M+ = 102 (4%) H H HO 28 (4%) 74 (0%) 15.0 1 18.0 3 26.0 1 27.0 12 28.0 4 29.0 27 31.0 57 39.0 5 41.0 26 42.0 3 43.0 6 44.0 1