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Spectroscopy Beauchamp 1

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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 neon-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 He-

highenergy R H

radical cation

+ + 2 e-EI mass spec

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 2000

25

50

75

100

percent relative intensity

masscharge

me=

85 = base peak57

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-hexanoneC12H16O , MW = 176

9185

57

27 58 86 92

119

39

27 628 229 2439 741 2642 143 150 151 355 357 9958 560 163 365 1177 285 10086 689 290 291 3692 6

176 7177 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.

me

B2r2

2V=

r = mVe

1B

m = masse = charge (usually +1)B = size of magnetic fieldr = radius of curvatureV = 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 (12C = 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 1H 1.00783 100.0 2H (D) 2.01410 0.015

carbon 12.01115 12C 12.00000 100.0 13C 13.00336 1.11

nitrogen 14.0067 14N 14.00307 100.0 15N 15.00011 0.37

oxygen 15.9994 16O 15.9949 100.0 17O 16.9991 0.04 18O 17.9992 0.20

fluorine 18.9984 19F 18.9984 100.0

silicon 28.086 28Si 27.9769 100.0 29Si 28.9765 5.06 30Si 29.9738 3.36

phosphorous 30.974 31P 30.9738 100.0

sulfur 32.064 32S 31.9721 100.0 33S 32.9715 0.79 34S 33.9679 4.43

chlorine 35,453 35Cl 34.9689 100.0 37Cl 36.9659 31.98

bromine 79.909 79Br 78.9183 100.0 81Br 80.9163 97.3

iodine 126.904 127I 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 C14H12, C13H8O and C12H8N2. 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 C15H17NO5, C16H21NNO4, and C17H25NO3. 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 CnHn+r. All molecular hydrocarbons have even mass molecular weights.

CH

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 lef t over hydrogen atoms = r

M13

n +=M = molecular weightn = number of CH units = quotientr = lef t over hydrogens = remainder

Possible hydrocarbon molecular formula = CnHn+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 CH4 (= 16) from the formula and substitute in the oxygen atom. If two oxygen atoms were present, we would subtract 2x(CH4) = C2H8 and so forth. Nitrogen-14 would substitute for CH2 and n nitrogen atoms would substitute for (CH2)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 CN C C C N

CnH2n+2Ox CnH2n+3N1CnH2n+4N2

(N is odd) (N is even)C = even massH = even massO = even massMW = even mass

C = even massH = odd massO = even massMW = odd mass

C = even massH = even massO = even massMW = 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 C4H10O. 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 C4H10O 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 13C atoms) 10 ways with hydrogen (2H = D) and 1 way with oxygen (17O). 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 13C or two 2H or one 13C and one 2H. The main contribution to the M+2 peak is the 18O isotope. Taken together, these three peaks would predict the indicated distribution for M+, M+1 and M+2 for this collection of atoms (C4H10O).



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

as a fraction = 1.000as 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 =1.11

101.11 (4 ways) = 0.0439

one 2H =0.015

100.015 (10 ways) = 0.0015

one 16O =0.04

101.24 (1 ways) = 0.0004

13C12C + 13C

2D1H + 2D

17O16O + 17O+ 18O

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 =1.11

101.114 x 32 x 1

two 2H =0.015

100.015

one 18O =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 92 x 1 = (2.25x10-8)(45 ways)

= 1 x 10-6 = 0.000001 = too small to consider

one 13C and one 2H =1.11

101.11 (4 ways) x0.015

100.015 (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 0.41CH4N3O 74.0355 2.33 0.22CH6N4 74.0594 2.70 0.03C2H2O3 74.0004 2.31 0.62C2H4NO2 74.0242 2.69 0.42C2H6NO 74.0480 3.06 0.23C3H6O2 74.0368 3.42 0.44C3H10N2 74.0845 4.17 0.07C4H10O 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 0.61CH4N3O 75.0320 2.70 0.43CH6N4 75.0798 3.45 0.05C2H2O3 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

(CH3-CH=CH2) and diazomethane (CH2=N=N). Can these two formulas (C3H6 vs CH2N2) 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 C2H5+ have fragment masses of approximately 29, yet CHO+ has a M+1 peak of

1.13% and M+2 peak of 0.20%, whereas C2H5+ 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 C2H5+.

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 (35Cl) to M+2 peak (37Cl)

M+ peak relative size

probability of 35Cl = 100100 + 32

(1 way) = 0.758

(assigned a referenced value of 100%)

M+2 peak relative size

probability of 37Cl = 32100 + 32

(1 way) = 0.242

percent of M+ peak = 0.2420.758

(100%) = 32% M+ M+1 M+2

100%

32%

one Br – comparison of M+ peak (79Br) to M+2 peak (81Br)


probability of 79Br = 100100 + 97

(1 way) = 0.508



probability of 81Br = 97100 + 97

(1 way) = 0.492


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

100% 97%



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


probability of 32S = 100100 + 0.79 + 4.43 (1 way) = 0.950



probability of 33S =


(100%) = 0.8%

M+ M+1 M+2

100%

4.4%


probability of 34S =


(100%) = 4.4%

0.79100 + 0.79 + 4.43 (1 way) = 0.008

4.43100 + 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


probability of 79Br = 0.508 (from above) probability of 35Cl = 0.758 (from above)(probability of 79Br)(probability of 35Cl) = (0.508) (0.758)(1 way) = 0.385



M+ M+2 M+4

100%

31%

129%

probability of 81Br = 0.492 (from above) probability of 37Cl = 0.242 (from above)(probability of 79Br)(probability of 37Cl)(1 way) = (0.508) (0.242)(1) = 0.123(probability of 81Br)(probability of 35Cl)(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 81Br)(probability of 37Cl)(1 way) = (0.492) (0.242)(1) = 0.119


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


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



M+ M+2 M+4

100%

10%

61%probability of 37Cl = 0.242 (from above)(probability of 35Cl)(probability of 37Cl)(2 ways) = (0.758) (0.242)(2) = 0.367


M+4 peak relative size(probability of 37Cl)(probability of 37Cl)(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 Br2. Use the probabilities from above.

Problem 6 - Calculate the relative intensities (as a percent) of M+, M+2, M+4 and M+6 for BrCl2 and Br2Cl. 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

heterolyticbond energy

C H

R-H

R H

homolyticbond

energy

Hfo(H ) = -52 heat of formatio of hydrogen

atom, common to all cycles

H-HH3C-H

CH3CH2-H(CH3)2CH-H

(CH3)3C-HCH2=CHCH2-H

C6H5CH2-H

Compound RadicalH (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- Hf

o(H )]

= energy to make R

1041059895

9286

ionizationpotential of R

R He-

HHf

o(H electron affinity) = -17

Hfo(R ) = + value

(see table)

Energy to formcarbocation

88

313227

193169

154186

165

-17-17

-52-52

I.P. E.A.(H) Hfo(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 109o (sp3) bond angles to 120o bond angles (sp2). 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

RR

R

more crowded as sp3 center = higher potential energy starting point with 3-4 larger groups around

tetrahedral carbon

less crowded as sp2with 3 groups around trigonal planar carbon is slightly more stable than it

would be if groups were smaller

C RR

R

R

Breaking a bond is a large uphill energy transformation,but less so with a sterically

crowded starting point, so Eais 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

HC CH2

H HH

a

b

c

d

e

f

g

high energy electron beam

2-methylbutane (isopentane)

CH3

H2C

H

HC CH2

H HH

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.

025 50 75 100

0

25

50

75

100

percentrelativeintensity

masscharge

me

=

29

41,42

43 = base peak

57 Many smallerpeaks not shown.

72

M+peak

isopentaneC5H12

75 eV

CH

CH3

H3C CH2

CH3

5715

4329

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

H2C

HC

CH3

a b c d

high energy electron beam

2,2,4-trimethylpentane radical cation

Possible fragmentations?

Energy to rupture bonds (eight calculations).

CH3 C

CH3

H3C

CH3

H2C

HC

CH3

a b c dCH3

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

025 50 75 100

0

25

50

75

100

percent relative intensity

masscharge

mZ

=

basepeak

2971

43

57

Many smaller peaks not shown.

M+ peak

octaneC8H18

75 eV

114

H3C

H2C

CH2

H2C

CH2

H2C

CH2

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+ = 3o > 2o > 1o > 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

proposedfragmentation

R

C4H9

probablerearrangement

C4H9

less stableprimary carbocation

more stabletertiary carbocation

can't tell whichmass = 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, CH2 = 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 13C 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% 13C 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 13C isotopes. If there were 10 carbons, the next mass peak would be expected to have 0.10x80% 8% size just based on the 13C 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 “CH2CH2“ (= 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 CH2),. 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. 3o R+ > 2o R+ > 1o R+ > H3C+). Also, alkanes can lose H2 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 %15.0 126.0 127.0 2028.0 429.0 2739.0 1240.0 241.0 4442.0 1543.0 10044.0 351.0 153.0 254.0 155.0 1156.0 1857.0 3458.0 269.0 270.0 1271.0 2072.0 184.0 785.0 2686.0 299.0 none

114.0 6115.0 1

M+ = 114C8H18

8529

7143

57

C8H18 C6H13 C5H11 C4H9 C3H7 C2H5 C1H3

M+ = 114 85 71 57 29 15

991557

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

7139

41

M+ = 11485

57

4229

28 5556

70

Only cations reach detector, so only the part with positive charge is observed at the detector. A positive chargeis written on all f ragments 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 fragmentmasses 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 10-5 second trip to the detector.

not observedin octane

octane - all alkane fragmentsare observed, except 99.

MW = 114

actual peaksin octane

M+

base

alkanepeaks

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 / 57MW = 114

(-RH)

possible alkylfragments

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 peaksin 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 = 11485

alkanepeaks

the basepeak 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 RRemaining sigma bondholds skeleton together.+ e-

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).

These alkenes all look similar.


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 CH2 x1, x2, …).

resonance stabilized carbocation

R' ionization R'R'

fragmentation

free radicalis 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 H2.

CR

H2CH

CH2R CH2

C

CH

CH2

mass = 42 (R = H)56 (R = CH3)70 (R= CH2CH3)84 (R = C3H7)

C

C

fragmentation

McLafferty-like rearrangement

RR

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

R2fragmentation is a

retro-Diels-Alder reactionR1

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%)43 (100%)69 (2%)

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

ORCH3

83 (0%)

97 (0%)

29 (27%)

15 (0%)

actual peaksfrom 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 -

65.0 166.0 -

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 -

65.0 166.0 -

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 -

65.0 166.0 -

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 -

65.0 266.0 -

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-4-octene trans-4-octenecis-3-octeneoctane

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, 99allylic fragments = 27, 41, 55, 69, 83, 97McLafferty fragments = 28, 42, 56, 70, 84, 98



Another Alkene Example (C7 alkene)

H

H

H

McLafferty rearrangementsallylic fragmentations

A pi bond can migrate through the skeleton to 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

allylicfragments

McLaffertyfragments

27 (26%)41 (97%)55 (68%)69 (31%)83 (4%)

28 (5%)42 (55%)56 (100%)70 (44%)

all peaks > 1%

Spectroscopy Beauchamp 24 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 CH2 x1, x2, …).

mass = 39 (R = H)53 (R = CH3)67 (R= CH2CH3)

Ralso

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.

RCH

CC

H

radical cation

fragmentation

R'

CH

CC

H

CH

CC

H

resonanceR R

R'

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

HC

CH

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

CH H

R H

Either fragment can be observed and both show an even mass.

RC

CH

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 rearrangementsallylic 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

C7H12M+ = 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

C7H12M+ = 96


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

C7H12M+ = 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 (RCH2-) 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

fragmentationCH2

R

lots ofresonance

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' = massH 91CH3 105C2H5 119HO 107H2N 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 HH

H

H

R

C

C

C

H HH

H

H

RH

Hor

can be on either fragment

R = massH 92CH3 106C2H5 120HO 108H2N 107

R = massH 28CH3 42C2H5 56C3H7 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.

R R



O

C

C

H

H

H

H

H

O

C

C

H

H

H

HH

Hor

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%)

C12H18M+ = 162


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

C12H18M+ = 162


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

C10H14M+ = 135


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 XR

fragmentation R

Free radicals are sucked away by the vacuum pump.

X = massCl 91Br 135I 183

X

R

fragmentationX R

Cations reach the detector, will see this mass.



X = massCl 63Br 107I 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

Clalkyl 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 fragmentsMcLafferty 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 10093.0 32 35Cl and 37Cl

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


27 (16%)41 (42%)63 (0%)

28 (3%)42 (10%)56 (5%)70 (3%)84 (1%)

C6H13Br = 165

1-bromorhexane

91.0 10093.0 32 35Cl and 37Cl

91.0 10093.0 32 35Cl and 37Cl

mass % mass %

Ialkyl branches 15 (1%) 29 (15%) 43 (100%) 57 (11%) 71 (0%) 85 (50%)

I

183 (0%)

I

107 (2%)

84 (4%), minus HI


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


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

I2-iodopropaneC3H7I = 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

OH

fragmentationR' O

HH

M-18

This can lead to alkene fragmentations.

OH OHOH

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 R1 ≠ R2 ≠ R3. (two ways with 2o alcohols). Often all are observed, when present.

OH C

R1

R2

R3

fragmentationOH C

R1

R2

R3

OH 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



4. Cyclic alcohols tend to show stronger M+ peaks than linear chains.

OHOH

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.

OH C H

H

OH 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

OH

M+ = 116C7H16O

98

31

71C5H11

HC

OH

45CH

OH

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 3o ROH, but is often observed (in ethers too).

OH

CH3 101

M+ = 116C7H16O

OHa 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


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 H2O.

R'H

OR

fragmentationR' O

RH

R = massH 18CH3 32C2H5 46C3H7 60

from either side

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

OR' C

R1

R2

R3

fragmentationOR' C

R1

R2

R3

OR' C R2

R3


radical cation

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

OR' C

R1

R2

R3

fragmentation

We only see the cations. The fragmentation could potentially occur from either side.radical cation

OR'C

R1

R2

R3

loss of alcoholfrom either side

15 (1%)

CH3

O

(-ROH) OH2

OO

C6H14O

46 (0%) 56 (24%)

59 (100%)87 (2%)

H2C

H2C

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%)HH

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 45.0 10 47.0 1 55.0 6 56.0 24 57.0 31 58.0 1 59.0 100 60.0 3 73.0 8 87.0 2 101.0 1 102.0 4

O

CH2

H

31 (57%)

not logical, but observed

ab

b

amass %

56 (24%), minus ROH28 (4%), minus ROHother alkene fragments

McLafferty allylic27 (12%)41 (26%)55 (6%)69 (0%)

28 (4%)42 (3%)56 (24%)70 (0%)84 (0%)

M+ = 102 (4%)

M+ = 102 (4%)



Thiols and Thioethers - Key Points

1. The M+2 peak with a single sulfur adds an extra 4.4% to this peak relative to the M+ peak (in addition to other M+2 contributions). Other than chlorine and bromine, this is the most significant M+2 contributor to common organic molecules.

2. Loss of H2S (M-34) is possible for thiols and RSH for sulfides (loss of CH3SH = (M-48)).

R'H

SR

fragmentationR' S

HR

R = massH 34CH3 48C2H5 62C3H7 76

M - (RSH mass)

This can lead to alkene fragmentations.

3. “Alpha” cleavage is possible because a resonance stabilized carbocation can form three possible ways. Often all are observed, when present.

SR C

R1

R2

R3

fragmentationSR C

R1

R2

R3

SR C R2

R3


radical cation

4. If a side chain has five or more atoms then cleavage is possible with ring formation (see the

halogens). Beta (β) cleavage is also reasonable.

R = massH 89CH3 103C2H5 117

S SR

fragmentation R


S

R

fragmentationS R




R R

RR

R = massH 61CH3 75C2H5 89


Example mass % mass %

SHalkyl branches 15 (1%) 29 (15%) 43 (48%) 57 (7%) 71 (0%) 85 (2%)

S

89 (3%)

S

61 (10%)

84 (16%), minus H2S


27 (16%)41 (35%)55 (35%)69 (25%)83 (1%)

28 (4%)42 (32%)56 (100%)70 (2%)84 (16%)

C6H14S = 118 120 (5.3%)

1-hexanethiol 26.0 1 27.0 16 28.0 4 29.0 15 35.0 2 39.0 9 40.0 2 41.0 35 42.0 32 43.0 48 44.0 2 45.0 4 46.0 2 47.0 15 48.0 1 53.0 2 54.0 3

55.0 35 56.0 100 57.0 7 59.0 2 60.0 2 61.0 10 62.0 1 69.0 25 70.0 2 83.0 1 84.0 16 85.0 2 89.0 3

118.0 30 119.0 2 120.0 1.6

HH

mass % mass %

Salkyl branches 15 (1%) 29 (50%) 43 (4%) 57 (0%) 71 (0%) 85 (0%)

S

89 (25%)

S

61 (38%)

56 (68%), minus H2S


27 (36%)41 (49%)55 (17%)69 (0%)83 (0%)

28 (9%)42 (4%)56 (68%)70 (0%)84 (0%)

C6H14S = 118 120 (4.8%)

butyl ethyl sulfide

HH 15.0 1 26.0 3 27.0 36 28.0 9 29.0 50 34.0 1 35.0 9 39.0 11 40.0 2 41.0 49 42.0 4 43.0 4 45.0 12 46.0 12 47.0 48 48.0 6 53.0 2 54.0 1

= M+

= M+2

55.0 17 56.0 68 57.0 17 58.0 3 59.0 6 60.0 6 61.0 38 62.0 47 63.0 20 75.0 100 76.0 8 77.0 5 89.0 25 90.0 3

103.0 2 118.0 56.4 119.0 4 120.0 2.7

H2S = 34 (1%)S

75 (100%)

CH3

S S(M-29) = 89 (25%)(M-57) = 89 (38%)

Phenols - Key Points 1. Phenols tend to have intense M+ peaks. (See below = 100% and 36%.)

2. Loss of CO with extensive rearrangement is common.

ROH

radical cation

fragmentationO

R

loss of carbon monoxide...?

R

C OR = massH 65CH3 79C2H5 93

R1

3. A hydroxy tropylium ion with no other substituents has a m/e = 107.

ROH

radical cation

fragmentationR = massCH3 107C2H5 121etc.

R1

O

HR'

Lots of resonance.



Examples mass % mass %

alkyl branches 15 (0%) 29 (0.8%) 43 (0.4%) 57 (0%) 71 (0%) 85 (0%)

65 (17%) 39 (14%)

C6H60 = 94 (100%) M+1 = (7%)

phenol

27.0 2 37.0 2 38.0 4 39.0 14 40.0 9 47.0 4 50.0 3 51.0 3 53.0 2 55.0 7 61.0 1

62.0 2 63.0 4 64.0 1 65.0 17 66.0 23

67.0 2 74.0 1 93.0 2 94.0 100 95.0 7

OH

mass % mass %

alkyl branches 15 (0.4%) 29 (0.4%) 43 (0.4%) 57 (0%) 71 (0%) 85 (0%)

39 (6%)

allylic R

27 (3%)41 (1%)55 (3%)69 (0%)83 (0%)

C6H60 = 122 (36%) M+1 = (3%)

p-ethylphenol

OH

27.0 3 38.0 1 39.0 6 41.0 1 50.0 3 51.0 5 52.0 3 53.0 2 55.0 3 62.0 1 63.0 2

65.0 3 77.0 13 78.0 3 79.0 2 91.0 4 94.0 1

103.0 2 107.0 100 108.0 8 121.0 3 122.0 36 123.0 3

R

R=H 65 (3%)R=CH3 79 (2%)R=C2H5 93 (1%)

OH

107 (100%)

OH

121 (3%)

Amines - Key Points

1. Amines often have weak or absent M+ peaks. An odd number of nitrogen atoms produces an odd

molecular ion peak.

H3C

H2C

NH

H

H3C

H2C

OH

Molecules made with C, H, S, O, halogens and an even number of nitrogen

atoms have even molecular masses.

Molecules made with an odd number of nitrogen atoms have odd molecular masses

because they have an odd number of hydrogens.

CnH2n+2OmCnH2n+2+NOm

2. Alpha cleavage is usually a major fragmentation pattern in a manner similar to alcohols and ethers.

NR' C

R1

R2

R3

fragmentation NR' C

R1

R2

R3

NR' C R2

R3

radical cation

R" R" R"resonance

The fragment mass depends on what is present in the "R" groups. If all R groups are "H" (H2N=CH2 ) then the mass will be 30, which shows up in almost every amine compound examined, even tertiary amines.

all R = H 30one CH3 44 C2H5 58etc.

mass


3. Loss of a branch at nitrogen is also possible in a manner similar to alcohols and ethers.

NR' C

R1

R2

R3

fragmentation NR' C R2

R3

radical cation

R" R"

The fragment mass depends on what is present in the "R" groups and which fragment retains the cation charge.

R1

4. Aromatic amines generally show intense M+ peaks.

NH2

radical cationodd mass

1. fragmentation2. rearrangement

R' = massH 106CH3 120C2H5 134etc.

R1

N

HR'

Lots of resonance.

R

HR' even mass

odd mass

Examples

n-isobutyl-sec-butylamineloss of amine

from either side

29 (18%)

CH3

NH

(-ROH)

NH

C8H19N

56

100 (18%)

N

ce

c

e

M+ = 129 (1%)

HH

NH2

CH2

30 (100%)

not logical, but observedand is even the base peak

ab

ba

mass %56 (6%), minus RNH2


27 (7%)41 (18%)55 (7%)69 (0%)

28 (8%)42 (4%)56 (6%)70 (2%)84 (2%)

H56

15.0 1 18.0 2 27.0 7 28.0 8 29.0 18 30.0 100 31.0 1 39.0 5 41.0 18 42.0 4 43.0 2 44.0 53 45.0 1 55.0 7 56.0 6 57.0 24 58.0 20 70.0 2 72.0 6 84.0 2 86.0 66 87.0 4

100.0 67 101.0 5 114.0 8 128.0 1 129.0 1

C2H5d

NH

114 (8%)

d

M+ = 129 (1%)

NH

C3H7

15 (1%)

86 (66%) 43 (2%)

29 (18%)43 (2%)57 (24%)71 (0%)85 (0%)

alkyl branches

n-isobutyl-sec-butylamine



15 (1%)

CH3NH

(-ROH)NH2 See alkene

fragmentations above.

NH

NH

C6H15N

45 (0%) 56 (3%)

58 (100%)

86 (2%)

H2C

H2C

CH3

c d c

43 (2%)

d

NH

e f

HN

NH

e

f

29 (8%) 72 (0%)

44 (10%) 57 (3%)

NH

M+ = 101 (9%) H

C6H15NCH2

NH2

30 (33%)

mass % 15.0 1 18.0 1 27.0 5 28.0 5 29.0 8 30.0 33 39.0 2 41.0 4 42.0 3 43.0 2 44.0 10 56.0 3 57.0 3 58.0 100 59.0 4 86.0 2 100.0 2 101.0 9

HH2N

28 (5%) 73 (0%)

NH4

18 (1%)

abb

a

butylethylamine

butylethylamine

butylethylamine

M+ = 101 (9%)

M+ = 101 (9%)

Carbonyl Compounds (aldehydes, ketones, esters, acids, amides, acid chlorides) - Key Points

1. M+ peaks are often observable (though they can be weak or absent). Several examples are provided below.

2. Alpha cleavage is possible from either side. Usually the more stable cation forms in greater amount. It is best to look for both possibilities.

radical cation

CR1

O

R2

R1 or R2 can be lost from aldehydes, ketones, acids, esters, amides, acid chlorides,etc.

CR1 O

C R2O

CR1 O

C R2O

An oxygen lone pair paritally fills in the loss of electrons at the carbocation site via resonance. This is a common fragmentation pattern for any carbonyl compound and can occur from either side, though some are more common than others.


3. Alpha cleavage can be followed by loss of CO (another -28). That would leave the side branches as observable peaks, plus any further fragment branches from those peaks.

CR1 O

C R2O

CR1 O

C R2O

loss of

C O

R1

R2

Subsequent loss of CO is possible after fragmentation, so not only can you see loss of an branch you can also see the mass of an branch.

4. McLafferty rearrangements are common with at least three carbons in a side chain. Cleavage occurs

between Cα and Cβ.

radical cation

CR1

O

C

C

C

H R

R

R

R

RR

CR1

O

C C

C

R

R

R

RRR

H Positive charge can be on either fragment, which typically has an even mass.

This is another common fragmentation pattern for carbonyl compounds (and other pi systems as well: alkenes, alkynes, aromatics, nitriles, etc.). If the pi bond has at least 3 additional nonhydrogen atoms attached and a hydrogen on the "gamma" atom, the branch can curve around to a comfortable 6 atom arrangement and the pi bond can pick up a hydrgen atom and cut off a fragment between the C and C positions. The positive charge can be seen on either fragment and usually the fragments have an even mass (unless there is an odd number of nitrogen atoms). The mass of either fragment depends on what "R"s are.

= alpha position = beta position = gamma position

The bottom line is there are several ways that carbonyl (C=O) functionality can fall apart. It is best to look for all possibilities. See the last example in this example list below (ketone).

Carbonyl Examples

Carboxylic Acids

OH

O

56 (8%)

28 (4%)42 (7%)70 (3%)

HO

H

60 (100%)

McLafferty

Loss of side chain, then CO (?)

17 (0.4%)99 (0.8%)

O

HO45 (100%)

a b

b

a

OOH

C

O

71 (2%)

15 (0.9%)29 (14%)43 (14%)57 (12%)71 (2%)85 (0.4%)99 (0.8%)

HO

OH

HO

71 (2%)

C6H12O2 = 116 (0%)

C6H12O2 = 116 (0%)

28 (5%)*

hexanoic acid

*28 could also be ethene

18.0 2 26.0 2 27.0 17 28.0 4 29.0 14 30.0 1 31.0 2 39.0 10 40.0 2 41.0 26 42.0 7 43.0 14 45.0 9 53.0 1 55.0 10 56.0 8 57.0 12 58.0 6 59.0 3 60.0 100 61.0 9 69.0 3 70.0 3 71.0 2 73.0 44 74.0 7 83.0 1 87.0 11

mass %

McLaffertyallylic

41 (26%)55 (10%)69 (3%)83 (1%)

alkyl branches

alkenes

OHO

(M-29) = 87 (11%)



Esters

28 (2%)*

OH

O

CH3O

H

74 (100%)

McLafferty


31 (2%)99 (19%)

O

CH3O59 (15%)

a b

b

aOH3CO

C

O

71 (10%)

15 (10%)29 (12%)43 (31%)57 (4%)71 (10%)85 (0.1%)99 (19%)

H3CO

71 (10%)

C7H14O2 = 130 (0.4%)

C7H14O2 = 130 (0.4%)

O

CH3O

methyl hexanoate


56 (2%)

28 (2%)42 (6%)70 (3%)

mass %

15.0 10 18.0 1 26.0 1 27.0 11 28.0 2 29.0 12 31.0 2 39.0 7 40.0 1 41.0 17 42.0 6 43.0 31 44.0 2 45.0 2 53.0 1 55.0 9 56.0 2 57.0 4 59.0 15 69.0 2 70.0 3 71.0 10 73.0 1 74.0 100 75.0 5 87.0 32 88.0 4 99.0 19

100.0 1 101.0 8

allylic

41 (17%)55 (9%)69 (2%)83 (0%)

McLafferty

alkyl branches

alkenes

OCH3O

(M-29) = 101 (8%)

Aldehyde

15.0 2 18.0 1 26.0 3 27.0 34 28.0 8 29.0 33 30.0 2 31.0 2 38.0 2 39.0 20 40.0 4 41.0 69 42.0 11 43.0 55 44.0 100

45.0 20 50.0 1 51.0 1 53.0 3 54.0 2 55.0 15 56.0 82 57.0 38 58.0 9 60.0 4 67.0 8 69.0 1 71.0 7 72.0 17 73.0 2 81.0 1 82.0 13 83.0 1

mass %

OH

O

H

H

C6H12O = 100 (0.4%) 44 (100%)

McLafferty


OH

H

1 (?)99 (0.4%)

O

H29 (33%)

a b

b

a OH

C

O

71 (7%)

C6H12O = 100 (0.4%)

15 (2%)29 (33%)43 (55%)57 (38%)71 (7%)85 (0.3%)99 (0.4%)

71 (7%)

hexanal

28 (8%)*


56 (82%)

28 (8%)42 (11%)70 (0%)

hexanal

alkyl branches

McLaffertyallylic

41 (69%)55 (15%)69 (1%)83 (1%)

alkenes

OH

(M-29) = 71 (7%)


Ketone

OH

OH

C7H14O = 114 (10%) 58 (91%)

McLafferty


15 (4%)99 (4%)

O

43 (100%)

a b

b

aO

CH3

C

O

71 (14%)

15 (4%)29 (9%)43 (100%)57 (2%)71 (14%)85 (3%)99 (4%)

OH

C7H14O = 114 (10%)

71 (14%)

2-heptanone


56 (2%)

28 (2%)42 (3%)70 (0%)

28 (2%)*

2-heptanone

mass %

alkyl branches

McLaffertyallylic

41 (12%)55 (5%)69 (0%)83 (0%)

15.0 4 18.0 2 27.0 9 28.0 2 29.0 9 39.0 7 40.0 1 41.0 12 42.0 3 43.0 100 44.0 2 45.0 1 53.0 1 55.0 5 56.0 2 57.0 2 58.0 91 59.0 15 71.0 14 72.0 4 85.0 3 99.0 4

113.0 2 114.0 10 115.0 1

alkenes

O

(M-29) = 85 (3%)

Amide

McLafferty


18 (2%)=NH4

99 (1%)98 (0.2%)

O

H2N44 (29%)

a b

b

a

ONH4

C

O

71 (2%)

15 (0%)29 (8%)43 (26%)57 (2%)71 (2%)85 (0%)99 (1%)

71 (2%)

O

H2N

OH

O

H2N

H

C6H13NO = 115 (0.6%)

H2N

59 (100%)

C6H13NO = 115 (0.6%)

hexanamide

56 (0%)

28 (2%)42 (4%)70 (0%)

28 (2%)*

*28 could also be ethenehexanamide

18.0 2 27.0 9 28.0 2 29.0 8 39.0 6 41.0 12 42.0 4 43.0 26 44.0 28 45.0 1 55.0 4 57.0 2 59.0 100 60.0 3 71.0 2 72.0 19 73.0 4 86.0 9 99.0 1

mass %alkyl branches

McLaffertyallylic

41 (12%)55 (4%)69 (0%)83 (0%)

18 could be NH4 or H2O

alkenes

OH2N

(M-29) = 86 (9%)



Full example for 2-methyl-4-heptanone

20 30 40 50 60 70 80 90 100 110 120

27

Mostly peaks greater than 5%of the base peak are shown.

57 = base

71

39

41M+ = 128

O

58

2-methylheptan-4-one

130

113100

86

43

42

29

28

57.0 100 58.0 27 59.0 1 69.0 2 70.0 1 71.0 70 72.0 3 85.0 72 86.0 11 113.0 6 128.0 23 129.0 2

27.0 17 28.0 2 29.0 17 39.0 11 40.0 2 41.0 36 42.0 5 43.0 73 44.0 3 53.0 1 55.0 2 56.0 1

C8H16OMW = 128

OHH

2 McLafferty possibilities

ba

C8H16O = 128 (23%)

O

28 (2%)

H

100 (0.4%)

OH

86 (11%)42 (5%)

a = only see the fragment that retains the positive charge

b = only see the fragment that retains the positive charge

a and b

a b

OLose left

branch

CO

c

c = only see the fragment that retains the positive charge

Lose CO

CO

57 (100%) 71 (70%)43 (73%)

C8H16O = 128 (23%)

c

28 (2%)

OLose right

branchC

Od

d = only see the fragment that retains the positive charge

Lose CO

CO

85 (72%)

reasonablemass peaks 128 100 86 85 71 57 43 42 28

C8H16O = 128 (23%)

d

43 (73%)57 (100%) 28 (2%)


Nitriles - Key Points

1. Usually have weak M+ peaks. An odd number of nitrogen atoms produces an odd molecular ion peak.

H3C

H2C

C

Alkynes made with C and H have even molecular masses.

Nitriles made with an odd number of nitrogen atoms have odd molecular masses

because they have an odd number of hydrogens.

CnH2n-2CnH2n-2+N

CH

H3C

H2C

CN

Compare.

C3H5NMW = 55

C4H6MW = 54

2. With side chains of three carbons or longer McLafferty rearrangements are possible.

CHC

CH

N

CCH

HNH


C

C

RR

oddmass

oddmass

radical cation

fragmentation

R R

R R

mass = 28 (all H)42 (1R = CH3)56 (1R= CH2CH3)70 (1R = C3H7)

evenmass

Nitrile McLafferty can cut off a fragment between the C and C positions. Either fragment can be observed (if the cation) and the one with the nitrogen atom will show an odd mass.

...or...

3. Alpha cleavage is possible.

CC

NR

radical cation

oddmass

fragmentation CC

N

R

The detector sees cations. Radicals are pumped away.

oddmass

evenmass



Example

McLafferty

Loss of side chain, then CN (?)

26 (5%)a

a

71 (1%)

15 (2%)29 (43%)43 (32%)57 (28%)71 (1%)85 (0%)

NH

NH

C6H11N = 97 (0.8%) 41 (100%)

NH

C6H11N = 97 (0.8%)

C

N

NH

Hperhaps...?

b 54 (82%)55 (42%)56 (4%)

43 (32%)42 (14%)41 (100%)

NHH

hexanenitrile

56 (4%)

28 (9%)42 (14%)70 (4%)

b

hexanenitrile

15.0 2 26.0 4 27.0 33 28.0 9 29.0 43 30.0 1.6 37.0 1 38.0 3 39.0 22 40.0 5 41.0 100 42.0 14 43.0 28 50.0 1 51.0 2 52.0 3 53.0 4 54.0 82 55.0 42 56.0 4 57.0 32 58.0 1 66.0 1 67.0 1 68.0 30 69.0 23 70.0 4 71.0 1 82.0 24 83.0 1 96.0 12 97.0 1

mass %

alkenes

allylic

41 (100%)55 (42%)69 (23%)83 (1%)

McLafferty

alkyl branches

27 (33%)

H2C CH

HCNH

or

similar masses

These are some of the more common organic functional group fragmentation patterns in EI mass spectroscopy. Most of the examples presented here are very simple monofunctional compounds. When more functional groups are present, more complexity is expected and it gets increasingly difficult to make definitive conclusions on the basis of mass spectroscopy. Even with simple monofunctional group compounds, we have seen that functional groups can change through rearrangements possible due to the high energy of ionization (e.g. alkanes alkenes). If you specialize in other specific patterns of functionality in your work, you will become familiar with useful mass spectral features of those groups. For us, the molecular weight is the primary information we seek from a mass spectrum, assisting us toward our main goal of determining organic structures from the available spectra.

A one page summary sheet showing many of the fragmentation patterns above is provided on the next page, and the following page shows common fragments and their extended variations. These two pages will explain most of what you will encounter as a burgeoning mass spectroscopist.


Common fragmentation patterns in mass spectroscopy (only cations are observed) 1. Fragment a branch next to a pi bond (α cleavage)

C C C

R

radical cation

pi bond of an alkene,alkyne or aromatic

C C C C C C

Pi electrons partially fill in loss of electrons at carbocation site via resonance. This is a common fragmentation pattern for alkenes, alkynes and aromatics.

Characteristic carbocation stability also applies.

3o R > 2o R > 1o R > CH3

We only see the cationic fragments. The radical fragments are lost to the vacuum.

R

2. Fragment a branch next to an atom with a lone pair of electrons

X C

R

radical cationadjacent lone pair ofan

oxygen or nitrogen atom

X C X C

X lone pair partially fills in loss of electrons at carbocation site via resonance. This is a common fragmentation pattern for any atom that has a lone pair of electrons (oxygen = alcohol, ether, ester; nitrogen = amine, amide; sulfur = thiol, sulfide, etc.) Alcohols often lose water (M-18), ethers can lose ROH, primary amines can lose ammonia (M-17), etc.

R

3. Fragment a branch next to a carbonyl (C=O) bond…and possible subsequent loss of carbon monoxide, CO

CR1 R2

O

CR1 O

C R2O

CR1 O

C R2O

CO

loss of

CO

loss of

R1

R2

R1 or R2 can be lost from aldehydes, ketones, acids, esters, amides...etc.

An oxygen lone pair partially fills in the loss of electrons at the carbocation site via resonance. This is a common fragmentation pattern for any carbonyl compound and can occur from either side, though some are more common than others.

Subsequent loss of CO is possible after fragmentation, so not only can you see loss of an a branch, you can also see the mass of an branch.

ab

b

a

4. McLafferty Rearrangement

This is another common fragmentation pattern for carbonyl compounds (and other pi systems as well: alkenes, aromatics, alkynes, nitriles, etc). If the pi bond has at least 3 additional nonhydrogen atoms attached and a hydrogen on the "gama" atom, the branch can curve around to a comfortable 6 atom arrangement and the pi bond can pick up a hydrogen atom and cut off a fragment between the C and C positions. The positive charge can be seen on either fragment and usually the fragments have an even mass (unless there is an odd number of nitrogen atoms in the observed fragment).

CR1 C

O

C

C

H

CR1 C

O

C

C

H The positive charge can be on either fragment, which typically have even masses (unless an odd number of N is present).

radical cation

= alpha position = beta position = gama positioncation fragment OR cation fragment

Knowing these few fragmentation patterns will allow you to make many useful predictions and interpretations in mass spectroscopy. Also loss of small molecules is common, producing an even mass if no nitrogen is present: H2O = 18, H2S = 34, CH3OH = 32, C2H5OH = 46, NH3 = 17, CH3CO2H = 62, HF = 20, HCl = 36/38, HBr = 80/82, etc. This can even include loss of an alkane equivalent (R branch plus H, 16, 30, 44, etc.) to leave behind an alkene cation that can also generate alkene fragments, which is shown later in the notes (McLafferty & allylic). Certain atoms generate characteristic M+2 peak patterns: 35Cl/37Cl = 75/25 ration, 79Br/81Br = 50/50 ratio, 32S/34S = 95/5 ratio. Any peak 1 amu larger than the one in front of it shows about 1% of the front peak for every carbon atom in the formula (e.g. C6 = M+ / M+1 ratio of 100% / 6%).



mass = 39 (R = H)53 (R = CH3)67 (R= CH2CH3)

RCH

R

mass = 41 (R = H)55 (R = CH3)69 (R= CH2CH3)83 (R=C3H7)

mass = 65 (R = H)79 (R = CH3)93 (R= CH2CH3)

R R

mass = 91 (R = H)105 (R = CH3)119 (R= CH2CH3)

CH3 = 15CH3CH2 = 29C3H7 = 43C4H9 = 57C5H11 = 71C6H13 = 85

CH2

HH

H

R

mass = 27

mass = 42 (R = H)56 (R = CH3)70 (R= CH2CH3)84 (R=C3H7)

CO R

mass = 29 (R = H)43 (R = CH3)57 (R= CH2CH3)71 (R = C3H7)85 (R = C4H9)99 (R = C5H11)105 (R = C6H5)45 (R= OH)59 (R= OCH3)44 (R= NH2)

C OH2N

C ORO

mass = 44

Loss of small molecules via elimination reactions.

H2O CH3OH C2H5OH NH3 CH3CO2HH2S HF HCl HBr

mass = 18 34 32 46 17 62 2036 (75%)38 (25%)

80 (50%)82 (50%)

A sampling of common and/or miscellaneous peaks that are often seen, (even when they don't make sense). Whatever the initial mass is, a series of masses increased by increments of 14 (CH2)n reveals additional "logical" fragment masses. Remember,we only see the cationic fragments.

CR

OH

CH2

McLafferty

mass =

44 (R = H)58 (R = CH3)72 (R = CH2CH3)86 (R = C3H7)

R CH2

R1

R2

HO R2

R1

variable mass,

(can sometimes see this fragment if it retains the cation charge)

Notice!even masses(without N)

McLafferty Rearrangement Possibilities

CR

H2CH

CH2

R CH2

R1

R2

HCH2

CH2

R1

R2

HC

H

CH2

R1

R2

HN

45 (R = H)59 (R = CH3)73 (R = CH2CH3)87 (R = C3H7)

mass =

CH2

R1

R2

H

H

H

CCH2

CHH

CCH2

NH

also works for

R CH2

mass = 77


mass =

R

92 (R = H)106 (R = CH3)120 (R= CH2CH3)134 (R = C3H7)



Similar Patterns - positive charge is written on both fragments to show that either fragment might be seen at the detector

=

HC

CH2

R

28 (R = H)42 (R = CH3)56 (R= CH2CH3)70 (R = C3H7)84 (R = C4H9)

mass =

R2

R1

R2

R1

R2

R1

R2

R1

60 (R = OH)74 (R = OCH3)59 (R = NH2)78 (R = Cl)


Very Brief Description of Various Mass Spec Techniques – There are other techniques others besides those mentioned below. If you need practical knowledge of the theory and instrumentation of these experimental techniques, you will need to consult specialty references or textbooks.

1. In electron impact (EI), vaporized sample is bombarded with a very high energy beam of

electrons at about 70 eV (1600 kcal/mole) knocking an electron out of a bonding orbital, forming a radical cation. EI is relatively inexpensive and additional information can be obtained from fragmentation patterns. However fragmentation can prevent seeing the molecular ion peak (parent peak), which may necessitate using another approach, such as CI (next).

2. Chemical ionization (CI) introduces a reagent gas in the source at higher concentration and the gas is ionized by electrons at 500 eV. The reagent gas acts as a strong acid to protonate a basic site in the molecule of interest (at much lower energy to minimize fragmentation). This adds some mass to the sample, such as +1 (proton) or +(mass of gas). The protonated reagent gas can also abstract a proton (forming M-1). Generally, one can see the molecular mass peak (+1) much more clearly using CI. However, the sample must be vaporized and thermally stable which limits many biological samples or high molecular weight samples. If EI-MS does not produce an M+ peak, we will provide a hypothetical CI mass peak (and always assume it represents M+1). If we have access to a proton and 13C NMR we can use those spectra to provide a proton and carbon count. Both IR and 13C can provide information about the functional groups that are present which will give us a clue about how many oxygen and nitrogen atoms are present. If any larger than expected M+2 peaks show up (molecular ion or in a fragment) we might gain information about chlorine, bromine or sulfur. Using such a combination approach could also lead us to a molecular formula.

3. In fast atom bombardment (FAB), a solution of the sample in a matrix of low volatility is bombarded with neutral fast heavy atoms (Xe, Ar at 7 kev). It is a good method for molecules up to 20 KDa (biological molecules), and one can sequence some proteins. However the matrix usually produces background peaks at nearly every mass. One can usually see ions at M+1 or M-1.

4. In electrospray (ES), a solution of the sample is sprayed at atmospheric pressure through a 2-5 kV potential and the resulting droplets are electrostatically charged. There is no matrix background, multicharged species, molecules up to 200 kDa can be analyzed. However the method is susceptible to contamination of ions in the mist solution and nonpolar molecules are not detected.

5. In matrix assisted laser desorption ionization (MALDI), ions are accelerated to an energy of 3kV for mass analysis. A matrix absorbs energy produced by a laser and there is minimal fragmentation with better resolution than ES and FAB, especially at high mass.

6. In field desorption (FD), a sample is deposited directly onto anode where a high electric field produces desorption and ionization. There are very few fragmentations and is a preferred method for synthetic polymers. However samples may begin to decompose before inserted to the direct inlet. It is not good for high sensitivity and biological samples and has poor reproducibility.



Mass Spec Problem Set Name ___________________________________ 1. If the molecular ion peak is 142, what molecular formula does the rule of 13 predict if the structure is a

hydrocarbon? What formula is predicted if there is one oxygen atom? Two oxygen atoms? Two nitrogen atoms? What is the degree of unsaturation for each possibility above (4 calculations)? Draw one structure for each possibility. What if the molecular ion peak is 143 (same questions)?

2. Both CHO+ and C2H5+ have fragment masses of approximately 29, yet CHO+ has a M+1 peak of 1.13%

and M+2 peak of 0.20%, whereas C2H5+ 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 C2H5+. Explain these

observations and show all of your work. Helpful data follow.

Average NuclideElement Atomic Mass (Relative Abundance) Mass H 1.00797 1H (100) 1.00783 H 2H (0.016) 2.01410 C 12.01115 12C (100) 12.00000 C 13C (1.08) 13.00336 O 15.9994 16O (100) 15.9949 O 17O (0.04) 16.9991 O 18O (0.20) 17.9992

3. What relative abundance would the characteristic M (let M be 100%), M+2, M+4, M+6 mass peaks have for: (a) tribromo, Br3 substituted alkane, (b) trichloro, Cl3, substituted alkane and (c) bromodichloro, BrCl2 substituted alkane? Show your work. You can use these approximate probabilities (P): P35Cl = 0.75, P37Cl = 0.25, P79Br = 0.50, P81Br = 0.50

4. Radical cations of the following molecules (e- + M M.+ + 2e-) will fragment to yield the indicated masses as major peaks. The molecular ion peak is given under each structure. The base peak is listed as 100%. Other values listed represent some relatively stable possibilities (hence higher relative abundance), or common fragmentations (expected), even if in low amount. For the fragments with arrows pointing at them, show what the fragment is and how it could form from the parent ion. This may be as easy as drawing a line between two atoms of a bond, or it may require drawing curved arrows to show how electrons move (e.g. McLafferty). Explain why each fragment is reasonable. This may involve drawing resonance structures or indicating special substitution patterns (3o R+ > 2oR+ > 1oR+ > CH3

+). If a fragment has an even mass and there is a pi bond, think McLafferty (unless an odd number of nitrogen atoms are present). Even masses can also be formed by elimination of a small molecule such as loss of water from an alcohol or loss of an alcohol from an ether or a retro-Diels-Alder reaction, etc. Make sure you show this. Peaks with arrows are expected from the functional group shown. Most of the other peaks should be explainable using the examples in the prior discussions. See how many you can explain.


M+ = 86

a.

H3C

C

CH2

CH3

m/e % base

CH3

H3C

27.0 17.2 29.0 33.6 41.0 49.1 42.0 5.6 43.0 100.0 55.0 11.3 56.0 28.0 57.0 98.3 71.0 76.7 72.0 4.5 86.0 <1.0

M+ = 114

b.

H3C

H2C

CH2

H2C

CH2

H2C

CH2

CH3

27.0 20.1 29.0 27.4 41.0 43.8 42.0 15.3 43.0 100.0 55.0 11.4 56.0 18.4 57.0 33.5 70.0 12.1 71.0 20.4 85.0 26.5 114.0 6.0

m/e % base

Is there a logical peak that is missing?

M+ = 84

m/e % basec.

27.0 10.0 41.0 49.5 42.0 24.7 43.0 11.2 56.0 100.0 69.0 35.4 84.0 17.5

No easy explanation for 56, but if ring opens and forms alkene, McLafferty might work.

H

?

27.0 20.8 41.0 68.2 42.0 31.4 43.0 100.0 56.0 49.8 69.0 16.9 84.0 11.7

M+ = 84

m/e % base

d.

M+ = 68

e. m/e % base

C

CH

CH2

H2C

H3C

27.0 32.9 29.0 24.4 39.0 54.9 40.0 61.2 41.0 22.7 42.0 22.3 53.0 44.0 67.0 100.0 68.0 15.3

Don't remove the sp C-H, there is a better spot to lose an H atom (resonance).

You might have to move the C=C around.

M+ = 120

f.

m/e % base

65.0 7.2 77.0 2.7 91.0 100.0 92.0 10.8 105.0 3.8 120.0 25.9

H2C

CH2

CH3

M+ = 74

g. m/e % base

27.0 32.7 29.0 16.1 31.0 83.4 41.0 65.6 42.0 31.6 43.0 59.3 55.0 14.1 56.0 100.0 57.0 5.9 74.0 <1.0

H3C

H2C

CH2

H2C

OH

56 is an even mass, but not McLafferty. A small molecule might help explain it.Use a bridging ring to make 105.



M+ = 74

h. m/e % base

H3C

CH

CH2

CH3

OH 27.0 9.8 28.0 51.5 29.0 6.0 31.0 16.8 41.0 11.7 43.0 9.2 45.0 100.0 56.0 1.5 59.0 20.5 74.0 <1.0

M+ = 88

i. m/e % base

H3C

H2C

CH2

O

CH2

CH3

27.0 23.0 28.0 7.9 29.0 34.9 31.0 100.0 42.0 4.1 43.0 39.8 59.0 98.3 73.0 3.3 88.0 25.7

28 and 42 are even, but not McLafferty. Think like "g", but "organic" water. 31 requires some drastic rearrangements.

Peak 31 is harder to explain, but common.

M+ = 73

j.

m/e % baseH3C

H2C

CH2

H2C

NH2

27.0 3.5 29.0 2.1 30.0 100.0 43.0 1.2 56.0 1.2 73.0 7.3

M+ = 100

k.m/e % base

H3C

C

O

CH2

H2C

CH2

CH3 27.0 8.2 29.0 14.8 43.0 100.0 57.0 15.8 58.0 49.8 85.0 6.4 100.0 8.0 43 is different than C3H7

+

Peak 30 dominates. Think of a small molecule elimination for peak 56.

M+ = 72

l.m/e % base

H

C

O

CH2

H2C

CH3

29 is different than C2H5+

27.0 73.5 29.0 54.8 41.0 69.1 43.0 75.3 44.0 100.0 57.0 23.3 72.0 53.6

M+ = 102

m. m/e % base

O

C

O

CH2

H2C

CH3

H3C

27.0 47.0 29.0 9.2 41.0 45.3 43.0 100.0 59.0 22.2 71.0 49.9 74.0 64.2 87.0 16.4 102.0 1.444 is an even mass, so...

74 is an even mass.

M+ = 88

n.m/e % base

HO

C

O

CH2

H2C

CH3

27.0 13.6 29.0 8.1 41.0 16.3 43.0 14.1 45.0 9.9 60.0 100.0 73.0 32.5 88.0 2.6

M+ = 87

o. m/e % base

H2N

C

O

CH2

H2C

CH3

27.0 26.6 29.0 26.1 41.0 53.4 43.0 32.2 44.0 66.3 59.0 100.0 71.0 8.0 72.0 19.2 87.0 2.9

Normally 59 would be even, but there is nitrogen present.

An even mass strikes again at 60 and 45 is not common,but expected here.

M+ = 69

p.m/e % base

C

CH2

H2C

CH3

N

27.0 28.6 29.0 66.3 40.0 3.8 41.0 100.0 42.0 4.0 54.0 1.2 69.0 0.2

Normally 41 would be even, but there is nitrogen present.

27.0 9.3 41.0 19.3 68.0 100.0 91.0 12.7 92.0 18.8 95.0 7.6 121.0 19.5 136.0 22.6

q.

M+ = 136

m/e % base

Two famous names goes with 68.


Problem 11 – On the following pages are 22 compounds (these are lettered A-V) from the 22 functional groups numbered below. Try to match each spectrum (A-V) to the class of functional group numbered 1-22, and then try to solve the exact structure of each compound. These are simple monofunctional group compounds. Explain the major peaks that helped decide on your structure. Why are these peaks formed in preference to others (what is the reason for their special stability)?

Classes of compounds

1. alkane 2. branched alkane 3. cycloalkane

4. alkene 5. alkyne 6. aromatic

7. fluorinated alkane 8. chlorinated alkane 9. brominated alkane

10. iodinated alkane 11. alcohol 12. ether

13. phenol 14. aldehyde 15. ketone

16. ester 17. acid 18. amine

19. amide 20. acid chloride 21. sulfide

22. thiol

A few hints are given with some of the spectra to help you match structures with the functional groups mentioned above. The mass of each peak is listed with its percent of the base peak. The IR spectra should also give you some functional group hints. Remember, not every wave number is interpretable.

Hint: No N or O. Explain peaks at 67, 53 and 39. Peak 54 associated with McLafferty.

M+ peak = 82 (very small)

10090

67 = base peak

Mass Spec - Only larger and/or significant peaks are shown.

20 30 80706050

40

8153

39

40

54

100

0

4000 500

= wavenumber = cm-1

3000 2000 1500 1000

1470

1385

25003500

50

%T

1375

2960-2850

2120

725

3310

650

m/e

mass percent

26 328 529 639 36 40 25 41 64 42 16 43 4950 651 852 4 53 15 54 27 55 365 5 67 100 (base)68 6 81 10 82 = M+

Major peaksSample A

0%

25%

50%

75%

100%

110

answer (remove) oct-1-yne

41

42

43



M+ = 96

10090

81 = base peak


20 30 80706050

5339

40

54

mass percent

Major peaksSample B

0%

25%

50%

75%

100%

110

41

67

me

27 629 439 1240 441 11 53 13 54 16 55 2865 467 42 68 36 77 579 11 81 100 (base)82 795 896 41 = M+97 4

100

0

4000 500

= wavenumber = cm-1

3000 2000 1500 1000

1470

1385

25003500

50

%T

1375

2960-2850

1685838

Hint: Strong M+ peak, easily lost branch explains 81 and two famous names are associated with 68.

1-methylcyclohexeneMW = 96

55 68

79

(remove)

180160

43 = base peak


20 40 1401201008060

mass percent

Major peaksSample C

0%

25%

50%

75%

100%

200me

(remove in book)

100

0

4000 500

= wavenumber = cm-1

3000 2000 1500 1000

1470

1385

25003500

50

%T

1250

2960-2850

640

MW = 164

560

4-methyl-1-bromopentane

Hint: (M+2) is helpful, as are 151/149, 109/107. Explain 85, 57 and 43. 15.0 1 27.0 13 28.0 1 29.0 8 38.0 1 39.0 11 40.0 2 41.0 44 42.0 42 43.0 100 44.0 3 53.0 2 55.0 5 56.0 12 57.0 9 69.0 31 70.0 2 83.0 1 84.0 2

(base)

= M+= M+2

mass percent

Major peaks

85.0 59 86.0 4 107.0 3 109.0 2 149.0 3 151.0 3 164.0 2 166.0 2

4142

5657

69

85

107109

149151

164166

= M+= M+2


180160

105 = base peak


20 40 1401201008060

mass percent

Major peaksSample D

0%

25%

50%

75%

100%

200me

(remove in book)

= M+

mass percent

Major peaksp-thiomethyltolueneMW = 138

100

0

4000 500

= wavenumber = cm-1

3000 2000 1500 1000

1440

1385

25003500

50

%T

1375

2960-2850 820

3050

3550

1520

Hint: (M+2) is helpful. There is no major peak at 91 for a reason, but 105 will substitute. 27.0 3 39.0 4 45.0 6 50.0 2 51.0 4 52.0 1 53.0 2 63.0 2 65.0 3 68.0 2 77.0 11 78.0 4 79.0 11 89.0 1 91.0 5 93.0 1 103.0 8 104.0 6 105.0 100

= M+2 (5.8%)

(base)

106.0 10 135.0 2 137.0 2 138.0 15.5 139.0 2 140.0 0.9

77 79138 = M+

140 = M+2

180160

94 = base peak


20 40 1401201008060

mass percent

Major peaksSample E

0%

25%

50%

75%

100%

200me

(remove in book) mass percent

Major peaks

77150 = M+

(base)

= M+

butoxybenzeneMW = 150

27.0 4.1 29.0 11.2 39.0 7.1 40.0 1.4 41.0 8.3 50.0 1.2 51.0 4.4 55.0 1.5 56.0 1.1 57.0 3.9 63.0 1.1 65.0 5.3 66.0 5.5 77.0 7.4 94.0 100.0 95.0 7.0 107.0 1.6 150.0 18.4 151.0 2.1

Hint: McLafferty can explain 94, though there is no carbonyl group, but there is an xygen. 107 is small, but more like what you would expect for this functional group. Regular peaks at 29 and 57.The IR peak at 1220 is important.

= wavenumber = cm-1

100

0

4000 5003000 2000 1500 1000

1380

25003500

50

%T

2960-2850 760

3040

1600 1500 6901250

5729 41 69 107



180160

91 = base peak


20 40 1401201008060

mass percent

Major peaksSample F

0%

25%

50%

75%

100%

200me


Major peaks

78

120 = M+

29 41 65 103105

(base)

= M+

27.0 2 39.0 4 41.0 2 50.0 1 51.0 4 63.0 2 65.0 7 77.0 3 78.0 6 79.0 1 89.0 1 91.0 100 92.0 12 103.0 1 105.0 4 120.0 26 121.0 3

100

0

4000 500

= wavenumber = cm-1

3000 2000 1500 100025003500

50

%T

2960-2850 740

30301610

1500

7001450

propylbenzeneMW = 120

Hint: The big base peak is a big clue.

160


20 40 1401201008060

mass percent

Major peaksSample G

0%

25%

50%

me


Major peaks

128 = M+

43 = base peak

100

0

4000 500

= wavenumber = cm-1

3000 2000 1500 100025003500

50

%T

2960-2850

1460 14101380

1720

3420

octan-3-oneMW = 128

Hint: There are a lot of peaks that could be explained, both large and small. Try 100, 86, 85, 71, 57, 43 and 29. McLafferty might help on some of these.

15.0 1 18.0 1 26.0 2 27.0 21 28.0 4 29.0 58 30.0 1 39.0 8 40.0 1 41.0 17 42.0 4 43.0 100 44.0 4 53.0 2 55.0 9 56.0 3 57.0 92 58.0 5 71.0 52

72.0 67 73.0 8 81.0 1 85.0 10 86.0 3 99.0 52 100.0 4 128.0 12 129.0 1

(base)

= M+

29

57

7172

85

99

180 200


160

72 = base peak


20 40 1401201008060

mass percent

Major peaksSample H

0%

25%

50%

75%

100%

me


Major peaks

128 = M+5515

100

0

4000 500

= wavenumber = cm-1

3000 2000 1500 100025003500

50

%T

2960-2850

1460 1340

1730

3420

2-ethylhexanalMW = 128

Hint: There are a lot of peaks that could be explained, both large and small. Try 100, 72, 57, 43 and 29. McLafferty might help on some of these.

28102700

27.0 15 28.0 1 29.0 24 39.0 9 40.0 1 41.0 34 42.0 4 43.0 40 44.0 2 53.0 2 54.0 3 55.0 11 56.0 4 57.0 82 58.0 4 67.0 2 68.0 1 69.0 2 70.0 1

71.0 5 72.0 100 73.0 5 81.0 1 82.0 3 85.0 3

100.0 1 128.0 1 = M+

(base)

27 29

41 43

57

180 200

160


20 40 1401201008060

mass percent

Major peaksSample I

0%

25%

50%

me


Major peaks

87 = M+

30 = base peak

100

0

4000 500

= wavenumber = cm-1

3000 2000 1500 100025003500

50

%T

2960-2850

1470 13801610

33703290

18.0 2 27.0 3 28.0 33 29.0 2 30.0 100 31.0 2 39.0 2 41.0 3 42.0 2 43.0 1 44.0 2 45.0 3 87.0 4

820

pentylamineMW = 87

Hint: Notice the odd mass peak and the really big peak at 30. Though small 29 and 43 are obvious.

= M+

(base)

27

180 200



160


20 40 1401201008060

mass percent

Major peaksSample J

0%

25%

50%

me


Major peaks

100 = M+

43 = base peak

(base)

= M+

29

57

71

72 85

99

100

0

4000 500

= wavenumber = cm-1

3000 2000 1500 100025003500

50

%T

2960-2850

14701380 720

heptaneMW = 100

Hint: Look how regular the peaks are. What is lost in each fragment…100, 85, 71, 57, 43, 29? 15.0 1 26.0 1 27.0 18 28.0 3 29.0 31 39.0 11 40.0 2 41.0 45 42.0 20 43.0 100 44.0 3 53.0 1 55.0 10 56.0 25 57.0 47 58.0 2 70.0 18 71.0 46

72.0 2 85.0 2 100.0 11

56

4142

70

180 200

160


20 40 1401201008060

mass percent

Major peaksSample K

0%

25%

50%


Major peaks

100 = M+

43 = base peak

(base)

= M+

2757

71

7255

4142 70

100

0

4000 500

= wavenumber = cm-1

3000 2000 1500 100025003500

50

%T

2960-2850

1460

1380 770

890

2-ethylpentaneMW = 100

Hint: You can barely see the M+ peak because…? 43 is big and 71, 57 and 29 are all there too.27.0 10

29.0 14 39.0 6 41.0 16 42.0 7 43.0 100 44.0 4 53.0 1 55.0 15 56.0 4 57.0 4 70.0 48 71.0 51 72.0 3

100.0 2

29

me

200180


160


20 40 1401201008060

mass percent

Major peaksSample L

0%

25%

50%


Major peaks

118 = M+

43 = base peak

(base)

me

200180

= wavenumber = cm-1

100

04000 5003000 2000 1500 1000

1380

25003500

50

%T

2960-2850 1470

15.0 2 27.0 25 29.0 6 39.0 16 41.0 2 42.0 65 43.0 100 47.0 47 61.0 35 75.0 13 76.0 50 89.0 92 103.0 4 118.0 63 119.0 5 120.0 3

= M+

= M+2 (4.8%)

47

61

76

89

103

dipropylsulfideMW = 118

Hint: The (M+2) at 120 is helpful, as are 103, 89, 43 and 29. Why is 89 so big and 103 so small?

120 = M+2 (4.8%)75

42

27

29

160


20 40 1401201008060

mass percent

Major peaksSample M

0%

25%

50%


Major peaks

43 = base peak

me

200180

212 = M+

= wavenumber = cm-1

100

04000 5003000 2000 1500 1000

1420

25003500

50

%T

2960-2850 1460

85

29

27.0 14 28.0 4 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

1370

57 155

1-iodohexaneMW = 212

Hint: Look at that big hole in the middle…then there's 85, 57, 43, 29.

(base)

= M+

86



160


20 40 1401201008060

mass percentMajor peaksSample N

0%

25%

50%

(remove in book) mass percentMajor peaks

56 = base peak

me

200180

= wavenumber = cm-1

29 M+ = 102is missing

15.0 1 18.0 3 27.0 15 28.0 3 29.0 20 31.0 24 39.0 8 41.0 36 42.0 43 43.0 59 45.0 3 53.0 2 55.0 49 56.0 100 57.0 7 69.0 25 70.0 3 71.0 2 73.0 1 83.0 2

84.0 9 102.0 0 M+ (missing)

43

4241

55

69

7071

84

100

0

4000 5003000 2000 1500 1000

1470

25003500

50

%T

1380

2960-2850 1060

3320

660

hexan-1-olMW = 102

Hint: If you look hard there is a tiny peak at 102 (=M+)…and then a gap of 18. Some familiar peaks at 57, 43 and 29 are helpful. There's a special reason that 31 is there…why?

1527

31

160


20 40 1401201008060

mass percentMajor peaksSample O

0%

25%

50%


88 = base peak

me

200180

29

M+ = 102is missing

4341 55

31

(base)

= M+

= wavenumber = cm-1

15.0 1 18.0 4 29.0 22 31.0 1 39.0 10 41.0 32 42.0 5 43.0 20 45.0 7 53.0 2 55.0 17 56.0 5 57.0 38 59.0 3 60.0 2 69.0 6 70.0 4 71.0 1

57

567071

101 115116

144 = M+

100

0

4000 5003000 2000 1500 100025003500

50

%T

12301710

3300-2500

29252864

1460

1380

940

73.0 92 74.0 5 87.0 21 88.0 100 89.0 5 101.0 18 115.0 11 116.0 14 144.0 0.5

hexanoic acidMW = 144

Hint: 144 is tiny, but important. (M-56) is big for a reason (McLafferty). Other helpful peaks are 73, 71, 57, 56, 45, 43, 29.

45

73

87


160


20 40 1401201008060

mass percentMajor peaksSample P

0%

25%

50%


85 = base peak

me

200180

(base)

73

= wavenumber = cm-1

100

0

4000 5003000 2000 1500 100025003500

50

%T

1250

17402850-2960

1460

1380

1180

15.0 2 27.0 24 28.0 8 29.0 29 31.0 4 39.0 10 41.0 38 42.0 19 43.0 40 55.0 7 56.0 5 57.0 57 59.0 7 60.0 34 61.0 29 73.0 19 84.0 1 85.0 100

1090

103

104 115

6061

57

43404129

27

M+ = 144is missing

propyl pentanoateMW = 144

.Hint: McLafferty can occur two ways, at 116 and 88. Other useful peaks are at 57, 43, 29 and 45 is there for a reason. By the way, the M+ peak is missing at 144

86.0 6 87.0 3 102.0 10 103.0 60 104.0 4 115.0 4

M+ = 144is missing15 31

160


20 40 1401201008060

mass percentMajor peaksSample Q

0%

25%

50%


me

200180

.

18

59 = base peak

(base)

= wavenumber = cm-1

M+ = 143

100

0

4000 5003000 2000 1500 100025003500

50

%T

16502850-2960

1460

1380

= M+

1005557

72

4142

4344

29

3360

3190

1630

720

640

heptanamideMW = 143

Hint: Has an odd M+ peak at 143. McLafferty can explain the base peak at 59 and other familiar peaks are 43 and 29. The peak at 44 can be explained too.

18.0 2 27.0 5 28.0 2 29.0 7 39.0 4 41.0 13 42.0 4 43.0 16 44.0 19 53.0 1 54.0 1 55.0 8 56.0 2 57.0 10 59.0 100 60.0 7 69.0 2 72.0 34

73.0 6 82.0 2 83.0 2 84.0 1 86.0 10 96.0 1 97.0 1 100.0 6 114.0 5 143.0 6



160


20 40 1401201008060

mass percentMajor peaksSample R

0%

25%

50%


me

200180

.

15

43 = base peak

(base)

= wavenumber = cm-1

M+ = 112

100

0

4000 5003000 2000 1500 100025003500

50

%T

1640

2850-2960

1470

1380

= M+

3080

910

990

15.0 1 27.0 25 28.0 5 29.0 35 39.0 28 41.0 82 42.0 66 43.0 100 53.0 8 55.0 99 56.0 87 57.0 19 69.0 44 70.0 86 71.0 12 83.0 34 84.0 22 85.0 2 112.0 20

84

69

5556

57

4142

7185

1-octeneMW = 112

Hint: Some important peaks are M+ at 112 and 41 has a special reason as does 56 (McLafferty-like, but there is no oxygen). Other familiar peaks are at 71, 57, 43 and 29.

293927

53

70

83

160


20 40 1401201008060

mass percentMajor peaksSample S

0%

25%

50%

(remove in book)

= M+

mass percentMajor peaks

me

200180

.

107 = base peak

3927

= wavenumber = cm-1

M+ = 122

100

0

4000 5003000 2000 1500 100025003500

50

%T

1620

2850-2960 1510

1450

3020

8301240

3330

121108

77

6555

p-ethylphenolMW = 122

Hint: The M+ peak is solid at 122…and look at that peak at 107 (think "91" plus a really good something extra). That's almost all there is.

27.0 3 38.0 1 39.0 6 41.0 1 50.0 2 51.0 4 52.0 2 53.0 2 55.0 2 62.0 1 63.0 2 65.0 3 77.0 13 78.0 3 79.0 2 91.0 4 94.0 1

103.0 2 107.0 100 108.0 8 121.0 3 122.0 36 123.0 3

(base)

91


160


20 40 1401201008060

mass percentMajor peaksSample T

0%

25%

50%


me

200180

.

42 = base peak

27

(base)

= wavenumber = cm-1

= M+2

M+ = 106M+2 = 108

100

0

4000 5003000 2000 1500 100025003500

50

%T

2850-2960 1470

1350

= M+

750

15.0 1 27.0 27 28.0 6 29.0 38 39.0 19 41.0 70 42.0 100 43.0 39 53.0 3 55.0 93 56.0 6 57.0 22 63.0 5 69.0 3 70.0 95 71.0 6 91.0 3 93.0 0.9 106.0 1.0 108.0 0.3

660

29 43

55

57

70

9193

1-chloropentaneMW = 106

Hint: The M+ peak barely shows at 106 (and 108 is 1/3 its size). 70 is really big because it lost a small molecule of… (36)? 71, 57, 43 and 29 are old familiar friends.

15

39

160


20 40 1401201008060

mass percentMajor peaksSample U

0%

25%

50%


me

200180

.

2729 5739

109 = base peak

(base)

= wavenumber = cm-1

100

0

4000 5003000 2000 1500 100025003500

50

%T

2850-2960

1510

1470

= M+

740 700

3080-30301380

980

110 = M+

918365 77

benzylfluorideMW = 110

Hint: M+ is big, but M-1 is bigger and that's unusual. However, it has lots of stabilization. There is a halogen present, but M+2 is not important. What does 91 remind you of?

27.0 2 28.0 2 31.0 1 39.0 8 44.0 1 45.0 2 50.0 5 51.0 8 57.0 5 62.0 3 63.0 6 65.0 4 77.0 2 81.0 2 83.0 12 89.0 4

91.0 8 92.0 1 107.0 2 108.0 1 109.0 100 110.0 55 111.0 4

314445

51



160


20 40 1401201008060

mass percentMajor peaksSample V

0%

25%

50%


me

200180

.

55 = base peak

benzoyl chorideMW = 134.6

(base)

= wavenumber = cm-1

= mass+2

100

0

4000 5003000 2000 1500 100025003500

50

%T

2860-2960

11301470

730

580

1380

960

99

91

15.0 4 27.0 59 29.0 44 39.0 36 41.0 84 42.0 43 43.0 97 53.0 5 55.0 100 56.0 27 57.0 45 60.0 13 65.0 2 69.0 10 70.0 23 71.0 40 77.0 4 78.0 34

= mass+2

1800

440

29

27

39

4142

43

53

5657

60 6970

71 78

80

105106107108

Hint: Peaks at 105 and 106 have M+2 about 1/3 their size. There is a really helpful band in the IR spectrum. There are the usual peaks at 29, 43, 57 and 71.

80.0 11 91.0 12 92.0 1 98.0 24 99.0 84

105.0 6 106.0 6 107.0 2 108.0 2

134.0 0 M+ (missing)

134.0 0 M+ (missing)

15 65

What Happens When There is More Than One Functional Group?

We have mostly looked at monofunctional groups to learn the main clues provided by each functional group toward our goal of determining organic structures. What if more than one functional group is present? We can pit two strongly stabilizing groups against one another and see what happens. Both benzyl (91 mass) and methyliminium (30 mass) carbocations are strongly stabilized and generate easily recognizable MS peaks. We have seen both individually earlier, but we will repeat them below for comparison.

H2N

H

HCH2

H2CNH2

15.0 1 18.0 3 26.0 1 27.0 4 28.0 8 29.0 2 30.0 100.0 31.0 2 39.0 2 41.0 5 42.0 3 43.0 2 44.0 1 56.0 1 58.0 2 59.0 9

M+ = 59 mass = 30mass = 91M+ = 120basepeak

27.0 2 39.0 4 41.0 2 50.0 1 51.0 4 63.0 2 65.0 7 77.0 3 78.0 6 79.0 1 89.0 1 91.0 100 92.0 11 103.0 1 105.0 4 120.0 26

basepeak

(M-29) (M-29)


One way we could pit these two groups against one another would be to look at the mass spectrum of benzyl amine. Do we lose the amine group, NH2, or do we lose the phenyl group, C6H5? Not surprisingly, we don’t lose either. Instead they work together to make an even more stable carbocation with mass of 106 (M-1).

NH2

M+ = 107

H2CNH2

mass = 30 (27%)

CH2

mass = 91 (15%)

27.0 2 28.0 11 29.0 5 30.0 27 32.0 1 38.0 2 39.0 6 41.0 1 50.0 6 51.0 11 52.0 4 52.5 2 53.0 3 62.0 1 63.0 3 65.0 5

74.0 2 75.0 1 76.0 2 77.0 19 78.0 12 79.0 35 80.0 3 89.0 4 90.0 2 91.0 15 92.0 2

103.0 2 104.0 5 105.0 1 106.0 100 107.0 60

basepeak

NH2

HH H

mass = 106 (100%)

imminiumion

benzylcarbocation

(tropylium ion)

both stabilizing

groups

(M-29) (M-16) (M-77)

Another way to compare these two groups would be to look at the spectrum of 2-phenylethylamine (phenethylamine). Do we lose benzyl or do we lose methylimminium? Here we find the methylimminium group is the preferred method of fragmentation, but the benzyl carbocation is still observable. As molecules get more complicated, so will their mass spectra. We will not emphasize such examples because there are other methods much more helpful to our goal of determining organic structures, namely 1H and 13C NMR spectroscopy.

NH2

H

H

H2CNH2

CH2

M+ = 121

mass = 30 (100%)mass = 30 (15%)

28.0 2 30.0 100 31.0 1 39.0 3 50.0 1 51.0 3 63.0 2 65.0 6 77.0 2 89.0 1 91.0 15 92.0 6 103.0 2 120.0 1 121.0 6

basepeak

benzylcarbocation

(tropylium ion)

imminiumion

NH2

H

H

mass = 120 (1%)(M-29) (M-91)(M-1)

Problem – Discuss the MS of the following compounds. How do they compare to those in the examples above?

HO

OH OH

M+ = 120basepeak

27.0 2 39.0 4 41.0 2 50.0 1 51.0 4 63.0 2 65.0 7 77.0 3 78.0 6 79.0 1 89.0 1 91.0 100 92.0 11

103.0 1 105.0 4 120.0 26

basepeak

basepeak(very

unusual) basepeak

15.0 1 26.0 2 27.0 10 28.0 4 29.0 7 30.0 2 31.0 100 32.0 2 33.0 1 39.0 4 41.0 7 42.0 12 43.0 2 45.0 2 57.0 1 59.0 16 60.0 10

18.0 1 26.0 1 27.0 6 28.0 2 29.0 4 31.0 3 37.0 2 38.0 3 39.0 11 40.0 1 41.0 1 43.0 1 49.0 1

62.0 3 63.0 6 64.0 2 65.0 7 74.0 3 75.0 2

74.0 1 77.0 5 78.0 4 79.0 1 89.0 3 90.0 1 91.0 100 92.0 60 93.0 4 103.0 4 104.0 4 122.0 30 123.0 3

27.0 1 31.0 4 38.0 1 39.0 8 41.0 1 50.0 3 51.0 6 52.0 2 62.0 1 63.0 4 64.0 1 65.0 15 66.0 1

50.0 10 51.0 21 52.0 6 53.0 6 53.5 2 54.0 1 61.0 1

76.0 2 77.0 49 78.0 11 79.0 100 80.0 10 89.0 6 90.0 9 91.0 17 92.0 2 105.0 4 107.0 68 108.0 99 109.0 8

M+ = 60

M+ = 108M+ = 122



Problem – Major peaks in mass spectra representing most organic groups are provided below. Explain as many peaks as seems reasonable. There is plenty of opportunity to practice your new mass spec skills. Alkanes (Also look for alkene fragments too.)

72.0 3 85.0 2 100.0 11

mass % mass %

C7H16M+ = 100

27.0 19 28.0 3 29.0 31 39.0 11 40.0 2 41.0 45 42.0 20 43.0 100

44.0 3 53.0 2 55.0 10 56.0 25 57.0 48 58.0 2 70.0 18 71.0 46

mass % mass % mass %

heptane

2-methylhexane

mass % mass %

C7H16M+ = 100

mass % mass % mass % 84.0 5 85.0 37 86.0 3 100.0 3

27.0 11 29.0 15 39.0 8 40.0 1 41.0 31 42.0 35 43.0 100 44.0 3

53.0 1 55.0 5 56.0 21 57.0 29 58.0 1 69.0 1 70.0 2 71.0 2

3-methylhexane

mass % mass %

C7H16M+ = 100

mass % mass % mass % 72.0 3.1 84.0 1.1 85.0 5.7 100.0 4.0

27.0 12.8 29.0 24.7 39.0 9.1 40.0 1.5 41.0 36.7 42.0 9.2 43.0 100.0 44.0 3.5

53.0 1.8 55.0 15.0 56.0 39.3 57.0 52.8 58.0 2.3 69.0 2.0 70.0 46.5 71.0 58.3

3-ethylpentane

mass % mass %

C7H16M+ = 100

mass % mass % mass % 55.0 14.9 56.0 3.6 57.0 4.5 70.0 48.4 71.0 50.6 72.0 2.8 100.0 1.7

27.0 9.9 29.0 14.3 39.0 6.4 41.0 16.1 42.0 6.8 43.0 100.0 44.0 3.5 53.0 1.4

2,3-dimethylpentane

mass % mass %

C7H16M+ = 100

mass % mass % mass % 71.0 38.2 72.0 2.2 84.0 1.0 85.0 12.6 100.0 3.0

27.0 14.8 28.0 1.3 29.0 27.0 39.0 10.8 40.0 2.0 41.0 52.5 42.0 26.0 43.0 100.0

44.0 3.4 53.0 2.2 55.0 9.8 56.0 99.8 57.0 76.2 58.0 3.3 69.0 1.9 70.0 10.8

3,3-dimethylpentane

mass % mass %

C7H16M+ = 100

mass % mass % mass % 86.0 1.3 100.0 0.0

27.0 9.3 29.0 11.5 39.0 6.4 41.0 15.7 42.0 2.7 43.0 100.0 44.0 3.4 53.0 1.8

55.0 11.2 56.0 1.3 57.0 6.1 69.0 1.4 70.0 18.0 71.0 69.7 72.0 4.1 85.0 19.3


2,4-dimethylpentane

mass % mass %

C7H16M+ = 100

mass % mass % mass % 85.0 18.5 86.0 1.3 100.0 0.0

15.0 1.0 27.0 10.6 29.0 14.0 39.0 7.6 40.0 1.4 41.0 33.5 42.0 23.9 43.0 100.0

44.0 3.3 53.0 1.0 55.0 3.2 56.0 35.3 57.0 71.7 58.0 3.2 69.0 2.4 71.0 1.1

2,2,3-trimethylbutane

mass % mass %

C7H16M+ = 100

mass % mass % mass % 83.0 1.0 85.0 34.4 86.0 2.3 100.0 0.0

15.0 1.5 27.0 8.4 29.0 15.6 39.0 9.3 40.0 1.6 41.0 42.1 42.0 2.7 43.0 62.6

44.0 2.1 53.0 1.6 55.0 5.1 56.0 58.3 57.0 100.0 58.0 4.6 59.0 2.8 69.0 2.4

C6H14M+ = 86hexane

mass % mass %mass % mass % mass % 86.0 10.0 15.0 1.3

26.0 1.7 27.0 22.7 28.0 4.7 29.0 42.5 30.0 1.0 38.0 1.2 39.0 14.8

40.0 3.1 41.0 72.5 42.0 42.4 43.0 80.9 44.0 2.8 51.0 1.2 53.0 2.2 54.0 1.2

55.0 12.1 56.0 70.5 57.0 100.0 58.0 4.7 69.0 8.5 70.0 2.6 71.0 11.0 84.0 3.2

C6H14M+ = 86

2-methylpentane

mass % mass %mass % mass % mass % 72.0 2.2 78.0 1.0 85.0 1.5 86.0 6.1

27.0 12.6 28.0 1.3 29.0 11.3 39.0 9.6 40.0 1.5 41.0 29.5 42.0 52.6 43.0 100.0

44.0 3.2 53.0 1.2 55.0 6.7 56.0 9.4 57.0 17.0 69.0 1.3 70.0 10.2 71.0 39.5

C6H14M+ = 86

3-methylpentane


27.0 13.3 28.0 2.2 29.0 39.1 39.0 9.2 40.0 1.2 41.0 53.4 42.0 3.6

43.0 25.4 53.0 1.8 55.0 6.7 56.0 76.7 57.0 100.0 58.0 4.5 70.0 1.6 71.0 5.7

C6H14M+ = 86

2,2-dimethylbutane

mass % mass %mass % mass % mass % 57.0 98.3 58.0 4.7 70.0 3.3 71.0 76.7 72.0 4.5 86.0 0.1

15.0 1.6 26.0 1.1 27.0 17.2 28.0 2.6 29.0 33.6 38.0 1.0 39.0 13.6 40.0 1.8

41.0 49.1 42.0 5.6 43.0 100.0 44.0 3.1 51.0 1.1 53.0 2.5 55.0 11.3 56.0 28.0

C6H14M+ = 86

2,3-dimethylbutane


27.0 13.7 28.0 1.9 29.0 7.5 39.0 9.0 40.0 1.5 41.0 27.4 42.0 87.0

43.0 100.0 44.0 3.5 53.0 1.4 55.0 5.2 56.0 1.5 57.0 2.3 71.0 19.2 72.0 1.1



Alkenes & Cycloalkanes

C7H14M+ = 98


1-heptene

69.0 31.1 70.0 44.2 71.0 2.5 83.0 4.2 98.0 13.8 99.0 1.1

15.0 1.1 18.0 1.2 26.0 2.0 27.0 25.7 28.0 4.9 29.0 55.9 30.0 1.2 38.0 1.8

39.0 30.5 40.0 4.9 41.0 96.8 42.0 54.9 43.0 15.9 50.0 1.5 51.0 2.3 52.0 1.0

53.0 6.5 54.0 7.6 55.0 67.6 56.0 100.0 57.0 30.7 58.0 1.3 67.0 2.3 68.0 3.6

C7H14M+ = 98


trans-2-heptene

71.0 1.1 81.0 1.2 83.0 4.2 98.0 43.4 99.0 3.6

15.0 1.4 26.0 2.2 27.0 26.3 28.0 3.7 29.0 22.0 38.0 2.1 39.0 27.6 40.0 4.8

41.0 74.3 42.0 19.3 43.0 20.5 50.0 1.8 51.0 3.1 52.0 1.4 53.0 9.9 54.0 11.6

55.0 100.0 56.0 90.4 57.0 9.9 65.0 1.3 67.0 5.0 68.0 3.6 69.0 48.1 70.0 16.9

C7H14M+ = 98


trans-3-heptene

79.0 1.3 81.0 1.7 83.0 8.3 97.0 2.0 98.0 80.2 99.0 6.1

27.0 13.5 28.0 3.1 29.0 12.8 38.0 1.1 39.0 18.4 40.0 3.7 41.0 95.3 42.0 23.7

43.0 18.0 50.0 1.2 51.0 2.4 52.0 1.0 53.0 7.7 54.0 8.3 55.0 82.5 56.0 98.8

57.0 14.2 65.0 1.8 67.0 8.5 68.0 5.2 69.0 100.0 70.0 28.3 71.0 1.6 77.0 1.0

C7H14M+ = 98


cis-2-heptene

69.0 49.4 70.0 17.6 71.0 2.6 81.0 1.2 83.0 4.5 98.0 43.3 99.0 3.9

15.0 1.4 26.0 1.7 27.0 24.8 28.0 4.9 29.0 23.4 38.0 1.6 39.0 23.0 40.0 4.3

41.0 87.8 42.0 22.8 43.0 27.5 44.0 1.0 45.0 1.1 50.0 1.5 51.0 2.4 52.0 1.0

53.0 7.6 54.0 9.7 55.0 79.3 56.0 100.0 57.0 18.4 65.0 1.0 67.0 4.0 68.0 3.6

C7H14M+ = 98


2-methy l-1-hexene

83.0 1.1 98.0 3.1

27.0 11.7 28.0 3.1 29.0 8.9 39.0 10.5 40.0 2.9 41.0 46.6 42.0 5.6 43.0 11.5

53.0 3.2 54.0 1.6 55.0 17.7 56.0 100.0 57.0 10.5 67.0 1.5 69.0 9.9 70.0 12.4

C7H14M+ = 98


3-methy l-3-hexene

98.0 35.9 99.0 3.2

15.0 1.4 26.0 1.1 27.0 15.5 28.0 3.3 29.0 12.2 38.0 1.5 39.0 18.9 40.0 3.4

41.0 82.1 42.0 6.1 43.0 8.2 50.0 1.7 51.0 3.2 52.0 1.4 53.0 8.2 54.0 2.2

55.0 69 56.0 19 57.0 2 65.0 2 67.0 7 68.0 3 69.0 100 70.0 13

72.0 1.2 73.0 1.0 77.0 1.0 79.0 1.4 81.0 2.5 83.0 14.9 84.0 1.1 85.0 1.5

C7H14M+ = 98


2,3-dimethy l-2-pentene

83.0 82.9 84.0 5.7 85.0 4.8 98.0 37.8 99.0 4.1

15.0 2.2 18.0 1.0 27.0 14.1 28.0 2.3 29.0 12.8 31.0 1.3 38.0 1.1 39.0 17.9

40.0 3.4 41.0 65.3 42.0 5.1 43.0 18.9 50.0 1.3 51.0 2.5 52.0 1.2 53.0 7.6

54.0 2.3 55.0 100.0 56.0 13.7 57.0 4.6 58.0 1.8 59.0 8.9 65.0 1.7 67.0 7.3

68.0 1.6 69.0 21.1 70.0 4.5 71.0 1.0 72.0 5.2 73.0 3.5 79.0 1.5 81.0 4.2


C7H14M+ = 98


cycloheptane

84.0 3.7 91.0 2.0 92.0 1.4 95.0 1.1 96.0 3.4 97.0 1.7 98.0 62.9 99.0 5.0

15.0 1.7 26.0 2.4 27.0 23.2 28.0 4.8 29.0 23.9 38.0 2.2 39.0 30.9 40.0 6.2

41.0 77.6 42.0 75.4 43.0 14.1 50.0 1.6 51.0 2.9 52.0 1.3 53.0 8.5 54.0 18.4

55.0 96.8 56.0 100.0 57.0 17.9 63.0 1.0 65.0 2.0 66.0 1.7 67.0 13.2 68.0 27.2

69.0 61.7 70.0 85.7 71.0 5.1 77.0 1.1 79.0 1.5 81.0 7.5 82.0 4.6 83.0 48.6

C7H14M+ = 98


methylcyclohexane

84.0 7.0 97.0 2.8 98.0 36.9 99.0 3.1

26.0 1.0 27.0 12.9 28.0 3.2 29.0 11.1 38.0 1.0 39.0 15.6 40.0 2.9 41.0 41.1

42.0 28.6 43.0 6.9 51.0 1.9 53.0 4.6 54.0 4.5 55.0 76.3 56.0 28.5 57.0 5.0

67.0 4.5 68.0 9.3 69.0 22.5 70.0 21.8 71.0 1.4 81.0 1.4 82.0 14.5 83.0 100.0

C7H14M+ = 98


ethylcyclopentane

83.0 8.2 98.0 12.3 99.0 1.0

15.0 1.0 26.0 1.4 27.0 13.7 28.0 2.6 29.0 12.5 38.0 1.2 39.0 18.6 40.0 3.5

41.0 63.7 42.0 41.3 43.0 7.0 50.0 1.0 51.0 1.7 53.0 5.1 54.0 5.3 55.0 47.9

56.0 44.7 57.0 7.4 65.0 1.2 67.0 10.0 68.0 66.4 69.0 100.0 70.0 56.1 71.0 3.2

C7H14M+ = 98


1,1,2,2-tet ramethylcyclopropane

69.0 2.4 81.0 1.4 83.0 88.4 84.0 6.0 98.0 18.4 99.0 1.4

27.0 9.6 28.0 2.3 29.0 8.3 38.0 1.1 39.0 18.0 40.0 3.0 41.0 48.6 42.0 2.7

43.0 17.4 51.0 1.6 53.0 3.9 55.0 100.0 56.0 11.2 57.0 2.0 65.0 1.1 67.0 3.7

Alcohols and ethers

C7H16OM+ = 116


1-heptanol

OH

73.0 2.0 83.0 7.4 98.0 5.6 116.0 0.0

15.0 1.3 18.0 3.2 26.0 1.2 27.0 20.2 28.0 4.6 29.0 27.2 31.0 25.6 39.0 11.3

40.0 2.4 41.0 62.6 42.0 48.5 43.0 66.7 44.0 2.7 45.0 3.6 53.0 2.5 54.0 5.6

55.0 65.7 56.0 95.5 57.0 23.4 67.0 1.8 68.0 13.3 69.0 51.6 70.0 100.0 71.0 6.3

C7H16OM+ = 116

mass % mass %mass % mass % mass %OH

2-heptanol

98.0 4.0 101.0 3.7 116.0 0.0

27.0 5.1 29.0 5.4 31.0 1.5 39.0 3.2 41.0 10.0 42.0 3.6 43.0 8.1 44.0 6.9

45.0 100.0 46.0 2.3 55.0 14.9 56.0 6.8 57.0 3.5 69.0 2.7 70.0 4.8 83.0 8.9

C7H16OM+ = 116


3-heptanol

OH

98.0 2.9 116.0 0.0

27.0 9.8 28.0 2.1 29.0 11.1 30.0 1.1 31.0 22.1 39.0 5.3 41.0 34.9 42.0 2.9

43.0 11.4 44.0 2.9 45.0 8.1 53.0 1.1 55.0 8.0 56.0 5.8 57.0 8.1 58.0 8.0

59.0 100.0 60.0 3.3 69.0 70.2 70.0 5.4 73.0 1.0 86.0 2.7 87.0 31.3 88.0 1.8



C7H16OM+ = 116


4-heptanol

OH 69.0 4.0 70.0 1.3 71.0 3.2 72.0 6.3 73.0 71.4 74.0 3.4 98.0 2.0 116.0 0.0

18.0 1.1 19.0 1.1 27.0 9.8 28.0 1.1 29.0 6.9 31.0 11.1 39.0 5.4 41.0 13.1

42.0 2.5 43.0 33.4 44.0 5.4 45.0 4.6 53.0 1.2 55.0 100.0 56.0 9.2 57.0 4.4

C7H16OM+ = 116


3-ethy l-3-pentanol

OH 67.0 1.2 69.0 28.1 70.0 2.7 83.0 1.6 87.0 100.0 88.0 5.4 98.0 2.8 116.0 0.0

18.0 1.7 27.0 8.5 28.0 1.5 29.0 13.9 31.0 6.1 39.0 4.2 41.0 24.9 42.0 1.5

43.0 12.5 45.0 73.7 46.0 1.7 53.0 1.9 55.0 8.2 56.0 2.3 57.0 11.2 59.0 2.5

C7H16OM+ = 116


1-methoxyhexane

O

57.0 3.4 69.0 13.2 70.0 2.4 83.0 1.3 84.0 19.4 85.0 1.8 116.0 0.0

15.0 2.0 27.0 5.9 28.0 1.9 29.0 8.2 31.0 1.8 33.0 5.2 39.0 4.5 41.0 14.1

42.0 13.1 43.0 13.3 45.0 100.0 46.0 2.4 47.0 1.3 54.0 1.3 55.0 16.4 56.0 53.7

C7H14OM+ = 114

mass % mass %mass % mass % mass %OH

cycloheptanol

85.0 5.6 86.0 4.1 95.0 2.9 96.0 22.6 97.0 2.0 113.0 1.1 114.0 2.1

15.0 1.3 18.0 1.4 26.0 1.0 27.0 11.9 28.0 2.8 29.0 12.0 30.0 1.1 31.0 5.4

39.0 12.7 40.0 2.8 41.0 25.0 42.0 14.8 43.0 10.3 44.0 22.5 45.0 5.3 51.0 1.1

53.0 4.8 54.0 15.9 55.0 23.1 56.0 4.4 57.0 100.0 58.0 9.1 66.0 2.0 67.0 19.8

68.0 38.6 69.0 4.0 70.0 13.0 71.0 14.9 72.0 6.3 81.0 34.2 82.0 2.5 83.0 1.4

113.0 1.4 114.0 2.8

C7H14OM+ = 114


4-methylcyclohexanol

OH 15.0 1.5 18.0 2.4 26.0 1.1 27.0 13.5 28.0 3.7 29.0 19.8 30.0 1.7 31.0 5.7

39.0 12.7 40.0 2.6 41.0 34.8 42.0 9.9 43.0 7.8 44.0 17.0 45.0 3.3 51.0 1.5

53.0 5.5 54.0 7.7 55.0 31.1 56.0 12.0 57.0 100.0 58.0 52.8 59.0 2.7 65.0 1.0

67.0 10.7 68.0 11.8 69.0 3.1 70.0 34.1 71.0 12.6 73.0 1.7 77.0 1.0 79.0 2.0

81.0 47.2 82.0 3.7 83.0 2.2 85.0 3.8 86.0 3.1 95.0 6.0 96.0 35.0 97.0 3.1

mass %

C7H14OM+ = 114


2-cyclopentylethanol

OH 71.0 1.9 79.0 1.5 81.0 23.9 82.0 2.2 83.0 7.5 95.0 3.0 96.0 4.0 114.0 0.0

18.0 1.3 27.0 10.6 28.0 2.5 29.0 8.4 31.0 11.0 39.0 13.4 40.0 3.8 41.0 40.7

42.0 9.6 43.0 4.8 44.0 5.8 45.0 2.4 51.0 1.2 53.0 7.5 54.0 12.3 55.0 36.9

56.0 7.0 57.0 7.3 58.0 1.0 65.0 1.2 66.0 7.6 67.0 99.6 68.0 100.0 69.0 15.4

C7H14OM+ = 114


3-butoxypropene

56.0 13.3 57.0 42.9 58.0 72.2 59.0 3.8 71.0 10.7 72.0 1.8 73.0 2.8 85.0 1.9

O 15.0 1.4 18.0 1.3 26.0 1.8 27.0 9.9 28.0 4.0 29.0 24.7 30.0 3.2 31.0 3.2

38.0 1.1 39.0 13.4 40.0 3.0 41.0 100.0 42.0 7.6 43.0 16.0 44.0 1.9 55.0 10.4

114.0 0.0


C8H18OM+ = 130


di-sec-butyl ether

O 112.0 1.0 115.0 1.0 130.0 0.2

15.0 1.2 27.0 8.0 28.0 2.0 29.0 23.8 31.0 2.9 39.0 5.2 41.0 28.8 42.0 1.7

43.0 4.7 44.0 1.3 45.0 100.0 46.0 2.2 53.0 1.0 55.0 8.3 56.0 6.8 57.0 84.5

58.0 3.8 59.0 29.5 60.0 1.0 73.0 5.4 83.0 9.7 97.0 1.1 101.0 39.1 102.0 2.6

C8H18OM+ = 130


di-isobutyl ether

O

55.0 2.0 56.0 4.7 57.0 100.0 58.0 4.6 59.0 1.7 73.0 1.0 87.0 8.2 130.0 2.6

27.0 4.0 29.0 12.4 31.0 1.0 39.0 4.2 41.0 19.3 42.0 2.2 43.0 3.6 45.0 2.0

Amines



Haloalkanes


1-chloropropane

Cl

C3H7ClM+ = 78.5(78 & 80)

63.0 4.7 65.0 1.4 78.0 2.7 80.0 0.8

15.0 1.3 18.0 1.3 26.0 3.9 27.0 31.9 28.0 12.9 29.0 40.6 36.0 1.2 37.0 2.3

38.0 3.1 39.0 11.4 40.0 3.1 41.0 23.2 42.0 100.0 43.0 13.7 49.0 4.2 51.0 1.3


2-chloropropane

Cl

78.0 9.0 80.0 3.1

C3H7ClM+ = 78.5(78 & 80)

15.0 1.7 26.0 2.4 27.0 29.3 36.0 1.2 37.0 1.5 38.0 2.4 39.0 8.5 40.0 2.1

41.0 21.8 42.0 6.6 43.0 100.0 44.0 3.6 62.0 1.8 63.0 17.0 65.0 5.7


1-chlorobutane

Cl

C3H7ClM+ = 92.6(78 & 80)

55.0 7.5 56.0 100.0 57.0 5.8 62.0 1.3 63.0 4.8 65.0 1.4

18.0 1.4 26.0 3.7 27.0 26.3 28.0 15.0 29.0 16.7 38.0 1.4 39.0 10.0 40.0 2.2

41.0 57.5 42.0 5.0 43.0 36.6 44.0 1.1 49.0 2.8 51.0 1.1 53.0 1.2



2-chloro-2-methylpropane

Cl

C3H7ClM+ = 92.6(90 & 92)

55.0 3.9 56.0 8.1 57.0 100.0 58.0 4.3 76.0 3.1 77.0 38.3 79.0 12.2

15.0 1.9 26.0 1.9 27.0 10.4 28.0 1.4 29.0 22.3 36.0 3.1 37.0 1.4 38.0 2.6

39.0 17.9 40.0 2.6 41.0 66.8 42.0 3.2 49.0 2.2 51.0 1.3 53.0 1.3


C6H13ClM+ = 120.6(120 & 122)

1-chlorohexane

Cl

91.0 100.0 93.0 32.0

15.0 1.3 18.0 3.2 26.0 2.2 27.0 27.0 28.0 5.0 29.0 32.0 31.0 1.4 38.0 1.2

39.0 16.7 40.0 2.9 41.0 59.0 42.0 44.7 43.0 72.0 44.0 2.5 49.0 3.0 51.0 1.4

53.0 3.7 54.0 3.9 55.0 81.1 56.0 56.5 57.0 14.7 62.0 1.2 63.0 4.7 65.0 1.5

67.0 2.7 69.0 22.1 70.0 2.1 71.0 2.8 82.0 1.2 83.0 1.3 84.0 4.2


C3H7BrM+ = 123

(122 & 124)

1-bromohexane

Br

107.0 1.5 109.0 1.2 122.0 8.6 124.0 8.3

15.0 1.6 26.0 2.4 27.0 25.6 28.0 1.6 29.0 4.0 37.0 1.2 38.0 2.2 39.0 9.0

40.0 2.1 41.0 31.0 42.0 7.7 43.0 100.0 44.0 3.5 93.0 1.1 95.0 1.0


C3H7BrM+ = 123

(122 & 124)

2-bromohexane

Br

41.0 31.9 42.0 4.0 43.0 100.0 44.0 3.6 107.0 1.1 109.0 1.0 122.0 5.9 124.0 5.7

15.0 1.4 26.0 2.2 27.0 27.4 28.0 1.1 37.0 1.2 38.0 2.1 39.0 9.9 40.0 1.9


C4H9BrM+ = 137

(136 & 138)

1-bromobutane

Br

136.0 7.9 138.0 7.8

15.0 1.5 26.0 5.7 27.0 29.4 28.0 13.3 29.0 40.9 38.0 2.0 39.0 14.8 40.0 2.2

41.0 64.5 42.0 3.2 43.0 3.9 50.0 1.6 51.0 1.4 53.0 1.5 55.0 7.2 56.0 16.4

57.0 100.0 58.0 4.6 79.0 1.0 81.0 1.0 93.0 1.4 95.0 1.3 107.0 3.7 109.0 3.6


C4H9BrM+ = 137

(136 & 138)

2-bromobutane

Br

58.0 4.4 107.0 1.2 109.0 1.0 136.0 0.6 138.0 0.6

15.0 1.4 26.0 3.9 27.0 18.6 28.0 5.3 29.0 40.9 38.0 1.8 39.0 14.3 40.0 1.7

41.0 53.2 42.0 2.9 50.0 1.6 51.0 1.5 53.0 1.6 55.0 4.9 56.0 8.0 57.0 100.0




C6H13BrM+ = 165.1(164 & 166)

1-bromohexane

Br

164.0 0.4 166.0 0.4

26.0 1.2 27.0 16.5 28.0 2.7 29.0 20.7 39.0 10.8 40.0 1.9 41.0 41.8 42.0 10.3

43.0 66.4 44.0 2.2 53.0 2.4 54.0 1.3 55.0 25.7 56.0 14.6 57.0 100.0 58.0 4.9

69.0 7.5 70.0 3.2 71.0 2.8 81.0 1.0 83.0 1.5 84.0 1.2 85.0 18.2 86.0 1.3

99.0 14.5 100.0 1.2 107.0 1.2 109.0 1.0 135.0 8.3 137.0 8.0


C3H7IM+ = 170.0

1-iodopropane

I

42.0 3.3 43.0 100.0 44.0 3.3 127.0 4.9 128.0 1.2 170.0 23.3

15.0 1.7 26.0 1.7 27.0 31.7 28.0 1.7 38.0 1.5 39.0 10.7 40.0 2.0 41.0 36.8

I+ = 127

HI+ = 128


C3H7IM+ = 170.0

2-iodopropane

I

42.0 3.6 43.0 100.0 44.0 3.4 127.0 5.8 128.0 1.7 170.0 24.3

15.0 1.1 26.0 1.4 27.0 27.7 28.0 1.6 38.0 1.6 39.0 11.5 40.0 2.0 41.0 35.8

I+ = 127

HI+ = 128


C5H11IM+ = 198.0

1-iodopentane

I

141.0 1.2 155.0 2.3 198.0 9.5

26.0 1.2 27.0 18.8 28.0 2.5 29.0 23.1 39.0 10.6 40.0 2.0 41.0 30.1 42.0 8.3

43.0 100.0 44.0 3.4 53.0 1.3 55.0 9.9 71.0 73.2 72.0 4.3 127.0 2.1 128.0 0.6

I+ = 127

HI+ = 128

Spectroscopy Beauchamp 72 Aldehydes and ketones

mass % mass %mass % mass % mass %O

H

propanal

58.0 85.0 59.0 12.1

14.0 2.0 15.0 4.7 18.0 5.2 25.0 2.4 26.0 16.4 27.0 59.0 28.0 90.8 29.0 100.0

30.0 7.8 31.0 5.6 37.0 2.4 38.0 2.3 39.0 4.0 40.0 1.7 56.0 1.7 57.0 7.3

C3H6OM+ = 58

C3H6OM+ = 58


propanone

59.0 3.1 14.0 2.9 15.0 23.1 26.0 3.5 27.0 5.7 28.0 1.2 29.0 3.1 37.0 1.8 38.0 2.2

39.0 4.2 40.0 1.0 41.0 2.0 42.0 9.1 43.0 100.0 44.0 3.4 57.0 1.7 58.0 63.8

C4H8OM+ = 72


H

butanal

60.0 2.7 71.0 5.4 72.0 53.6 73.0 2.7

14.0 1.7 15.0 5.9 18.0 1.2 26.0 8.2 27.0 73.5 28.0 19.6 29.0 54.8 30.0 1.2

31.0 2.6 32.0 1.3 37.0 3.1 38.0 5.3 39.0 27.3 40.0 3.5 41.0 69.1 42.0 9.4

43.0 75.3 44.0 100.0 45.0 3.2 50.0 1.1 53.0 1.1 54.0 2.3 55.0 1.5 57.0 23.3

C4H8OM+ = 72


H

pentanal

15.0 2.4 26.0 5.6 27.0 54.9 28.0 13.1 29.0 31.5 31.0 1.5 37.0 3.4

38.0 5.7 39.0 28.7 40.0 3.9 41.0 86.7 42.0 14.0 43.0 100.0 44.0 7.4 45.0 1.1

50.0 1.0 53.0 1.5 55.0 3.3 56.0 1.8 57.0 6.9 71.0 4.4 72.0 92.3 73.0 5.9

C4H8OM+ = 72


butanone

41.0 1.1 42.0 4.1 43.0 100.0 44.0 2.6 57.0 8.0 72.0 22.1 73.0 1.0

14.0 1.2 15.0 6.6 18.0 1.3 26.0 2.6 27.0 8.9 28.0 1.3 29.0 18.8 39.0 1.6


C5H10OM+ = 86

O

H

pentanal

71.0 1.8 73.0 1.1 85.0 1.2 86.0 1.1

15.0 3.9 26.0 3.4 27.0 28.2 28.0 10.5 29.0 52.3 30.0 2.2 31.0 1.9 38.0 1.3

39.0 12.3 40.0 2.2 41.0 41.0 42.0 11.1 43.0 18.7 44.0 100.0 45.0 12.2 53.0 1.6

55.0 4.4 56.0 2.2 57.0 19.8 58.0 31.4 59.0 1.3 60.0 2.9 67.0 1.1 68.0 1.0


C5H10OM+ = 86

O

H

3-methylbutanal

87.0 1.2 15.0 4.2 26.0 2.8 27.0 41.9 28.0 4.0 29.0 46.3 30.0 1.0 31.0 1.3 37.0 2.3

38.0 5.0 39.0 38.7 40.0 5.7 41.0 89.8 42.0 24.6 43.0 93.4 44.0 100.0 45.0 19.6

50.0 2.5 51.0 2.4 53.0 5.7 55.0 6.2 56.0 3.6 57.0 37.0 58.0 81.4 59.0 3.1

60.0 2.7 67.0 1.9 68.0 2.3 69.0 2.6 71.0 36.0 72.0 1.7 85.0 2.8 86.0 11.7




C5H10OM+ = 86

O

H

3-methylbutanal

86.0 6.3 87.0 3.2

15.0 2.3 18.0 2.3 26.0 4.5 27.0 31.1 28.0 9.4 29.0 100.0 30.0 4.0 31.0 1.7

37.0 1.1 38.0 1.9 39.0 18.4 40.0 2.7 41.0 92.4 42.0 5.8 43.0 11.9 44.0 3.8

45.0 3.0 50.0 1.4 51.0 1.5 53.0 3.6 55.0 9.4 56.0 7.8 57.0 95.8 58.0 60.0

59.0 2.7 67.0 1.1 69.0 1.3 70.0 2.3 71.0 4.9 73.0 1.8 74.0 11.1 85.0 1.8


C5H10OM+ = 86

O

H

2,2-dimethylpropanal

86.0 18.1 87.0 3.1

15.0 2.3 18.0 3.0 26.0 1.7 27.0 16.4 28.0 3.2 29.0 51.5 30.0 1.1 32.0 1.0

37.0 1.1 38.0 2.5 39.0 19.4 40.0 2.6 41.0 83.5 42.0 8.8 43.0 26.6 50.0 1.2

51.0 1.3 53.0 2.1 55.0 5.5 56.0 2.8 57.0 100.0 58.0 5.2 59.0 1.5 71.0 1.6


C5H10OM+ = 86

O

2-pentanone

42.0 4.0 43.0 100.0 44.0 2.3 58.0 10.3 71.0 11.0 86.0 20.2 87.0 1.2

15.0 4.8 26.0 1.3 27.0 10.5 28.0 1.3 29.0 1.9 38.0 1.2 39.0 6.3 41.0 11.9


C5H10OM+ = 86

O

3-pentanone

43.0 1.6 55.0 1.3 56.0 3.7 57.0 100.0 58.0 3.4 86.0 21.2 87.0 1.2

26.0 2.5 27.0 12.4 28.0 4.3 29.0 59.4 30.0 1.4 39.0 1.8 41.0 2.0 42.0 1.8


C5H10OM+ = 86

O

3-methyl-2-butanone

51.0 1.1 57.0 3.7 71.0 6.9 86.0 22.6 87.0 1.0

14.0 1.6 15.0 9.8 26.0 3.1 27.0 19.3 28.0 3.1 29.0 3.5 37.0 1.4 38.0 2.9

39.0 16.3 40.0 2.1 41.0 26.2 42.0 4.9 43.0 100.0 44.0 2.4 45.0 1.3 50.0 1.3


H

hexanal

C6H12OM+ = 100

83.0 1.0 100.0 0.4

15.0 2.2 18.0 1.0 26.0 2.7 27.0 33.9 28.0 8.1 29.0 33.0 30.0 1.6 31.0 1.8

38.0 1.9 39.0 20.1 40.0 3.8 41.0 69.1 42.0 10.8 43.0 55.1 44.0 100.0 45.0 19.5

50.0 1.0 51.0 1.3 53.0 2.9 54.0 2.3 55.0 15.3 56.0 82.0 57.0 38.1 58.0 9.0

60.0 3.6 67.0 8.1 69.0 1.4 71.0 6.7 72.0 16.7 73.0 1.8 81.0 1.2 82.0 12.8


H

2-methylpentanal

C6H12OM+ = 100

72.0 1.0 74.0 1.7 100.0 2.0

15.0 1.0 18.0 1.0 26.0 1.5 27.0 17.6 28.0 2.1 29.0 21.4 30.0 2.5 38.0 1.0

39.0 11.4 40.0 1.9 41.0 27.2 42.0 5.2 43.0 100.0 44.0 3.4 53.0 1.9 55.0 10.3

56.0 1.7 57.0 12.1 58.0 90.1 59.0 3.4 67.0 1.3 69.0 1.3 70.0 1.2 71.0 15.3



C6H12OM+ = 100

O

H

2-ethylbutanal

88.0 1.5 100.0 3.1

26.0 1.2 27.0 14.9 28.0 2.0 29.0 22.2 39.0 10.9 40.0 1.3 41.0 27.5 42.0 4.9

43.0 100.0 44.0 5.5 53.0 2.7 54.0 1.6 55.0 12.6 56.0 2.9 57.0 15.3 58.0 3.8

59.0 1.2 67.0 2.6 69.0 1.7 70.0 2.9 71.0 34.2 72.0 48.1 73.0 3.1 82.0 4.9


C6H12OM+ = 100

O

2-hexanone

85.0 6.4 100.0 8.0

15.0 3.6 18.0 1.9 26.0 1.1 27.0 8.2 28.0 2.0 29.0 14.8 39.0 5.6 41.0 14.1

42.0 3.1 43.0 100.0 44.0 2.4 55.0 1.4 57.0 15.8 58.0 49.8 59.0 3.1 71.0 5.4


C6H12OM+ = 100

O

3-hexanone

101.0 2.0 15.0 2.4 18.0 1.8 26.0 2.9 27.0 27.6 28.0 7.5 29.0 53.0 30.0 1.1 32.0 1.6

38.0 1.0 39.0 7.9 40.0 1.2 41.0 20.3 42.0 3.6 43.0 100.0 44.0 3.4 53.0 1.0

55.0 2.4 56.0 1.8 57.0 84.9 58.0 3.1 71.0 54.0 72.0 6.1 85.0 2.9 100.0 28.6


C6H12OM+ = 100

O

4-methyl-2-pentanone

67.0 1.9 72.0 1.3 85.0 17.7 86.0 1.0 100.0 19.0 101.0 1.4

15.0 5.1 27.0 7.2 28.0 1.7 29.0 11.0 31.0 1.1 39.0 8.3 40.0 1.3 41.0 19.2

42.0 3.1 43.0 100.0 44.0 2.6 55.0 1.4 56.0 1.4 57.0 24.9 58.0 42.6 59.0 3.5


C6H12OM+ = 100

O

3-methyl-2-pentanone

67.0 1.6 71.0 2.4 72.0 50.7 73.0 2.2 85.0 8.3 100.0 17.6 101.0 2.4

15.0 2.9 26.0 1.0 27.0 8.0 28.0 2.2 29.0 33.6 39.0 7.1 41.0 41.8 42.0 4.0

43.0 100.0 44.0 4.4 45.0 2.5 53.0 1.6 55.0 5.1 56.0 23.5 57.0 67.9 58.0 3.3


heptanal

O

H

C7H14OM+ = 114

97.0 2.1 114.0 1.5

15.0 2.0 26.0 2.1 27.0 30.7 28.0 7.4 29.0 40.2 30.0 1.5 31.0 1.9 38.0 1.2

39.0 19.2 40.0 3.3 41.0 66.7 42.0 53.0 43.0 84.0 44.0 100.0 45.0 21.9 51.0 1.5

53.0 4.0 54.0 7.6 55.0 58.6 56.0 9.7 57.0 46.5 58.0 6.4 59.0 1.0 60.0 1.1

67.0 8.8 68.0 17.5 69.0 7.1 70.0 93.7 71.0 23.7 72.0 8.5 74.0 2.0 81.0 19.2

82.0 2.2 83.0 3.2 84.0 1.2 85.0 4.1 86.0 15.4 87.0 1.3 95.0 2.2 96.0 14.4

mass %


2,3-dimethylpentanal

O

H

C7H14OM+ = 114

mass % 59.0 4.3 69.0 1.5 70.0 1.5 74.0 2.5 85.0 9.9 91.0 1.3 97.0 1.5

27.0 9.8 28.0 3.4 29.0 25.0 30.0 2.7 39.0 6.9 40.0 1.2 41.0 31.1 42.0 1.9

43.0 70.9 44.0 2.4 45.0 1.2 53.0 1.9 55.0 7.2 56.0 7.9 57.0 31.8 58.0 100.0




2-heptanone

O

C7H14OM+ = 114

115.0 1.0 15.0 4.2 18.0 1.5 27.0 8.9 28.0 2.0 29.0 8.7 39.0 6.7 40.0 1.0 41.0 11.6

42.0 3.0 43.0 100.0 44.0 2.4 45.0 1.4 53.0 1.0 55.0 5.1 56.0 1.5 57.0 1.6

58.0 90.6 59.0 14.8 71.0 14.0 72.0 3.9 85.0 3.3 99.0 4.1 113.0 1.7 114.0 9.5


3-heptanone

C7H14OM+ = 114

59.0 1.2 71.0 5.6 72.0 32.9 73.0 2.0 85.0 43.3 86.0 2.6 114.0 14.0 115.0 1.1

15.0 1.2 26.0 2.5 27.0 22.8 28.0 8.2 29.0 79.1 30.0 1.8 39.0 8.8 40.0 1.0

41.0 33.7 42.0 2.8 43.0 15.2 53.0 1.5 55.0 3.6 56.0 2.8 57.0 100.0 58.0 4.4


4-heptanone

C7H14OM+ = 114

72.0 3.8 86.0 1.4 99.0 2.0 114.0 14.4 115.0 1.2

15.0 1.5 26.0 1.0 27.0 15.9 28.0 1.1 29.0 2.9 39.0 6.3 40.0 1.0 41.0 17.5

42.0 2.6 43.0 100.0 44.0 3.5 55.0 2.2 57.0 1.4 58.0 6.7 70.0 1.2 71.0 84.7


5-methyl-3-hexanone

C7H14OM+ = 114

115.0 1.5 15.0 1.4 26.0 1.3 27.0 13.3 28.0 2.6 29.0 44.8 30.0 1.0 39.0 8.0 40.0 1.2

41.0 25.7 42.0 3.0 43.0 15.1 53.0 1.0 55.0 1.6 56.0 1.7 57.0 100.0 58.0 5.0

69.0 1.2 71.0 1.5 72.0 14.0 73.0 1.0 85.0 41.3 86.0 2.8 99.0 4.3 114.0 20.4


5-methyl-2-hexanone

C7H14OM+ = 114

59.0 12.4 71.0 9.6 72.0 1.2 81.0 4.8 85.0 1.4 86.0 1.2 99.0 2.3 114.0 4.3

15.0 4.6 18.0 1.0 27.0 9.3 28.0 1.6 29.0 7.1 39.0 7.0 40.0 1.0 41.0 13.3

42.0 2.1 43.0 100.0 44.0 2.4 53.0 1.3 55.0 4.1 56.0 3.8 57.0 14.8 58.0 50.2


4-nonanone

C9H18OM+ = 142

71.0 59.5 72.0 3.4 86.0 15.3 87.0 2.5 99.0 34.3 100.0 3.4 142.0 5.2

15.0 1.1 27.0 16.0 28.0 1.7 29.0 11.5 39.0 7.0 40.0 1.4 41.0 18.7 42.0 4.2

43.0 100.0 44.0 3.8 53.0 1.2 55.0 5.5 57.0 2.5 58.0 34.4 59.0 1.5 70.0 1.5

Spectroscopy Beauchamp 76 Esters


C8H16O2

M+ = 144

ethyl hexanoate

O

O 98.0 1.399.0 51.9

100.0 3.5101.0 25.8102.0 3.6115.0 8.0116.0 1.0117.0 5.0144.0 1.6

15.0 1.926.0 1.827.0 22.328.0 4.029.0 49.830.0 1.431.0 1.539.0 8.7

40.0 1.741.0 22.342.0 10.143.0 61.044.0 2.245.0 17.453.0 1.355.0 12.4

56.0 2.657.0 2.559.0 1.060.0 38.961.0 23.669.0 5.570.0 22.971.0 26.1

72.0 1.573.0 23.374.0 4.183.0 1.387.0 6.888.0 100.089.0 5.897.0 2.0




C8H16O2

M+ = 144

isopropyl 2-methylbutanoate

O

O144.0 0.615.0 1.0

26.0 1.127.0 13.028.0 5.729.0 15.731.0 1.038.0 1.339.0 12.0

40.0 2.741.0 43.442.0 9.643.0 100.044.0 3.345.0 5.053.0 1.055.0 4.2

56.0 12.457.0 92.658.0 4.159.0 10.069.0 1.473.0 2.674.0 19.585.0 66.8

86.0 3.787.0 11.0102.0 10.5103.0 32.0104.0 1.8116.0 21.3117.0 1.4129.0 5.0

Carboxylic acids



144.0 0.1O

OH

2,2-dimethylhexanoic acid

C8H16O2

M+ = 144

18.0 1.527.0 8.928.0 1.029.0 11.931.0 1.139.0 8.440.0 1.441.0 27.3

42.0 3.443.0 25.244.0 1.145.0 2.553.0 2.255.0 11.556.0 4.157.0 100.0

58.0 4.459.0 5.560.0 1.069.0 3.370.0 3.871.0 1.273.0 19.274.0 1.3

83.0 2.887.0 7.388.0 67.989.0 3.199.0 28.5100.0 2.2101.0 6.2115.0 2.0

Alkynes Acid chlorides Nitriles Anhydrides Amides



Amides

27.0 8.7 29.0 8.4 39.0 5.5 41.0 11.6 43.0 25.8 44.0 28.5 59.0 100.0 72.0 18.7 86.0 9.3

hexanamide

NH2

15.0 6.5 27.0 6.4 28.0 5.6 29.0 8.3 30.0 19.0 42.0 6.6 43.0 27.6 44.0 32.9 58.0 100.0 72.0 15.1 100.0 5.9 115.0 33.5

N,N-diethylethanamide

27.0 12.3 28.0 6.9 29.0 40.0 30.0 100.0 41.0 12.6 44.0 30.5 57.0 51.6 58.0 7.2 74.0 11.1 86.0 22.8 87.0 25.0 100.0 14.5 129.0 13.2

N-butylpropanamide

M+ = 115M+ = 115

M+ = 129

27.0 25.1 28.0 10.4 29.0 100.0 30.0 69.7 42.0 5.4 44.0 72.0 46.0 9.1 56.0 8.6 57.0 63.2 72.0 62.1 86.0 9.6 100.0 5.4 101.0 92.6 102.0 5.6

N-ethylpropanamide

M+ = 101

O

N

O

NH

O

NH

O

C

M+ = 111

heptanenitrile 1-heptyne 27.0 20.6 28.0 6.7 29.0 22.5 39.0 17.8 41.0 87.3 42.0 12.0 43.0 60.2 54.0 55.3 55.0 50.2 56.0 8.2 57.0 11.2 68.0 23.1 69.0 14.7 71.0 5.2 82.0 100.0 83.0 59.4 96.0 14.3 110.0 7.8 111.0 0.7

N

27.0 18.4 29.0 45.7 39.0 29.8 40.0 11.7 41.0 70.6 42.0 8.0 51.0 5.9 53.0 17.9 54.0 35.4 55.0 51.0 56.0 26.1 57.0 28.4 65.0 7.1 67.0 44.0 68.0 30.2 79.0 10.6 81.0 100.0 82.0 7.3 95.0 9.4 96.0 1.0

2-heptyne 3-heptyne 27.0 39.9 28.0 6.7 29.0 8.6 39.0 50.8 40.0 7.6 41.0 67.6 42.0 7.2 43.0 25.6 50.0 6.3 51.0 11.5 52.0 8.6 53.0 46.9 54.0 81.8 55.0 22.3 56.0 8.2 65.0 9.9 66.0 5.5 67.0 43.3 68.0 42.5 77.0 5.3 79.0 13.8 81.0 100.0 82.0 7.6 95.0 5.3 96.0 18.0

M+ = 96 M+ = 96 M+ = 96

27.0 23.3 29.0 13.9 39.0 43.1 40.0 11.5 41.0 83.6 42.0 10.5 50.0 6.3 51.0 11.7 52.0 7.3 53.0 48.8 54.0 24.6 55.0 26.5 56.0 5.0 63.0 5.2 65.0 21.3 66.0 11.1 67.0 100.0 68.0 29.2 77.0 9.2 79.0 32.4 81.0 92.6 82.0 6.4 95.0 6.9 96.0 69.6

Spectroscopy Beauchamp 80 Aromatics

M+ = 162

hexylbenzene

27.0 5.2 29.0 5.6 39.0 5.5 41.0 7.6 43.0 16.6 65.0 8.8 78.0 6.4 91.0 100.0 92.0 95.1 93.0 7.7 105.0 11.2 133.0 5.4 162.0 33.2

27.0 6.0 39.0 5.4 41.0 13.1 43.0 33.4 65.0 5.2 66.0 5.2 77.0 7.2 79.0 5.6 91.0 25.7 105.0 30.2 119.0 41.8 133.0 6.5 147.0 100.0 148.0 14.4 162.0 36.4

m-diisopropylbenzene p-diisopropylbenzene

41.0 9.3 43.0 19.2 77.0 5.5 91.0 19.2 105.0 21.8 117.0 7.5 119.0 30.8 131.0 5.0 147.0 100.0 148.0 12.4 162.0 33.1

M+ = 162

M+ = 162

spectroscopy beauchamp 1psbeauchamp/pdf_book/ms_chapter.pdf · value of 100% and all other...

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