Unit V. Mass spectroscopy. [08]

5.1 Introduction

5.2 Principle of mass spectroscopy

5.3Mass spectrometer - schematic diagram

5.4 Types of ions produced in mass spectrum

5.5 Fragmentation patterns of- alkanes, alkenes, aromatic hydrocarbons, alcohols, phenols,

amines and carbonyl compounds

5.6 McLafferty rearrangement

5.7 Applications-

5.1. Introduction :

Mass spectrometry is a powerful analytical technique that is used to identify unknown

compounds, to quantify known compounds, and to elucidate the structure and chemical

properties of molecules. In Mass spectroscopy, the compound in gas phase is separated in the

form of positive ions according to their mass to charge (m/z) ratio. As the sample cannot be

recovered, this is a destructive technique unlike all other spectroscopic techniques such as UV,

IR or NMR.

Mass spectrometry (MS) is an analytical technique for the determination of the elemental

composition of a sample or molecule. It is also used for elucidating the chemical structures of

molecules, such as peptides and other chemical compounds.

According to Physics / General Physics Mass spectrometer is an analytical instrument in

which ions, produced from a sample, are separated by electric or magnetic fields according to

their ratios of mass to charge. A record is produced (mass spectrum) of the types of ion present

and their relative amounts.

Simply, Mass Spectrometer (MS) is a kind of machine which uses an analytical technique

to measure the mass to charge ratio of ions. This analytical technique is also known as Mass

spectrometry. An ion is an atom or group of atoms which have lost one or more electrons,

making them positively charged.

Mass spectrometry has number of applications in various fields. Biotechnology industry

uses mass spectrometry to assay and sequence proteins, oligonucleotides, and polysaccharides.

The pharmaceutical industry uses mass spectrometry in all phases of the drug development

process, from lead compound discovery and structural analysis, to synthetic development and

combinatorial chemistry, and to pharmacokinetics and drug metabolism. In health clinics around

the world, mass spectrometry is used in testing blood and urine for everything from the presence

and levels of certain compounds that are “markers” for disease states, including many cancers, to

detecting the presence and quantitative analysis of illicit or performance-enhancing drugs.

Environmental scientists rely on mass spectrometry to monitor water and air quality, and

geologists use mass spectrometry to test the quality of petroleum reserves.

To date, five Nobel Prizes have been awarded for work directly related to mass

spectrometry: J. J. Thomson (Physics, 1906) for “theoretical and experimental investigations on

the conduction of electricity by gases”; F. W. Aston (Chemistry, 1922) for “discovery, by means

of a mass spectrograph, of isotopes, in a large number of non-radioactive elements”; W. Paul

(Physics, 1989) “for the development of the ion trap technique”; and most recently J. B. Fenn

and K. Tanaka (Chemistry, 2002) “for the development of soft desorption ionization methods for

mass spectrometric analyses of biological macromolecules.”

5.2 Principle of mass spectroscopy

“The basic principle of mass spectrometry (MS) is to generate ions from either inorganic

or organic compounds by any suitable method, to separate these ions by their mass-to-charge

ratio (m/z) and to detect them qualitatively and quantitatively by their respective m/z and

abundance. The analyte may be ionized thermally, by electric fields or by impacting energetic

electrons, ions or photons. The ions can be single ionized atoms, clusters, molecules or their

fragments or associates. Ion separation is effected by static or dynamic electric or magnetic

fields.” Although this definition of mass spectrometry dates back to 1968 when organic mass

spectrometry was in its infancy, it is still valid.

In mass spectroscopy, electrons are emitted from a hot filament usually tungsten, i.e. a

light bulb. These electrons are accelerated towards an electrode so that their energy > 60 eV. The

organic substance is bombarded with this beam of electrons. When the energy of electrons is

increased to attain the critical value of 8-15 eV, This energy is equal to the ionization potential of

the molecule, results in the ionization of the molecule as follows.

M (g) + e- M+(g) +2e-

Where M(g) is the gaseous molecule which on bombardment of electrons, emit an

electron from the lowest energy molecular orbital and gives rise to an ion M+ which is called

parent ion or molecular ion. This is called ionization.

Ionisation: Ionization is the removal of an electron to give an atom with an overall +ve

charge. The amount of energy required to do this (ionization energy) depends on how tightly the

nucleus binds it’s electrons.

If the energy of bombarding electrons is increased to 50-80 eV, the parent ion formed

will have excess energy spread over the molecule as a whole. But within the short time of about

ao-10 to 10-3 secs, the excess energy is localized at certain bonds in the molecule. This results

into the breaking of the bond to produce the pair of fragments of the molecule. The fragments

may be a radical (m2.) plus a cation. ( m1

+) or the fragments can be a small molecule (m2)and a

radical cation (m1+.)

These fragment ions may in turn break decompose further to form more number of ions.

This is called fragmentation pattern. The occurrence of the fragmentation pattern is expressed in

terms of relative abundance. i.e the height of the peak is in accordance with the total number of

that cation. Most formed cation thus gives most intense peak.

Eg. An organic compound A..B when ionizes (gives out one electron and becomes a

cation), produces a molecular ion. (Mc)

M+. or Mc

A..B + e- 2e- + A.+B

M

[100]

However these cations are further fragmented by variety of ways and produce fragment

cations (Fc)and radical cations (Rc)in different number.

A+ + B.

M+. or Mc

A.+B

Fc radical………………………………………………………………….(1)

A. + B+

M+. or Mc

A.+B

radical Fc………………………………………………………………….………(2)

A + B+.

M+. or Mc

A.+B

Rc………………………………………………………………………..(3)

A+. + B

M+. or Mc

A.+B

Rc…………………………………………………………………..(4)

Such fragmentation of almost every bond is possible. The kind of ions produced most

forms an intense peak in the spectrum and hence its abundance is considered as 100 and is

considered as base peak. The abundance of other ions is compared with this intensity and

expressed as relative abundance. However the abundance of every fragment cation and radical

cation varies with the structure of the organic compound.

Eg. Methane molecule can show the following fragmentation pattern. The number in the

square bracket is relative abundance.

CH4

M+ (m/z=16)

e- beam of 70 eV

ionisation CH4

+.+2e-+ e- [100]

methane

CH4+. + [87]

Mc

fragmentationCH3

+

Fc (m/z=15)

H.

radical

CH4+. + [very less]

M+

fragmentationCH3

.

Fc (m/z=1)

H+

CH4+. + [less]

M+

fragmentation

Rc (m/z=2)

CH2 H2+.

+ e-CH3+ + [18.3]CH2

+.fragmentation

Rc (m/z=14)

H.

radical

+ e- + [2.9]C+fragmentation

Fc (m/z=12)

CH+ H.

radical

Naturally carbon occurs as isotopes 12C and 13C. Depending upon their abundance the

molecules with 13C also undergo ionization as follows with different m/z ratio.

M+ (m/z=17)

e- collision

ionisation 13CH4

+.+2e-+ e- [11]

methane

13CH4

Thus bombardment of an organic molecule with high energy electrons produces a

complex mixture of different kind of positive ions such as molecular cation, fragmentation

cation, radical cations along with neutral radicals and few negative ions too. Bigger the organic

molecule, greater is the number of bonds present in it and more complex is the fragmentation

pattern. The abundance of each of the species varies with the structure of the organic molecule

and the energy of the electron beam used for collision.

Each cation had unique m/z ratio. When these cations are passed through a variable

magnetic field, they are deflected to different degree. Lower the mass, easier is the deflection.

Hence the ions get separated and produce an ionic current in proportion to their relative

abundance. This current is recorded in the mass spectrometer and is seen as a peak in mass

spectrum. However, the neutral radicals and other fragmentation molecules cannot be detected in

mass spectrometry.

A mass spectrum is a graph of m/z ratio to the relative abundance of the cations. Eg.

Mass spectrum of methane will appear as shown in Fig. 5.1

Fig. 5.1 : Mass spectrum of methane molecule.

Due to the presence of different isotopes of C, H, N, O and halogens, some additional

peaks are possible. Thus because of natural abundance of 13C and 2H, there are small peaks at

one unit mass higher than the parent peak. These are called (M+1) peak. If both the higher

isotope are present in the same molecule then a peak at two units mass higher than the parent

peak is possible. It is called (M+2) peak. For chlorine and bromine, the (M+2) peaks are more

intense due to more abundance of their isotopes naturally.

5.3Mass spectrometer - schematic diagram

Generally, a typical Mass Spectrometer consists of three parts: an ion source, a mass

analyzer, and a detector. The function of the ion source is to produce ions from the sample.

Finally, the mass spectrum is generated after all the data have been collected.

The ionizer converts a portion of the sample into ions. There is a wide variety of ionization

techniques, depending on the phase (solid, liquid, gas) of the sample and the efficiency of

various ionization mechanisms for the unknown species. An extraction system removes ions

from the sample, which are then targeted through the mass analyzer and onto the detector. The

function of the Mass Analyzer is to separate ions with different mass-to-charge ratios. The

differences in masses of the fragments allow the mass analyzer to sort the ions by their mass-to-

charge ratio. Then the numbers of different ions are detected by the detector. The detector

measures the value of an indicator quantity and thus provides data for calculating the abundances

of each ion present. The output, i.e. mass spectrum, is an intensity vs. m/z (mass-to-charge ratio)

graph, from which the Chemists are able to draw some clues about the ions.

1. Ion source : The ion source is the part of the mass spectrometer that ionizes the material under

analysis (the analyte). The ions are then transported by magnetic or electric fields to the mass

analyzer. Techniques for ionization have been key to determining what types of samples can be

analyzed by mass spectrometry. Electron ionization and chemical ionization are used

for gases and vapours.

The compound may be ionized in several ways:

(i)Electron ionization: bombardment with electrons.

(ii)Chemical ionization: a reagent gas (e.g.,CH4, NH3, isobutane) is ionized by electron impact.

These ions react with the sample molecules to produce ions which are called 'ions of the

molecular species.

(iii)Electrospray ionization :The sample solution is sprayed across a high potential difference (a

few kilovolts) from a needle into an orifice in the interface. Heat and gas flows are used to

desolvate the ions existing in the sample solution. Electrospray ionization often produces

multiply charged ions with the number of charges tending to increase as the molecular weight

increases.

(iv)The atmospheric pressure chemical ionization: contains a heated vaporizer that aids in the

vaporization of the compound.

(v)MALDI (matrix assisted laser desorption ionization): Matrix: The compound is dissolved in a

solution containing an excess of the matrix compound which absorbs at a laser wavelength. The

solution is placed on the laser target. A pulse of UV radiation of this mixture produces a plasma

from the vaporization of both the matrix and the compound.

2. Mass analyzer: There are many types of mass analyzers, using either static or dynamic fields,

and magnetic or electric fields. Each analyzer type has its strengths and weaknesses. Many mass

spectrometers use two or more mass analyzers for tandem mass spectrometry.

There are several important analyser characteristics. The mass resolving power is the

measure of the ability to distinguish two peaks of slightly different m/z. The mass accuracy is the

ratio of the m/z measurement error to the true m/z. Mass accuracy is usually measured

in ppm or milli mass units. The mass range is the range of m/z amenable to analysis by a given

analyzer. The linear dynamic range is the range over which ion signal is linear with analyte

concentration. Speed refers to the time frame of the experiment and ultimately is used to

determine the number of spectra per unit time that can be generated.

3. Detector: The final element of the mass spectrometer is the detector. The detector records

either the charge induced or the current produced when an ion passes by or hits a surface. In a

scanning instrument, the signal produced in the detector during the course of the scan versus

where the instrument is in the scan (at what m/z) will produce a mass spectrum, a record of ions

as a function of m/z.

Typically, some type of electron multiplier is used, though other detectors including

Faraday cups and ion-to-photon detectors are also used. Because the number of ions leaving the

mass analyzer at a particular instant is typically quite small, considerable amplification is often

necessary to get a signal. Microchannel plate detectors are commonly used in modern

commercial instruments.

Schematic diagram of the mass spectrometer is shown in Fig. 5.2.

Fig. 5.2: Mass spectrometer

Working of MS:

Sample preparation: Mass spectrometry can analyze gases, liquids, and solid samples. For

a compound to be analyzed in a mass spectrometer, it must be in the gaseous state. Gases and

liquids are introduced directly into the vacuum chamber through a syringe. Solid samples can be

placed on the tip of a rod, which is then inserted into the vacuum chamber through a vacuum

tight seal. The solid is sublimed or vaporized in the presence of heat. If a sample is impure, it

must be separated into its individual components prior to analysis by mass spectrometry.

Separation is necessary since a mixture will generate many overlapping peaks, and a pure

compound may generate many peaks. Most commonly, the mixture is separated by gas

chromatography. A sample is injected into the gas chromatograph. A chromatogram is generated

that shows the individual components. An individual component is then selected for analysis

with the mass spectrometer. For a compound to be analyzed in a mass spectrometer, it must be in

the gaseous state. Solids and liquids are vaporized with heat. The vapours are then bombarded

with a beam of electrons.

Mass Spectrometry can be combined with gas chromatography to analyze mixtures of

compounds. GC separates the components of the mixture. Each component is analyzed by the

Mass Spectrometer. In modern MS, the integrated equipment is available. Though it isn’t part

of the mass spectrometer, but liquid chromatograph or gas chromatograph (i.e. LC, GC) are often

directly coupled to the ion source and sold with the instrument. Hence, the name LCMS or

GCMS.

The positive ions are accelerated through the accelerating plates. The magnetic field

separates the ions by their mass-to-charge ratio (m/z). By varying the magnetic field, the

abundance of each mass is detected. A detector records the mass-to-charge ratio and this

information is given as a mass spectrum.

The overall working of mass spectrometer is given below.

1. A little amount of a compound, typically one micromole or less is evaporated. The

vapours are passed into the ionization chamber where a pressure is maintained of about 10-7

mbar.

2. The vapour molecules are now ionized by an electron-beam. A heated cathode, the

filament, produces this beam. Ionization is achieved by loss of valence electrons, mainly positive

ions are produced.

3. The positive ions are forced out of the ionization chamber by a small positive charge

(several Volts) applied to the repeller opposing the exit-slit (A). After the ions have left the

ionization chamber, they are accelerated by an electrostatic field (A>B) of several hundreds to

thousands of volts before they enter the analyzer.

4. The separation of ions takes place in the analyzer at a pressure of about 10-8 mbar.

This is achieved by applying a strong magnetic field perpendicular to the motional direction of

the ions. The fast moving ions then will follow a circular trajectory, due to the Lorenz

acceleration, whose radius is determined by the mass/charge ratio of the ion and the strength of

the magnetic field. Ions with different mass/charge ratios are forced through the exit-slit by

variation of the accelerating voltage (A>B) or by changing the magnetic-field force.

5. After the ions have passed the exit-slit, they collide on a collector-electrode. The

resulting current is amplified and registered as a function of the magnetic-field force or the

accelerating voltage.

5.4 Types of ions produced in mass spectrum

When the sample organic compound is bombarded with electrons, there occurs ionisation

of the molecule, causing formation of molecular ion, which further gives rise to different ions.

These types of ions in mass spectrometry are as follows:

1. Molecular ion: ion derived from the neutral molecule by loss or gain of an electron. M is the

molecular ion (un-fragmented molecule minus one electron) made up of “isotopes with the

lowest mass numbers (1H, 12C, etc)”

The molecular ion has the same empirical formula as the corresponding neutral molecule.

The neutral and its molecular ion only differ by one (or more) electron(s). A singly charged

molecular ion can be a positive radical ion, M+•. This is also called odd electron ion. The mass

of this ion corresponds to the sum of the masses of the most abundant isotopes of the various

atoms that make up the molecule. The symbolism M+• does not mean one added electron. The

radical symbol is added to the molecular ion only to indicate a remaining unpaired electron after

ionization. Eg. Methane molecule is an even electron species which gives out one electron and

forms an odd electron molecular ion CH4+•

CH4

M+ (m/z=16)

e- beam of 70 eV

ionisation CH4

+.+2e-+ e-

methane

The peak for molecular ion will most likely the highest peak with the largest m/z. When

z=1 (charge is usually +1), M provides the mass (in amu) of the compound with the lowest mass

isotopes. Eg. We get the M peak for methane at m/z =16.

Similarly, we expect the M peak of CH3CH2CH2Br should be at

(3)(12)+(7)(1)+(1)(79)=122. This is verified by the mass spectrum above. Where 12,1 and 79

are the isotopic masses of C,H and Br respectively and not the atomic weight, of each atom.

2. Ion of the molecular species: In chemical ionisation method, the ions are formed by a

chemical reaction. If an ion originates from the molecule by the abstraction of a proton [M−H]−

or by a hydride abstraction [M−H]+, or by interaction with a proton or a cation to form a

protonated molecule [M+H]+ or a cationized molecule [M+Cat]+. These ions allow the molecular

weight to be deduced. [The ambiguous and obsolete terms ‘quasimolecular ion’ or

‘pseudomolecular ion’ should be avoided. ]

3. Precursor ion: ion that decomposes or changes its charge, yielding a product ion. Also named

parent ion. Eg. Molecular ion M+• [CH4+• in the following reaction] is the parent ion or

precurser ion if it decomposes further to give rise to the product ions.

CH4+. + [87]

fragmentationCH3

+

Fc (m/z=15)

H.

radicalprecurser ion/ parent ion

4. Fragment ion: ion derived from the decomposition of a precursor ion. When the precursor ion

is an odd electron ion like molecular ion M+•, it can give rise to fragment ion which can be an

even electron fragment ion and a radical.

Eg. CH3+ in the following reaction.

CH4+. + [87]

Mc

fragmentationCH3

+

Fc (m/z=15)

H.

radical

5. Metastable ion: Some of the fragment ions or molecular ions are very short lived. These ions

are fragmented spontaneously even in the absence of the magnetic field.

Ionization happens on the low femtosecond timescale, direct bond cleavages require

between some picoseconds to several tens of nanoseconds, and rearrangement fragmentations

usually proceed in much less than a microsecond. Finally, some fragment ions may even be

formed after the excited species has left the ion source giving rise to metastable ion dissociation.

The ion residence time within an electron ionization ion source is about 1 μs.

[Non-decomposing molecular ions and molecular ions decomposing at rates below about 105 s–1

will reach the detector without fragmentation and are therefore termed stable ions.

Consequently, ions dissociating at rates above 106 s–1 cannot reach the detector. Instead, their

fragments will be detected and thus they are called unstable ions. A small percentage, however,

decomposing at rates of 105–106 s–1 will just fragment on transit through the mass analyzer;

those are termed metastable ions.]

6. Adduct ion: Ion formed by direct combination of two separate molecular entities, usually an

ion and a neutral molecule, in such a way that there is change in connectivity promoted by

intermolecular binding, but no loss of atoms within the precursor entities. In Matrix-Assisted

Laser Desorption/Ionization, Solutions containing metal salts, e.g., from buffers or excess of

noncomplexed metals, may cause a confusingly large number of signals due to multiple

proton/metal exchange and adduct ion formation. i.e., [M+H]+, [M+Na]+ and [M+K]+ showing

peaks at (M+1), (M+23) and (M+39) are observed respectively. This makes the recognition of

those peaks straightforward and effectively assists the assignment of the molecular weight.

7. Radical ion :(synonymous with odd-electron ion): When a neutral molecule is hit by an

energetic electron with electronvolts (eV) of kinetic energy, some of the energy of the electron is

transferred to the neutral molecule. If the electron, in terms of energy transfer, collides very

effectively with the neutral, the amount of energy transferred can effect ionization by ejection of

one electron out of the neutral, thus making it a positive radical ion:

If it was an even-electron molecule, the molecular ion is formed which is called a radical cation

or an odd-electron (open-shell) ion, e.g., for methane we obtain ion containing an unpaired

electron.

CH4

M+ (m/z=16)

e- beam of 70 eV

ionisation CH4

+.+2e-+ e-

methane

Similarly, the odd electron molecular ion can also give radical ion and a neutral molecule.

Eg. CH2+• is a radical ion or odd electron ion is the following reaction.

In the rare case the neutral precursor is a radical, the ion created by electron ionization would be

even-electron, e.g., for nitric oxide:

8. Cluster ion: Ion formed by combination with two or more molecules either of the same

species or of different species. These are the ions formed by a multi-component atomic or

molecular assembly of one or more ions with atoms, ions or molecules. Cluster refers to an

associate of more atoms, molecules or ions of the same species, sometimes associated to one

other species. Such ions are not common in organic chemistry. e.g., Cesium iodide on fast atom

bombardment (FAB) mass spectrometry yields cluster ions of the general formula [Cs(CsI)n]+

9. Distonic ion: These are radical ions in which the charge and radical sites are formally located

on different atoms in the molecule.

A distonic ion is a positive radical ion, which would formally arise by ionization of a

zwitterion or a diradical, by isomerization or fragmentation of a classical molecular ion, or by

ion-molecule reactions. Consequently, distonic ions have charge and radical at separate atoms.

10. Isotopic ion: Most elements occur naturally as a mixture of isotopes. Any ion containing one

or more of the less abundant, naturally occurring isotopes of the elements that make up its

structure is an isotopic ion. Eg, 13CH4+• in the following reaction is an isotopic ion.

M+ (m/z=17)

e- collision

ionisation 13CH4

+.+2e-+ e- [11]

methane

13CH4

The presence of significant amounts of heavier isotopes leads to small peaks that

have masses that are higher than the parent ion peak.

• M+1 = a peak that is one mass unit higher than M+

For 13C, 15N, and 33S the isotopic ion peaks are observed significantly as the M+1 peak.

M+1 corresponds to the molecular ion with mass one higher than M due to presence of “one

atom that is a heavier isotope”

Eg. For 1-bromopropane, the M peak is at 122, so the M+1 peak should be at

122+1=123.

• M+2 = a peak that is two mass units higher than M+

For 34S, 37Cl, and 81Br the isotopic ion peaks are observed significantly as the M+2 peak.

M+2 corresponds to the molecular ion with mass two units higher than M due to presence of

“one atom that is a heavier isotope by 2”

Eg. For 1-bromopropane, the M peak is at 122, so the M+2 peak should be at 122+2=124

If intensity of Molecular ion M is considered to be 100%, then the intensity of M+2 peak

is at the proportion in accordance with the natural abundance of the isotope.

Eg. For molecule with Br , the intensity of M+2 peak is almost equal to that of M+ peak.

(50.5% 79Br & 49.5% 81Br, [Eg. Mass spectrum of bromomethane CH3Br. (Fig. 5.3)

Fig.5.3: Mass spectrum (MS)of bromomethane CH3Br

Similarly, for molecule with Cl, the intensity of the M+2 peak is 1/3rd of the M+ peak. Eg. Mass

spectrum of 2-chloro-2-methyl propane (Fig. 5.4)

Fig. 5.4: Mass spectrum of 2-chloro-2-methyl propane

For molecule with S, the intensity of M+2 peak is 4% to that of M+ peak.

11. Multiply charged ion: If the electron in terms of energy transfer collides effectively with the

neutral, the energy transferred can exceed the ionization energy (IE) of the neutral. Depending on

the analyte and on the energy of the primary electrons, doubly charged and even triply charged

ions may be observed.

Eg. when the ionisation is brought about by electrospray ionization, such ions containing several

positive charges can be observed. As for these ions mass is same but z=1,2,3 etc, they differ in

the m/z ration and appear at different position in the spectrum. These ions will be detected at

lower m/z than the corresponding singly-charged ion of the same mass. Isotopic distributions

remain unaffected as far as the relative intensities are concerned. As the distance between

isotopic peaks is reduced to 1/z u, the charge state can readily be assigned.

12. Isobaric ion: These are the ions of the same nominal mass. Eg. the molecular ions of

nitrogen, N2+•, carbon monoxide, CO+•, and ethene, C2H4+•, have the same nominal mass of

28 u, i.e., they are so-called isobaric ions.

5.5 Fragmentation patterns of- alkanes, alkenes, aromatic hydrocarbons, alcohols, phenols,

amines and carbonyl compounds:

In mass spectrometry, fragmentation is the dissociation of energetically unstable

molecular ions formed from passing the molecules in the ionization chamber of a mass

spectrometer. The fragments of a molecule cause a pattern in the mass spectrum used to

determine structural information of the molecule.

A molecule hit with the electron either becomes a molecular ion or breaks apart into

fragments. The impact of the stream of high energy electrons often breaks the molecule into

fragments, commonly a cation and a radical. Some of these fragmentation patterns are fairly

simple. Others involve additional reactions or rearrangements.

Fragmentation in Mass Spectrum helps in deducing the correct structure of the molecule.

However with certain general rules the fragmentation occurs differently in different compounds.

We are going to discuss the fragmentation patterns of- alkanes, alkenes, aromatic hydrocarbons,

alcohols, phenols, amines and carbonyl compounds,

General rules:

1. Lone pair electrons from a heteroatom are more easily displaced than bonding electrons.

2. Electrons in pi-bonds are more easily displaced than those in sigma bonds.

3. Cleavage is formed at branch sites that lead to more substituted carbocations and radicals.

4. Bonds break to give the most stable cation. Stability of the radical is less important.

5.The bond breaking can be done either by homolytic or heterolytic cleavage.

Fragmentation patterns of Alkanes:

1. The relative height of the M+ peak is greatest for straight-chain molecules and decreases as

the branching increases.

2. The relative height of the M+ peak decreases with chain length for a homologous series.

3. Cleavage is favoured at alkyl-substituted carbons, with the probability of cleavage increasing

as the substitution increases.

4. The fragmentation pattern contains clusters of peaks 14 mass units apart (which represent loss

of (CH2)nCH3). Fragmentation often splits off simple alkyl groups results in the characteristic

peaks at ( M+ - 15) due to loss of methyl group, at ( M+ - 29) due to loss of ethyl group, at ( M+

- 43) due to loss of propyl group, at ( M+ - 57) due to loss of butyl group.

Eg. Hexane (C6H14) with MW = 86.18 shows peaks at m/z = 86, 71, 57, 43, 29 and 15.

The spectrum and the fragmentation can be explained as follows (Fig. 5.5)

Fig.5.5: Fragmentation of Hexane (C6H14)and mass spectrum

5. Branched alkanes tend to fragment forming the most stable carbocations. In branched alkanes,

the carbocation stability show the following trend:

Benzylic > Allylic > Tertiary > Secondary >> Primary

“Stevenson’s Rule” Most Stable carbocation is formed favourably over the least stable one.

Hence in mass spectrum of isohexane (2-methyl pentane), the most prominent and

intense peak is observed at m/z=43 due to most stable tertiary carbocation. (Fig. 5.6)

Fig. 5.6: Mass spectrum of 2-methyl pentane and fragmentation

Similarly in mass spectrum of isooctane (2, 2, 4-trimethylpentane), since tertiary

carbocations are the most stable of the saturated alkyl carbocations, this cleavage is particularly

favorable and accounts for the intense fragment peak at m/z = 57. (Fig. 5.7)

Fig. 5.7: Mass spectrum of 2, 2, 4-trimethylpentane (isooctane).

Fragmentation patterns of Alkenes:

The mass spectra of most alkenes show distinct molecular ion peaks. Naturally, the mass

of the molecular ion should correspond to a molecular formula with an index of hydrogen

deficiency equal to at least one. Apparently, electron bombardment removes one of the electrons

in the π bond, leaving the carbon skeleton relatively undisturbed.

1. When alkenes undergo fragmentation processes, the resulting fragment ions have formulas

corresponding to [CnH2n]+ and [CnH2n−1]+.

2. Usually there is cleavage called α-cleavage, as the bond α to the double bond is cleaved to

form the allylic carbocation which is more stable than the other carbocations possibly formed.

Hence, fragmentation typically forms resonance stabilized allylic carbocations.

Eg. In 2-hexene, the allylic carbocation is formed by following reaction.

Hence a base peak is observed at m/z=55. (Fig. 5.8)

Fig. 5.8: Mass spectrum of 2-hexene

3. It is sometimes difficult to locate double bonds in alkenes since they migrate readily. The

position of the double bond result in different spectra.

• 1-alkene (terminal alkene): The allyl carbocation (m/z = 41) is an important fragment in

the mass spectra of terminal alkenes and forms via an allylic α-cleavage.

Eg. Mass spectrum of 1-pentene : Besides the peak at 41 due to allylic carbocation, the fragment

at m/z = 55 is from loss of a methyl radical. This fragment is the base peak in the spectra of the

cis and trans 2- pentene isomers since loss of the methyl group distal to the alkene creates an

allylic cation that is resonance stabilized. A large fragment at m/z = 42 in the spectrum of 1-

pentene. This ion is formed via loss of ethylene through a McLafferty-type rearrangement of the

molecular ion. (Fig. 5.9)

Fig. 5.9: Mass spectrum of 1-pentene.

• 2-alkene: The allyl carbocation (m/z = 55) is an important fragment in the mass spectra of

terminal alkenes and forms via an allylic α-cleavage.

Eg. Mass spectrum of 2-pentene : Besides the base peak at 55 due to allylic carbocation, the

fragment at m/z = 41 and 42 is from loss of a methyl radical. This fragment is the base peak in

the spectra of the cis and trans 2- pentene isomers since loss of the methyl group distal to the

alkene creates an allylic cation that is resonance stabilized. A large fragment at m/z = 42 in the

spectrum of 2-pentene. This ion is formed via loss of ethylene through a McLafferty-type

rearrangement of the molecular ion. (Fig. 5.10)

Fig. 5.10 Mass spectrum of Z-2-pentene

Fragmentation patterns of Aromatic hydrocarbons:

The mass spectra of most aromatic hydrocarbons show very intense molecular ion

peaks. Aromatics may also have a peak at m/z = 77 for the benzene ring. As is evident from the

mass spectrum of benzene, fragmentation of the benzene ring requires a great deal of energy.

Fig. MS of benzene (Fig. 5.11)

Fig. 5.11:Mass spectrum of benzene

Such fragmentation is not observed to any significant extent. In the mass spectrum of

toluene, loss of a hydrogen atom from the molecular ion gives a strong peak at m/z = 91.

Although it might be expected that this fragment ion peak is due to the benzyl carbocation

(C6H5CH2+), isotope-labeling experiments suggest that the benzyl carbocation actually

rearranges to form the aromatic delocalized tropylium ion (C7H7+, Fig. 5.12). When a benzene

ring contains larger side chains, a favored mode of fragmentation is cleavage of the side chain to

form initially a benzyl cation, which spontaneously rearranges to the tropylium ion. When the

side chain attached to a benzene ring contains three or more carbons, ions formed by a

McLafferty rearrangement can be observed.

Fig. 5.12:Mass spectrum of toulene

• Molecular ion peaks are strong due to the stable structure. Eg. Mass spectrum of

Naphthalene (C10H8) with MW = 128.17 (Fig. 5.13)

Fig. 5.13:Mass spectrum of Naphthalene (C10H8)

The mass spectra of the xylene (Fig. 5.14) isomers show a medium peak at m/z = 105,

which is due to the loss of a hydrogen atom and the formation of the methyl tropylium ion. More

importantly, xylene loses one methyl group to form the tropylium (m/z = 91).

Fig. 5.14: Mass spectrum and fragmentation of ortho-xylene.

The mass spectra of ortho-, meta-, and para-disubstituted aromatic rings are essentially

identical. (Fig. 5.14 and 5.15) As a result, the substitution pattern of polyalkylated benzenes

cannot be determined by mass spectrometry. The formation of a substituted tropylium ion is

typical for alkyl-substituted benzenes.

Fig. 5.15: Mass spectrum of meta-xylene

In the mass spectrum of isopropylbenzene, (Fig. 5.16) a strong peak appears at m/z =

105. This peak corresponds to loss of a methyl group to form a methyl-substituted tropylium ion.

Fig. 5.16: Mass spectrum of isopropylbenzene (cumene).

The tropylium ion has characteristic fragmentations of its own. The tropylium ion can

fragment to form the aromatic cyclopentadienyl cation (m/z = 65) plus ethyne (acetylene). The

cyclopentadienyl cation in turn can fragment to form another equivalent of ethyne and the

aromatic cyclopropenyl cation (m/z = 39) (Fig. 5.17). Formation and fragmentation of the

tropylium ion further gives characteristic peaks at for tropylium ion (91), methyl tropylium

ion(105), cyclopentadienyl cation (65), methyl cyclopentadienyl cation(79), cyclopropenyl

cation (39), methyl cyclopropenyl cation (53) respectively as given in Fig. 5.17

Fig. 5.17: Formation and fragmentation of the tropylium ion.

In the mass spectrum of butylbenzene (Fig. 5.18), a strong peak due to the tropylium ion

appears at m/z = 91. When the alkyl group attached to the benzene ring is a propyl group or

larger, a McLafferty rearrangement is likely to occur, producing a peak at m/z = 92. Indeed, all

alkyl benzenes bearing a side chain of three or more carbons and at least one hydrogen on the γ-

carbon will exhibit a peak at m/z = 92 in their mass spectra from the McLafferty rearrangement.

Fig. 5.18: Mass spectrum of butylbenzene.

Other aromatic benzyl compounds also fragment at the benzylic carbon, forming a

resonance stabilized benzylic carbocation (which rearranges to the tropylium ion at m/z=91)

(Fig. 5.19)

Fig. 5.19: Mass spectrum of benzyl bromide

• Other aromatic compounds also fragment at the benzene carbon, showing a peak at

m/z = 77 for the benzene ring forming a resonance stabilized benzene carbocation

(Fig. 5.20)

Fig. 5.20: Mass spectrum of nitrobenzene

Fragmentation pattern in alcohols:

• The intensity of the molecular ion peak in the mass spectrum of a primary or

secondary alcohol is usually rather low, and the molecular ion peak is often entirely

absent in the mass spectrum of a tertiary alcohol.

• The most important fragmentation reaction for alcohols is the loss of an alkyl group

via α-cleavage. Common fragmentations of alcohols are α-cleavage adjacent to the

hydroxyl group and dehydration. The mass spectrum of straight-chain pentanol

isomers, 1-pentanol (Fig. 5.22), 2-pentanol (Fig. 5.23), and 3-pentanol (Fig. 5.24) all

exhibit very weak molecular ion peaks at m/z = 88

• A second common mode of fragmentation involves dehydration. The importance of

dehydration increases as the chain length of the alcohol increases., it is quite weak in

the other pentanol isomers.

• May lose hydroxyl radical or water giving peak at M+ - 17 or M+ - 18

• Commonly lose an alkyl group attached to the carbinol carbon which is forming an

oxonium ion.

• Primary alcohols: Usually has prominent peak at m/z = 31 corresponding to

H2C=OH+ Eg. 1-butanol (Fig. 5.21)

Fig. 5.21: Mass spectrum of 1-butanol

• 1-pentanol:

• In the spectrum of 1-pentanol (Fig. 5.22), the peak at m/z = 31 is due to the loss of a butyl

group to form an H2C=OH+ ion.

• While the fragment ion peak resulting from dehydration (m/z = 70) is very intense in the

mass spectrum of 1-pentanol

• Alcohols containing four or more carbons may undergo the simultaneous loss of both

water and ethylene. This type of fragmentation is not prominent for 1-butanol but is

responsible for the base peak at m/z = 42 in the mass spectrum of 1-pentanol

Fig. 5.22: Mass spectrum of 1-pentanol.

• Secondary alcohol: Fragmentation depends on the position of the hydroxyl group and

hence the alkyl radicals removed.

• 2-Pentanol :

• 2-Pentanol (Fig. 5.23) loses either a propyl group to form the CH3CH=OH+ fragment at

m/z = 45 or a methyl radical to form the relatively small peak at m/z = 73 corresponding

to CH3CH2CH2CH=OH+.

Fig. 5.23: Mass spectrum of 2-pentanol.

• 3-Pentanol :

• 3-Pentanol loses an ethyl radical to form the CH3CH2CH=OH+ ion at m/z =59. The

symmetry of 3-pentanol means there are two identical α-cleavage paths, making the peak

corresponding to that ion even more prevalent.

Fig. 5.24: Mass spectrum of 3-pentanol.

Tertiary alcohol: While the molecular ion in the mass spectrum of the tertiary alcohol

2-methyl-2-butanol (Fig. 5.25) is entirely absent. As discussed earlier, the largest alkyl group is

most readily lost. 2-Methyl-2-butanol (Fig. 5.25) undergoes α-cleavage to lose a methyl radical

two different ways, creating a considerable size peak at m/z = 73 in addition to the peak at m/z =

59 corresponding to the (CH3)2C=OH+ ion formed by loss of an ethyl radical.

Fig. 5.24: Mass spectrum of 2-methyl-2-butanol

Fragmentation pattern in phenols:

The mass spectra of phenols usually show strong molecular ion peaks. In fact, the

molecular ion at m/z = 94 is the base peak in the MS of phenol (Fig. 5.25). Favored modes of

fragmentation involve loss of a hydrogen atom to create an M – 1 peak (a small peak at m/z =

93), loss of carbon monoxide (CO) to produce a peak at M – 28 (m/z = 66), and loss of a formyl

radical (HCO•) to give a peak at M – 29. In the case of phenol itself, this creates the aromatic

cyclopentadienyl cation at m/z = 65. In some cases, the loss of 29 mass units may be sequential:

initial loss of carbon monoxide followed by loss of a hydrogen atom.

Fig. 5.24: Mass spectrum of phenol

The mass spectrum of ortho-cresol (2-methylphenol) exhibits a much larger peak at M–

1 (Fig. 5.25) than does unsubstituted phenol. Note also the peaks at m/z = 80 and m/z = 79 in the

o-cresol spectrum from loss of CO and formyl radical, respectively.

Fig. 5.25: Mass spectrum of o-cresol

Fragmentation pattern in amines:

When nitrogen is present in any organic compound, it has odd molecular mass and

hence its M+ peak appears at odd number. This is applicable for the compounds having odd

number of nitrogens.

In amines, there occurs α-cleavage to form an iminium ion. The intensity of the

molecular ion of aliphatic amines decreases regularly with increasing molecular weight.

Primary amines: In the mass spectra of primary amines the methylene immonium ion,

CH2=NH+, m/z 30, resulting from α--cleavage either represents the base peak or at least is the by

far most abundant of the immonium ion series.

NHCH2NH2

+

m/z=30

+

m/z=16+14n C(n-1) H(n-1)2

Secondary amines: In the mass spectra of secondary amines the bonds of the alpha carbon break

in two ways to produce two peaks by elimination of two alkyl radicals. The larger radical is

removed preferably hence forms a base peak. Eg. In Mass spectrum of N-propyl -1-butanamine

(Fig. 5.26) the bonds of the alpha carbon break in two ways to produce two peaks at 72 and 86

by elimination of propyl and ethyl radical respectively. The larger radical is removed preferably

hence the peak at 72 is a base peak.

Fig. 5.26: Mass spectrum of N-propyl -1-butanamine

Diethyl amine undergoes C-N bond breaking to give ethyl amine cation on m/z=44 as given in

Fig. 5.27

Fig. 5.27 Mass spectrum of Diethyl amine

Tertiary amines: In the mass spectra of secondary amines the bonds of the alpha carbon break

in two ways to produce three peaks by elimination of three alkyl radicals. The larger radical is

removed preferably hence forms a base peak. Eg. The mass spectrum of N-ethyl-N-methyl-

propanamine, C6H15N, shows the molecular ion peak at m/z 101 (Fig. 5.28). The primary

fragmentations of the molecular ion may well be explained in terms of the α-cleavage and

accordingly, the peaks at m/z 72 and 86 can be assigned as immonium fragment ions due to ethyl

and methyl loss, respectively. Methylene imminium ion, CH2=NH+, m/z 30, resulting from α--

cleavage is also seen in the spectrum.

Fig. 5.28Mass spectrum of N-ethyl-N-methyl-propanamine

Fragmentation pattern in carbonyl compounds:

Carbonyl compounds are aldehydes and ketones. They both contain oxygen with two lone

pairs of electrons which produce oxonium radical ion on bombardment of electrons to give rise

to molecular ions.

The fragmentation involves formation of acylium ion RC≡O+ on further removal of

hydrogen radical and an alkyl cation with mass(M+-29) by removal of carbon monoxide from

the acylium ion.

R

O

R H

O

R H

O

R+

....

..+.

..

M+ M+-1

+

M+-29aldehyde

Example : in mass spectrum of hydro cinnamaldehyde, one can see M+ peak at 134 and peaks at

133(M+-1), 105 (M+-29) and 91(M+-29-14) due to fragments removed as H, CHO & CH2CHO

(Fig. 5.29)

Fig. 5.29: Mass spectrum of hydro cinnamaldehyde C6H5CH2CH2CHO

While in ketones, the fragmentation involves formation of acylium ion RC≡O+ on removal of

alkyl radical. The peaks can give information about the chain size of the two alkyl radicals.

R

O+

R R'

O

R R'

O

R'

O+

....

..+.

M+ M+-R'

+

M+-Rketone

+

Eg. In the mass spectrum of 2-pentanone, molecular peak is observed at 86 and the two other

prominent peaks at 71 and 43 are due to removal of propyl and methyl radicals respectively.

(Fig. 5.30)

Fig. 5.30: Mass spectrum of 2-pentanone

5.7 McLafferty rearrangement :

Some fragment ions cannot be explained by the simple cleavage of bonds; they result

from intramolecular rearrangements, such as migration of H• to (or abstraction of H• by) a

heteroatom. Organic compounds having the γ carbon containing a hydrogen undergo the γ-H

shift with β-cleavage, commonly known as McLafferty rearrangement reaction. It is the most

prominent rearrangement reaction. McLafferty rearrangement was first described by Fred McLafferty in 1956 and is one of

the most predictable fragmentations, next to the simple α-cleavage. In the McLafferty

rearrangement, a hydrogen atom on a carbon 3 atoms away from the radical cation of an alkene,

arene, carbonyl, or imine (a so-called γ-hydrogen) is transferred to the charge site via a six-

membered transition state, with concurrent cleavage of the sigma bond between the α and β

positions of the tether. This forms a new radical cation and an alkene with a π bond between

what were the original β and γ carbons. For simplicity, the mechanism of the McLafferty

rearrangement is usually drawn as a concerted process, as in Fig. 5.31. There is experimental

evidence, however, that the fragmentation is in fact stepwise, and as a general rule

fragmentations that involve breaking more than one bond are probably stepwise.

Fig. 5.31 McLafferty rearrangement

Requirements for the McLafferty rearrangement: i) the atoms Z and Y can be

carbons or heteroatoms, ii) Z must be connected by a double bond, iii) at least one γ-hydrogen is

available that iv) is selectively transferred to B via a six-membered transition state, v) causing

alkene loss upon cleavage of the β-bond.

The distance between the γ-hydrogen and the double-bonded atom must be less than 1.8

× 10–10 m [87,88] and the Cγ–H bond must be in plane with the acceptor group.

The McLafferty rearrangements occurs in molecular ions of various substrates as follows.

1-alkene (terminal alkene): 1-Alkenes with γ carbon containing a hydrogen undergo

McLafferty rearrangement. Eg. Mass spectrum of 1-pentene: Besides the peak at 41 due to allylic

carbocation, a large fragment at m/z = 42 in the spectrum of 1-pentene is formed via loss of

ethylene through a McLafferty-type rearrangement of the molecular ion. (Fig. 5.9)

HH H+.

+.

+

+.

m/z=42

Aldehydes: The peak at m/z 44 in the mass spectrum of butanal can be explained by C2H4 loss

from the molecular ion.

Aromatic compounds: When the side chain attached to a benzene ring contains three or more

carbons, ions formed by a McLafferty rearrangement can be observed. Eg. In the mass spectrum

of butylbenzene (Fig. 5.18), a McLafferty rearrangement is likely to occur, producing a peak at

m/z = 92. Indeed, all alkyl benzenes bearing a side chain of three or more carbons and at least

one hydrogen on the γ-carbon will exhibit a peak at m/z = 92 in their mass spectra from the

McLafferty rearrangement.

Esters: Esters of higher alkanoic acids with γ-carbon with hydrogen atom undergo McLafferty

rearrangement to give peak at m/z=74 for the fragment cation produced as follows.

Carboxylic acids: The mass spectra of carboxylic acids and their derivatives are governed by

both α-cleavage and McLafferty rearrangement. McLafferty rearrangement can only occur from

butanoic acid and its derivatives onwards. Analogous to aliphatic aldehydes, the same fragment

ions are obtained for a homologous series of carboxylic acids, provided they are not branched at

the α-carbon. Thus, highly characteristic fragment ions make their recognition straightforward.

In all aliphatic carboxylic acids, the prominent peak at m/z=60 is due to McLafferty

rearrangement.

Table 5.1 gives information about the characteristic peaks due to McLafferty rearrangement.

These peaks when appear in the MS of the sample, help to identify the molecule.

Table 5.1: Frequent product ions of the McLafferty rearrangement.

5.8 Applications:

Mass spectrometry has both qualitative and quantitative uses. These include

identifying unknown compounds, determining the isotopic composition of elements in a

molecule, and determining the structure of a compound by observing its fragmentation. MS is

now in very common use in analytical laboratories that study physical, chemical, or biological

properties of a great variety of compounds. Some of the applications are given below.

1. To know the most abundant ion:

The Base Peak is the peak with the greatest intensity (usually set to 100% relative

abundance) in the mass spectrum, corresponding to the most abundant ion. (M and the base peak

are only the same if the molecular ions pass the detector without breaking into fragments).

2. To get the Molecular mass of the sample molecule

M is the molecular ion (un-fragmented molecule minus one electron) made up of

“isotopes with the lowest mass numbers (1H, 12C, etc)” This peak will most likely be indicated

on an exam, but is usually the peak with the largest m/z. When we assume that z=1 (charge is

usually +1), M provides the mass (in amu) of the compound with the lowest mass isotopes. To

find the m/z for the M peak of a compound, multiply the isotopic mass (not the atomic weight!)

of each atom by the number of atoms of each in the formula.

EX: We expect the M peak of CH3CH2CH2Br should be at (3)(12)+(7)(1)+(1)(79)=122.

This is verified by the mass spectrum. Note that we use the isotopic mass, not the atomic weight,

of each atom.

Table 5.2 gives some frequently found atoms and their isotopic masses, relative

abundance and atomic weights.

Table 5.3: isotopes and their relative abundance

Element Isotopic mass

Relative abundance

Exact Isotopic mass

Average Atomic mass

C 12 13

100 1.1

12.000000 13.003355

12.0108

H 1 2

100 0.0115

1.007825 2.014101

1.00795

Cl 35 37

100 31.96

34.968853 36.965903

35.4528

Br 79 81

100 97.28

78.918338 80.916291

79.904

S 32 33 34 36

100 0.80 4.52 0.02

31.972071 32.971459 33.967867 35.967081

32.067

3. To find the number of carbon atoms in the molecule.

Since M+1 and M+2 are always less intense than M, they can never be the base peak.M+1 is

next-door to M on the mass spectrum, and corresponds to the molecular ion with mass one

higher than M due to presence of “one atom that is a heavier isotope”

EX: For 1-bromopropane, the M peak is at 122, so the M+1 peak should be at 122+1=123

Only 13C, 15N, and 33S contribute significantly to the M+1 peak, 13C is most important

c) If intensity of Molecular ion M is considered to be 100%, then the intensity of M+1

peak is used to calculate the number of the carbon atoms present in the molecule.

f) If M is scaled to 100%, then the M+1 intensity helps to determine the number of

carbons in the compound by formula given below:

Note: One has to be careful to avoid math errors here! Note about rounding:

If remainder is 0.4 or less, then round down

If remainder is 0.7 or more, then round up

if remainder is between 0.4-0.7, then consider two different carbon counts for formula candidates. Some of these may be rejected

later based on other data.

Problem: If the relative abundance of the M+1 peak (relative to M) is 4.95%, how many carbons

are there in the formula?

ANSWER: 4.95/1.1 = 4.5 There are 4 or 5 carbons!

4. To determine the presence of an isotopic element as Cl/Br/S etc.

M+2 is a peak that is two mass units higher than M+. It is next-door to M+1 on the mass

spectrum, and has a mass two higher than M. The elements that make significant contributions to

this peak are 34S, 37Cl, and 81Br. If intensity of Molecular ion M is considered to be 100%, then

the intensity of M+2 peak is at the proportion in accordance with the natural abundance of the

isotope. i.e The M:M+2 ratio indicates the presence of S(100:4.4), Cl (100:31.9), or Br(100:97.2)

in the compound.

Eg. For molecule with Br, the intensity of M+2 peak is almost equal to that of M+ peak as the

relative abundance of the two isotopes is almost equal (50.5% 79Br & 49.5% 81Br)(relative

abundance=100:97.28)

Hence, M and M+2 peaks for 1-bromopropane are almost the same height, indicating that Br is

present in the molecule.[Eg. Mass spectrum of bromomethane CH3Br. (Fig. 5.3)

Similarly, for molecule with Cl, the intensity of the M+2 peak is 1/3rd of the M+ peak.

Eg. Mass spectrum of 2-chloro-2-methyl propane (Fig. 5.4)

For molecule with S, the intensity of M+2 peak is 4% to that of M+ peak.

Presence of iodine is characterised by peak at m/z=127.

Presence of more than 1 halogens and sulphur in a molecule can be known from the presence of

M+4 peaks at respective abundance as given below.

5. Getting the Molecular Formula from the Mass Spectrum:

• The number of carbon atoms is determined from M+1 peak as given above.

• The presence of Cl/Br/S is determined from M+2 peak as given above.

• The Nitrogen Rule states that if m/z for M is odd, then the molecular formula must have

an odd number of nitrogens. If m/z for M is even, then the molecular formula must have

an even number of nitrogens (this includes 0).

EX: For 1-bromopropane, m/z for M=122. The even number is in accordance with the

even number of nitrogens in the formula (zero).

• The Hydrogen Rule states that the maximum number of hydrogens in the molecular

formula is 2C+N+2. In the formula, C: no. of carbons, N: no. of nitrogens

EX: For CH3CH2CH2Br, there are three carbons, so the max no. of hydrogens is

2(3)+2=8

• Subtract the isotopic masses of the known atoms from the mass of the entire compound

(m/z of M peak). The remaining value represents the amu left over for the other atoms.

See whether it is oxygen (16 amu), nitrogen (14 amu each), or another carbon (12 amu).

Lower number may be attributed to the equal number of hydrogens simply.

6. To determine the Number of Rings or Unsaturations

An aliphatic hydrocarbon has a formula CnH2n+2. Each ring or unsaturation that is

present in a hydrocarbon decreases the number of hydrogen atoms by two units. Let x be

the number of hydrogen atoms observed; then, if Ni is the number of rings and unsaturations,

we have

For example, benzene has formula C6H6:

Benzene indeed contains one ring and three unsaturations (three double bonds).

The presence of an oxygen or a sulfur atom in the molecule does not modify these

numbers, as we can see when comparing CH4 with CH3OH and CH3SH, or CH3CH3 with

CH3OCH3 and CH3SCH3.

Each halogen atom ‘replaces’ a hydrogen atom and decreases the number of hydrogen

atoms by an equal number, as is shown by CH4 and CH2Cl2.

Each nitrogen or phosphorus atom increases the number of hydrogen atoms of a

compound by one unit: compare CH4 with CH3NH2 and CH3PH2 as an example.

Let n be the number of carbon atoms, and nX and nN be those of halogens and nitrogen

(or phosphorus), respectively. The following rule can be deduced, where Ni is the number of

rings or unsaturations and x is the number of hydrogen atoms that are found experimentally:

7. To confirm the molecular formula of the compound: The possibilities for elemental

composition determination can often be restricted by using isotopic abundance data. For

example, C10H20 and C8H12O2, both show M+ peak at 140. But, C10H20 produce peaks at mass

141 which is 11% of the M peak and C8H12O2 produce peaks at mass 141which is 8.8%of the M

peak with mass 140. From this information, the number of carbon atoms is calculated and found

to be different in C10H20 and C8H12O2. This is the result of a different statistical probability of

having 13C isotopes. These two elemental compositions can thus be distinguished in a mass

analysis, which helps in confirming the molecular formula.

8. To determine the Low-Mass Fragments and Lost Neutrals:

The molecular ion fragments and produces ions and neutrals that are not observed in the

spectrum. The mass of the neutral product can be deduced from the difference between the mass

of the parent ion and that of the observed ionic fragment. One can deduce much information

concerning the elemental composition from the masses of the neutrals or of the low-mass

fragments. In the case of lost neutral fragments, it is important to take into account the fact that a

fragment can result from several successive steps starting from the molecular ion.

Table 5.3 gives such lost neutrals

Table 5.3 : Common fragments lost

9. To determine the functional group : The general behaviour of the fragmentation pattern of

different organic compounds helps us to identify the functional group present in the compound.

Table 5.4 gives common fragment ions for the functional groups present in the compound.

Table 5.4 common fragment ions

Summary: Mass spectrometry is a powerful analytical technique that is used to identify unknown compounds, to quantify known compounds, and to elucidate the structure and chemical properties of molecules. In Mass spectroscopy, the compound in gas phase is separated in the form of positive ions according to their mass to charge (m/z) ratio. Exercises:

Objective type questions

1. The Mass spectrometer is an analytical instrument in which ions, produced from a

sample, are separated by electric or magnetic fields according to their----------

a. ratios of mass to charge b. ratios of charge to atomic weight

c. mass of the proton d. ratios of charge to mass.

2. In mass spectroscopy, the organic substance is bombarded with the beam of--------

a. electromagnetic radiations b. infra red radiations

c. protons d. electrons

3. The energy equal to the --------of the molecule, results in the ionization of the molecule

a. molar mass b. ionization potential

c. potential energy d. size

4. The gaseous molecule on bombardment of electrons, emit an electron from the --------

energy molecular orbital

a. highest b. moderate

c. zero d. lowest

5. The gaseous molecule on bombardment of electrons, gives rise to an ion M+ which is

called ------------.

a. parent ion b. molecular ion

c. radical cation d. All a,b,c

6. The ---------- cannot be detected in mass spectrometry.

a. neutral radicals b. fragmentation molecules

c. negative ions d. All a,b,c

7. A mass spectrum is a graph of ---------------of the cations..

a. energy absorbed to strength of magnetic field

b. mass to intensity

c. atomic mass to energy absorbed

d. m/z ratio to the relative abundance

8. Because of natural abundance of 13C and 2H, there are small peaks at ------ than the parent

peak

a. two units mass higher b. one unit mass higher

c. lower d. less intensity

9. For a compound to be analyzed in a mass spectrometer, it must be in the -----------state

a. solid b. liquid

c. purest d. gaseous

10. Fragmentation of alkane often splits off simple alkyl groups results in the characteristic

peaks at --------due to loss of methyl group

a. ( M+ - 14) b. ( M+ +1)

c. ( M+ - 15) d. ( M+ + 15)

11. Branched alkanes tend to fragment forming the most stable---------------.

a. radicals b. carbocations

c. carbanions d. molecules

12. In mass spectra of straight chain alkanes there are peaks at ------mass units apart from

each other.

a. 14 b. 15

c. 2 d. 1

13. Allyl carbocation (m/z = 55) is an important fragment in the mass spectra of ----------

a. straight chain alkanes b. terminal alkenes

c. aromatic compounds d. aldehydes

14. Benzyl carbocation (m/z = 91) is an important fragment in the mass spectra of ----------

a. straight chain alkanes b. terminal alkenes

c. aromatic compounds d. aldehydes

15. α-Cleavage and dehydration are the common modes of fragmentations of ---------.

a. phenols b. aldehydes

c. alcohols d. ketones

16. Favored modes of fragmentation involve--------- in phenols..

a. loss of a hydrogen atom to create an M – 1 peak

b. loss of carbon monoxide (CO) to produce a peak at M – 28

c. loss of a formyl radical (HCO•) to give a peak at M – 29

d. All a,b,c

17. Favored modes of fragmentation involve--------- in aldehydes..

a. loss of a hydrogen atom to create an M – 1 peak

b. loss of a formyl radical (HCO•) to give a peak at M – 29

c. loss of carbon monoxide (CO) to produce a peak at M – 28

d. both a & b,

18. Organic compounds having the γ carbon containing a hydrogen undergo the ----------------

-----------commonly known as McLafferty rearrangement reaction

a. α-H shift with β-cleavage, b. β-H shift with α-cleavage,

c. γ-H shift with α-cleavage, d. γ-H shift with β-cleavage,

19. The Base Peak is the peak with the ------------in the mass spectrum

a. lowest intensity

b. greatest intensity

c. greatest mass

d. greatest intensity and mass both

20. To find the m/z for the M peak of a compound, multiply the ---------of each atom by the

number of atoms of each in the formula

a. atomic weight, b. isotopic mass,

c. atomic number d. valency,

21. intensity of M+1 peak is used to calculate the ----------present in the molecule.

a. nature of the carbon atoms

b. number of the carbon atoms

c. number of the total atoms

d. number of the isotopic atoms

22. If M and M+2 peaks for any compound are of almost the same height, indicating that -----

------------ is present in the molecule

a. Chlorine, b. bromine,

c. Oxygen d. sulphur

23. If M+2 peak for any compound is 1/3 of the M peak, indicating that ----------------- is

present in the molecule

a. Chlorine, b. bromine,

c. Oxygen d. sulphur

24. Prsence of peak at m/z=127, indicates that ----------------- is present in the molecule

a. Chlorine, b. bromine,

c. iodine d. sulphur

25. Presence of more than 1 halogens and sulphur in a molecule can be known from the

presence of -----------------peaks.

a. M+1

b. M+2

c. M+3

d. M+4

[Ans: 1.a; 2.d; 3.b; 4.d; 5.d; 6.d; 7.d; 8.b; 9.d; 10.c; 11.b; 12.a; 13.b; 14.c; 15.c; 16: d,

17:d ; 18:d ; 19:b ; 20:b; 21:b; 22:b; 23:a; 24: c; 25: d ]

Long answer type questions:

1. Describe principles of Mass spectrometry.

2. Write a brief note on instrumentation and working in mass spectrometry.

3. What are the different types of ions produced in mass spectrometry with suitable

examples.

4. Explain Fragmentation patterns of-

a) alkanes,

b) alkenes,

c) aromatic hydrocarbons,

d) alcohols,

e) phenols,

f) amines

g) carbonyl compounds

5. Describe the McLafferty rearrangement with suitable examples.

6. Explain how to determine the molecular mass and formula from mass spectrum? .

7. Explain how will you determine the presence of chlorine, bromine and iodine in

the molecule from mass spectrum?

8. How the number of carbon atoms and nitrogen atoms present in the molecule is

determined from mass spectrum?

9. Describe different applications of mass spectrometry.