MASS SPECTROSCOPY

Ionization potential:

The energy required to remove an electron from an atom or molecule is called ionization potential. Its unit is eV.  Most organic compounds have ionization potentials ranging between 8 to 15 eV. The energy require for removing one electron from the neutral patent molecule is usually 10 eV. However, a beam of electrons does not produce ions with high efficiency until the beam of electrons striking the steam of molecules has a potential of from 50 to 70 eV. 

Base peak:

The most intense peak in the mass spectrum is called base peak. Its intensity is highest and counted as 100%. The intensity of other peaks are determined by comparing with it. Base peak appears either due to resonance stabilization or for the formation of the ion through various common pathways. 

Instrumentation:

The instrument needed to produce the mass spectrum of a compound consists of the following parts: -----

Ion source

Mass analyzer

Ion detector.

Ion source:

In the ionization chamber, the molecules of the sample are exposed to bombardment with energetic electrons, produced from an electrically heated tungsten filament having energy of about 70eV. Due to bombardment the molecules generally lose one electron to form a parent ion radical.

The energy required for removing one electron from the neutral parent molecule is called ionization potential, which is usually 10eV. With this much energy no ions are formed. But if the energy of the bombarding electrons is around 70eV then the additional energy is consumed in fragmenting the parent ion. This results in the formation of fragment ions or the daughter ions.

A repeller plate having a positive electrical potential, directs the newly created positive ions towards a series of accelerating plates. A large potential difference, ranging from 1 to 10 kilovolts, applied across these accelerating plates produces a beam of rapidly traveling positive ions.

Most of the sample molecules are not ionized at all. These are continuously drawn off by the vacuum pumps which are connected to the ionization chamber. Some of the molecules are converted to the negative ions that are absorbed by the repeller plate. Small portion of the positive ions which are formed may have a charged greater than one, are accelerated in the same way as the single charged positive ions.

Mass analyzer:

After acceleration of the positive ion from ionization chamber ions are then entered into the mass analyzer. Here the fragment ions are differentiated on the basis of their mass to charge (m/e) ratio. The positive ions travel through whole of the analyzer portion of the mass spectrometer with high velocity and are separated according to their m/e ratio.

The positive ions travel in a circular path through 1800 under a magnetic field H. Suppose, an ion having a charge “e” is accelerated through a voltage “V”, then: -------

                        The kinetic energy of the ions is: ½ mv2

                        The potential energy of the ions is: eV

            Here,

                        V = potential applied

                        v = velocity of the ions after acceleration.

                                    ½ mv2 = eV ---------------------------------------- (1)

It may be noted that a massive ion will travel, slowly in a circular path compared to the lighter fragment. In passing a magnetic field H, any ion will develop two anti-parallel force of attraction such as:-----

                        Centripetal force = Hev

                        Centrifugal force = mv2/r.

            Hence, from Newton’s second Law of motion: -------

                                                Hev = mv2/r         

or, He = mv/r                 

or v = Her/m

                        Putting the value of “v” in equation (1), we get: -----

                                    eV = ½ m (Her/m)2

                                          = ½ m ( H2e2r2/m2

                                          = H2e2r2/2m

                                    or, V = H2er2/2m

                                    or, m/e = H2r2/2V ------------------------------ (2)

From this equation, it is clear that at a given magnetic field strength and accelerating voltage, the ions of m/e will follow a circular path of radius r. the ions of various m/e values reach the collectors, amplified and recorded. The mass spectrum can be obtained either by: -----

Changing H at constant V

Changing V at constant H

When magnetic field is varies, the method is called magnetic scanning and electric voltage scanning is called if potential is varied at constant field strength H.

Ion detector:

The ions which are separated by the analyzer are detected and measured electrically or photographically. The ions pass through the collecting slit one after the other and fall on the detector. The spectrum is scanned by going up the scale. The ion currents are amplified using a direct current amplifier. The spectrum is recorded by using a fast scanning oscillograph.

Limitation:

The prime limitation is that the resolving power is limited by initial spread of translational energy of ion leaving the source. This problem is overcomed by passing the ions through electric field prior to the magnetic field. 

Molecular ion peak or molecular ion:

The electron bombardment with energy 10 – 15 eV usually removes one electron from the molecule of the organic compound in the vapor phase. It results in the formation of molecular ion. The molecular ion represents the actual molecular weight of the original molecule. It is frequently symbolized by (M+(). It is positively charged molecule with an unpaired electron.

Important features:

The mass of the molecular ion gives the molecular mass of the sample.

Molecular ion is located at the high mass region of the spectrum.

The stability of the molecular ion decides it s relative abundance.

The peak intensity of the molecular ion differs from one compound to another. Sometimes molecular ion peak is not formed which means that the rate of decomposition of molecular ion is too high for its detection.

The rate of decomposition of the molecular ion increases with the molecular size in the homologous series.

The molecular ion peak in aromatic compounds is relatively much intense due to the presence of ( electron system.

Conjugated olefins show more intense molecular ion peak as compared to the corresponding non conjugated olefins with the same number of un-saturation.

Un-saturated compounds give more intense peak as compared to the saturated or the cyclic compounds.

The relative abundance of the saturated hydrocarbon is more than the corresponding branched chain compounds with the same no. of carbon atoms.

The substituents groups like (- OH, - OR, - NH2 etc) lower the ionizational potential but increase the relative abundance in case of aromatic compounds. Also the group like (-NO2

, - CN etc) increase the ionizational potential but decrease the relative abundance of the aromatic compounds.

Absence of molecular ion peak in the mass spectrum means that the compound is highly branched or tertiary alcohol. Primary or secondary alcohol gives very small molecular ion peak.

In case of chloro or bromo compounds, isotope peaks are also formed along with the molecular ion peak. In case of bromo compounds, M+ and (M+ +2) peaks are formed in the intensity ratio 1:1. In case of chloro compounds, M+ and (M+ +2) peaks are formed in the intensity ration 1:3.   

Types of molecular ion peak:

There are two types of molecular ion peak: ------

Strong and intense molecular ion peak

Weak and not discernible molecular ion peak.

Strong and intense molecular ion peak: Generally 90% organic compounds give this peak. This peak represents the molecular structure of this compound. Generally conjugated compounds such as aromatic, hetero aromatic give this peak due to the presence of more ( (pie) electron. Straight chain ketone, ester, acid, aldehydes, amide, halide and ether also give this peak.      

Weak and not discernible molecular ion peak: Generally 10% organic compounds give this peak. Generally aliphatic alcohol, amine, nitrite, nitro compounds, nitrile and branched hydrocarbon compounds give this peak.

Causes:

The molecular ion is less stable than fragment ions.

Elimination of neutral molecule from the molecular ion. (e.g. elimination of one mole H2O from alcohol give M – 18 peak which is more intense than molecular ion.)

Elimination of a methyl group from molecular ion. (e.g. elimination of methyl group (- CH3) from branched chain molecule give M – 15 peak which is more intense than molecular ion.)

If impurities are present in molecule.

Life span of molecular ion is less than 10-5.

Overcome:

Through the bombardment of low energy electron.

The ion molecular reaction: If the pressure of the ionization chamber is increased, the ion-molecular reaction is also increased. As a result the molecular ion receives an H+ ion and gives M+1 peak. Thus the intensity of M+1 peak is increased. So the previous peak of M+1 is molecular ion (M+() peak.

Field ionization method: If the charge of repeller plate is decreased the expulsion of molecular ion (M+() is also decreased. So the life span of molecular ion is increased and leads to form M+1 peak.

Nitrogen rule: Through the application of “nitrogen rule” it is helpful to recognize the molecular ion peak from the molecular ion cluster. This rule holds for all compounds containing C, H, O, N, S and halogen.      

“A molecule of even numbered molecular mass must contain no nitrogen atom or an even number of nitrogen atoms. An odd numbered molecular mass requires an odd number of nitrogen atoms. ”

Confusion arises about molecular ion:

Confusion about molecular ion peak arises due to:

Isotopic peak

Impurity peak

Traced peak

Isotopic peak: Molecules which occur in nature don’t occur as isotopically pure species. Virtually all atoms have heavier isotopes which occur in varying natural abundance. Most often the isotopes occur at one or two mass units above the mass of the normal atom. Therefore, besides looking for the molecular ion (M+() peak, one also attempts to locate the M+1 and M+2 peaks. The relative abundance of these M+1 and M+2 peaks can be used to determine the molecular formula of the substance being studied.

Impurity peak: If the weight of the impurity is more than original compound, then trace peak occur, and the molecular ion peak is not necessarily the extreme right peak.  

Trace peak: Sometimes, trace peak also occurs, if the compound is unstable. Un-stability occurs due to the elimination of H2O, HCN and H2S etc. from the compound and if the weight of the impurity is more than original compound. When trace peaks occur, then the molecular ion peak is not necessarily the extreme right peak. 

Metastable ions or peaks:

The ion which gives an abnormal flight path on its way to the detector by receiving energy from the uncharged portion of the original ion during its acceleration in ionization chamber in spite of having lower life times (10-6 sec.) is termed as metastable ion.

Formation of metastable ion:

Let consider that m1+ is the parent ion and m2+ is the daughter ion. If the reaction m1+ ( m2+ takes place in the source, then the daughter ion, m2+ may travel the whole analyzer region and is recorded as m2+ ion. On the otherhand, if the transition m1+ to m2+ occurs after the source exists and before arrival at the collector, then metastable ion is formed. It is frequently symbolized by m*. The position of the metastable ion in mass spectroscopy may determine by the following equation: ------

This equation is called parent – daughter relationship and expressed the relationship between m*, m1+and m2+

  Important features:

The parent-daughter relationship is possible only for those compounds whose fragmentation occur in acceleration region of the ion chamber.

The presence of actual metastable ion (m*) prove that, the fragmentation process (m1 ( m2) is completed in one steps.

They frequently may appear at nonintegral m/e value.

These are much broader than the normal peaks

These are of relatively low abundance

The m/e value of m* is always less than the m/e values of parent and daughter ion.

The theoretical m/e value of m* is 0.1-0.4 mass unit less compared the actual m/e value in mass spectrum.

The life span of metastable ion (m*) is 10-6 sec.

Metastable peak (m*) has a distance below m2+ on the mass scale. The distance is approximately similar to the distance that m2+ lies below m1+.

Example: Consider the formation of metastable peaks in the spectrum of p-amino anisol. The parent ion appears at m/e 123. Suppose the fragmentation of parent ion into daughter ion (due to the loss of methyl radical, i.e. loss of 15 mass units) takes place in acceleration region. The position of the metastable peak can be calculated as follows: 

The position of the metastable peak: -------

  

We see that the position of metastable peak (94.8) below m2+ ion (108) is approximately the same as m2+ ion is below m1+ (123) on the mass scale.

  Importance of metastable peaks:

The metastable peaks in the mass spectrum greatly contribute in structure elucidation.

The presence of metastable peaks at the expected positions of the suspected compounds lends weight to its structure.

Metastable ion peaks can be used to prove a proposed fragmentation pattern or to aid in the solution of structure proof problems.

Question: Write down the consequences of incident of the electron beam of the following energies on the sample molecule. 

10 – 15 eV

(70 eV

( 15eV but ( 70eV.

Answer:

10 – 15 eV:

Generally this energy is equal to ionization potential of organic compounds. 10 -15 eV electronic bombardments on a molecule causes loss of an electron and results in formation of molecular ion.

( 15eV but ( 70eV:

Upon the application of this energy molecular ion is fragmented into small ions (cations), radicals and neutral fragment may be produced during this process. 

(70 eV:

This energy causes loss of two electrons from molecular ion and results in the formation of doubly charged cations.

Question: Write a short note on Mclafferty Rearrangement.

Answer:

Mclafferty rearrangement:

            Mclafferty rearrangement occurs in compounds having double bond and ( hydrogen. It involves the migration of (-hydrogen atom followed by the cleavage of a (-bond. This rearrangement proceeds through a six-membered transition state and leads to the elimination of a neutral molecule.

Question: Show the molecular ion cluster for the following compound.

Answer:

The compound contains two chlorine (Cl). Chlorine has two isotopes: ----

35Cl (lighter isotope)

37Cl (higher isotope)

Abundance in nature is: ----  35Cl :  37Cl = 75.53 : 24.47 = 3 : 1

Molecular ion cluster of a compound is determined by binomial expansion rule:  (a +b)n. now we apply this rule for the compound                 we have:

            a = Relative abundance of lighter isotope 35Cl

            b = relative abundance of heavier isotope 37Cl 

            n = no. of halogen atoms (in this case chlorine) = 2

                       

(a + b)n

(3 + 1)2 = 32 + 2.3.1 + 12 = 9 : 6 : 1