Introduction:
The production of a molecular ion is frequently followed via its dissociation or fragmentation into ions and neutral species of the lower mass, which in turn might dissociate further. Fragmentation patterns are characteristic of specific molecular structures and can point out the presence of specific functional groups, therefore providing helpful information on the structure and identity of the original molecule. The points of cleavage in a molecule are found out by individual bond strengths all through the structure and additionally, molecular rearrangements and recombination can take place.
Fragmentation patterns are an invaluable help in the interpretation of mass spectra and in the recognition or confirmation of the structural characteristics.
Interpretation of the Mass Spectrum:
Mass spectra can be hard to interpret, due to the complexity of the fragmentation. Usually, one identified the peak because of the molecular ion (M+) or MH+ ion and uses this as the reference point. The mass differences between this peak and other peaks are found out and the likely nature of the fragments lost from the reference peak deduced from the reference tables. One can work the other manner round and consult tables that point out the most likely composition of the fragment. The peaks on either side of the main peak might be made because of the presence of isotopes in a fragment. These peaks should be taken into account whenever making any deductions, and initially it is best to consider just the most abundant peaks. Unfortunately, the great diversity of fragments generated in a mass spectrum signifies that these deductions are of limited value on their own. However taken in conjunction by other experimental proof can be of considerable use either in recognizing or finding out the structure of a compound.
Some Rules employed in the Interpretation of Mass Spectra:
1) The nitrogen rule defines that compounds having an even numbered RMM should contain zero or an even number of nitrogen atoms and those having an odd numbered RMM should contain an odd number of nitrogen atoms.
2) The unsaturated sites rule gives a means of computation of the number of double-bond equivalent in the molecule from the formula:
No of C atoms + 1/2(no of N atoms) - 1/2(no of H atoms + halogen atoms) + 1
For illustration: C7H7ON, the formula provides 7 + 0.5 - 3.5 - 1 = 5 double bond equivalents. This corresponds to benzamide, C6H5CONH2, the aromatic ring being counted as the three double bonds plus one for the ring.
3) The intensity of molecular ion peak reduces with increasing chain in length in the spectra of a homologous series of compounds and by increased branching of the chain.
4) Double bonds are cyclic structures tend to stabilize the molecular ion, saturated rings losing side chains at the α-position.
5) Alkyl-substituted aromatic rings (that is, benzyl group) undergo rearrangement to form a tropylium cation C7H7+ (see below) providing a prominent peak at m/z = 91.
6) Small neutral molecules like CO, C2H4, C2H2, H2O and NH3 are often lost throughout fragmentation.
7) The C-C bond adjacent to a heteroatom (N, O and S) is often cleaved leaving the charge on the fragment having the heteroatom (Y), whose nonbonding electrons give resonance stabilization example:
CH3-CH2-Y+•-R → (-CH3•) → CH2=Y+-R ↔ •CH2-Y-R
8) McLafferty rearrangements:
McLafferty rearrangement takes place in carbonyl compounds example:
HCH2-CH2-CH2-CO+•-OR ↔ C2H2 + CH2=C(O+•)-
OR
A neutral molecule of ethene is lost in the procedure.
The mass spectrum of toluene (that is, methyl benzene) is illustrated below (figure shown below). The spectrum displays a strong molecular ion at m/z = 92, small m + 1 and m +2 peaks, a base peak at m/z = 91 and an assortment of the minor peaks m/z =65 and below.
Fig: Mass Spectrum of Toluene
The molecular ion again symbolizes loss of an electron and the peaks above the molecular ion are due to isotopic abundance. The base peak in toluene is because of loss of a hydrogen atom to form the relatively stable benzyl cation. This is thought to experience rearrangement to form the extremely stable tropylium cation, and this strong peak at m/z = 91 is a hallmark of compounds having a benzyl unit. The minor peak at m/z = 65 symbolizes loss of a neutral acetylene from the tropylium ion and the minor peak beneath this occur from more complex fragmentation.
a) Alkanes:
Simple alkanes tend to experience fragmentation via the initial loss of a methyl group to form a (m-15) species. This carbocation can then experience stepwise cleavage down the alkyl chain, expelling neutral two-carbon units (that is, ethane). Branched hydrocarbons form more stable secondary and tertiary carbocation, and these peaks will tend to control the mass spectrum.
Fig: Alkanes tend to undergo fragmentation
b) Aromatic Hydrocarbons:
The fragmentation of the aromatic nucleus is fairly complex, producing series of peaks having m/z = 77, 65, 63 and so on. While these peaks are difficult to explain in simple terms, they do form a pattern (the aromatic cluster) that becomes recognizable with experience. If the molecule includes a benzyl unit, the main cleavage will be to produce the benzyl carbocation that rearranges to form the tropylium ion. Expulsion of acetylene (that is, ethyne) from this produces a characteristic m/z = 65 peak.
c) Aldehydes and Ketones:
The predominate cleavage in aldehydes and ketones is loss of one of the side chains to produce the substituted oxonium ion. This is an extremely favorable cleavage and this ion frequently represents the base peak in the spectrum. The methyl derivative (CH3C ≡ O+) is generally referred to as the acylium ion.
Fig: Predominate cleavage in aldehydes and ketones
The other common fragmentation noticed in the carbonyl compounds (and in nitriles and so on) comprises the expulsion of neutral ethane through a process termed as the McLafferty rearrangement.
d) Esters, acids and amides:
As with aldehydes and ketones, the main cleavage observed for these compounds comprises expulsion of the 'X' group, as illustrated below to form the substituted oxonium ion. For carboxylic acids and unsubstituted amides, characteristic peaks at m/z = 45 and 44 are as well often noticed.
Fig: Cleavage observed for Esters, acids and amides
e) Alcohols:
Moreover to losing a proton and hydroxyl radical, alcohols tend to loose on the α-alkyl groups (or hydrogen) to form the oxonium ions illustrated below. For primary alcohols, this produces a peak at m/z = 31; secondary alcohols produced peaks with m/z = 45, 59, 73 and according to substitution.
Fig: Alcohols tend to lose α-alkyl groups
f) Ethers:
Following the trend of alcohols, ethers will fragment often via loss of an alkyl radical, to form a substituted oxonium ion.
g) Halides:
Organic halides fragment having simple expulsion of the halogen, as illustrated below. The molecular ions of chlorine and bromine having compounds will illustrate multiple peaks because of the fact that each of these exists as two isotopes in relatively high abundance. Thus for chlorine,
the 35Cl-37Cl ratio is roughly 3.08:1 and for bromine, the 79Br-81
Br ratio is 1.02:1. The molecular ion of a chlorine-containing compound will have two peaks, separated by two mass units, in the ratio ~ 3:1, and a bromine-containing compound will have two peaks, again separated by two mass units, having approximately equivalent intensities.
The lists given above are by no means exhaustive and represent just the simplest and most common fragments seen in the mass spectrum.
Examples of Mass Spectra Interpretation:
a) The figure below includes the mass spectra data of octane and 2,2,4-trimethylpentane. Octane is a saturated straight chain therefore the spectra is characterized via clusters of peaks 14 mass units (CH2 groups) apart, as successive C-C bonds all along the chain are cleaved in various molecules. Octane consists of a base peak at m/z = 43 due to the CH3CH2CH2+ fragment ion and a small molecular ion peak at m/z (rule 3).
Table: Commonly lost Fragments from a Molecular Ion
Fig: Mass Spectrum of 2,2,4-Trimethylpentane
Branching of the chain modify the relative intensities of the clusters, as illustrated by the spectrum of the isomeric 2,2,4-trimethylpentane that consists of a base peak at m/z = 57 due to the (CH3)3C+ fragment ion, and no significant m/z = 71, 85 or molecular ion peak (rule 3).
b) The spectrum of methylbenzene (figure below) typifies alkyl-substituted aromatic compounds, having a base peak corresponding to the tropylium ion, C7H7+, at m/z = 91 and a large molecular ion peak at m/z = 92 (rules 4 and 5).
Fig: Mass Spectrum of Methylbenzene
Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)
Expand your confidence, grow study skills and improve your grades.
Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.
Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology. Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.
Free to know our price and packages for online Chemistry tutoring. Chat with us or submit request at [email protected]
www.tutorsglobe.com offers chemical thermodynamics homework help, chemical thermodynamics assignment help, online tutoring, chemical thermodynamics solutions and instant help in projects by online chemistry tutor's help.
Crystal Diffraction-I tutorial all along with the key concepts of Bragg formulation of diffraction by a crystal, Statement of Bragg's law, Von Laue formulation of diffraction by a crystal, Diffraction of crystals by electrons and neutrons
line profiles tutorial all along with the key concepts of transition, transition rules in absence of magnetic field, degeneracy of atomic levels, absorption lines, electron lifetime and levels, lorentzian line profile, doppler line broadening, gaussian profile, bell-shaped profile
Theory and lecture notes of Public Goods all along with the key concepts of public goods, Free Rider Problem, Optimal Output Decision of Public Good, Externalities and Public Goods. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Public Goods.
microbial interactions tutorial all along with the key concepts of neutralism, mutualism, commensalism, supply of nutrients, amensalism, prey-predator relationship and competition
www.tutorsglobe.com offers reactions of substituted benzenes homework help, reactions of substituted benzenes assignment help, online tutoring assistance, organic chemistry solutions by online qualified tutor's help.
Habitat Studies tutorial all along with the key concepts of Animal Sampling Methods, Short vegetation and litters, Tall vegetation or grasses, Plant Sampling Methods and Soil Sampling Methods
tutorsglobe.com capital structure assignment help-homework help by online financial management tutors
A case of historic US (United States) use of the identical word Radiophone presents in the 1960s. This word is employed to explain push-to-talk two-way equipment and systems.
boost your grades and future with reliable wellness and lifestyle analysis assignment help. we deliver a++ solutions within deadline!
venetian lines, ghost image, only plain raster, negative picture, over loading, picture ok, low volume, distorted audio, hum audio, sound bars in picture, noisyaudio.
www.tutorsglobe.com offers Selling and Distribution Overheads homework help, assignment help, case study, writing homework help, online tutoring assistance by accounting tutors.
www.tutorsglobe.com offers answering questions to factors affecting price elasticity, economics free tutorial and assignment help - homework help.
tutorsglobe.com selection assignment help-homework help by online breeding experiments tutors
tutorsglobe.com bronchitis assignment help-homework help by online respiration tutors
1959605
Questions Asked
3689
Tutors
1470187
Questions Answered
Start Excelling in your courses, Ask an Expert and get answers for your homework and assignments!!