This essay has been submitted by a student. This is not an example of the work written by our professional essay writers.
The Mass Spectrometer. In order to measure the characteristics of individual molecule, a mass spectrometer converts them to ions so that they can be moved about & manipulated by external electric fields. The three essential functions of mass spectrometer & its associated components are-
1. A small sample is ionized, usually to loss of cations by loss of electrons: The ion source
2. The ions are separated & sorted according to their mass & charge: The mass Analyser
3. The separated ions are then measured & then displayed on a chart:
THE MASS SPECTRUM
In the mass spectrometer, molecules are bombarded with a beam of energetic electrons. The molecules are ionized & broken up into many fragments, some of which are positive ions. Each kind of ion has a particular ratio of mass to charge, or m/e ratio. For most of the ions, the charge is 1, so m/e is simply the mass of the ion
C (CH3)4-------> 2e_ + (C5H12)+ m/e=72
The set of ions is analyzed in such a way that a signal is obtained for each value of m/e that is represented; the intensity of each signal reflects the relative abundance of the ion producing the signal. The largest peak is called Base peak; its intensity is taken as 100, & the intensities of the other peaks are expressed relative to it. A plot or even a list showing the relative intensities of signals at the various m/e values is called a mass spectrum, & is highly characteristic of a particular compound.Compare,for example,the spectra of three isomers shown below-
N, N-Diethyl methylamine
The greater the number of physical properties measured, the stronger the evidence. Now a single mass spectrum amounts to dozens of physical properties, since it shows relative abundances of dozens of different fragments. If we measure the mass spectrum of an unknown compound and find it to be identical with the spectrum of a previously reported compound of known structure, then we can conclude that almost beyond the shadow of doubt the two compounds are identical.
The mass spectrum helps to establish the structure of a new compound in several different ways: it can give exact molecular weight; it can give a molecular formula or at least narrow the possibilities to a very few; and it can indicate the presence in a molecule of certain structural units.
If an electron is removed from the parent molecule, there is produced M+, the molecular ion or(a parent ion ) ,whose m/e value is, of course, the molecular weight of compound. Sometimes the M+ peak is the base peak, and is easily recognized; often, though, it is not the base peak it may even be very small and considerable work is required to locate it. Once identified, it gives the most accurate weight obtainable.
M + e_ -----> M+ + 2e_
molecular ion (parent ion), m/e=mol wt
We might at first think that the M+ peak would be the peak of highest m/e value. This is not so, however. Most elements occur naturally as several isotopes; generally the lightest one greatly predominates, and the heavier ones occur to lesser extent. Table the relative abundances of several heavy isotopes.
ABUNDANCE OF SOME HEAVY ISOTOPES
Heavy Abundance relative to isotope isotope of lowest at. wt
The molecular weight that one usually measures and works with is the sum of the average atomic weights of the elements, and reflects the presence of these heavy isotopes. But this is not true of the molecular weights obtained from the mass spectrum; here, the M+ peak is due to molecules containing only the commonest isotope of each element.
Consider Benzene, for example. The M+ peak, m/e 78, is due only to ions of formula C6H6+. There is a peak at m/e 79, the M+1 peak, which is due to C513CH6+ and C6H5D+. There is an M+2 peak at m/e 80, due to C413C2H6+, and C513CH5D+, and C6H4D2+. Now, because of low natural abundance of most heavy isotopes, these isotopic peaks are generally much less intense depends on which elements they are due to. In the case of benzene, the M+1 and M+2 peaks are, respectively, 6.75% and 0.18% as intense as the M+ peak. (Table shows us, however, that a monochloro compound would have an M+2 peak about one third as intense as the M+ peak, and a monobromo compound would have M and M+2 peaks of about equal intensity.)
It is the isotopic peak that makes us possible for us to determine the molecular formula of the compound. Knowing the relative natural abundances of isotopes, one can calculate for any molecular formula the relative intensity to be expected each isotopic peak: M+1, M+2, etc. The results of such calculations are available in tables. Consider, for example, a compound for which M+ is 44. The compound might be (among other less likely possibilities) N2O, CO2, C2H4O, or C3H8. By use of table, we clearly could pick out the most likely formula from the mass spectral data.
CALCULATED INTENSITIES OF ISOTOPIC PEAKS
M M+1 M+2
N2O 100 0.8 0.2
CO2 100 1.16 0.4
C2H4O 100 1.91 0.01
C3H8 100 3.37 0.04
Finally, study of compounds of known structure is beginning to reveal the factor that determines which fragments a particular structure is likely to break into. In this we can find much that is familiar to us: the preferential formation of carbocations that we recognize as being relatively stable ones; elimination of small stable molecules like water, ammonia, and carbon monoxide. Under the energetic conditions, extensive arrangement can occur, complicating and interpretation; But here, too, patterns are emerging. The direction of rearrangements is, as we would expect, towards more stable ions. As this knowledge accumulates, the process is reversed: from the kind of fragmentation as unknown compound gives, its structure is deducted.