MASS SPECTROMETRY OVERVIEW

Mass Spectrometry

Mass spectrometry is a powerful analytical technique used to quantify known materials, to identify unknown compounds within a sample, and to elucidate the structure and chemical properties of different molecules. The complete process involves the conversion of the sample into gaseous ions, with or without fragmentation, which are then characterized by their mass to charge ratios (m/z) and relative abundances. This technique basically studies the effect of ionizing energy on molecules. It depends upon chemical reactions in the gas phase in which sample molecules are consumed during the formation of ionic and neutral species. Basic Principle A mass spectrometer generates multiple ions from the sample under investigation, it then separates them according to their specific mass-to-charge ratio (m/z), and then records the relative abundance of each ion type. The first step in the mass spectrometric analysis of compounds is the production of gas phase ions of the compound, basically by electron ionization. This molecular ion undergoes fragmentation. Each primary product ion derived from the molecular ion, in turn, undergoes fragmentation, and so on. The ions are separated in the mass spectrometer according to their mass-to-charge ratio, and are detected in proportion to their abundance. A mass spectrum of the molecule is thus produced. It displays the result in the form of a plot of ion abundance versus mass-to-charge ratio. Ions provide information concerning the nature and the structure of their precursor molecule. In the spectrum of a pure compound, the molecular ion, if present, appears at the highest value of m/z (followed by ions containing heavier isotopes) and gives the molecular mass of the compound. Components The instrument consists of three major components: Ion Source: For producing gaseous ions from the substance being studied. Analyzer: For resolving the ions into their characteristics mass components according to their mass-to-charge ratio. Detector System: For detecting the ions and recording the relative abundance of each of the resolved ionic species. In addition, a sample introduction system is necessary to admit the samples to be studied to the ion source while maintaining the high vacuum requirements (~10-6 to 10-8 mm of mercury) of the technique; and a computer is required to control the instrument, acquire and manipulate data, and compare spectra to reference libraries. Figure: Components of a Mass Spectrometer With all the above components, a mass spectrometer should always perform the following processes: Produce ions from the sample in the ionization source. Separate these ions according to their mass-to-charge ratio in the mass analyzer. Eventually, fragment the selected ions and analyze the fragments in a second analyzer. Detect the ions emerging from the last analyzer and measure their abundance with the detector that converts the ions into electrical signals. Process the signals from the detector that are transmitted to the computer and control the instrument using feedback. Analysis of Biomolecules using Mass Spectrometry Mass spectrometry is fast becoming an indispensable field for analyzing biomolecules. Till the1970s, the only analytical techniques which provided similar information were electrophoretic, chromatographic or ultracentrifugation methods. The results were not absolute as they were based on characteristics other than the molecular weight. Thus the only possibility of knowing the exact molecular weight of a macromolecule remained its calculation based on its chemical structure. The development of desorption ionization methods based on the emission of pre-existing ions such as plasma desorption (PD), fast atom bombardment (FAB) or laser desorption (LD), allowed the application of mass spectrometry for analyzing complex biomolecules. Analysis of Glycans Oligosaccharides are molecules formed by the association of several monosaccharides linked through glycosidic bonds. The determination of the complete structure of oligosaccharides is more complex than that of proteins or oligonucleotides. It involves the determination of additional components as a consequence of the isomeric nature of monosaccharides and their capacity to form linear or branched oligosaccharides. Knowing the structure of an oligosaccharide requires not only the determination of its monosaccharide sequence and its branching pattern, but also the isomer position and the anomeric configuration of each of its glycosidic bonds. Advances in glycobiology involves a comprehensive study of structure, bio-synthesis, and biology of sugars and saccharides. Mass spectrometry (MS) is emerging as an enabling technology in the field of glycomics and glycobiology. Analysis of Lipids Lipids are made up of many classes of different molecules which are soluble in organic solvents. Lipidomics, a major part of metabolomics, constitutes the detailed analysis and global characterization, both spatial and temporal, of the structure and function of lipids (the lipidome) within a living system. Many new strategies for mass-spectrometry-based analyses of lipids have been developed. The most popular lipidomics methodologies involve electrospray ionization (ESI) sources and triple quadrupole analyzers. Using mass spectrometry, it is possible to determine the molecular weight, elemental composition, the position of branching and nature of substituents in the lipid structure. Analysis of Proteins and Peptides Proteins and peptides are linear polymers made up of combinations of the 20 amino acids linked by peptide bonds. Proteins undergo several post translational modifications, extending the range of their function via such modifications. The term Proteomics refers to the analysis of complete protein content in a living system, including co- and post-translationally modified proteins and alternatively spliced variants. Mass Spectrometry has now become a crucial technique for almost all proteomics experiments. It allows precise determination of the molecular mass of peptides as well as their sequences. This information can very well be used for protein identification, de novo sequencing, and identification of post-translational modifications. Analysis of Oligonucleotides Oligonucleotides (DNA or RNA), are linear polymers of nucleotides. These are composed of a nitrogenous base, a ribose sugar and a phosphate group. Oligonucleotides may undergo several natural covalent modifications which are commonly present in tRNA and rRNA, or unnatural ones resulting from reactions with exogenous compounds. Mass spectrometry plays an important role in identifying these modifications and determining their structure as well as their position in the oligonucleotide. It not only allows determination of the molecular weight of oligonucleotides, but also in a direct or indirect manner, the determination of their sequences. A mass spectrometer allows us to determine the mass of a molecule such as a peptide from within a heterogeneous mixture. To do this a mass spectrometer uses electromagnetic fields in order to control the movement of charged molecules (ions) within the instrument. Mass Spectrometer Overview These ions need to be in the gas-phase in order to move freely within the instrument in response to the imposed electromagnetic fields. To allow for largely uninhibited movement of the gas-phase ions within the instrument mass spectrometers are contained within a vacuum chamber that has a vacuum generated within it by rotary and turbo-molecular vacuum pumps. This removes the ambient air from the instrument and depending on the type of mass analyser can span a wide range of vacuum strengths. For heterogeneous samples some form of pre-fractionation is used prior to mass spectrometry to reduce the complexity and this is shown below as the inlet. For peptides this is usually reverse phase liquid chromatography. From the inlet the molecules need to become charged and in the gas-phase prior to entrance to the mass spectrometer. This takes place in the ion source where different techniques may be used to generate gas-phase ions. To become positively charged molecules (peptides) gain protons to become positively charged. For peptides previously separated by reverse phase chromatography this is usually electrospray ionisation (ESI). Once within the mass spectrometer the movement of the ions is controlled based upon two of their physical parameters: their charge state (z) and their mass (m). Therefore the mass spectrometer does not measure mass directly but measures a combination of these two parameters the mass-to-charge ratio (m/z). The separation of ions in the mass spectrometer takes place with the mass analyser(s). There are many different types of mass analysers which function based on different physical principles but these devices separate ions with different m/z to allow for their separate detection. The detector within a mass spectrometer is most commonly an electron multiplier were the collision of ions with a charged anode leads to a cascade of increasing number of electrons which can be detected by an electrical circuit connected to a computer (data storage system). However many other types of detectors exists which may be destructive or non-destructive such as the detection system of the orbitrap mass analyser. This data will lead to the generation of a mass spectrum which shows a graph of the relative intensity of all ions detected in the mass spectrometer within a defined period of time versus their m/z. The most intense peak in a mass spectrum is called the “base peak”. The software which displays the mass spectrum will allow the user to zoom in on individual regions or peaks to appreciate the entire dataset in each spectrum.

An outline of what happens in a mass spectrometer Atoms and molecules can be deflected by magnetic fields - provided the atom or molecule is first turned into an ion. Electrically charged particles are affected by a magnetic field although electrically neutral ones aren't. The sequence is : Stage 1: Ionisation The atom or molecule is ionised by knocking one or more electrons off to give a positive ion. This is true even for things which you would normally expect to form negative ions (chlorine, for example) or never form ions at all (argon, for example). Most mass spectrometers work with positive ions.
	Note:  All mass spectrometers that you will come across if you are doing a course for 16 - 18 year olds work with positive ions. Even if a few atoms in a sample of chlorine, for example, captured an electron instead of losing one, the negative ions formed wouldn't get all the way through the ordinary mass spectrometer. But it has been pointed out to me that there is work being done on negative ion mass spectrometers, although they use a different ionisation technique.My thanks to Professor John Todd of the University of Kent for drawing this to my attention.
Stage 2: Acceleration The ions are accelerated so that they all have the same kinetic energy. Stage 3: Deflection The ions are then deflected by a magnetic field according to their masses. The lighter they are, the more they are deflected. The amount of deflection also depends on the number of positive charges on the ion - in other words, on how many electrons were knocked off in the first stage. The more the ion is charged, the more it gets deflected. Stage 4: Detection The beam of ions passing through the machine is detected electrically. A full diagram of a mass spectrometer Understanding what's going on The need for a vacuum It's important that the ions produced in the ionisation chamber have a free run through the machine without hitting air molecules. Ionisation The vaporised sample passes into the ionisation chamber. The electrically heated metal coil gives off electrons which are attracted to the electron trap which is a positively charged plate. The particles in the sample (atoms or molecules) are therefore bombarded with a stream of electrons, and some of the collisions are energetic enough to knock one or more electrons out of the sample particles to make positive ions. Most of the positive ions formed will carry a charge of +1 because it is much more difficult to remove further electrons from an already positive ion. These positive ions are persuaded out into the rest of the machine by the ion repeller which is another metal plate carrying a slight positive charge.
	Note:  As you will see in a moment, the whole ionisation chamber is held at a positive voltage of about 10,000 volts. Where we are talking about the two plates having positive charges, these charges are in addition to that 10,000 volts.
Acceleration The positive ions are repelled away from the very positive ionisation chamber and pass through three slits, the final one of which is at 0 volts. The middle slit carries some intermediate voltage. All the ions are accelerated into a finely focused beam. Deflection Different ions are deflected by the magnetic field by different amounts. The amount of deflection depends on: the mass of the ion. Lighter ions are deflected more than heavier ones. the charge on the ion. Ions with 2 (or more) positive charges are deflected more than ones with only 1 positive charge. These two factors are combined into the mass/charge ratio.Mass/charge ratio is given the symbol m/z (or sometimes m/e). For example, if an ion had a mass of 28 and a charge of 1+, its mass/charge ratio would be 28. An ion with a mass of 56 and a charge of 2+ would also have a mass/charge ratio of 28. In the last diagram, ion stream A is most deflected - it will contain ions with the smallest mass/charge ratio. Ion stream C is the least deflected - it contains ions with the greatest mass/charge ratio. It makes it simpler to talk about this if we assume that the charge on all the ions is 1+. Most of the ions passing through the mass spectrometer will have a charge of 1+, so that the mass/charge ratio will be the same as the mass of the ion.
	Note:  You must be aware of the possibility of 2+ (etc) ions, but the vast majority of A'level questions will give you mass spectra which only involve 1+ ions. Unless there is some hint in the question, you can reasonably assume that the ions you are talking about will have a charge of 1+.
Assuming 1+ ions, stream A has the lightest ions, stream B the next lightest and stream C the heaviest. Lighter ions are going to be more deflected than heavy ones. Detection Only ion stream B makes it right through the machine to the ion detector. The other ions collide with the walls where they will pick up electrons and be neutralised. Eventually, they get removed from the mass spectrometer by the vacuum pump. When an ion hits the metal box, its charge is neutralised by an electron jumping from the metal on to the ion (right hand diagram). That leaves a space amongst the electrons in the metal, and the electrons in the wire shuffle along to fill it. A flow of electrons in the wire is detected as an electric current which can be amplified and recorded. The more ions arriving, the greater the current. Detecting the other ions How might the other ions be detected - those in streams A and C which have been lost in the machine? Remember that stream A was most deflected - it has the smallest value of m/z (the lightest ions if the charge is 1+). To bring them on to the detector, you would need to deflect them less - by using a smaller magnetic field (a smaller sideways force). To bring those with a larger m/z value (the heavier ions if the charge is +1) on to the detector you would have to deflect them more by using a larger magnetic field. If you vary the magnetic field, you can bring each ion stream in turn on to the detector to produce a current which is proportional to the number of ions arriving. The mass of each ion being detected is related to the size of the magnetic field used to bring it on to the detector. The machine can be calibrated to record current (which is a measure of the number of ions) against m/z directly. The mass is measured on the 12C scale.
	Note:  The 12C scale is a scale on which the 12C isotope weighs exactly 12 units.
What the mass spectrometer output looks like The output from the chart recorder is usually simplified into a "stick diagram". This shows the relative current produced by ions of varying mass/charge ratio. The stick diagram for molybdenum looks lilke this: You may find diagrams in which the vertical axis is labelled as either "relative abundance" or "relative intensity". Whichever is used, it means the same thing. The vertical scale is related to the current received by the chart recorder - and so to the number of ions arriving at the detector: the greater the current, the more abundant the ion. As you will see from the diagram, the commonest ion has a mass/charge ratio of 98. Other ions have mass/charge ratios of 92, 94, 95, 96, 97 and 100. That means that molybdenum consists of 7 different isotopes. Assuming that the ions all have a charge of 1+, that means that the masses of the 7 isotopes on the carbon-12 scale are 92, 94, 95, 96, 97, 98 and 100.

The origin of fragmentation patternsThe formation of molecular ions When the vaporised organic sample passes into the ionisation chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion - or sometimes the parent ion.
	Note:  If you aren't sure about how a mass spectrum is produced, it might be worth taking a quick look at the page describing how a mass spectrometer works.
The molecular ion is often given the symbol M+ or  - the dot in this second version represents the fact that somewhere in the ion there will be a single unpaired electron. That's one half of what was originally a pair of electrons - the other half is the electron which was removed in the ionisation process. Fragmentation The molecular ions are energetically unstable, and some of them will break up into smaller pieces. The simplest case is that a molecular ion breaks into two parts - one of which is another positive ion, and the other is an uncharged free radical.
	Note:  A free radical is an atom or group of atoms which contains a single unpaired electron.More complicated break-ups are beyond the scope of A'level syllabuses.
The uncharged free radical won't produce a line on the mass spectrum. Only charged particles will be accelerated, deflected and detected by the mass spectrometer. These uncharged particles will simply get lost in the machine - eventually, they get removed by the vacuum pump. The ion, X+, will travel through the mass spectrometer just like any other positive ion - and will produce a line on the stick diagram. All sorts of fragmentations of the original molecular ion are possible - and that means that you will get a whole host of lines in the mass spectrum. For example, the mass spectrum of pentane looks like this:
	Note:  All the mass spectra on this page have been drawn using data from the Spectral Data Base System for Organic Compounds (SDBS) at the National Institute of Materials and Chemical Research in Japan.They have been simplified by omitting all the minor lines with peak heights of 2% or less of the base peak (the tallest peak).
It's important to realise that the pattern of lines in the mass spectrum of an organic compound tells you something quite different from the pattern of lines in the mass spectrum of an element. With an element, each line represents a differentisotope of that element. With a compound, each line represents a different fragment produced when the molecular ion breaks up.
	Note:  If you are interested in the mass spectra of elements, you could follow this link.
The molecular ion peak and the base peakIn the stick diagram showing the mass spectrum of pentane, the line produced by the heaviest ion passing through the machine (at m/z = 72) is due to the molecular ion.
	Note:  You have to be a bit careful about this, because in some cases, the molecular ion is so unstable that every single one of them splits up, and none gets through the machine to register in the mass spectrum. You are very unlikely to come across such a case at A'level.
The tallest line in the stick diagram (in this case at m/z = 43) is called the base peak. This is usually given an arbitrary height of 100, and the height of everything else is measured relative to this. The base peak is the tallest peak because it represents the commonest fragment ion to be formed - either because there are several ways in which it could be produced during fragmentation of the parent ion, or because it is a particularly stable ion. Using fragmentation patternsThis section will ignore the information you can get from the molecular ion (or ions). That is covered in three other pages which you can get at via the mass spectrometry menu. You will find a link at the bottom of the page. Working out which ion produces which line This is generally the simplest thing you can be asked to do. The mass spectrum of pentane Let's have another look at the mass spectrum for pentane: What causes the line at m/z = 57? How many carbon atoms are there in this ion? There can't be 5 because 5 x 12 = 60. What about 4? 4 x 12 = 48. That leaves 9 to make up a total of 57. How about C4H9+ then? C4H9+ would be [CH3CH2CH2CH2]+, and this would be produced by the following fragmentation: The methyl radical produced will simply get lost in the machine. The line at m/z = 43 can be worked out similarly. If you play around with the numbers, you will find that this corresponds to a break producing a 3-carbon ion: The line at m/z = 29 is typical of an ethyl ion, [CH3CH2]+: The other lines in the mass spectrum are more difficult to explain. For example, lines with m/z values 1 or 2 less than one of the easy lines are often due to loss of one or more hydrogen atoms during the fragmentation process. You are very unlikely to have to explain any but the most obvious cases in an A'level exam. The mass spectrum of pentan-3-one This time the base peak (the tallest peak - and so the commonest fragment ion) is at m/z = 57. But this isn't produced by the same ion as the same m/z value peak in pentane. If you remember, the m/z = 57 peak in pentane was produced by [CH3CH2CH2CH2]+. If you look at the structure of pentan-3-one, it's impossible to get that particular fragment from it. Work along the molecule mentally chopping bits off until you come up with something that adds up to 57. With a small amount of patience, you'll eventually find [CH3CH2CO]+ - which is produced by this fragmentation: You would get exactly the same products whichever side of the CO group you split the molecular ion. The m/z = 29 peak is produced by the ethyl ion - which once again could be formed by splitting the molecular ion either side of the CO group. Peak heights and the stability of ions The more stable an ion is, the more likely it is to form. The more of a particular sort of ion that's formed, the higher its peak height will be. We'll look at two common examples of this. Examples involving carbocations (carbonium ions)
	Important!  If you don't know what a carbocation (or carbonium ion) is, or why the various sorts vary in stability, it's essential that you follow this link before you go on.Use the BACK button on your browser to return quickly to this page.
Summarizing the most important conclusion from the page on carbocations: Order of stability of carbocations primary < secondary < tertiary
	Note:  The symbol "<" means "is less than". So what this is saying is that primary ions are less stable than secondary ones which in turn are less stable than tertiary ones.
Applying the logic of this to fragmentation patterns, it means that a split which produces a secondary carbocation is going to be more successful than one producing a primary one. A split producing a tertiary carbocation will be more successful still. Let's look at the mass spectrum of 2-methylbutane. 2-methylbutane is an isomer of pentane - isomers are molecules with the same molecular formula, but a different spatial arrangement of the atoms. Look first at the very strong peak at m/z = 43. This is caused by a different ion than the corresponding peak in the pentane mass spectrum. This peak in 2-methylbutane is caused by: The ion formed is a secondary carbocation - it has two alkyl groups attached to the carbon with the positive charge. As such, it is relatively stable. The peak at m/z = 57 is much taller than the corresponding line in pentane. Again a secondary carbocation is formed - this time, by: You would get the same ion, of course, if the left-hand CH3group broke off instead of the bottom one as we've drawn it. In these two spectra, this is probably the most dramatic example of the extra stability of a secondary carbocation. Examples involving acylium ions, [RCO]+ Ions with the positive charge on the carbon of a carbonyl group, C=O, are also relatively stable. This is fairly clearly seen in the mass spectra of ketones like pentan-3-one. The base peak, at m/z=57, is due to the [CH3CH2CO]+ ion. We've already discussed the fragmentation that produces this.
	Note:  There are lots of other examples of positive ions with extra stability and which are produced in large numbers in a mass spectrometer as a result. Without making this article even longer than it already is, it's impossible to cover every possible case.Check past exam papers to find out whether you are likely to need to know about other possibilities. If you haven't got past papers, follow the link on the syllabuses page to find out how to get hold of them.
Using mass spectra to distinguish between compounds Suppose you had to suggest a way of distinguishing between pentan-2-one and pentan-3-one using their mass spectra. pentan-2-one CH3COCH2CH2CH3 pentan-3-one CH3CH2COCH2CH3 Each of these is likely to split to produce ions with a positive charge on the CO group. In the pentan-2-one case, there are two different ions like this: [CH3CO]+ [COCH2CH2CH3]+ That would give you strong lines at m/z = 43 and 71. With pentan-3-one, you would only get one ion of this kind: [CH3CH2CO]+ In that case, you would get a strong line at 57. You don't need to worry about the other lines in the spectra - the 43, 57 and 71 lines give you plenty of difference between the two. The 43 and 71 lines are missing from the pentan-3-one spectrum, and the 57 line is missing from the pentan-2-one one.
	Note:  Don't confuse the line at m/z = 58 in the pentan-2-one spectrum. That's due to a complicated rearrangement which you couldn't possibly predict at A'level.
The two spectra look like this: Computer matching of mass spectra As you've seen, the mass spectrum of even very similar organic compounds will be quite different because of the different fragmentations that can occur. Provided you have a computer data base of mass spectra, any unkown spectrum can be computer analysed and simply matched against the data base.

Using a mass spectrum to find relative formula massThe formation of molecular ions When the vaporised organic sample passes into the ionisation chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion.
	Note:  If you aren't sure about how a mass spectrum is produced, it might be worth taking a quick look at the page describing how a mass spectrometer works.
The molecular ion is often given the symbol M+ or  - the dot in this second version represents the fact that somewhere in the ion there will be a single unpaired electron. That's one half of what was originally a pair of electrons - the other half is the electron which was removed in the ionisation process. The molecular ions tend to be unstable and some of them break into smaller fragments. These fragments produce the familiar stick diagram. Fragmentation is irrelevant to what we are talking about on this page - all we're interested in is the molecular ion.
	Note:  If you are interested in a detailed look at fragmentation patterns you could follow this link.
Using the molecular ion to find the relative formula mass In the mass spectrum, the heaviest ion (the one with the greatest m/z value) is likely to be the molecular ion. A few compounds have mass spectra which don't contain a molecular ion peak, because all the molecular ions break into fragments. That isn't a problem you are likely to meet at A'level. For example, in the mass spectrum of pentane, the heaviest ion has an m/z value of 72.
	Note:  This mass spectrum has been drawn using data from the Spectral Data Base System for Organic Compounds (SDBS) at the National Institute of Materials and Chemical Research in Japan.It has been simplified by omitting all the minor lines with peak heights of 2% or less of the base peak (the tallest peak).
Because the largest m/z value is 72, that represents the largest ion going through the mass spectrometer - and you can reasonably assume that this is the molecular ion. The relative formula mass of the compound is therefore 72.
	Note:  This assumes that the charge on the ion is 1+. That's always the case when you are interpreting these mass spectra.
Finding the relative formula mass (relative molecular mass) from a mass spectrum is therefore trivial. Look for the peak with the highest value for m/z, and that value is the relative formula mass of the compound. There are, however, complications which arise because of the possibility of different isotopes (either of carbon or of chlorine or bromine) in the molecular ion. These cases are dealt with on separate pages.
	Note:  The presence of the carbon-13 isotope in a molecular ion causes a little peak 1 unit to the right of the M+ peak. This is called the M+1 peak.The presence of a chlorine atom in a compound causes two peaks in the molecular ion region - the M+ peak and the M+2 peak depending on whether the particular molecular ion contains a chlorine-35 or chlorine-37 isotope. Bromine creates a similar problem. Follow these links if you are interested - or explore them later via the mass spectrometry menu.
Using a mass spectrum to find a molecular formulaSo far we've been looking at m/z values in a mass spectrum as whole numbers, but it's possible to get far more accurate results using a high resolution mass spectrometer. You can use that more accurate information about the mass of the molecular ion to work out the molecular formula of the compound. Accurate isotopic masses For normal calculation purposes, you tend to use rounded-off relative isotopic masses. For example, you are familiar with the numbers: 1H 1 12C 12 14N 14 16O 16 To 4 decimal places, however, these are the relative isotopic masses: 1H 1.0078 12C 12.0000 14N 14.0031 16O 15.9949 The carbon value is 12.0000, of course, because all the other masses are measured on the carbon-12 scale which is based on the carbon-12 isotope having a mass of exactly 12. Using these accurate values to find a molecular formula Two simple organic compounds have a relative formula mass of 44 - propane, C3H8, and ethanal, CH3CHO. Using a high resolution mass spectrometer, you could easily decide which of these you had. On a high resolution mass spectrometer, the molecular ion peaks for the two compounds give the following m/z values: C3H8 44.0624 CH3CHO 44.0261 You can easily check that by adding up numbers from the table of accurate relative isotopic masses above. A possible exam question A gas was known to contain only elements from the following list: 1H 1.0078 12C 12.0000 14N 14.0031 16O 15.9949 The gas had a molecular ion peak at m/z = 28.0312 in a high resolution mass spectrometer. What was the gas? After a bit of playing around, you might reasonably come up with 3 gases which had relative formula masses of approximately 28 and which contained the elements from the list. They are N2, CO and C2H4. Working out their accurate relative formula masses gives: N2 28.0062 CO 27.9949 C2H4 28.0312 The gas is obviously C2H4. In an exam, you would hope that - apart from the most simple cases - you would be given the possible formulae to work from. Trying to work out all the possible things which might add up to the value you want is quite time-consuming - and it's easy to miss an important possibility!

What causes the M+1 peak?What is an M+1 peak? If you had a complete (rather than a simplified) mass spectrum, you will find a small line 1 m/z unit to the right of the main molecular ion peak. This small peak is called the M+1 peak. In questions at this level (UK A level or its equivalent), the M+1 peak is often left out to avoid confusion - particularly if you were being asked to find the relative formula mass of the compound from the molecular ion peak. The carbon-13 isotope The M+1 peak is caused by the presence of the 13C isotope in the molecule. 13C is a stable isotope of carbon - don't confuse it with the 14C isotope which is radioactive. Carbon-13 makes up 1.11% of all carbon atoms. If you had a simple compound like methane, CH4, approximately 1 in every 100 of these molecules will contain carbon-13 rather than the more common carbon-12. That means that 1 in every 100 of the molecules will have a mass of 17 (13 + 4) rather than 16 (12 + 4). The mass spectrum will therefore have a line corresponding to the molecular ion [13CH4]+ as well as [12CH4]+. The line at m/z = 17 will be much smaller than the line at m/z = 16 because the carbon-13 isotope is much less common. Statistically you will have a ratio of approximately 1 of the heavier ions to every 99 of the lighter ones. That's why the M+1 peak is much smaller than the M+ peak. Using the M+1 peakWhat happens when there is more than 1 carbon atom in the compound? Imagine a compound containing 2 carbon atoms. Either of them has an approximately 1 in 100 chance of being 13C. There's therefore a 2 in 100 chance of the molecule as a whole containing one 13C atom rather than a 12C atom - which leaves a 98 in 100 chance of both atoms being 12C. That means that the ratio of the height of the M+1 peak to the M+ peak will be approximately 2 : 98. That's pretty close to having an M+1 peak approximately 2% of the height of the M+ peak.
	Note:  You might wonder why both atoms can't be carbon-13, giving you an M+2 peak. They can - and do! But statistically the chance of both carbons being 13C is approximately 1 in 10,000. The M+2 peak will be so small that you couldn't observe it.
Using the relative peak heights to predict the number of carbon atoms If there are small numbers of carbon atoms If you measure the peak height of the M+1 peak as a percentage of the peak height of the M+ peak, that gives you the number of carbon atoms in the compound. We've just seen that a compound with 2 carbons will have an M+1 peak approximately 2% of the height of the M+ peak. Similarly, you could show that a compound with 3 carbons will have the M+1 peak at about 3% of the height of the M+ peak. With larger numbers of carbon atoms The approximations we are making won't hold with more than 2 or 3 carbons. The proportion of carbon atoms which are 13C isn't 1% - it's 1.11%. And the appoximation that a ratio of 2 : 98 is about 2% doesn't hold as the small number increases. Consider a molecule with 5 carbons in it. You could work out that 5.55 (5 x 1.11) molecules will contain 1 13C to every 94.45 (100 - 5.55) which contain only 12C atoms. If you convert that to how tall the M+1 peak is as a percentage of the M+ peak, you get an answer of 5.9% (5.55/94.45 x 100). That's close enough to 6% that you might assume wrongly that there are 6 carbon atoms. Above 3 carbon atoms, then, you shouldn't really be making the approximation that the height of the M+1 peak as a percentage of the height of the M+ peak tells you the number of carbons - you will need to do some fiddly sums!

The effect of chlorine or bromine atoms on the mass spectrum of an organic compoundCompounds containing chlorine atoms One chlorine atom in a compound
	Note:  All the mass spectra on this page have been drawn using data from the Spectral Data Base System for Organic Compounds (SDBS) at the National Institute of Materials and Chemical Research in Japan.With one exception, they have been simplified by omitting all the minor lines with peak heights of 2% or less of the base peak (the tallest peak).
The molecular ion peaks (M+ and M+2) each contain one chlorine atom - but the chlorine can be either of the two chlorine isotopes, 35Cl and 37Cl. The molecular ion containing the 35Cl isotope has a relative formula mass of 78. The one containing 37Cl has a relative formula mass of 80 - hence the two lines at m/z = 78 and m/z = 80. Notice that the peak heights are in the ratio of 3 : 1. That reflects the fact that chlorine contains 3 times as much of the 35Cl isotope as the 37Cl one. That means that there will be 3 times more molecules containing the lighter isotope than the heavier one. So . . . if you look at the molecular ion region, and find two peaks separated by 2 m/z units and with a ratio of 3 : 1 in the peak heights, that tells you that the molecule contains 1 chlorine atom. You might also have noticed the same pattern at m/z = 63 and m/z = 65 in the mass spectrum above. That pattern is due to fragment ions also containing one chlorine atom - which could either be 35Cl or 37Cl. The fragmentation that produced those ions was:
	Note:  If you aren't sure about fragmentation you might like to have a look at this link.
Two chlorine atoms in a compound
	Note:  This spectrum has been simplified by omitting all the minor lines with peak heights of less than 1% of the base peak (the tallest peak). This contains more minor lines than other mass spectra in this section. It was necessary because otherwise an important line in the molecular ion region would have been missing.
The lines in the molecular ion region (at m/z values of 98, 100 ands 102) arise because of the various combinations of chlorine isotopes that are possible. The carbons and hydrogens add up to 28 - so the various possible molecular ions could be: 28 + 35 + 35 = 98 28 + 35 + 37 = 100 28 + 37 + 37 = 102 If you have the necessary maths, you could show that the chances of these arrangements occurring are in the ratio of 9:6:1 - and this is the ratio of the peak heights. If you don't know the right bit of maths, just learn this ratio! So . . . if you have 3 lines in the molecular ion region (M+, M+2 and M+4) with gaps of 2 m/z units between them, and with peak heights in the ratio of 9:6:1, the compound contains 2 chlorine atoms. Compounds containing bromine atoms Bromine has two isotopes, 79Br and 81Br in an approximately 1:1 ratio (50.5 : 49.5 if you want to be fussy!). That means that a compound containing 1 bromine atom will have two peaks in the molecular ion region, depending on which bromine isotope the molecular ion contains. Unlike compounds containing chlorine, though, the two peaks will be very similar in height. The carbons and hydrogens add up to 29. The M+ and M+2 peaks are therefore at m/z values given by: 29 + 79 = 108 29 + 81 = 110 So . . . if you have two lines in the molecular ion region with a gap of 2 m/z units between them and with almost equal heights, this shows the presence of a bromine atom in the molecule.

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palawan, philippines

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• Palawan - Wikipedia, the free encyclopedia

  en.wikipedia.org/wiki/Palawan
  

  
  

  Palawan (Tagalog pronunciation: [pɐˈlawan]), officially the Province of Palawan(Filipino: Lalawigan ng Palawan), is an island province of the Philippines that ...

  ‎Puerto Princesa - ‎Mimaropa - ‎El Nido, Palawan - ‎Palawan (island)

  

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1. Points of interest
   
   

   

   Puerto Princesa Subterran...

   

   Coron Island

   

   Honda Bay, Philippines

   

   Tubbataha Reef

   Calamian Islands
2. Destinations
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   Puerto Princesa

   

   El Nido, Palawan

   

   Coron

   

   Busuanga, Palawan

   

   Taytay, Palawan

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 Mass Spectrometry 

            A.  Theory
                        1.  Molecules are bombarded with a beam of  high energy electrons.
                        2.  The molecules are ionized and fragmented into many smaller pieces.
                        3.  Most of the particles carry a +1 charge.
                                    a)  This means that their mass/charge  ratio, the m/e value, is the mass of the ion.
                                    b)  For the case of the molecular cation, the m/e value is the molecular weight of the compound.
                                                i)  This is referred to as the M⁺ peak, and M⁺ is the molecular ion or parent ion.
                        4.  The different ions produced is sensed by a detector, which sends the information to a recorder which
                                    plots the signals in relationship to their intensities.
                                    a)  The largest signal is referred to as the base peak and is given a relative intensity of 100%.
                                                i)  The M⁺ is not necessarily the base peak.
                                                ii)  The M⁺ peak can sometimes be extremely small.
                                    b)  The intensities of the other peaks are then expressed relative to the base peak.
                                    c)   The mass spectrum is highly characteristic of a given compound and can be used to prove the
                                                identity of a compound or used to determine the structure of a compound.
                        5.  The mass spectrum helps in structure determination by
                                    a)  giving a molecular weight for the compound.
                                    b)  indicating certain structural units existing in the molecule, and
                                    c)  sometimes can give a molecular formula for the compound.

            B.  The M + 1 Peak
                        1.  The M⁺ peak is not the peak of highest m/e value.
                        2.  This is because of the existence of isotopes.
                        3.  Since about 1.1% of all carbon atoms is C-13, some of the molecular ions have a m/e value 1 unit
                                    higher than the parent peak.
                                    a)  The (M + 1) peak can also be due to the presence of deuterium, although the % abundance of ²H
                                                is only   0.015%.
                        4.  In the case of benzene, it would be expected that the (M + 1) peak would be about 6.6% as intense as
                                    the M⁺ peak.
                                    a)  Since there are six carbons in benzene, the % intensity expected is 6 x (1.1%).
                        5.  In compounds containing chlorine a significant (M + 2) peak would be observed since the
                                     % abundance of ³⁷Cl (2 amu's more massive than the more abundant ³⁵Cl) is about 24.23%.
                        6.  In compounds containing Br, an (M + 2) peak of about equal intensity as the parent peak is observed
                                    since the two isotopes of Br, ⁷⁹Br and ⁸¹Br, exist in a 100:98 [~1:1] ratio.
                                    a)  “Twin” peaks of about equal intensity but two units different in mass is a good indication of Br in
                                                 the compound. 

            C.  Example

                        1.  Neopentane has been analyzed by mass spectrometry and found to break into several fragments
                                    including the following.
                                    a)  Notice that the fragments formed are not obvious fragmentation products of neopentane.
                                                i)  Rearrangements obviously must occur.
                                                ii)  Stability of carbocations is a driving force for rearrangements in mass spectrometry.
                                    b)  Only about 10^-6 gram of sample is needed for the experiment.

                         2.  Fragmentation patterns and types of   rearrangements are well-known and are a key to the full
                                    utilization of mass spectrometry as a tool for structure determination.