So what does this mass spectrometer thing look like?

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1 Mass Spectrometry: What is it? Mass spectrometry is an analytical tool that makes use of the mass to charge ratio of particles to determine the molecular formula of a compound. In the lab, this tool can be immensely powerful in making sure what you make is actually what you want to make. The Theory Behind It So what does this mass spectrometer thing look like? As was mentioned before, mass spectrometry uses mass to charge ratios to determine molecular weights. To go from a sample of substance to something that can be analyzed, the mass spectrometer must first vaporize the sample and ionize it (give it a charge). In our context, this ionization occurs by slamming the gas particles of our sample with a beam of electrons. This beam will inevitably knock off an electron from the gas particles and form positively charged ions. Now, to keep everything simple, we want to ensure that every particle has only a +1 charge. Remember that this is because we are using the mass to charge ratio, so if the charge is a constant 1, then determining mass is made

2 that much simpler. So in order to ensure that things don t get too ionized (we only want +1 charged particles), we can put magnets that will accelerate the molecules as soon as they are ionized, as shown in the diagram. The accelerated particles are then directed into a magnetic field that will cause its flight trajectory to curve. The degree of curvature is dependent on how much mass each molecular ion has. The more mass an ion has, the less curvature its trajectory will experience. The ions then slam into a detector and its mass can be recorded. Reading the Spectra: How does one go about reading all the lines on the graph? In order to read the graph, we must first understand the basics of the graph itself. The x-axis of our graph is the mass-to-charge ratio (m/z for short), and because the charge of the ions is 1, our mass-tocharge ratio will be simply mass divided by 1, or more simply put our mass. The y-axis on the other hand is the relative ion abundance of our molecule at different weights. Relative abundance means that the abundances are based off of the most prevalent ion, not how many ions there are in total. Now that we know what the graph means, how do we find M, M+1 and M+2 from our spectra? See below.

3 What are M, M+1, and M+2? M is simply defined as the mass of the molecule with the lowest isotopes for each atom that comprises this molecule. M+1 is the molecular mass plus one. M+2 is you get the point.

4 What do we use M, M+1, and M+2 for? M is used to give us the molecular weight of the molecule. M+1 is used to determine the number of carbons within the molecule, and M+2 can be used to find the presence of sulfur, chlorine, or bromine. How do we determine this information from M+1 and M+2? In order to utilize M+1 and M+2 to determine information about the molecule, we must first set M as 100% abundance and set M+1 and M+2 as relative percentages of M. We do this to so that our analysis of the M, M+1, and M+2 abundances are parallel with the universal relative abundances, and from this parallel we can estimate the number of specific atoms within the molecule. The abundance of carbon-12 in nature is %, where as carbon-13 has an abundance of 1.107%. We set carbon-12 as 100% by dividing 1.107% by % and get a relative abundance of approximately 1.12% for carbon-13. Therefore, we can assume that the M+1 abundance present in the molecule is due to the carbon-13. For each carbon that is present in the molecule, the M+1 abundance goes up proportionally by 1.1%. Therefore, by dividing our M+1 value by 1.1% we are able to approximate the number of carbons present in the molecule. A similar concept applies for finding the amount of chlorine, sulfur, and bromine from M+2. The top slide of page 101 of the lecture supplement numerates the relative ratios of these atoms Determining Molecular Formula So now that we ve taken into account the relative isotope abundances, we can begin to determine the structure of our molecule. The best place to start with this is to locate the M+1 peak and use that to determine the approximate number of carbons in the structure. However, nitrogen and sulfur must also be kept in mind when looking at M+1 because the isotopes nitrogen-15 and sulfur-32 have a relative abundance of about 0.4% and 0.76%, respectively. Now we can examine the M+2 peak, assuming that there is a significant enough one. Referring to the relative abundance charts, we know that a sulfur atom is present if the M+2 peak is at about 4.4% intensity relative to the M peak. A chlorine atom is present if this peak is at about 32-33% intensity. Lastly, a bromine atom is present if the M and M+2 peaks are at a 1:1 ratio. Remember, presence of fluorine and iodine cannot be determined via mass spectrometry. If these atoms are present, Dr. H will let us know that they re there. With all this, we just need a few more bits of information before proposing some formulas

5 The Nitrogen Rule The nitrogen rule is a handy bit of information to keep in mind when proposing possible formulas. The basic idea is that any formula with an even mass number will have an even number of nitrogen atoms (0, 2, 4, ). On the other hand, if the mass number is odd, then the number of nitrogen atoms will be odd (1, 3, 5, ). This rule helps us limit the number of possible formulas that we can propose. The Hydrogen Rule We learned earlier that the general formula for a maximum number of hydrogen atoms on an acyclic alkane with n carbons could be given as 2n+2. We can use this information to help rule out formulas in mass spectrometry. Remember that any double bond or ring structure will decrease the number of hydrogen atoms by 2. This is illustrated below: Hexane: n = 6 carbons, 2n + 2 = 14 hydrogens. 2-Hexene: 6 carbons, but only 12 hydrogens. Presence of double bond reduces number of hydrogens by two

6 Cyclohexane: 6 carbon, but only 12 hydrogens. Presence of ring structure reduces number of hydrogens by two. When taking into account the number of hydrogens, we also have to consider that each nitrogen will increase the maximum number of possible hydrogens by 1. So if there is one nitrogen in a 4 carbon formula, then the maximum possible number of hydrogens becomes 11 instead of 10 (as predicted by the 2n + 2 formula). With all of these considerations, we get the general formula: Max number of hydrogens + halogens = 2C + N +2 This is referred to as the hydrogen/halogen rule. Putting It All Together So with everything covered up until this point, we have taken into account the number of carbons and the presence or absence of sulfurs, chlorines, bromines, fluorines, and iodines. We can subtract these masses from the total mass of the molecule (M) to get the mass that the nitrogens, hydrogens, and oxygens account for, and with that, formulas can be proposed. Just remember to check the validity of formulas with the nitrogen and hydrogen/halogen rules. Implications? We ve gone through all this trouble simply to learn to derive possible formulas from looking at mass spectra. Admittedly, this gives us a good deal of information regarding the composition of atoms within the molecules, but is that all? Of course not! With the molecular formula, it is possible to reveal more about the structure. Here, we introduce the concept of the double bond equivalent (DBE). This is essentially a unit defined as two hydrogens less than the maximum number of hydrogens possible in the structure. For example, two DBE s means that there are four less hydrogens than possible in the formula. If you remember back to the bit about the hydrogen rule, you ll see that each DBE implies the presence of either a double bond or a ring. When it comes to determining structure, this bit of information can be extremely useful. So to calculate DBE s, we can use a variation on the hydrogen/halogen rule formula. DBE = C H/2 + N/2 + 1

7 Now just a few more reminders. DBE s can only come in whole and positive numbers. Any negative or fraction DBE is an indication that you should probably recheck your formula and recalculate the DBE. Best of luck! References bih=683&tbm=isch&tbnid=sim1w9cg74feym%3a&imgrefurl=http%3a%2f%2fandromeda.ru tgers.edu%2f%7ehuskey%2f335f11_lec.html&docid=kqsqvbl- U5n4cM&imgurl=http%3A%2F%2Fnewark.rutgers.edu%2F%7Ehuskey%2Fimages%2Fmass_spe ctrum2.jpg&w=800&h=600&ei=xuert4zgjkqwiqkg8ydfag&zoom=1&iact=hc&vpx=753&vp y=154&dur=963&hovh=194&hovw=259&tx=182&ty=127&sig= &pag e=1&tbnh=144&tbnw=246&start=0&ndsp=13&ved=1t%3a429%2cr%3a3%2cs%3a0%2ci% 3A78