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Lecture Comments (26)

1 answer

Last reply by: Professor Starkey
Thu Aug 4, 2016 4:37 PM

Post by Adel Althaqafy on August 3 at 09:55:45 PM

Hi Dr
thank you for explain mass spectroscopy and do you have any articles that by PDF to read it or something is short like summarize about mass spectroscopy
many thanks  

1 answer

Last reply by: Professor Starkey
Sun Oct 18, 2015 11:04 AM

Post by Jinhai Zhang on October 17, 2015

Prof. Starkey:
I have a question about the McLafferty mechanism should I use the double-headed arrow or single-headed arrow, because I google this, the web gives me a single headed arrow instead of double headed, and the mechanism is a little bit different from yours?

1 answer

Last reply by: Professor Starkey
Wed Sep 16, 2015 12:16 AM

Post by Tamrat Regasa on September 15, 2015

how can i sketch mass spectrum for 2 phenyl butane?

1 answer

Last reply by: Professor Starkey
Mon Jan 5, 2015 11:45 PM

Post by Rene Whitaker on January 5, 2015

I am still not understanding how you determine the M+ peak in chloroethane (at 20:52) and bromobutane, other than you know the compound you are looking at and so know at about what amu you should be looking at.  For example, at 20:52, if you did not know the compound was chloroethane, how would you know the peak labeled at M+ is not the base peak (since it is at 100% relative abundance)?

1 answer

Last reply by: Professor Starkey
Sun Oct 26, 2014 12:53 AM

Post by Datevig Daghlian on October 24, 2014

Professor Starkey,

  Thank you very much for your lectures! I am currently a high school student and have taken AP Chemistry and am very interested in O. Chemistry. Would you recommend I watch your lectures on O. Chemistry or should I hold off till I get to University Chemistry? Thank you!

George D.  

3 answers

Last reply by: Professor Starkey
Thu Sep 11, 2014 10:07 AM

Post by Brandon West on September 9, 2014

Do you have a systematic way of doing mass spec like you have for HNMR and IR. I really like the systematic approach you have for HNMR. In my organic class my professor will give me all 3(HNMR or c13NMR, IR, and mass spec) and I will use all three to determine the structure.

1 answer

Last reply by: Professor Starkey
Thu Jul 17, 2014 4:17 PM

Post by Kim Tran on July 17, 2014

Professor Starkey, is there anyway I can open the lecture slides in powerpoint? Thank you

1 answer

Last reply by: Professor Starkey
Tue Mar 18, 2014 12:15 AM

Post by saima khwaja on March 17, 2014

Professor Starkey,

Do you do any lectures on Organometallic Compounds?

1 answer

Last reply by: Professor Starkey
Thu Feb 13, 2014 12:17 AM

Post by Jude Nawlo on February 12, 2014

Going back to the mass spectra of aromatic compounds section: If we see 91 amu on the MS, then do we assume that our compound has the "base" with the double bond on the CH2 or does it not matter, considering the benzylic resonance forms coexist? Awesome lecture, thank you again!

1 answer

Last reply by: Professor Starkey
Fri Jan 24, 2014 10:45 AM

Post by Udoka Ofoedu on January 24, 2014

You are the best . Thank God he brought me here

1 answer

Last reply by: Professor Starkey
Fri Oct 25, 2013 10:05 PM

Post by Saif Al-Wahaibi on October 25, 2013

For the McLafferty problem, where would the positive charge be in C4H8O so that it would be observed in the mass spectrometer?

1 answer

Last reply by: Professor Starkey
Wed Sep 11, 2013 11:05 AM

Post by Ramin Sadat on September 10, 2013

Thank you for the Mass Spectrometry lecture. I have been waiting for this! Very Helpful!

Mass Spectrometry

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  1. Intro
    • Introduction to Mass Spectrometry
    • Obtaining a Mass Spectrum
    • The Components of a Mass Spectrum
    • What is the Mass of a Single Molecule
    • Other Isotopes of High Abundance
    • Isotopic Abundance can be Calculated
    • Determining Molecular Formula from High-resolution Mass Spectrometry
    • Fragmentation of various Functional Groups
    • Mass Spectra of Alkanes
    • Mass of Common Fragments
    • Mass Spectra of Alkanes
    • Branched Alkanes
    • Mass Spectra of Alkenes
    • Mass Spectra of Aromatic Compounds
    • Mass Spectra of Alcohols
    • Mass Spectra of Ethers
    • Mass Spectra of Amines
    • Mass Spectra of Aldehydes & Ketones
    • McLafferty Rearrangement
    • Mass Spectra of Esters
    • Mass Spectrometry Discussion I
    • Mass Spectrometry Discussion II
    • Mass Spectrometry Discussion III
    • Mass Spectrometry Discussion IV
    • Mass Spectrometry Discussion V
    • Mass Spectrometry Discussion VI
    • Intro 0:00
    • Introduction to Mass Spectrometry 0:37
      • Uses of Mass Spectrometry: Molecular Mass
      • Uses of Mass Spectrometry: Molecular Formula
      • Uses of Mass Spectrometry: Structural Information
      • Uses of Mass Spectrometry: In Conjunction with Gas Chromatography
    • Obtaining a Mass Spectrum 2:59
      • Obtaining a Mass Spectrum
    • The Components of a Mass Spectrum 6:44
      • The Components of a Mass Spectrum
    • What is the Mass of a Single Molecule 12:13
      • Example: CH₄
      • Example: ¹³CH₄
      • What Ratio is Expected for the Molecular Ion Peaks of C₂H₆?
    • Other Isotopes of High Abundance 16:30
      • Example: Cl Atoms
      • Example: Br Atoms
      • Mass Spectrometry of Chloroethane
      • Mass Spectrometry of Bromobutane
    • Isotopic Abundance can be Calculated 22:48
      • What Ratios are Expected for the Molecular Ion Peaks of CH₂Br₂?
    • Determining Molecular Formula from High-resolution Mass Spectrometry 26:53
      • Exact Masses of Various Elements
    • Fragmentation of various Functional Groups 28:42
      • What is More Stable, a Carbocation C⁺ or a Radical R?
      • Fragmentation is More Likely If It Gives Relatively Stable Carbocations and Radicals
    • Mass Spectra of Alkanes 33:15
      • Example: Hexane
      • Fragmentation Method 1
      • Fragmentation Method 2
      • Fragmentation Method 3
    • Mass of Common Fragments 37:07
      • Mass of Common Fragments
    • Mass Spectra of Alkanes 39:28
      • Mass Spectra of Alkanes
      • What are the Peaks at m/z 15 and 71 So Small?
    • Branched Alkanes 43:12
      • Explain Why the Base Peak of 2-methylhexane is at m/z 43 (M-57)
    • Mass Spectra of Alkenes 45:42
      • Mass Spectra of Alkenes: Remove 1 e⁻
      • Mass Spectra of Alkenes: Fragment
      • High-Energy Pi Electron is Most Likely Removed
    • Mass Spectra of Aromatic Compounds 49:01
      • Mass Spectra of Aromatic Compounds
    • Mass Spectra of Alcohols 51:32
      • Mass Spectra of Alcohols
    • Mass Spectra of Ethers 54:53
      • Mass Spectra of Ethers
    • Mass Spectra of Amines 56:49
      • Mass Spectra of Amines
    • Mass Spectra of Aldehydes & Ketones 59:23
      • Mass Spectra of Aldehydes & Ketones
    • McLafferty Rearrangement 1:01:29
      • McLafferty Rearrangement
    • Mass Spectra of Esters 1:04:15
      • Mass Spectra of Esters
    • Mass Spectrometry Discussion I 1:05:01
      • For the Given Molecule (M=58), Do You Expect the More Abundant Peak to Be m/z 15 or m/z 43?
    • Mass Spectrometry Discussion II 1:08:13
      • For the Given Molecule (M=74), Do You Expect the More Abundant Peak to Be m/z 31, m/z 45, or m/z 59?
    • Mass Spectrometry Discussion III 1:11:42
      • Explain Why the Mass Spectra of Methyl Ketones Typically have a Peak at m/z 43
    • Mass Spectrometry Discussion IV 1:14:46
      • In the Mass Spectrum of the Given Molecule (M=88), Account for the Peaks at m/z 45 and m/z 57
    • Mass Spectrometry Discussion V 1:18:25
      • How Could You Use Mass Spectrometry to Distinguish Between the Following Two Compounds (M=73)?
    • Mass Spectrometry Discussion VI 1:22:45
      • What Would be the m/z Ratio for the Fragment for the Fragment Resulting from a McLafferty Rearrangement for the Following Molecule (M=114)?

    Transcription: Mass Spectrometry

    Hi; welcome back to Educator.0000

    Today, we are going to talk about another spectroscopy analytical tool that is called mass spectrometry.0001

    Now, mass spectrometry does not involve the absorption of light, like we had in infrared (IR) spectroscopy; in NMR, we had the absorption of energy from radio frequencies.0007

    Because it is not an absorption of light, we don't call it a spectroscopy, even though that word is used a lot; the correct term is mass spectrometry--a slight difference in the word there.0021

    We usually call it "mass spec" for short.0034

    And, as the name implies, it has something to do with the mass of a molecule; so what we can use mass spec to learn about the structure of a molecule is: we can determine its molecular mass.0038

    That is a great analytical tool: let's say you have isolated an unknown compound from a natural product, and you want to characterize--you want to learn something about it.0050

    One thing you could do is find out how much it weighs--what is its molecular mass?0060

    And, if you have a high-resolution mass spec--if you can find out very, very precisely what its molecular mass is--you can actually learn what its molecular formula is.0064

    Instead of doing an elemental analysis to determine its molecular formula, we can learn that information from the mass spec.0075

    In addition, it tells us some structural information; we are going to maybe figure out how the molecule is put together when we look at the mass spectrum of that molecule.0081

    There is not quite as much structural information as we might get from the proton NMR, where it tells you exactly which carbons are connected to which carbons, and so on; but when we use all of these spectroscopic techniques in conjunction with each other, all looking at every single avenue for a given unknown compound, that is when they can kind of feed into each other and build on each other, and we come up with a precise structure.0094

    Another very useful tool is to use this in conjunction with gas chromatography--GC is something we use to separate a mixture of compounds.0123

    So, if you can shoot an unknown mixture into a GC and have them spread out (like column chromatography) into their different components, and then, at the end, instead of just having a detector that says, "Oh, something came out; there is an organic molecule coming out," and indicating the presence of that, if it feeds into a mass spectrometer, then not only do you know that a component has exited the column, but then you take a mass spec of that component and find out what its molecular weight is, and again, something about its structural information.0133

    So, this is an extremely powerful analytical tool; it is called GC-MS for gas chromatography and mass spectrometry together; and that is just very routinely used in the analysis of mixtures.0164

    How does mass spectrometry work--how do you go about achieving a mass spectrum?0180

    First of all, let's assume we have a neutral molecule, just represented by these little shapes here; and we have this sample: we are going to inject it into the GC, and it is going to be vaporized.0186

    The first thing we do is: we heat it up and turn it into a gas.0197

    OK, then it is going to be blasted with a beam of electrons, and it is going to ionize it.0200

    This is known as EI, electron ionization; and what happens when you hit a molecule with a beam of electrons: it causes an electron to be ejected from the compound.0210

    An electron is going to be ejected from each molecule, so now, we have an unpaired electron; so it's going to be, now, a radical.0228

    So remember, electrons come in pairs: they are either bonded pairs, or they are non-bonded pairs; they are either lone pairs, or they are part of a double bond or a single bond or a triple bond...0237

    Electrons always come in pairs; so if you remove an electron from a neutral molecule, you now have a radical somewhere--an unpaired electron.0246

    Furthermore, you have a positive charge, because that electron you removed had a negative charge.0253

    So, if it started out neutral, the molecule now has a positive charge.0257

    What we get are radical cations: radical cations are the species that are formed in mass spectrometry.0262

    And this molecule is still intact--it is still a complete molecule--and there is essentially no change in mass.0270

    Remember, an electron has an insignificant mass compared to the neutrons and the protons in the molecule.0279

    So, although we have imparted a charge on the molecule, or caused a charge to form, we have not changed the mass.0284

    And so, we will analyze this to get the mass of the parent; but then, what happens is: this is a very high-energy environment, and so fragmentation occurs.0293

    What happens is: these molecules break apart, one way or another, to split up into radicals and cations.0306

    We are going to get some fragments that are cations, and some fragments that are radicals; we are going to get a mixture, then, of both charged and uncharged particles.0314

    OK, then this passes through a magnetic field and causes a deflection of the charged compound (the uncharged compounds don't bend, and continue on to the detector, but the charged compounds do).0323

    And it separates them based on their mass-to-charge ratio; so we call that the m/z; that is read as "mass-to-charge."0339

    Usually, the charge is +1...almost always, the charge is +1; so, essentially, what we are doing is: by bending it through this magnetic field, we are separating the fragments by their masses.0350

    They are starting out all together; they break apart, and then they spread out, and they get separated by their mass.0364

    We have these various charged particles now, as a result, and the lower-mass fragment is going to travel at a faster rate than the higher-mass fragment.0372

    And this one that is the complete molecule, intact, that has not undergone any fragmentation--we describe this as M+; that is called the molecular ion, because it is the complete molecule that has just been ionized by removing an electron.0382

    That is called the molecular ion.0401

    Let's take a look at an example of a mass spectrum; this is what a mass spectrum looks like.0405

    Down on the x-axis, we have our mass-to-charge ratio; so again, remember, our charge is usually +1, so that is, we are showing the various masses of the different fragments.0412

    And over on the y-axis, we have the relative intensity: this is the abundance.0424

    So, what we are seeing here is a histogram: how many of each mass was recorded--was formed in the mass spec and then recorded in the spectrum?0430

    These various charged fragments (remember, everything we see in a mass spec represents a charged ion, a +1)--these are going to hit the detector; it records the frequency of each mass, and therefore, the taller the peak, the larger the signal; that means it is a more abundant fragment.0442

    OK, so we look at this, and we see lots of peaks here, little peaks, little, tiny, tiny peaks, even; and here is our biggest peak.0462

    The biggest peak we set to a value of 100; so this is a relative intensity--it is compared to the fragment that you have the most of; and we describe that as the base peak.0469

    So, the base peak is the tallest peak in the mass spec; it is defined as 100%, and it sets the scale for everything else.0482

    Everything else is relative: this has about a 30%, compared to the base peak, and so on.0490

    Now, why do we have a lot of this fragment?0496

    Now, the way you read a mass spec is: you can literally just count over from these labeled hash marks--count over to see what number, what mass, we have.0500

    Here is 40, so this is 41, 42, 43; so this is a mass-to-charge of 43, and that is our base peak.0510

    Why do we have so much of that fragment--why did that occur so much?--it must be a very stable fragment.0521

    If this is a stable fragment--a stable cation or a radical cation--then, if it's very stable, then it is more likely to be formed, and it is going to be showing up in a higher abundance.0530

    What we do is: we look over wherever we see the highest mass; that is typically our molecular ion, so we call that the M+.0545

    And I say it's almost always the highest mass, because sometimes our molecules are so huge or so unstable that, as soon as they ionize, they fragment right away, and you never see...none of the full molecular ion makes its way to the detector.0555

    So, it's possible that no molecular ion is evident, but...well, it depends on what you are studying, but typically, in the problems we are working on, you will see a molecular ion, and it's going to be the highest one.0570

    So here, we have the molecular ion; it is at mass-to-charge of 100 in this case; and that is the mass, given in atomic mass units.0583

    That is how we would find out, looking at the mass spectrum of this compound--we would say, "Oh, we know it has a molecular mass of 100 there."0594

    A few other things we want to point out: the mass-to-charge of 43, our most stable fragment--if you compare that to the molecular ion of 100, you see that, if we have only 43 mass units remaining, that means it broke off a fragment that was the difference of 57.0605

    So, another way that you can describe this is: you can call this M-57.0626

    There are 43 mass units remaining, and we know that a break occurred in the molecule to remove 57 mass units from it; so there a couple ways that you can describe a peak that way.0632

    And, kind of like IR, which had a lot of peaks--a lot of stretches throughout the spectrum--we don't try and analyze every peak; we just pick out the major ones.0648

    OK, and the same is going to be true for mass spec: you are going to see lots and lots of fragments; molecules can just break apart in all sorts of ways; but we are going to look for significant peaks.0658

    And typically, when you are given a mass spec, you will be asked to discuss certain peaks, and that is what we'll be doing today.0667

    Now, this is interesting: we have our molecular ion here at 100, but if you look very, very carefully (this arrow is off a little bit--sorry), you will see that we have a second little peak--very tiny--at 101.0676

    We would describe that peak as M+1, because it has 1 atomic mass unit higher than the parent.0695

    Why is that--what would cause a remember, this is what is interesting about mass spec: we are recording the mass of single molecules and single fragments.0705

    So, if we are looking one molecule at a time, what would cause a molecule that has a mass of 100 to sometimes weigh 101--to sometimes have 101 grams per mole, or something like that, when we are looking at the molecular mass?0715

    Well, let's think about a single molecule and consider how we calculate the mass, or how we determine the mass.0735

    OK, so if we have methane (CH4), the mass of that will be the mass of a carbon atom (and that is 12) and the mass of four hydrogens (those are 1 each--4 times 1); so we are going to get 16.0742

    AMU are the units we are dealing with--atomic mass units.0760

    We would expect, if we ran a mass spec on methane, to find our molecular ion, our M+, peak at 16.0763

    OK, but remember, we have some carbons that are not C-12, but instead C-13; so what does it mean to be an isotope?--it means that you have a different number of neutrons in your molecule.0772

    So, carbon-12 has 6 protons (that is what makes it carbon) and 6 neutrons: we add those together, and that is how we get 12.0785

    But about 1% (so not many, but about 1%) of carbon atoms exist as C-13; so they have an extra neutron.0794

    When that 1 molecule out of every 100--when that one molecule comes through the mass spec, we are going to see that the carbon does not weigh 12; it weighs 13; plus, we have our four hydrogens; and it's going to be 17 amu's.0804

    And so, that is the molecule that is going to give rise to this M+1 peak.0820

    And because C-13 occurs about 1%, at 1 out of every 100 carbon atoms, this M+1 peak is going to be only about 1% of the relative intensity of the M+ peak.0828

    And that is pretty small, because our molecular ion peak is usually not a very large peak, because fragmentation usually occurs quite readily.0845

    So, it's usually a small peak; so we are looking at 1%, so it might be very, very small.0855

    OK, but what if we had two carbons?--let's think about: what would the molecular ion look like for ethane, C2H6?0861

    OK, well, now we have CH3, CH3; if we kind of have a bag full of ethane molecules, and we pull one out, one of them might have...most of them are going to have both of these as C-12; so let's say, for every 100 of these molecules, we are going to have one where there is a C-13 on the first molecule--on the first carbon atom.0869

    And we are going to have about 1% of them that will have a C-13 on the second carbon atom--so approximately 1 for each of would more precisely be 1 for every 200; we would get one of these and one of these.0904

    OK, so what that tells us, though: if we have two carbons, then our M+1 is about 2% of the M+, and so on.0920

    So actually, this is something that someone who is really studying mass spec and knows it very well--that person is going to look at the size of that M+1 peak, and that is going to tell us something about how many carbons there are in that molecular ion.0936

    That is a little bit of an analytical tool we can have.0953

    Sometimes, it can be big; if it's a very large molecule, that M+1 peak might be significant, because we have really increase...let's say you have 30 carbons: you have really increased the chances, the likelihood, that one of those carbons is going to be a C-13.0956

    And every time we have one of those molecules, they are always going to fall in that same slot, where it's one more than the molecular ion is.0970

    OK, so the take-home message is that the number of carbon atoms affects the relative height of the M+1 peak; it is usually very small, but it is going to be there.0979

    OK, well, what other isotopes are there that have high abundance?--1% is not very significant, and so it is not going to ever give rise to a really significant peak.0991

    But chlorine and bromine are two atoms that have...each of them has two isotopes that are relatively high-abundance.1002

    Unlike carbon, which is almost always just C-12, when you look at chlorine--when you look at the periodic table, you will see that it's 35.45; that is the mass that is assigned to chlorine.1013

    Well, you can't have a partial proton or a partial neutron in the structure, so of course your atomic mass units...your molecular weight is based on averages--averages of all the isotopes.1030

    But, since we are looking at single molecules and single atoms in mass spec, we need to know about the single isotopes that exist.1042

    And so, the way that, mathematically, we get this average of about 35 and 1/2 is: about 76% of the chlorine atoms are the 35 isotope, and about 24% are a 37 isotope.1051

    So, this is now going to be +2; so the naturally-occurring isotope of chlorine has two extra neutrons, compared to the lower-mass one.1066

    And so, what we are going to see is about a 3:1 ratio--if you have three-quarters...75% versus 25%, it's about a 3:1 ratio of M to M+2.1076

    We are going to have our molecular ion; if we have one chlorine, we are going to have our molecular ion, and then, +2 from that, we are going to have a peak that is about 1/3 of the size.1091

    And now, again, that is a big clue to the chemist looking at the spectrum, to see, "Wow, it looks like we have a chlorine in our molecule," because of this significant M+2 peak.1102

    OK, bromine is another example where it has an M+2; the two masses of bromine are either 79 or 81.1113

    And they are almost exactly equal--almost 1:1--so when you look at the mass of bromine, you see it's 80, but there is no bromine atom that weighs 80.1122

    It either weighs 79, or it weighs 81: because they are nearly equal and we average them together, we get the average atomic mass number of 80.1132

    OK, so what does a bromine look like?--again, very obvious when we see it in the mass spec: every fragment that contains a bromine has about a 1:1 ratio of M to M+2.1143

    We are going to get little doublets of peaks for every fragment that has a bromine in it.1157

    And so, both chlorine and bromine have a significant M+2 peak; so let's look at it one at a time.1162

    Here we have a mass spec; this is a chloroethane, so it does have a chlorine in it; now, how can we tell that it has a chlorine in it, even if we didn't know the structure of it?1168

    Well, we go to find our molecular ion, and we see that we have a lot of peaks up here; it is going to be very...this can't be our molecular ion, and then somehow we have two less from that.1177

    It is this large peak; actually, in this case, the molecular ion is also the base peak; this is going to be our M+--we'll label this as our M+, and this one is M+2.1191

    You can see that, if this one is that high, then this one is about three times higher than can go a little higher; it's about three times higher than that.1205

    We'll label this one as our M+.1216

    You see the difference there: the evidence here that we have chlorine is these two peaks; the lower-mass one is about three times higher than the higher-mass one.1221

    And they are two units apart: you can count that over, see?--1, 2.1233

    Are there any other fragments in this mass spec that also have chlorine?1238

    We look, and do we see that pattern again?--well, here it is at 49; I'm sorry, we could look at this, and we could say we have 49 and 51.1243

    And again, chlorine has a mass of 35 of 37, so this fragment is big enough to still contain a chlorine atom.1255

    And so, this also must still have a chlorine, because we see that 3:1 ratio here, as well.1264

    That can tell us something about our fragments that we have.1277

    OK, now let's take a look at our bromine: this is bromobutane, and this has a molecular ion of 137.1283

    We look for 137, and here it's very, very small peaks, but what do we see?--we don't see just one peak; we see two, and they are almost exactly the same abundance.1289

    This is what we would label as M (the lower-mass one), and we describe the higher one as M+2.1299

    We describe it with reference to the lower-mass isotope; the other one is the M+2.1307

    Any other bromine fragments?--I think I see a lot of little doubled peaks--this one and this one; those fragments must contain a bromine; and fragments contain bromine.1315

    And what about this one?--here we have another doublet: one is at 79; one is 81.1332

    What do we think that is?--well, remember, 79 and 81--that is the actual mass of bromine itself; so here, we are seeing the bromine atom as the cation; so this is 79Br as a cation, and this is 81Br as a cation.1339

    There is a lot of evidence here that we have a bromine in our structure.1359

    If we have multiple isotopes, we can actually calculate--based on their known abundance, we can calculate what the ratio of the peaks would be.1370

    And of course, the computers can predict this as well.1380

    What if we had dibromomethane, CH2Br2?--let's think about what the pattern would be there.1383

    So, what are the possible masses that we have for the CH2 group and the two bromines?1389

    The CH2 is going to be 14; again, we will assume that, every time we see carbon, it is carbon-12, because the likelihood of it being a C-13 is so small; OK, so we will ignore C-13, moving forward.1396

    This will weigh 12 for the carbon and 2 for the hydrogen, so that is 14.1410

    And how about the bromines?--well, the bromines can either be the lower isotope--they could both be the lower isotope, so that is a total amu of 172.1415

    Or we could, again, have a 14 for the CH2, and we can have...this first bromine can be the higher-weighing isotope, 81 (so that would be 2 more, so this would be 174).1427

    Or, instead, we could have the second bromine (if we are kind of labeling them #1 and #2)--we could say, "Well, the first one is a 79, and the second one is an 81"; this would still weigh 174.1445

    And then, of course, the final possibility is that both of these bromines are 81; and remember, this is like flipping a coin--bromine is like flipping a coin, because it's a 1:1 ratio--it's just as likely to get a 79 as it is an 81.1458

    So, what would this final mass be?--we started at 172, and then we added 2 and another 2; so this is 176.1475

    If you flip these coins--if you pull the molecules of CH2Br2 out of a bag, what are the odds that you are going to pull out a molecule that weighs 172?1483

    Well, out of every 4 molecules you pick out, one of them is going to be a 172; one of them is going to be a 176; but you are twice as likely to take a dibromomethane molecule that weighs 174, because there are two different ways that you can reach that same amu.1494

    OK, so what are the peaks going to look like?--they are going to be 1:2:1, 1 to 2 to 1; the middle mass is going to be twice as frequent, twice as abundant, as the lower mass and the higher mass.1520

    OK, and so, here is an example: this is the mass spec of dibromomethane, the exact molecule we are looking at.1538

    We predicted 172, 174, 176, and here it is: here is our 172; now, how do you count that?1543

    Now, again, this is 160 and 200; so that means this is 170, 80, 190; so this is 170, and then 171, 172; so this 172, we would describe as M or M+; this tall peak that is twice as high is M+2, so that is what we get when just one of the bromines is the higher isotope; and then, back to the same ratio: we have M+4, because both of our bromines had the higher isotope.1552

    We get a 1:2 ratio, and so on, and so on; if you have multiple bromines and chlorines, you can calculate those, and you can use the computer to predict those peaks, as well.1585

    But these ratios of molecular ion peaks really tell us something about how many halogens we have.1597

    Bromine and chlorine are the most significant isotopes that we are going to be encountering in mass spec.1603

    Now, if we do a high-resolution mass spec--a very, very precise mass spec where we can measure very precisely what the mass of the molecular ion is--we can actually determine a molecular formula.1615

    That is because carbon-12 is defined as exactly 12 amu; that is the standard; that is the reference point; that is what we set to be 12, that number 12.1629

    And everything else is relative to that; so hydrogen, we know, is 1, and nitrogen, 14, oxygen, 16.1640

    We are going to keep a periodic table nearby when we are working on mass spec problems; we are also going to keep a calculator nearby when we are working on mass spec problems, because we are going to be a lot of math, adding up fragments, molecular masses, and so on.1646

    OK, but these are the numbers that we are probably pretty familiar with.1658

    But oxygen is 16 (I'm sorry, did I say 15?)--oxygen is 16, but the actual number is 15.994914; it is not an exact number.1661

    And so, if you take your atomic mass unit out to 4 decimal places, then you will actually see a difference, based on the molecular formula.1671

    If we find that a molecule has a molecular mass of 98 grams per mole, what are the different formulas you can have for 98?1684

    There are a lot of different ways to come up with 98.1693

    So, we have all of these different options: 4, 5, 6, 7 carbons, and so on--some nitrogens, some oxygens--but when we carry it out to very precise masses, you see that they do differ, and the mass spec, then, will be able to differentiate between one formula and another.1697

    This is a really great tool for analysis: knowing exactly what your molecular formula is.1713

    OK, so what we are going to talk about--we have kind of talked about the theory of mass spec and how it works, and the idea that our molecules kind of fragment apart, and we get this spectrum out; what does the molecular ion look like?1724

    OK, so what we are going to look at next is: we are going to go functional group by functional group, and we are going to look at their structures, and we are going to ask, "Well, what are typical fragmentation patterns that we can have, based on a given functional group?"1736

    This is going to, now, tell us something about, "OK, we have this molecule; I don't even know how much it weighs"; but by studying how the molecule breaks apart, we can help put those pieces back together and learn something about the molecular structure, on how those atoms are arranged.1751

    If we consider that our radical cation that we started with is going to cleave, and we are going to get radicals; we are going to get cations...we need to talk a little bit about the relative stability of cations and radicals.1768

    OK, so let's just look at this example: we have a carbocation (a methyl carbocation) and a methyl radical; and here is the question: which do you think is going to be more stable--the carbocation or the radical?1782

    Now, both of these are unstable; we see them as only fleeting intermediates in reaction mechanisms that we have studied, so neither of these is a stable species, because both of them are missing an octet.1795

    Remember, carbon is happy when it has four bonds; it wants to have an octet.1808

    But who is less stable, and who is more stable?1814

    Well, if you think about it, the carbocation has only 6 total electrons around it, where the radical has 7 total electrons around it; this is actually (the radical is) closer to an octet.1817

    So, it is not quite as unstable; so this is more stable--it is not quite as unstable as a carbocation.1840

    The carbocation is missing a full bond; it has only 6 electrons; it is missing so many electrons that it has a formal charge; it has a positive charge; this is going to be less stable.1849

    For example, if we think about the mechanisms we have seen, if we ever wanted to form a methyl carbocation or even a primary carbocation, we would say "no way."1859

    That reaction doesn't happen, because this carbocation is so unstable.1869

    OK, but on the other hand, when we look at radical mechanisms like free radical halogenation, you can do that on methane; you can form a methyl radical or primary radical.1873

    It is not as stable as a more substituted radical, but it is doable.1882

    OK, so if we ever have to make a choice, we are going to find that putting the carbocation in the most stable position is going to be the best choice.1887

    OK, and so, the fragmentation we are going to have is the best one that gives relatively stable (remember, none of them are super stable, but the more stable the carbocation is, the more stable the radical is, the better)...1898

    And what is a very stable radical or carbocation?--well, the most stable is going to be tertiary; tertiary is the most stable, and we have the same reason for both.1912

    If we have a carbocation here or a radical here, this is electron deficient; and so, the more carbon groups you put on here, the better, because they can donate electron density and take away that deficiency.1927

    A tertiary is the best carbocation and the best radical; that is more stable than a secondary; it is more stable than a primary; it is more stable than a methyl.1939

    The methyl cation or radical (C+ carbocation or radical--we are looking at both of those)--this is the least stable.1951

    We need to keep that in mind as we are working problems today, because we are going to be counting on that.1964

    Remember, another way to stabilize a carbocation or a radical (besides being tertiary) is: what if we had resonance stabilization?1970

    If there is a way to delocalize that charge or that radical onto multiple sites, that would be very stable.1976

    So remember, we could have allylic or benzylic; those would also be very, very stable carbocations or radicals.1982

    We'll keep an eye out for those, too.1994

    OK, so let's take a look at alkanes: if we were to take an alkane and do a mass spec, what would we find?1998

    So, for example, hexane--just n-hexane, straight chain: if we inject this into the mass spec, it gets hit by the electron beam and shoots an electron out, and now we get a radical cation.2003

    Now, what electron got removed?--we can't tell: all of the electrons here are in σ bonds, are in single bonds; so there is no easy way to draw a single bond that isn't two electrons--that is only one electron.2016

    Usually, what we do is: we just put a bracket around the whole species, and we make it a radical cation.2030

    This is just telling us that somewhere in the structure, we are missing an electron, and therefore, the structure overall has a positive charge.2038

    OK, so this is a radical cation; because it's the complete molecule, we call that M+; and hexane has a mass of 86, so we would expect a parent ion--the molecular ion--at 86.2046

    OK, but then the molecule can break apart; it can undergo fragmentation.2059

    And where can it fragment?--it can fragment anywhere you want.2062

    If we break this bond, for example, it is going to give two fragments: one of these carbons is going to end up as a cation, and the other one is going to end up as a radical.2065

    And you have a choice of one or the other--either we could have this butyl group being the carbocation with an ethyl radical, or we could have the butyl radical and the ethyl carbocation.2076

    OK, so that cleaving can occur in either direction, with the electron going to the left or the right.2088

    OK, so what is that going to show us in the mass spec?--in the mass spec and in the mass spectrum, we are going to see the charged species only.2092

    We are going to observe these carbocations: we can see this peak at mass-to-charge of 57; so 57...we'll get to know some of these numbers; we are going to see them pretty regularly, and so we'll get to know them.2101

    57 tells me it's a 4-carbon unit, so it's a butyl group; and you could also describe this as M-29; 29 is an ethyl group, so this fragment that is remaining has lost an ethyl group from the parent, and there is that ethyl group.2117

    OK, or it could fragment in the other direction, so that the peak we see is 29; and we would describe that as M-57, because it's the parent minus a butyl group.2134

    OK, so there are a few different ways we could describe that.2144

    What is another way that we could break it up?--we could break it up right in the middle here, and now that breaks the molecule symmetrically; so no matter which way this breaks, we get the same two results: we get a propyl radical and a propyl carbocation.2147

    What is going to show up here?--the carbocation is going to show up, and the mass-to-charge of this is 43, which is also M-43.2161

    We have a propyl group remaining, and we had a propyl group that was removed.2170

    OK, or we could break off one of these end carbons and fragment, and when we do that, we can either end up with the methyl radical or the methyl carbocation.2176

    And so, the two fragments that we will see would be mass-to-charge of 15 (that is the methyl group), or mass-to-charge of 71 (remember, we started with 86, so to get to 71, we lost 15 amu, and 15 is what we get for a methyl group, carbon-12 and three hydrogens, so 15 is a number we'll see a lot, too).2186

    OK, so what do we expect to find in the mass spec of hexane?2211

    We would expect to find all of these peaks: the molecular ion, and then peaks at 57, 29, 43, 71, and 15.2217

    OK, and let's see what we have here...actually, before we get to the spectrum, let me just share with you some of these common fragments.2226

    You can find lists like this in any textbook--most Organic textbooks; any spectroscopy textbooks, for sure, will have a more complete list than this one.2235

    And it is a simple list of: for each mass number, what is a common way that you can have a combination of atoms to add up to that?2247

    So, rather than you having to do the work all the time, you can just use this table to help you identify them very quickly.2259

    For example, we saw that 15 is a methyl group; so either you have lost a methyl group, or the methyl group is remaining.2268

    You can look at these as either the radical portion or the cation portion, but all of these are just fragments that would be cut off from the whole structure.2276

    That is a methyl; this is an ethyl; so when you go from a methyl to an ethyl, you added a CH2.2288

    And so, that is 14 mass units, and so we are going to be seeing that; as we go from an ethyl to a propyl, then, we add 14 to 29.2297

    Here it is--here is our propyl.2309

    This is a methyl and an ethyl and a propyl; so these numbers are going to be ones that we will probably see a lot: 15, 29, 43, and then when you go to butyl, that is another 14, so that is 57.2315

    I wrote out C4H9; it doesn't matter how those four carbons are arranged--they could be in a straight chain; they could be branched--however they are; it is still going to be a butyl group; it is still going to have a mass of 57; and so on.2334

    You see some common things here that have some carbonyls in there; we might see those.2347

    So anyway, a table like this is pretty handy to have access to and to use; so you can search online for some tables like this: just Google mass spec common fragments, and you will find some examples there, too, if you don't have a textbook handy.2352

    OK, so now, let's take a look at that hexane spectrum: here are the peaks that we predicted that we would find, and so let's see: where is 15?2369

    This is 20, 30, 40; so this is 10, so this is 15--mass-to-charge of 15 (we can bring it in here).2379

    OK, we said we would find...and notice, 29, then, is an extra CH2; is 30; so 29 is one below that.2392

    That is there; 43--this is 40, so 41, 42, 43--it's this peak here.2402

    57--55, 56, 57--so these are all pretty significant peaks, right?--pretty highly abundant peaks.2410

    71, down here: this tiny little peak is 71.2418

    And 86 was our molecular ion.2424

    I'm sorry, this isn't a peak; this is just the edge of the spectrum--sorry; that is a little misleading.2429

    OK, so in fact, we see (and remember, this is our molecular ion at 86) all the peaks that we predicted; there are more peaks than what we predicted; so again, we are not going to try and worry about every peak in the spectrum--just the ones that we're interested in.2438

    But when you look at the ones that we predicted, there are two that are very, very small--15 and 71 were not very abundant.2454

    Can we explain that?--that would be something that is useful to explain.2463

    How do we get 15 or 71?--we started at 86, so 15 means that we broke this bond, and we saw this methyl carbocation; so CH3+ would be mass-to-charge of 15.2466

    And how did we get 71?--well, 71 is M-15, so that is where we lost a methyl unit--a methyl radical; and what was remaining was the charged...the remaining 5 carbons, so a pentyl group is 71.2487

    Oh, this is a carbocation--excuse me.2513

    This is our mass-to-charge of 71, and this is our mass-to-charge of 15; now clearly, both of those are very low-occurring fragmentations, so can we see why that is--why that is not as good as some of the others?2515

    Well, think about the carbocation just formed: you just formed a methyl carbocation; that is not very stable.2531

    And in order to get this primary carbocation--this primary carbocation is stable, but what did you have to lose in order to get there?2538

    You had to cleave a methyl radical.2544

    So, this fragmentation, no matter which way it goes, produces a relatively unstable fragment; and therefore, it's less likely to occur in that position.2547

    CH3+ and CH3 radical are relatively unstable; so that is the answer to our question of, "Why are the peaks so small?"2558

    These are relatively unstable, so this fragmentation is relatively rare.2576

    It is going to be less abundant; very good.2589

    What if we had a branched alkane, instead of just a hexane itself, but we had this 2-methylhexane (is that still 6 carbons...yes, 2-methylhexane)?2593

    We have a base peak; it is telling that the base peak is at mass-to-charge of 43, and it's asking to explain why that would be.2605

    Now, remember what the base peak is: what does it mean to be the base peak?2614

    That means it is the most abundant fragment; we set this to 100%; this is the tallest peak in the spectrum.2618

    So, why would this be an exceptionally high peak?2629

    Well, remember, 43...if we go back to our little table here, 43 is a propyl group--is 3 carbons; let's just say it's 3 carbons, 36 plus the hydrogens, where 57 is a 4-carbon chain.2635

    So, we have these seven carbons; we are breaking it somehow into a 3-carbon unit and a 4-carbon unit; and which is the one that has the charge?2655

    Remember, mass-to-charge: whatever we see in the spectrum is the thing that has the positive charge, so we're going to get a 3-carbon carbocation by losing a 4-carbon radical.2664

    Where can we break this molecule that would be a favorable disconnection--a favorable fragmentation that would give a stable, a stable 3-carbon carbocation?2674

    Well, if we go to the site of the branching, if we were to cleave it in this position, we would get a 3-carbon chain and a 4-carbon chain; and the 3-carbon carbocation would be a secondary carbocation.2689

    That is relatively stable, compared to all of the other carbocations; if you broke it in any other position, you would be getting a primary or methyl carbocation; but in this one, we would get a secondary carbocation.2711

    We have removed this 4-carbon radical that was 57, and what we are left with is the 3-carbon mass-to-charge of 43.2721

    And so, yes, we would actually predict that to be a favorable fragmentation; in fact, in this spectrum, we find that it is the base peak.2734

    Let's take a look at some other functional groups: what if we have functional groups in the molecule, like an alkene--where would we expect the molecule to cleave, preferentially (more likely to cleave)?2743

    OK, once again, we are going to start with the same process: we start with our parent compound; we remove one electron from it; and we end up with a radical cation--the radical cation.2755

    We just put the entire structure in brackets to make things easy; we could just put the whole thing in brackets and put a plus radical--radical cation.2765

    OK, now where is it going to cleave?2774

    One possibility is allylic cleavage: if we look at this carbon that is next to the π bond (remember, a carbon next to the π bond we call an allylic carbon, so this is the allylic carbon), we would describe this as allylic cleavage.2776

    And if we break it there to give a carbocation, we would break off this ethyl group; it gets lost as a radical; and if we put the carbocation in this position, what is a good thing about having a carbocation in this position?2798

    It is an allylic carbocation; that is one of the best carbocations we have, because it has resonance.2811

    What does the resonance look like?--we can bring this π bond over and move the positive charge to another location; that is the best resonance that we, the best stabilization that we have: it's when you can delocalize the positive charge and move it to another atom.2816

    OK, so what we are going to look for, as usual, is a fragmentation that is going to result in stable fragments, stable carbocations, and this gives us an opportunity for an allylic one--very, very stable.2833

    Now, sometimes you are asked for a mechanism, rather than just this vague dotted line saying, "Oh, we're going to break here."2846

    If you were asked for a mechanism--a reaction mechanism--to show how the molecule cleaves, well, we can do that for an alkene, because really, if you are going to remove an electron from this system, it is probably going to be one of the high-energy π electrons that gets ejected.2854

    And so, instead of just using the brackets like this, we can redraw the π bond as a radical cation.2871

    And now, when we do that radical cation, we come back to our allylic; this was allylic to the π bond; this is the bond that we want to cleave; and we can now show a mechanism for that cleavage, where we break it homolytically.2879

    One electron goes to the ethyl radical we are losing; the other electron pairs up with the radical; and again, we are losing our ethyl dot, our ethyl radical.2896

    And what we are left with is our resonance-stabilized allylic carbocation.2910

    You could see, we jumped right to this second resonance form that we had here.2918

    So sometimes, you will be asked for a mechanism for the fragmentation, and it would be a radical mechanism like this, and we would start with an explicit radical cation, rather than just kind of using the brackets.2922

    What if we have an aromatic compound, like something with a benzene ring in it?2943

    OK, now again, let's look at this molecular ion; it's a radical cation; where could we break it?2946

    Well, just like we looked for the allylic carbon as a good place to break from, this is called the benzylic carbon; and so, breaking here, we would describe this as a benzylic fragmentation.2953

    And that would be a good place to break the molecule up, because it will allow us to put the carbocation in the benzylic position.2972

    What fragment did we just lose?--remember, our radical cation is going to be cleaving into a radical and a cation.2979

    We just lost this 2-carbon...just like in the last problem, it's a 2-carbon radical, this ethyl radical.2988

    And what remains is this benzylic carbocation.2994

    Now, why is it good to have a benzylic carbocation?--because just like being allylic, this is resonance-stabilized, and we can delocalize the positive charge...and so on; etc.3000

    We could end up using all three of these π bonds that are in the benzene ring.3018

    So, this is another very, very stable carbocation to have; this benzylic resonance is very stable.3022

    And in fact, what happens sometimes when you have this benzyl cation--it can rearrange within the hydrogen mass spectrum environment; this benzylic carbocation can rearrange to this one, known as the tropylium carbocation.3035

    And this is still an aromatic system; it has 2, 4, 6 π electrons; it is aromatic.3053

    And so, when we see this peak of 91, it is either the benzylic phenyl CH2...phenyl CH2 is 91, or it may have rearranged to be this tropylium cation.3063

    It doesn't really matter what the actual structure of the fragment is, but it tells us that we had a phenyl group with a CH2 as part of the molecule, initially.3077

    When we take a look at alcohols, alcohols are going to undergo a type of fragmentation described as α cleavage; and α means it is on the very first carbon, so this is the carbon that is bearing the functional group--this is the α carbon.3094

    And so, breaking at that carbon is described as α cleavage.3110

    So, if I break off this ethyl group again...if I remove this ethyl group, it gives me the carbocation in this position, in the α position, and why is that a good place to put a carbocation?3115

    Once again, we get that resonance: any time you can come up with a resonance-stabilized carbocation, that is going to be a significant fragmentation in the mass spec.3130

    So, how does this resonance work?3138

    Well, in this case, with the oxygen, it shares its lone pairs and delocalizes the positive charge onto the oxygen.3139

    So, once again, we have a resonance-stabilized carbocation.3148

    And so, α cleavage is the place we are going to look for, for alcohols.3160

    Now again, this is another one where we can show a mechanism for that cleavage, because when we have a lone pair of electrons in a molecule, again, those are the highest-energy electrons, so those are the ones that are most easily removed when we initially do the ionization.3165

    Instead of drawing it with this bracket, we can actually remove one of the lone pair electrons on the oxygen; so this oxygen, now, has a positive charge.3185

    1, 2, 3, 4, 5: oxygen has 5; it wants 6, so this is an O+.3194

    And now, to do our α cleavage, this is the bond we break, and we break it homolytically.3200

    One electron gives us the ethyl radical we are expecting; the other electron pairs up with the existing radical.3207

    And we end up with our resonance-stabilized carbocation.3218

    And again, this kind of brings us right to the better resonance form that has a filled octet; so alcohols are another one that we can show a radical mechanism on how that fragmentation occurs.3225

    What else can we have for alcohols?--we can have α cleavage as one thing to look for; another possibility is to lose water from the molecule.3237

    In other words, M-18 may be a peak that is observed.3248

    Now, when you look at this structure, and you think, "Well, sure: I could break this bond, C-O,"--but this is not 18; this is only 17, so this would not be the correct thing: an oxygen is 16+1.3252

    So, losing water means that, first, a rearrangement must have occurred so that the oxygen gets another hydrogen on it.3267

    Then there is a cleavage, so that the fragment that is lost is H2O.3275

    So again, why is it favorable?--because you are losing a very, very stable molecule, but the mechanism is a little more complex.3279

    If you see an M-18, it might tell you that you have an alcohol, as well.3287

    What if we have an ether instead of an alcohol?3296

    We're going to do the same sort of thing: we are going to do an α cleavage, and this is the carbon that has (let me flip my page--sorry) the oxygen on it, so that is the α carbon.3298

    This is the bond we are breaking; so we would end up with a carbocation at this position.3318

    Or, now, because an ether has two carbon groups attached to it, we could look at the other α carbon; this is our α carbon, and so we break the bond from that α carbon.3327

    Remember, our goal is to get the carbocation at the same carbon that the oxygen is attached to.3338

    We can either lose this carbon group or this carbon group; we would get both of these peaks that we would expect, and why is this a stable peak?--because being next to the oxygen, we could have resonance stabilization.3345

    OK, so with an ether, we have more than one α cleavage option available to us.3359

    It is also possible that you can cleave right at the oxygen, the carbon-oxygen bond; so, in that case, you would lose an O-R radical (like a methoxy, ethoxy, propoxy, something like that--whatever that alkyl group is, plus the oxygen), leaving behind the other carbocation.3367

    The mass of this would be the mass of the parent, minus the O-R group (methoxy, ethoxy, propoxy, something like that).3390

    So, that is another possibility; so again, we have our α cleavage, and then we can also--less common, but we can also break at the C-O bond.3398

    Amines are going to be doing very much the same thing we saw for oxygens or for alcohols: we still call it α cleavage.3411

    Now, where would the α cleavage occur?--here is the carbon with the nitrogen, so that is the α carbon, and that is where we want to put the positive charge; so that is where we would be breaking the molecule.3419

    In this case, we would be clipping off this propyl group.3430

    This propyl radical would be lost, and we would end up with this CH2NH2.3434

    OK, and why is that a good place to fragment--why is that favorable?--because, just like oxygen, this nitrogen has a lone pair.3443

    It is, in fact, even better than oxygen; nitrogen loves sharing its lone pair and doing resonance, so we once again have a resonance-stabilized carbocation.3449

    So, α cleavage is very good for amines, just like with the oxygens.3463

    Now, another interesting thing that comes from amines is something known as the nitrogen rule.3471

    Now remember, nitrogen has three bonds in its stable form.3476

    Carbon, oxygen...they have even numbers of bonds (2 or 4), but nitrogen has three; so every time you have a nitrogen, it requires an extra hydrogen overall in the structure, to fill that bond.3481

    When you look at a molecular mass, almost all molecular masses are even; they end in an even number (0 or 2, 4, 6, 8).3498

    So, if you see an odd number mass, that tells you that you must have a nitrogen in your structure.3508

    And so, that is an interesting tool--a little trick that we call the nitrogen rule.3516

    So, if it is odd, then there is an odd number of nitrogens; so you have 1 nitrogen, or maybe you have 3 nitrogens--because, if you had 2 nitrogens, then that adds one hydrogen and it adds another hydrogen; we are back to an even number.3520

    The rule is that, if you have an odd molecular mass, you either have one or three or five or seven have some odd number of nitrogens.3532

    And if you have an even molecular mass, you either have no nitrogens, or you have an even number of nitrogens (2, 4, 6, and so on).3539

    That is another tool that chemists have when they are learning something about a molecular mass: you will see, right away--there is a hint whether or not you have a nitrogen in your structure.3547

    What about carbonyl-containing compounds?3564

    Let's look at aldehydes and ketones.3566

    OK, now these still undergo what is described as α cleavage, but it looks a little different than we saw for oxygen or nitrogen compounds--ethers and alcohols--because it occurs right at the carbonyl.3569

    This is what we usually describe as the α carbon--next to a carbonyl, we call those α carbons.3583

    We can deprotonate α carbons and make enolates and that sort of thing.3590

    And so, the α cleavage is actually going in the other direction; it is cleaving between the α carbon and the carbonyl.3593

    And when we do that, we lose the R. here; this radical gets cleaved, and we get the carbocation at the carbonyl.3600

    This is described as an acylium ion, and why is this a good ion to have?3610

    Well, once again, we are next to lone pairs of electrons, and any time you have a vacancy next to lone pairs, we can have resonance.3615

    We can redraw that carbocation to be an O+, and it's resonance stabilized.3625

    So, aldehydes and ketones cleave on either side of the carbonyl; so this mass-to-charge would be the parent minus this R group, or we could cleave in the other way.3631

    So, if we call this A and we call this B, for B, we would have this acylium ion that is resonance stabilized.3644

    You don't always have to show the resonance; I'm just trying to make sure that you are comfortable with it and have enough experience with it that you recognize it as a good carbocation.3657

    And we also form our prime dot in this case; so we have lost an R prime; so the mass-to-charge for this acylium ion would be M minus R prime, whatever that R' group may be...methyl, ethyl, propyl, butyl--it could be a massive carbon chain--whatever it is, your molecular ion is going to weigh that much less.3666

    Now, there is another interesting rearrangement--there is another interesting fragmentation that can occur for ketones or aldehydes or anything containing a carbonyl.3691

    It is described as a McLafferty rearrangement.3701

    This is what it looks like: you are going to have your molecular ion; so you start with your parent compound; and what happens is: you kind of wrap around the molecule so that one of these hydrogens is now 6 atoms away--1, 2, 3, 4, 5, 6 atoms away.3703

    And, when you have the possibility for a 6-membered transition state, those mechanisms are always very favorable, or can be very favorable, because it is very easy for the molecule to achieve the 6-membered ring because there is no ring strain.3722

    What happens is that this π bond reaches up...we are going to form a bond here between the oxygen and the hydrogen.3736

    We use the π bond to form that bond; we break this C-H bond to form a π bond; and now, this carbon would have 5 bonds, so we have to break a bond there, and we break this carbon-carbon bond and move it over to be a π bond.3745

    It is a 6-electron system; we call these pericyclic reactions, and we have cyclic transition states--all of the electrons flowing around in the ring.3760

    And it ends up breaking (see, notice, we just cleaved the molecule between those two carbons)--we break between the α and the β carbons of the carbonyl-containing compound, and we get two fragments.3769

    Now, this fragment is a neutral molecule; we just created this neutral alkene, so that just kind of hits the walls when it goes through the magnetic deflection; this kind of falls off--we don't see it in the detector.3782

    But this is the fragment we see now.3798

    This is the charge that is observed.3802

    Now, we just made an enol; this enol would very likely tautomerize to the ketone; but the structure of the fragment doesn't matter so much--what matters is the mass and how we got there.3807

    What we would do is: if you are ever asked to do a McLafferty rearrangement, it really helps to actually draw out the intermediate that is formed, the intermediate transition state that you are getting, because you will notice: it is not just a matter of breaking this carbon-carbon bond; you also transfer this hydrogen from this carbon chain to this carbon chain.3823

    There is a 1-unit difference in this original carbonyl portion that remains.3846

    What do we expect from an ester?--an ester can do a few things: it can undergo α cleavage, so just like any carbonyl, it can undergo α cleavage.3857

    We can lose this R. to give an acylium ion.3868

    It can undergo a McLafferty rearrangement, so one of these sides can wrap around and form the 6-membered ring and cleave that way between the α and β carbons.3873

    Or, it can lose RO.; so just kind of like we saw for ethers, that also corresponds to an α cleavage here.3884

    You could kind of treat it like a ketone; it can break on either side of the carbonyl, or it can possibly undergo McLafferty rearrangement.3893

    Let's try some problems, now that we have had an introduction to mass spec and thinking about the various fragmentation patterns that are possible.3903

    Let's try a few problems.3914

    OK, so for example, "For the given molecule, mass of 57"...that is know how I knew I had a typo?--why can this molecule not have a mass of 57?--because that is an odd mass, and there are no nitrogens in the structure.3916

    Look at that: I just realized that I have a typo (and see, I corrected it in my notes, but not on my slide).3934

    So, actually, it has a mass of 58, which you can calculate if you need to (so again, always have that calculator handy); but a lot of times, you will be given the mass.3940

    The question is: do you expect the more abundant peak to be mass-to-charge of 15 or mass-to-charge of 43?3948

    If we use our cheat sheet, do we get to 15?3956

    15 is a CH3, so that would be a CH3 carbocation; and 43 is actually 3 carbons; so 15 is one carbon, and 43 is three carbons.3963

    Remember, you had 36, and then you add in the extra 7 hydrogens.3975

    So, this is a 4-carbon...I'm sorry, 3-carbon (I just said 3 and wrote 4); so we are going to be breaking this molecule up, this 4-carbon molecule, into 3 carbons and 1 carbon.3981

    If this is the only place we can break it to do that, how do we get these two different fragments as a result?3995

    Well, remember, let's start with this as a radical cation, because that is the only way it is going to cleave.4003

    Let's assume it is a radical cation; and now, we can break it so that the propyl group ends up with the radical and the methyl is the cation, or the propyl group is the carbocation and the methyl is the radical.4011

    OK, so the question is: which of these do you expect to be more likely?4030

    They are both forming unstable methyl species, but which one is even less stable?--this methyl group is less stable than the radical.4035

    We would expect that this primary carbocation is more stable, even though it had to kick off a methyl radical; that is OK--that is better than having this methyl carbocation.4050

    This is less stable.4069

    We expect...who do we expect to be more abundant?--the more stable; this has the mass-to-charge of 43; it is going to be more abundant.4072

    This is the better fragmentation.4085

    OK, very good.4091

    OK, now we have a molecular ion of 74; so again, we can kind of assume this is a radical cation, so that that is our 74 for the M+.4096

    And the same thing--we are asking, "Who is more abundant: 31, 45, or 59, and how do we get those?"4112

    OK, so now we have to think about our structure and think about how to break it apart.4120

    59...a mass-to-charge of 59: what size fragment have we lost?4125

    We started at 74; so it looks like this is -15...4134

    In fact, these are good problems for you to try, if you want to pause the lecture and try each of these on your own, before I do the discussion of it.4139

    This is really good practice because, based on what we have had so far, you really should be able to work on these problems on your own.4147

    The mass-to-charge of 59 is -15; so that means it cleaved like this to leave a peak of 59.4154

    45, mass-to-charge of 45, is M-29; so 29 is our ethyl group, so that means we cleaved off this 2-carbon unit.4167

    And the mass-to-charge of 31 we could describe as -43, which is a 3-carbon unit, so now we are breaking it like this.4184

    Which of these three bonds are we breaking?4201

    And so, you think about an alcohol, and you say, "Which of these is more likely--is there one that is going to be more favorable than another?" and you think about the carbocations that we form, either at this carbon or this carbon or this carbon.4206

    Which is going to be the most favorable carbocation?--it's going to be the one where the carbocation is at the same carbon as the oxygen.4221

    So remember, we described this as α cleavage; we said α cleavage is very stable; it is very favorable, because the carbocation we would get is resonance-stabilized.4228

    OK, and so, not only answering the question we just asked (which do you expect to be most abundant?) the mass-to-charge of 43 we would expect to be the most abundant; but your answer should extend beyond: "It's most abundant, because it's α cleavage, and that is what I learned."4251

    You want to look at the other possibilities and show that that is actually the most stable carbocation.4274

    If we had this carbocation--it's just a random primary carbocation, not very stable--and if you took off the methyl group, we would have this carbocation; these are just primary carbocations and no resonance stabilization.4280

    You want to demonstrate why the α cleavage is preferred over the other possible fragmentations.4295

    OK, here is one more to try: Explain why the mass spectra of methyl ketones typically have a peak of mass-to-charge of 43.4305

    So, provide the structure of this fragment and a mechanism for its formation.4313

    First of all, what does it mean to be a methyl ketone?4318

    A ketone has 2 R groups on either side, and a methyl ketone tells us that one of those sides is a methyl group.4322

    And what fragmentation do we expect as a common one for ketones?4332

    Well, for ketones, we can break on either side here; and so, if we take a look at cleaving in this position, that would cleave off whatever that...4338

    This is why it is common: it doesn't matter what that other carbon group is, because it gets lost in this fragmentation, and what remains is this unit.4350

    Is that 43?--well, we can check our cheat sheet if we have it; otherwise, we could do our old-fashioned...we have an oxygen; we have carbon times 2; we have hydrogen times 3.4362

    We really have to be able to do this math, either in our head or with a calculator if we are allowed.4373

    This is 16; this is 24; and this is 3; and we have 43.4379

    OK, so don't hesitate to do this math and do the calculation; you don't want to make an assumption or think it's "close enough"; you really want to verify that.4386

    This certainly would have a mass-to-charge of 43, and that is the fragment that is there.4395

    What if we had to show a mechanism for this?4401

    Well, if we consider the molecular right now, this kind of assumed that this was the radical cation; so remember, this fragmentation isn't something that happens with neutral, stable molecules.4403

    This only happens after we ionize the molecule; we form a radical cation; it is a very high-energy species; and then, that is how we break it into a radical and a cation.4417

    So, make sure you are starting with the proper species.4427

    What we can show is: instead of using the brackets, we can show one of the lone pairs of electrons on the oxygen being removed.4431

    In that case, it's the oxygen that is the radical cation.4445

    And so, what does my mechanism look like?--this is the bond that I'm breaking, so we are always going to break that homolytically.4451

    We are going to lose the R.; we lose the R., and then this electron pairs up with the radical that was here; so that brings us, right now, to a triple bond.4456

    We call this an acylium ion.4470

    That is the way that we can show the mechanism for this α cleavage; we call this α cleavage for a ketone.4476

    OK, so now we have a mass spectrum of this given molecule; the mass is 88.4489

    Account for the peaks at mass-to-charge of 45 and mass-to-charge of 57.4497

    So, mass-to-charge of 45 is going to be...well, we started at 88, so this is M-43; so 43 is our 3-carbon unit.4502

    And mass-to-charge of 57 is M-31; now, 31 isn't a number that is common to me, but 57 is; this tells me that it is a 4-carbon unit, a butyl group.4523

    This is a propyl group; this is a butyl group.4539

    Whether you are looking at the mass, trying to learn something from the mass of the fragment--the cation...sometimes that works, and you can find a connection there and where to disconnect, but sometimes it's looking at the mass of the fragment that is lost, the part that is lost; you are looking for a recognizable thing.4541

    Here I see we have a 3-carbon chunk being lost; so here is my propyl group (1, 2, 3); that looks like a good disconnection here.4562

    If we call this a and this would be disconnection a; and mass-to-charge of 57 means we are left with this 4-carbon chain; we are left with this butyl group, so this must be disconnection b.4571

    In other words, these are both common peaks we would expect for the mass spec of an ether.4590

    This one is minus...losing an OR., and so this is the fragment that is left here.4600

    And this is the α cleavage; so here is our α carbon, and we are doing α cleavage.4609

    For the 45, we have lost the propyl group, so we have done our α cleavage, and that is resonance-stabilized.4618

    And for b, we are remaining with our 4-carbon unit; so this is -OR.; we have lost a methoxy; and for a, we have done our α cleavage.4638

    And, in this case, the α cleavage has lost a propyl group of 43.4664

    OK, so it just asks for a count for the peaks; add those two masses; it doesn't say to show a complete mechanism or something, so we can just kind of do this work in our heads.4670

    And keep in mind that if you don't have a table that adds up 45, that you can do the math here, of your 24 and 16 and another 5; and this is 4 carbons and 6, 7, 8, 9 hydrogens.4685

    OK, so make sure you do the math to double-check that.4701

    OK, this is another common type of problem we have in mass spec; it's "How would you use mass spectrometry to distinguish between the following two compounds?"4707

    They both have the same mass; they both have a molecular mass of 73; and the question is: how is their mass spec going to differ?4717

    They both have molecular ion peak; but what other peaks are we going to expect, that are abundant, for one or the other?4728

    OK, and this comes from knowing what the common fragmentations are for your different functional groups.4735

    So, for an amine, what we are going to do--where would this amine typically break up--the most common place for this to break up?4740

    You are thinking, "Where is the best place for me to break a bond so I result in a carbocation that is going to be a stable carbocation?"4746

    So remember, we call this α cleavage, and it is going to be at the carbon attached to the nitrogen.4757

    We are going to break here to put a positive charge on the carbon with the nitrogen: α cleavage to give this resonance-stabilized cation.4767

    OK, so that is what we would predict for butylamine; this is methyl propylamine, and where can that break?4785

    Well, here is the carbon; it can lose a hydrogen, so we might expect that there might be a significant M-1 peak here, but let's look at the carbon-carbon bond that we are breaking.4794

    Here is our α carbon; here is our α cleavage in this case; and we would expect to form this carbocation.4806

    And again, why there?--because it's resonance-stabilized.4818

    We can draw both of those forms.4824

    What would we see in the mass spec?--well, over here we have lost a propyl group--that is M-43 (that is one way you can kind of calculate it based on the molecular ion).4833

    This has a mass-to-charge of 30; nitrogen is 15, and carbon is I adding that up right?4849

    Mass-to-charge of 30...oh, and we are missing a hydrogen; so the reason that this is not even...the M+1 rule with the even and odd numbers (this is something that is interesting to point out) is: that is for whole, complete molecules.4860

    OK, when we are looking at mass-to-charge ratios of fragments, by definition, a fragment means it's not complete octets; there is somewhere that someone is missing a bond; so in this case, then, a nitrogen-containing fragment is going to be evenly numbered, because we are also missing, let's say, a hydrogen or something like that.4876

    So anyway, mass-to-charge of 30 makes sense here--it's M-43--where in this case, we lost an ethyl group, which is 29.4895

    And when we calculate this mass-to-charge, we get 44.4907

    When we take the mass spec of these two amines, we expect to find a significant peak for butylamine at a mass-to-charge of 30, and a significant peak in the methyl propylamine at mass-to-charge of 44.4913

    Not only could we distinguish between those, sometimes we are actually given the spectra and asked which is which--you know, "match them up," kind of like you might have with an IR problem--here are the IR spectra; here are your possible structures; match them up and provide evidence to defend your choice--how did you make your choice?4927

    OK, so here we said (this is how we would use mass spec), "and provide the structures and the mass-to-charge values for the significant fragments expected."4946

    You not only want to have "Hey, this should have a peak at 30, because of α cleavage," but you show what that structure looks like; and the same here.4956

    OK, finally, let's try a McLafferty rearrangement, because that is kind of a unique phenomenon we have.4966

    And so, we start with 114, and there are two questions here: one is "what is the fragment resulting from a McLafferty rearrangement?" and then there is a second question: "What accounts for the peak at mass-to-charge of 57?"4976

    We can do either one we want; let's look at the 57.4997

    So, what 57 is actually a butyl group, and we started at 114, so it's actually -57; so we have either lost a butyl group or we still have a butyl group--you can kind of look at it that way.5003

    And so, where could this molecule cleave that would be both reasonable for the functional group...5017

    It's always very important--we don't want to just start hacking the molecule apart in whatever way suits us; it has to be something that is logical for that functional group.5026

    And so, what we saw for ketones is that it can break on either side of the carbonyl; so if we break on this side, then that would lose a butyl group, and that would give this acylium ion (carbocation--I just draw it as the other resonance form); and this has a mass-to-charge of 57.5035

    So, this underwent α cleavage to form 57.5060

    So it turns out that a butyl group has the same mass as an ethyl group plus a carbonyl.5064

    Again, that is where that table kind of comes in handy; you can see a few different options for how to get to a certain mass--how to achieve a certain mass that you are looking for.5071

    OK, so that is the mass-to-charge of 57 fragment; now, how about this McLafferty rearrangement?5081

    What we want to do is: we want to look around; this oxygen is going to look to either side, and it is going to find a hydrogen that is 6 atoms away; we have to form a 6-membered ring.5088

    So, if this is 1, 2, 3, 4, 5, is going to be taking one of the hydrogens from this end carbon, and that hydrogen is going to be transferring it over.5100

    You can try and do the mechanism with the structure drawn out, just in the straight chain; and maybe with practice you can get there; but let's draw the mechanism with the 6-membered ring first, to get some practice with that.5113

    Again, if you want to pause this and try in on your own, this is a great one to try on your own, rather than just see the solution.5131

    We're going to draw our CH2, and then a CH with a methyl on it, and then a CH2, and then there is that hydrogen that is 6 atoms away.5136

    If we want to number our carbons, 1, 2, 3, 4, just to keep track of everyone--1, 2, 3, 4...5147

    OK, any time you are redrawing a structure, doing rotation, something like that, numbering your carbons is a really good way of keeping track and making sure we don't lose anyone.5155

    OK, so that is the bond that is formed; what does the mechanism look like?--we could show this carbonyl grabbing onto that hydrogen; it forces this bond to break; it forces that bond to break.5163

    OK, so where is the cleavage that occurred?--it's right here, and it's right here; between the α and the β carbon is this bond you are always going to break with McLafferty rearrangement.5175

    OK, so what do we have on the left-hand side of the molecule?5185

    We have...I'm sorry, this is a CH3, CH, double bond, CH2; this is a neutral molecule that is not observed.5192

    OK, but what is, I'm sorry; you know what, I keep making this mistake.5203

    Before I can do a McLafferty, I have to have a radical cation; I have to have a radical cation before I can do my cleavage, the α cleavage, before I can do a rearrangement; I need to have a radical cation.5213

    So, this clips off as a neutral molecule; it is still this radical cation that remains.5227

    This is what is going to be observed--the enol is the fragment that is going to be observed in the mass spectrum--of course, as the rearranged ketone.5235

    This is C4 (1, 2, 3, 4) H8 (2, 4, 5, 6, 7, 8) O....C4H8O; so we can do that math, and we get mass-to-charge of 72.5246

    That is another peak that we would expect to find in this molecule, thanks to the McLafferty rearrangement.5264

    We have had an introduction to mass spec, just kind of the theory about it and how we might use it to learn something about a molecule's structure.5270

    And so, hopefully, you can find a good book to work on some extra problems until you get the hang of it and be able to handle it.5278

    A lot of times in more advanced classes, you will have spectroscopy problems where you need to come up with a structure, and you will be given all the spectroscopic information about that compound.5285

    "Here is the molecular weight; here is the IR; here is the NMR; here is the mass spec."5296

    And so now you will know how to take that mass spec information and really work with it and come up with some structural information.5302

    Thanks so much for joining me, and I hope to see you again really soon.5312