IR spectroscopy (review) Unit 1 Mass spectroscopy NMR UV/Vis spectroscopy IR spectroscopy (review) 1

Mass Spectroscopy Mass Spectroscopy involved the following steps: The vaporization, ionization and fragmentation of a compound. The separation of the cationic fragments based on their mass to charge ratio (m/z). The relative abundance of the different fragments is determined. The more stable an ion is the more abundant the ion will be. A spectrum is generated that shows the relative abundance of each ion as a function of increasing m/z.

Mass Spectroscopy A mass spectrum shows the relative abundance of fragments vs. the mass to charge (m/z) ratios of the fragments. As a general rule the charge on each fragment is +1. As a result the m/z is the mass of the fragment. The mass spectrum for 2-butanol is given below. All MS were taken from: http://sdbs.db.aist.go.jp (National Institute of Advanced Industrial Science and Technology, 8/22/13).

Mass Spectrometers There are numerous Mass Spectrometer designs. Magnetic –Deflection Cycloidal Isotope-ratio Double-Focusing Time-of-Flight Quadrupole Etc. The different designs have various advantages and disadvantages. Such as the size of the instrument and the resolution of the instrument.

Mass Spectrometer – the following diagram is typical of a magnetic deflection (sector) mass analyzer.

Ion source In the ion source the sample is vaporized. The vapor phase molecules are then bombarded with a beam of electrons (typically 70 eV). This bombardment with electron causes the molecule to lose an electron from a bond and become a radical cations and is called the molecular ions (M+).

Fragmentation During the electron bombardment bonds in the molecules are also broken, producing a cation fragment and a radical fragment.

Mass analyzer/Detector The cation fragments that are generated in the ion source are accelerated into the mass analyzer section of the mass spec where they will be separated based on their m/z. The detector then generates an output signal that is proportional to the number of each different type of ion. 2-butanol molecular ion is 74 You are responsible for learning how to interpret mass spectra.

Interpretation of Mass spectra Where to start? The first thing that you should do is make a list of the “main peaks” in the spectra. If the mass (m/z) of a peak indicates a specific structure is present then also draw the structure. (To do this second part you need to know what fragments are commonly seen in mass spectra.)

Fragmentation of alkanes + . CH3CH2CH2CH3 + e  [CH3CH2CH2CH3] + 2 e M+ with m/z = 58 + . . [CH3-CH2-CH2-CH3]  CH3CH2 + +CH2CH3 radical fragment (not detectable) carbocation fragment m/z = 29 + . . [CH3-CH2-CH2-CH3]  CH3 + +CH2CH2CH3 radical fragment (not detectable) carbocation fragment m/z = 43 + . . [CH3-CH2-CH2-CH3]  CH3+ + CH2CH2CH3 carbocation fragment m/z = 15 radical fragment (not detectable)

Fragmentation of alkanes 29 . CH3-CH2-CH2-CH3  CH3CH2 + +CH2CH3 radical fragment (not detectable) carbocation fragment m/z = 29 . CH3 + +CH2CH2CH3 43 radical fragment (not detectable) carbocation fragment m/z = 43 CH3-CH2-CH2-CH3 15 . CH3+ + CH2CH2CH3 carbocation fragment m/z = 15 radical fragment (not detectable) Which is the more stable cation?

Fragmentation of alkanes The relative abundances of fragments is determined by the stabilities of the fragments (i.e. more stable fragments are more easily formed and therefore will be produced with higher abundances). The most common fragment is assigned a relative abundance of 100 and will be known as the base peak in the mass spectrum. All other peaks are reports as a percentage of the base peak. The stability of the carbocation fragment is more important than the stability of the radical fragment when consider how a molecule will fragment. . CH3CH2CH2CH2 + +CH3 15 CH3-CH2-CH2-CH2-CH3 m/z = 15 57 . CH3CH2CH2CH2+ + CH3 m/z = 57

Fragmentation of alkanes . CH3CH2CH2CH2 + +CH3 15 CH3-CH2-CH2-CH2-CH3 m/z = 15 57 . CH3CH2CH2CH2+ + CH3 m/z = 57 The m/z = 57 fragment is more common since a 1 carbocation is more stable than a methyl carbocation. M+

Molecular ion The molecular ion (M+) gives you the molecular mass of the compound being analyzed. The molecular ion will be the highest molecular weight ion in the spectrum. There will be peaks at higher m/z than M+ because of the presents of various isotopes. However, these peaks are generally small. There are cases where identification of the M+ ion can be tricky because it is very small or not present. Note: I will give you the value of M+ if the peak is not observed in the spectra. .

Interpretation of Mass spectra For Hydrocarbons: The first data that goes into the leak list are M+ and the base peak. Then you list any peaks that indicate common alkyl groups. It is common not to observe some fragments. (Why?) Simply analysis of the data can identify missing fragments. Calculate the mass of the fragment that was lost to give a particular peak in the spectra. M+ - m/z (of observed peak) = mass of lost fragment

Fragmentation- base peak 43 . CH3-CH2-CH2-CH2-CH3  CH3CH2 + +CH2CH2CH3 The base peak in this case is a product of a fragmentation that produces a 1 carbocation and a 1 radical (a 1 radical is more stable than methyl radical). base peak M+

Identification of M+ and base peak N,N-dimethylacetamide

Identification of M+ and base peak Benzoic acid

Identification of M+ and base peak 2,3-dimethyl-1-butanol M+ = 102 ↓

Identification of M+ and base peak Methyl iodide

Fragmentation of alkanes Indicate the fragmentation that would produce the base peak for 2-methylbutane, 2-methylpentane and 3-methylpentane. Hint: Draw the structure first.

Fragmentation of alkanes You need to know the expected m/z ratios for common alkyl fragments: 1 C alkyl fragment = 15 (CH3 = 12 + 3(1) = 15) 2 C’s = 29 (15 + 14) (CH2 = 12 + 2(1) = 14) 3 C’s = 43 (29 + 14) 4 C’s = 57 (43 + 14) 5 C’s = 71 (57 + 14) Note: Other groups could give the above m/z ratios. i.e. m/z of 43 could be either a propyl cation or an acylium ion CH3CO+.

Fragmentation . CH3-CH2-CH2-CH2-CH3  CH3CH2 + +CH2CH2CH3 M  43 = 29

Fragmentation- Resonance effects Since the stabilities of the fragments affect how common the fragments will be, you should recognize that fragmentations producing resonance-stabilized fragments will be favored. + . . [CH2=CH-CH2-CH3]  [CH2=CH-CH2+  +CH2-CH=CH2] + CH3 resonance-stabilized allylic carbocation fragment m/z = 41 (add to your list of fragments to know) allylic position benzylic position (charge is delocalized into ring)

Fragmentation-benzylic group 91 Be sure you recognize that a major peak at m/z = 91 as likely indicating a benzyl group. This is a common base peak. If the ring has additional substituents the m/z would be higher. What would you see if there was a Cl or CH3 on the ring? tropylium ion

Fragmentation-benzene ring A fragment with a m/z= 77 indicates the presence of a mono substituted benzene. This m/z is not as common of fragment as a benzylic group. What is the base peak? 77

Determining molecular formula A procedure for generating possible chemical formulas from mass spectrum data is called the “Rule of Thirteen”. Find the molecular ion (M+). This is the molecular weight of the compound. Determine if there are any heteroatoms in the cpd. Subtract the mass of each heteroatom from M+ . This will give you the mass of the hydrocarbon portion of the molecule (CnHm). To determine the value of n, divide the hydrocarbon mass of the molecule by 13. The result is n. The value of m is the sum of n plus any remainder from the division done in step 4. The formula for the compound will be CnHm plus any heteroatoms.

Determining molecular formula Determine the molecular formula for the following. Hydrocarbon M+ = 92 Hydrocarbon M+ = 142 Hydrocarbon M+ = 128 Cpd containing bromine M+ = 122 Cpd containing nitrogen M+ = 107 Cpd containg oxygen M+ = 72

Sample problems (hydrocarbons) tetradecane

Sample problems (hydrocarbons) tetralin

Sample problems (hydrocarbons) butylbenzene

Sample problems (hydrocarbons) O-xylene

Sample problems (hydrocarbons) cumene

Sample problems (hydrocarbons) 3,5-dimethylheptane M+ = 128 ↓

Fragmentation- heteroatoms The presence of heteroatoms (atoms other than C and H) in a compound can be determined by looking for specific patterns in the mass spectra. The common heteroatoms that are found in organic compounds are oxygen (O), nitrogen (N) and the halogens (Cl, Br and I).

Heteroatoms- bromine Most compounds contain the natural isotopic abundance of each element in the compound. For bromine the natural isotope distribution is: Br  50.5% = 79Br & 49.5% = 81Br This distribution of Br isotopes means that for fragments containing Br, a little over half will contain 79Br and a little less than half will contain 81Br. This means that the presence of Br is signaled by two peaks of about the same height within 2 mass units (or m/z units) of one another. The molecular ion containing 79Br is designated M+ (slightly taller peak); the molecular ion containing 81Br is designated M+2.

The M+ peak represents the whole molecule – no fragment should be bigger (unless it contains a heavier isotope). To identify M+ look to the far right of the spectrum.

Fragmentation- chlorine The natural distribution of Cl isotopes: 75.5% 35Cl , 24.5% 37Cl Therefore fragments containing 35Cl will be 3x as common as those containing 37Cl. A molecular ion peak containing the more common 35Cl (designated M+) will be 3x taller than the molecular ion peak containing 37Cl (designated M+2)

What is the m/z of the base peak? If we subtract the mass of base peak and 35Cl from the M+ peak, what’s left?

Fragmentation- iodine The only stable isotope of iodine is 127I. Iodine radicals are relatively stable and break off easily leaving an M – 127 fragment. Therefore a gap of 127 between the M+ peak and the next significant peak signals the presence of I.

Sample problems (halogens) 1-chloropentane

Sample problems (halogens) 1-bromo-4-methylpentane

Sample problems (halogens) 1-bromohexane

Sample problems (halogens) iodobenzene

Fragmentation- (oxygen) Alpha cleavage of alcohols Note: structure of fragment = all resonance forms 31 .. + . CH3CH2CH2-CH2OH  CH3CH2CH2 + [+CH2-OH  CH2=OH] .. .. m/z = 31 (you should know)  position What is the M+?

Fragmentation- (oxygen) alcohols Alcohols often lose water to form a stable alkene radical cation fragment. This loss of water is so common, M+ peaks of alcohols often do not show up in mass spectra. If M+ is not observed in a spectra I will give you the value. The alkene radical cation can undergo further cleavages (allylic cleavage being most common).

Fragmentation- (oxygen) ethers Ethers also undergo alpha cleavage: R-CH2-O-R’  R· + [+CH2-O-R’  CH2=O-R’] .. .. + ..  position Ethers also cleave adjacent to the O (along with a proton transfer) to lose an alkyl group: .. + + R-CH2-O-R’  [R-CH-OH  R-CH=OH] + ·R’ .. .. What in the m/z of this cation?

Fragmentation-(oxygen) carbonyl cpds Ketones and aldehydes commonly cleave adjacent to the carbonyl groups to produce acylium ions: An acylium ion m/z= 43 Notice that the mass of the different acylium ions correspond to the mass of the alkane cations that have one carbon longer chains. How would you know that the compound had the carbonyl?

Sample problems (oxygen) 2-butanol

Sample problems (oxygen) 2-methyl-1-propanol

Sample problems (oxygen) 2-phenyl-1-propanol

Sample problems (oxygen) 2-pentanone

Sample problems (oxygen) 2-methyl-3-hexanone

Sample problems (oxygen) 1-phenyl-2-butanone

Fragmentation-alpha cleavage of amines Resonance stabilized cation .. 30 + . CH3CH2CH2-CH2NH2  CH3CH2CH2 + [+CH2-NH2  CH2=NH2] m/z = 30 (know)  position base peak M+

Fragmentation- amines Show the structures for the alpha cleavage fragments of the following amine and predict their m/z ratios. MW = 115 There are two alpha bonds where fragmentation can occur.

Fragmentation- amines Show the structures of the base peak fragment in the mass spec of N-methylbutylamine.

Fragmentation- Predict the most common fragments of 2-pentanol, 2-pentylamine and 2-pentanone in a mass spectrum. What do you want to do first?

Fragmentation- nirtogen Most molecules (and therefore M+’s) have even m/z ratios. Notable exceptions are molecules containing N which has an odd-numbered valence (3 vs. C’s 4). Molecules containing odd numbers of N’s therefore have odd m/z ratios. You can use M+ peaks with an odd m/z ratio as an indicator for the presence of N. N-ethylpropylamine

Interpretation of mass spectra Procedure for interpreting mass spectra. Identify M+ and base peak. Are there any heteroatoms? Is there an aromatic ring in the cpd? Are there any common alkyl groups in the cpd? Use the rule of thirteen to determine the number of carbons and hydrogens in the cpd. Determine the possible molecular formula for the cpd. Determine a structure for the cpd that is consistent with the above information.

Interpretation of mass spectra Factors to be aware of when interpreting mass spectra. Impurities in a sample can produce peaks that are irrelevant. This is particularly important when you are trying to determine the m/z of M+. Pecks from impurities are more of a problem when running CG-MS analysis where cross contamination can be a problem. Background scans can be used to identify extraneous pecks. Many compound are easily fragmented and the molecular (parent) ion is not observed. The mass spectrum of pure cpds oven have pecks that are hard to identify. These pecks are the result of complex fragmentation or complex rearrangements. Do not dwell on peaks that are not readily identifiable.

CHEM 2124 – Mass Spectra – Example 1 (3-chloropropyl)benzene 62

CHEM 2124 – Mass Spectra – Example 2 3-chlorohexane (molecular ion peak not present, hard to distinguish Cl) 63

CHEM 2124 – Mass Spectra – Example 3 1-bromo-4-methylpentane 64

CHEM 2124 – Mass Spectra – Example 4 2-phenylethylamine 65

CHEM 2124 – Mass Spectra – Example 5 1-iodopentane 66

CHEM 2124 – Mass Spectra – Example 6 3-methyl-1-butanol 67

CHEM 2124 – Mass Spectra – Example 7 2-methylpentane 68

CHEM 2124 – Mass Spectra – Example 7 N,N-dimethylacetamide 69