Mass Spectrometry

3 The GC-MS => A mixture of compounds is separated by gas chromatography, then identified by mass spectrometry.

Only Cations are Detected

How the mass spectrometer works: 1Sample is vapourised in an evacuated chamber M (l) or M (s) → M (g) 2.The gaseous molecules are then bombarded with a stream of high energy electrons (50-70 eV) to give a radical cation (M + , a molecular ion) in the ionization chamber M (g) + e - → M +  (g) + 2e - Mass Spectroscopy

Ionization to Radical Cation Molecular Ion (m + )

Definitions Molecular ion(or parent )peak - The ion obtained by the loss of one electron from the molecule (M + ) Base peak - The most intense peak in the MS, assigned 100% intensity Radical cation - positively charged species with an odd number of electrons Fragment ions - Lighter cations (and radical cations) formed by the decomposition of the molecular ion. These often correspond to stable carbocations. Isotopic peaks: due to presence of isotopes of C, H, O, N,... in the sample molecule and appear around the main peaks. m/z - mass to charge ratio

The resulting mass spectrum is a graph of the mass of each cation vs. its relative abundance. The peaks are assigned an abundance as a percentage of the base peak. –the most intense peak in the spectrum The base peak is not necessarily the same as the parent ion peak. Background

9 Example =>

Isotopes Mass spectrometers are capable of separating and detecting individual ions even those that only differ by a single atomic mass unit. As a result molecules containing different isotopes can be distinguished. This is most apparent when atoms such as bromine or chlorine are present ( 79 Br : 81 Br, intensity 1:1 and 35 Cl : 37 Cl, intensity 3:1) where peaks at "M" and "M+2" are obtained. The intensity ratios in the isotope patterns are due to the natural abundance of the isotopes. "M+1" peaks are seen due the the presence of 13 C in the sample.

11 Molecules with Heteroatoms Isotopes: present in their usual abundance. Hydrocarbons contain 1.1% C-13, so there will be a small M+1 peak. If Br is present, M+2 is equal to M +. If Cl is present, M+2 is one-third of M +. If I is present, peak at 127, large gap. If N is present, M + will be an odd number. If S is present, M+2 will be 4% of M +. =>

12 Isotopic Abundance => 81 Br

14 Mass Spectrum with Chlorine =>

15 Mass Spectrum with Bromine =>

Bromomethane

17 Mass Spectrum with Sulfur =>

18 Mass Spectra of Alkanes More stable carbocations will be more abundant. =>

Octane, m + = 114 m-29 m-43 m-57 m-71 Base peak m+m+

n-Decane

Effect of Branching in Hydrocarbons

Isooctane, no molecular ion

2-Methylpentane

2-Chloropropane

1-Bromopropane

Chapter 1226 Mass Spectra of Alkenes Resonance-stabilized cations favored. =>

Mass Spectra of Alkynes

–Fragment easily resulting in very small or missing parent ion peak ( M + may not be visible). –May lose hydroxyl radical or water M or M –Commonly lose an alkyl group attached to the carbinol carbon forming an oxonium ion. 1 o alcohol usually has prominent peak at m/z = 31 corresponding to H 2 C=OH + Mass Spectra of Alcohols

MS for 1-propanol M+M+ M SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/28/09) Mass Spectra of Alcohols

1-Butanol m-18

Methanol

3-Methyl-1-penten-3-ol m/z = 71 m + =100

4-Methyl-1-penten-3-ol M + =100 m/z = 57

Amines –Odd M + (assuming an odd number of nitrogens are present) –a-cleavage dominates forming an iminium ion Mass Spectra of Amines

Ethers –  -cleavage forming oxonium ion –Loss of alkyl group forming oxonium ion –Loss of alkyl group forming a carbocation Mass Spectra of Ethers

MS of diethylether (CH 3 CH 2 OCH 2 CH 3 ) Mass Spectra of Ethers

Aldehydes (RCHO) –Fragmentation may form acylium ion –Common fragments: M for M for Mass Spectra of Aldehydes

MS for hydrocinnamaldehyde M + = SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/28/09) Mass Spectra of Aldehydes

Ketones –Fragmentation leads to formation of acylium ion: Loss of R forming Loss of R’ forming Mass Spectra of Ketones

MS for 2-pentanone M+M+ SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/28/09) Mass Spectra of Ketones

Esters (RCO 2 R’) –Common fragmentation patterns include: Loss of OR’ –peak at M + - OR’ Loss of R’ –peak at M + - R ’ Mass Spectra of Esters

M + = SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/28/09) Mass Spectra of Esters

Examples of Mass Spectra of Carbonyl compounds

3-Pentanone m+m+ m-29 base

Cyclohexanone

Chloroacetone

McLafferty Rearrangement for carbonyl compounds

2-Octanone

Decanoic Acid

Methyl Octanoate

Mass Spectra of Aromatics

–Fragment at the benzylic carbon, forming a resonance stabilized benzylic carbocation (which rearranges to the tropylium ion) M+M+ Mass Spectra of benzene

Aromatics may also have a peak at m/z = 77 for the benzene ring. M + =

Toluene m+m+ m-1

(3-Chloropropyl)benzene

Propylbenzene M+M+ M-29

Isopropylbenzene M+M+ M-15

n-Butylbenzene

McLafferty Rearrangements in Alkyl Benzenes

Benzamide

p-Chloroacetophenone

2,4-Dimethoxyacetophenone

High Resolution Mass Spectrometry Determination of Molecular Formula

Isotope Ratios Can Help to Determine Molecular Formula Relative intensities (%) MFMW M M+1M+2 CO N C 2 H

The Nitrogen Rule The nitrogen rule states that an odd number of nitrogen atoms will form a molecular ion with an odd mass number. An even number of nitrogen atoms (or none at all) will produce a molecular ion with an even mass number. This occurs because nitrogen has an odd-numbered valence. Examples: C 6 H 5 CH 2 NH 2 MW = 107 H 2 NCH 2 CH 2 NH 2 MW = 60

The “Rule of Thirteen” can be used to identify possible molecular formulas for an unknown hydrocarbon, C n H m. –Step 1: n = M + /13 (integer only, use remainder in step 2) –Step 2: m = n + remainder from step 1 Rule of Thirteen

Example: The formula for a hydrocarbon with M + =106 can be found: –Step 1: n = 106/13 = 8 (R = 2) –Step 2: m = = 10 –Formula: C 8 H 10 Rule of Thirteen

Rule of 13 Heteroatom Equivalents ElementCH Equivalent ElementCH Equivalent 1 H 12 C 31 PC2H7C2H7 16 OCH 4 32 SC2H8C2H8 14 NCH 2 16 O 32 SC4C4 16 O 14 NC2H6C2H6 35 ClC 2 H FCH 7 79 BrC6H7C6H7 28 SiC2H4C2H4 127 IC 10 H 7

Candidate molecular formulas for M +. = /13 = 8, R = 4  C 8 H 12 O = CH 4 ; therefore, C 8 H 12 – CH 4 + O = C 7 H 8 O or C 8 H 12 – 2(CH 4 ) + 2O = C 6 H 4 O 2 or C 8 H 12 – C 6 H 7 + Br = C 2 H 5 Br Calculate three candidate molecular formulas for C 10 H 18.