Mass Spec - Fragmentation



Similar documents
13C NMR Spectroscopy

HOMEWORK PROBLEMS: IR SPECTROSCOPY AND 13C NMR. The peak at 1720 indicates a C=O bond (carbonyl). One possibility is acetone:

Determining the Structure of an Organic Compound

for excitation to occur, there must be an exact match between the frequency of the applied radiation and the frequency of the vibration

ALCOHOLS: Properties & Preparation

For example: (Example is from page 50 of the Thinkbook)

MOLECULAR REPRESENTATIONS AND INFRARED SPECTROSCOPY

Suggested solutions for Chapter 3

Electrophilic Aromatic Substitution Reactions

How to Quickly Solve Spectrometry Problems

Experiment 11. Infrared Spectroscopy

Molecular Models Experiment #1

Survival Organic Chemistry Part I: Molecular Models

Carboxylic Acid Derivatives and Nitriles

Chapter 5 Classification of Organic Compounds by Solubility

electron does not become part of the compound; one electron goes in but two electrons come out.

Organic Functional Groups Chapter 7. Alcohols, Ethers and More

Chapter 6 An Overview of Organic Reactions

4/18/ Substituent Effects in Electrophilic Substitutions. Substituent Effects in Electrophilic Substitutions

Q.1 Draw out some suitable structures which fit the molecular formula C 6 H 6

passing through (Y-axis). The peaks are those shown at frequencies when less than

Infrared Spectroscopy 紅 外 線 光 譜 儀

But in organic terms: Oxidation: loss of H 2 ; addition of O or O 2 ; addition of X 2 (halogens).

Typical Infrared Absorption Frequencies. Functional Class Range (nm) Intensity Assignment Range (nm) Intensity Assignment

How to Interpret an IR Spectrum

CHEM 51LB EXP 1 SPECTROSCOPIC METHODS: INFRARED AND NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

Unit Vocabulary: o Organic Acid o Alcohol. o Ester o Ether. o Amine o Aldehyde

These instructions are for a classroom activity which supports OCR A Level Chemistry A.

Electrophilic Aromatic Substitution

EXPERIMENT 1: Survival Organic Chemistry: Molecular Models

Alcohols An alcohol contains a hydroxyl group ( OH) attached to a carbon chain. A phenol contains a hydroxyl group ( OH) attached to a benzene ring.

Identification of Unknown Organic Compounds

Chemistry 1110 Organic Chemistry IUPAC Nomenclature

Solving Spectroscopy Problems

Chapter 13 Spectroscopy NMR, IR, MS, UV-Vis

INFRARED SPECTROSCOPY (IR)

Molecular Formula Determination

ammonium salt (acidic)

Symmetric Stretch: allows molecule to move through space

Acids and Bases: Molecular Structure and Acidity

Ultraviolet Spectroscopy

m/z

Chapter 11. Free Radical Reactions

IDENTIFICATION OF ALCOHOLS

CHM220 Addition lab. Experiment: Reactions of alkanes, alkenes, and cycloalkenes*

Chemistry Notes for class 12 Chapter 13 Amines

Figure 8. Example of simple benzene naming with chlorine and NO 2 as substituents.

AROMATIC COMPOUNDS A STUDENT SHOULD BE ABLE TO:

Mass Spectrometry. Overview

Writing a Correct Mechanism

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Benzene Benzene is best represented as a resonance hybrid:

IR Summary - All numerical values in the tables below are given in wavenumbers, cm -1

DETERMINACIÓN DE ESTRUCTURAS ORGÁNICAS (ORGANIC SPECTROSCOPY) IR SPECTROSCOPY

ORGANIC CHEM I Practice Questions for Ch. 4

Physicochemical Properties of Drugs

Organic Spectroscopy. UV - Ultraviolet-Visible Spectroscopy. !! nm. Methods for structure determination of organic compounds:

Resonance Structures Arrow Pushing Practice

ORGANIC CHEMISTRY I PRACTICE PROBLEMS FOR BRONSTED-LOWRY ACID-BASE CHEMISTRY

Suggested solutions for Chapter 7

Chapter 22 Carbonyl Alpha-Substitution Reactions

CH 102 Practice Exam 2 PCC-Sylvania

Question Bank Organic Chemistry-I

Infrared Spectroscopy

methyl RX example primary RX example secondary RX example secondary RX example tertiary RX example

IR Applied to Isomer Analysis

Conjugation is broken completely by the introduction of saturated (sp 3 ) carbon:

Organic Spectroscopy

NOMENCLATURE OF ORGANIC COMPOUNDS 2010, 2003, 1980, by David A. Katz. All rights reserved.

CHE334 Identification of an Unknown Compound By NMR/IR/MS

Laboratory 22: Properties of Alcohols

1. The functional group present in carboxylic acids is called a A) carbonyl group. B) carboxyl group. C) carboxylate group. D) carbohydroxyl group.

Lecture Overview. Hydrogen Bonds. Special Properties of Water Molecules. Universal Solvent. ph Scale Illustrated. special properties of water

Chapter 18: Organic Chemistry

Electrophilic Aromatic Substitution

Organic Chemistry Tenth Edition

UNIT (9) CARBOXYLIC ACIDS, ESTERS, AMINES, AND AMIDES

Chapter 2 Polar Covalent Bonds; Acids and Bases

Everything You Need to Know About Mechanisms. First rule: Arrows are used to indicate movement of electrons

12.4 FUNCTIONAL-GROUP INFRARED ABSORPTIONS

17.2 REACTIONS INVOLVING ALLYLIC AND BENZYLIC RADICALS

Unit 2 Review: Answers: Review for Organic Chemistry Unit Test

IUPAC System of Nomenclature

Copyright 2010 Pearson Education, Inc. Chapter Fourteen 1

CHEM 203 Exam 1. KEY Name Multiple Choice Identify the letter of the choice that best completes the statement or answers the question.

Benzene and Aromatic Compounds

RESONANCE, USING CURVED ARROWS AND ACID-BASE REACTIONS

Introduction to Biodiesel Chemistry Terms and Background Information

ORGANIC COMPOUNDS IN THREE DIMENSIONS

CHEM 211 CHAPTER 16 - Homework

Chapter 4 Lecture Notes

Molecular Models in Biology

Homolytic vs. Heterolytic Fragmentation

REACTIONS OF AROMATIC COMPOUNDS

partial positive an acid is a hydrogen ion donor, or proton donor base is a hydrogen ion acceptor, or proton acceptor acidic protons acid base

Aromaticity and Reactions of Benzene

ACID and BASES - a Summary

Chapter 10. Conjugation in Alkadienes and Allylic Systems. Class Notes. B. The allyl group is both a common name and an accepted IUPAC name

The dipolar nature of acids

Transcription:

Mass Spec - Fragmentation An extremely useful result of EI ionization in particular is a phenomenon known as fragmentation. The radical cation that is produced when an electron is knocked out of a neutral closed-shell molecule in EIMS initially possesses a lot of energy. Energy sufficient to break chemical bonds: radical cation will usually break into a neutral radical and a cation. It is also possible for a neutral closed shell fragment (such as water) to fall off. H 3 C H 1 electron H 3 C H 2 electrons H 3 C H CH 2 H H H 3 C H CH 3 H

Predicting Fragmentation A considerable volume of literature regarding the fragmentation reactions of certain molecules is available. Using this knowledge, we can predict how a given molecule might fragment. But first a few conventions.

Conventions It is always best to localize the cation and radical, obeying the rules of Lewis structures (i.e. don t put more than 8 electrons on carbon). For example, for the three structures above, A is discouraged in favour of B or C. This convention is not always possible to obey (alkanes).

More on Conventions When a molecule is ionized by EIMS, must decide which electron to remove. Always best to remove an electron from a π-bond, or from a heteroatom lone pair. It will become more clear which is the better choice when we look at individual examples. The bond breaking can be done either by homolytic or heterolytic cleavage. H 3 C CH 2 R H 3 C CH 2 Br 1 e EI CH 3 CH 2 R EI CH 3 CH 2 Br 2 e CH 3 H 2 C R CH 3 CH 2 Br Homolytic Heterolytic

Fragmentation Notes - Most fragments are even-electron cations. These split to make more even-electron cations. - The probability of cleaving a given bond is related to the bond strength, and to the stability of the fragments formed. In particular, cations like to rearrange or decay into more stable cations. - Remember that for a ring system, at least 2 bonds must break for the ring to fragment. In addition, there are 10 general rules to keep in mind when predicting the most likely ions to be formed for a given molecule.

Fragmentation Guidelines 1. The relative height of the M peak is greatest for straightchain molecules and decreases as the branching increases. 2. The relative height of the M peak decreases with chain length for a homologous series. 3. Cleavage is favoured at alkyl-substituted carbons, with the probability of cleavage increasing as the substitution increases. These rules mostly arise from the fact that carbocation and radical stability show the following trend: Most Stable Benzylic > Allylic > Tertiary > Secondary >> Primary Least Stable Stevenson s Rule At the point of breakage, the larger fragment usually takes the radical to leave the smaller cation.

Fragmentation Guidelines 4. Double bonds, cyclic structures, and especially aromatic rings will stabilize the molecular ion and increase its probability of appearance. 5. Double bonds favor allylic cleavage to give a resonance stabilized allylic carbocation, especially for cycloalkenes. EI 6. For saturated rings (like cyclohexanes), the side chains tend to cleave first leaving the positive charge with the ring. R R

Fragmentation Guidelines 7. Unsaturated rings can also undergo retro-diels-alder reactions to eliminate a neutral alkene. EI 8. Aromatic compounds tend to cleave to give benzylic cations, or more likely tropylium cations. R R EI m/z 91

Fragmentation Guidelines 9. C-C bonds next to heteroatoms often break leaving the positive charge on the carbon with the heteroatom. 10. Cleavage is often favoured if it can expel small stable molecules like water, C, NH 3, H 2 S, etc. In addition to bond fragmentation, various intramolecular rearrangements can take place to give sometimes unexpected ions.

Rearrangements ne common type of rearrangement in MS is the McLafferty rearrangement which takes place in compounds that contain a carbonyl group. Y H H Many other rearrangements are possible, even some that are not well understood and are considered random. H 3 C CH 3 CH 3 CH 3 CH 2 CH 3

Chem 325 Mass Spectra of Various Classes of rganic Compounds

1. Saturated Hydrocarbons General rules 1-3 apply well to hydrocarbons. Rearrangements are common but usually do not give intense peaks. The M peak of a straight-chain hydrocarbon is always visible, but decreases in intensity as the molecule gets larger. The fragmentation pattern is characterized by peaks separated by 14 mass units (a CH 2 group). The most intense peaks are the C 3 C 5 fragments.

Branched Hydrocarbons The MS for branched saturated hydrocarbons are similar, except certain fragments become more prominent.

CI-MS For some types of compounds the M peak is very weak or not observable at all! Due to EI ionization being a hard ionization : so much internal energy is given to the molecular ion that extensive fragmentation immediately results. How to measure the molecular mass?? Chemical Ionization (CI).

Chemical Ionization Ion source filled with relatively high pressure reagent gas, example isobutane i-c 4 H 10, ammonia, or methane CH 4 Electron impact ionizes the CH 4, high pressure ensures many ion-molecule collisions and reactions to produce CH 5, which acts as a protonating agent when it collides with sample molecule M e CH 4 CH 5 CH 5 M MH CH 4 (MC 2 H 5 ), (MC 3 H 5 ) Can also use i-c 4 H 10, NH 3, etc. soft ionization : much less fragmentation, can see molecular ion

Cyclic Hydrocarbons Cyclic hydrocarbons show a much more intense M ion (Rule 4) Two bonds must break to form fragments.

2. Alkenes It is usually easy to see the M peak of alkenes in EIMS. In acyclic alkenes, the double bond freely migrates in the fragments, so it can be difficult to determine the double bond location, but for cyclic alkenes it is easier. Cleavage usually happens at allylic bonds (Rule 5).

3. Aromatic Hydrocarbons Aromatic hydrocarbons usually show strong M peaks (Rule 4). Aromatic rings are stable and have a lower tendency to fragment. Alkyl substituted benzenes often give a strong peak at m/z 91 due to benzylic cleavage (Rule 8).

MS of n-butylbenzene m/z 91

4. Alcohols Alcohols fragment very easily secondary and primary alcohols show very weak M peaks, and tertiary alcohols often do not show M at all. MW is often determined by derivatization or CI-MS. The C-C bond nearest the H is frequently the first bond to break. Thus primary alcohols often show a prominent peak at 31 m/z. H H H m/z = 60 m/z = 31

Secondary Alcohols Secondary alcohols cleave in the same way, often showing a prominent CHR-H peak. H m/z = 74 H m/z = 45 H H m/z = 74 m/z = 59

Alcohols Sometimes the hydrogen R 2 CH-H in 1 and 2 alcohols cleaves rather than an alkyl group. The result is an M-1 peak. H H H M m/z = 74 M-1 m/z = 73

Tertiary Alcohols Tertiary alcohols cleave in a similar fashion to give CRR-H fragments. A H B m/z = 116 (not observed) A B H m/z = 59 H m/z = 101

More on Alcohols Alcohols can lose a molecule of water to show a sometimes prominent M-18 peak. It is especially noticeable for primary alcohols. H H H H M m/z = 102 M-18 m/z = 84

Benzylic Alcohols Benzylic alcohols fragment much differently from aliphatic alcohols. Benzylic cleavage happens as expected (Rule 8). Benzyl alcohol fragments via the following pathway: H H -H -C H m/z = 108 m/z = 107 m/z = 79 m/z = 77 H H H H H H

Benzylic Alcohols Benzylic alcohols will usually lose water (M-18). The M-18 peak is especially strong for molecules for which loss of water is mechanistically straightforward. H H m/z = 122 m/z = 104

Phenols Phenols often show peaks at 77 m/z resulting from formation of phenyl cation, and peaks resulting from loss of C (M-28) and CH (M-29) are usually found in phenols. H 94 M-29 M-28

Phenols If alternative attractive cleavage pathways are available for phenols, the molecule will often take that path. Molecular ions will fragment along the path of least resistance!

Example: ctane m/z = 114

Example: 4-methyl-2-hexene m/z = 98

Example: 3-methyl-1-pentanol m/z = 102 H

Example: 4-methyl-2-pentanol H m/z = 102

Example: 2-phenylethanol m/z = 122 H

Example: 1-phenylethanol H m/z = 122

5. Ethers Cleavage happens in two main ways: 1. Breakage of the C-C bond next to (like alcohols) CH 3 A CH 3 B m/z = 88 A B m/z = 73 CH 3 or m/z = 59 m/z = 29

Ethers Cleavage happens in two main ways: 2. C- bond cleavage with the charge on the C fragment. CH 3 CH 3 m/z = 88 m/z = 57

Aromatic Ethers M is usually strong. MS is similar to phenols both form phenoxyl cation (m/z 93) and associated daughters.

6. Ketones Ketones usually give strong M peaks. A major fragmentation pathway involves α-cleavage to give an acylium ion. A A m/z = 71 B m/z = 100 B m/z = 57

6. Ketones The carbonyl-containing fragment can also take the radical. A B A m/z = 29 m/z = 100 B m/z = 43

6. Ketones For ketones with longer chains, the McLafferty rearrangement often leads to strong peaks. H H m/z = 128 m/z = 58

6. Aromatic Ketones M is evident. Primary cleavage is α to the carbonyl to give a strong ArC peak (m/z 105 when Ar = Ph). This will lose C to give the phenyl cation (m/z 77).

7. Aldehydes Aldehydes show weak but discernable M peaks. Major pathways are α-cleavage and McLafferty Rearrangement. H H A B A M-1 m/z = 85 H C M m/z = 86 H H B M-R m/z = 29 H m/z = 44 C

7. Aromatic Aldehydes are similar to aromatic ketones. M is strong, and M-1 (αcleavage to carbonyl) is also strong to give the ArC ion (m/z 105 for Ar = Ph). Loss of C from this ion is common to give m/z77 phenyl cation. H -C M m/z = 106 M-1 m/z = 105 m/z = 77

8. Carboxylic Acids (Aliphatic) M is weak, and not always visible. A characteristic m/z 60 peak is often present due to the McLafferty Rearrangement. Bonds α to carbonyl also frequently break to give M-H and M-C 2 H peaks. H M m/z = 88 m/z = 71 m/z = 43 H H m/z = 60

Aromatic Acids M is very prominent. Common peaks are loss of H (M-17) and loss of C 2 H (M-45). H 3 C H H 3 C H 3 C M m/z = 136 M-17 m/z = 119 M-45 m/z = 91

Aromatic Acids If an ortho hydrogen-bearing group is present, loss of water (M-18) is visible as well. H H H H 2 H 2 M m/z = 136 M-18 m/z = 118

9. Aliphatic Esters M is usually distinct. The most characteristic peak is due to the McLafferty rearrangement. H H M m/z = 144 m/z = 88

Aromatic Esters M is usually prominent, unless R chain is long. The base peak is loss of R. m/z = 164 m/z = 105 C m/z = 77

Aromatic Esters The McLafferty rearrangement give the corresponding acid. A more complicated rearrangement often gives a prominent acid1 peak. H H m/z = 164 m/z = 122 H H m/z = 123

10. Amines Aliphatic monoamines have odd numbered and weak M peaks. Most important cleavage is usually breakage of the C-C bond next to the C-N bond. NH 2 3 H m/z = 73 NH 2 NH 2 NH 2 m/z = 72 m/z = 58 m/z = 44

10. Amines The base peak in nearly all primary amines comes at 30 m/z. m/z = 101 NH 2 NH 2 m/z = 30

Aromatic Amines M is intense (why?). An NH bond can be broken to give a moderately intense M-1 peak. H N H - H N H m/z = 93 m/z = 92

Aromatic Amines A common fragmentation is loss of HCN and H 2 CN to give peaks at 65 and 66 m/z. H N H m/z = 93

Aromatic Amines Alkyl substituted aromatic amines typically show breakage of the C-C bond next to the C-N to give a strong peak at 106 when Ar = Ph. H N H N m/z = 121 m/z = 106

11. Amides Primary amides give a strong peak at m/z 44 due to breakage of the R-CNH 2 bond. NH 2 NH 2 m/z = 115 m/z = 44

11. Amides Aliphatic amides M is weak but discernible. For straightchain amides more than 3 carbons, McLafferty gives the base peak at m/z 59. H H m/z = 115 NH 2 m/z = 59 NH 2

Any familiar features? Aromatic Amides

Aromatic Amides M is strong. Loss of NH 2 to make PhC (105) followed by loss of C to make the phenyl cation (77). NH 2 m/z = 121 m/z = 105 m/z = 77

12. Nitriles M are weak or absent for aliphatic nitriles. Loss of the α- hydrogen can give a weak M-1 peak. C N C N H m/z = 69 m/z = 68

Nitriles Base peak is usually 41 due to a rearrangement like the McLafferty. This has limited diagnostic value since (C 3 H 5 ) has the same mass. H H H N C NH C CH 2 m/z = 69 m/z = 41

13. Nitro compounds Aliphatic nitro compounds have weak odd M. The main peaks are hydrocarbon fragments up to M-N 2. m/z = 29 N 2 m/z = 131 m/z = 43 m/z = 57 m/z = 71 m/z = 85

Aromatic Nitro Compounds M is strong. Prominent peaks result from loss of N 2 radical to give an M-46 peak. Also prominent is M-30 from loss of N. N - 2 m/z = 123 M-46 m/z = 77 N M-30 m/z = 93

Example: 4-methyl-2-hexene m/z = 98 69 41 29 83 98

Example: 3-methyl-1-pentanol 56 m/z = 102 H 69 31 45 57 84 102

Example: 4-methyl-2-pentanol 45 H m/z = 102 43 69 57 87 84 102

Example: 2-phenylethanol H 91 m/z = 122 122 31 77 104

Example: 1-phenylethanol H 79 107 m/z = 122 43 77 91 105 122