nuclear magnetic resonance and mass spectrometryhome.konkuk.ac.kr/~parkyong/classes/ch09.pdf ·...

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Created byProfessor William Tam & Dr. Phillis Chang

Chapter 9

Nuclear MagneticResonance and Mass

Spectrometry

Copyright © 2016 by John Wiley & Sons, Inc. All rights reserved.

1. Introduction


Spectroscopy – the study of the interaction of light with matter

Spectroscopy provides information about molecular structure.

Methods we will explore:• Nuclear Magnetic Resonance (NMR)

Spectroscopy• Mass Spectrometry (MS)• Infrared (IR) Spectroscopy (Section 2.15)

v Electromagnetic spectrum

cosmic & g-rays X-rays ultraviolet visible infrared micro-

waveradio-wave

1Å = 10-10m1nm = 10-9m1mm = 10-6m

l: 0.1nm 200nm 400nm 800nm 50mm

X-RayCrystallography

UV IR NMR


2. Nuclear Magnetic Resonance(NMR) Spectroscopy

v The nuclei of some isotopes, such as 1H and 13C, behave as magnets.

v When 1H or 13C atoms are placed in a magnetic field and irradiated with electromagnetic energy, some frequencies are absorbed (“magnetic resonance”)

v A graph that shows the characteristic energy absorption frequencies and intensities for a sample in a magnetic field is called a nuclear magnetic resonance (NMR) spectrum


2


Example of an NMR spectrum 1. The number of signals in the spectrum tells us how many different sets of protons there are in the molecule

2. The position of the signals in the spectrum along the x-axis tells us about the magnetic environment of each set of protons arising largely from the electron density in their environment


3. The area under the signal tells us about how many protons there are in the set being measured

4. The multiplicity (or splitting pattern) of each signal tells us about the number of protons on atoms adjacent to the one whose signal is being measured


2A. Chemical Shiftv The position of a signal along the x-axis of

an NMR spectrum is called its chemical shift

v The chemical shift of each signal gives information about the structural environment of the nuclei producing that signal

v Counting the number of signals in a 1H NMR spectrum indicates, at a first approximation, the number of distinct proton environments in a molecule


3


v Normal range of 1H NMR chemical shifts

15 -10d ppm

"upfield" (more shielded)"downfield" (deshielded)

(high field strength)

(low field strength)


Chemical shift depends on a given nucleus’s magnetic environment.

Magnetic environment is affected by such factors as electron density.

v Reference compoundTMS = tetramethylsilane

as a reference standard (0 ppm)Reasons for the choice of TMS as reference

t Resonance position at higher field than most other organic compounds

t Unreactive and stablet Volatile and easily removed from

sample (B.P. = 28oC)

MeSi MeMeMe


v NMR solvent● Normal NMR solvents should not

contain hydrogen● Common solvents

t CDCl3t C6D6

t CD3OD

t CD3COCD3 (d6-acetone)

4

v The 300-MHz 1H NMR spectrum of 1,4-dimethylbenzene


2B. Integration of Signal Areas

Integral Step Heightsv The area under each signal in a 1H

NMR spectrum is proportional to the number of hydrogen atoms producing that signal

v It is signal area (integration), not signal height, that gives information about the number of hydrogen atoms


O

Ha HaHb

HbHbR

HbHa

2 Ha 3 Hb


2C. Coupling (Signal Splitting)v Coupling is caused by the magnetic

effect of nonequivalent hydrogen atoms that are within 2 or 3 bonds of the hydrogens producing the signal

v The n+1 rule● Rule of Multiplicity:

If a proton (or a set of magnetically equivalent nuclei) has n neighbors of magnetically equivalent protons, its multiplicity is n + 1


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v Examples

Hb C C ClHaHb

Hb Ha

Ha: multiplicity = 3 + 1 = 4 (a quartet)

Hb: multiplicity = 2 + 1 = 3 (a triplet)

(1)

Cl C C ClHbHa

Cl Hb

Ha: multiplicity = 2 + 1 = 3 (a triplet)

Hb: multiplicity = 1 + 1 = 2 (a doublet)

(2)

Copyright © 2016 by John Wiley & Sons, Inc. All rights reserved. Copyright © 2016 by John Wiley & Sons, Inc. All rights reserved.

v Examples

Note: All Hb’s are chemically and magnetically equivalent.

Hb C C BrHaHb

Hb

Ha: multiplicity = 6 + 1 = 7 (a septet)

Hb: multiplicity = 1 + 1 = 2 (a doublet)

(3)

HbHb

Hb


v Pascal’s Triangle● Use to predict relative intensity of

various peaks in multiplet● Given by the coefficient of

binomial expansion (a + b)n

singlet (s) 1doublet (d) 1 1triplet (t) 1 2 1quartet (q) 1 3 3 1quintet 1 4 6 4 1sextet 1 5 10 10 5 1

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v Pascal’s Triangle

● For

● For

Br C C BrHbHa

Cl Cl

Due to symmetry, Ha

and Hb are identicalÞ a singlet

Cl C C BrHbHa

Cl Br

Ha ≠ Hb

Þ two doublets

© 2014 by John Wiley & Sons, Inc. All rights reserved.

3. How to Interpret Proton NMRSpectra

1. Count the number of signals to determine how many distinct proton environments are in the molecule (neglecting, for the time being, the possibility of overlapping signals)

2. Use chemical shift tables or charts to correlate chemical shifts with possible structural environments


3. Determine the relative area of each signal, as compared with the area of other signals, as an indication of the relative number of protons producing the signal

4. Interpret the splitting pattern for each signal to determine how many hydrogen atoms are present on carbon atoms adjacent to those producing the signal and sketch possible molecular fragments

5. Join the fragments to make a molecule in a fashion that is consistent with the data


v Example: 1H NMR (300 MHz) of an unknown compound with molecular formula C3H7Br


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v Three distinct signals at ~ d3.4, 1.8and 1.1 ppmÞ d3.4 ppm: likely to be near an

electronegative group (Br)


d (ppm): 3.4 1.8 1.1

Integral: 2 2 3


d (ppm): 3.4 1.8 1.1

Multiplicity: triplet sextet triplet

2 H's on adjacent C

5 H's on adjacent C

2 H's on adjacent C


Complete structure:

BrCH2

CH2

CH3

• 2 H's from integration

• triplet


• sextet


• triplet

most upfield signalmost downfieldsignal


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4. Shielding and Deshielding of Protons: More About Chemical Shift


Protons absorb at different NMRfrequencies depending on the electrondensity around them and the effects oflocal induced magnetic fields.

The magnetic field associated with a spinning

proton

The spinning proton

resembles a tiny bar magnet


© 2014 by John Wiley & Sons, Inc. All rights reserved. © 2014 by John Wiley & Sons, Inc. All rights reserved.

9


© 2014 by John Wiley & Sons, Inc. All rights reserved. Copyright © 2016 by John Wiley & Sons, Inc. All rights reserved.

• The applied magnetic field causes s electrons to circulate in a way that induces a local magnetic field.

• The hydrogen of a C-H bond experiences a net smaller magnetic field than the applied field.

• The proton is said to be shielded from the applied magnetic field.

Protons of Hydrogen Atoms in Alkyl C-H Groups

The chemical shift for hydrogens of unsubstituted alkanes is typically in the range of d 0.8 – 1.8.

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© 2014 by John Wiley & Sons, Inc. All rights reserved. Copyright © 2016 by John Wiley & Sons, Inc. All rights reserved.

• Electronegative groups draw electron density away from nearby hydrogen atoms.

• Electronegative groups diminish the shielding of protons by circulating selectrons.

• The proton is said to be deshielded from the applied magnetic field.

Protons of Hydrogens Near Electronegative Groups

The chemical shift for hydrogens bonded to a carbon bearing an oxygen or halogen atom is typically in the range of d 3.1 – 4.0.


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12

Electronegativity

Complete structure:

BrCH2

CH2

CH3


• triplet


• sextet


• triplet

most upfield signalmost downfieldsignal


● If we were to consider only the relative electronegativities of carbon in its three hybridization states, we might expect the following order of protons attached to each type of carbon:

(higher frequency) sp < sp2 < sp3 (lower

frequency)© 2014 by John Wiley & Sons, Inc. All rights reserved.

● In fact, protons of terminal alkynes absorb between d 2.0 and d 3.0, and the order is

(higher frequency)

sp2 < sp < sp3 (lower frequency)


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v Shielding and deshielding by circulation of p electrons • The p electrons in alkenes, alkynes, and aromatic rings also

circulate as to generate an induced, local magnetic field

• Whether shielding or deshielding occurs depends on the location of the protons in the induced magnetic field.

Protons of Hydrogen Atoms Near p Electrons

Protons of Hydrogen Atoms Near p ElectronsThe hydrogens of benzene absorb at d 7.27.

Hydrogens bonded to substituted benzene rings have chemical shifts in the range of d6.0 – 8.5.

The chemical shift of alkene hydrogens is typically in the range of d 4.0 – 6.0.

The chemical shift of an alkyne hydrogen is typically in the range of of d 2.5 – 3.1.


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● e.g.

Hd Hc

Hb

Ha

d (ppm)Ha & Hb: 7.9 & 7.4 (deshielded)Hc & Hd: 0.91 – 1.2 (shielded)


● Aldehydes

OR

H

Electronegativity effect + Anisotropy effectÞ d = 8.5 – 10 ppm (deshielded)


v Two or more protons that are in identical environments have the same chemical shift and, therefore, give only one 1H NMR signal

v Chemically equivalent protons are chemical shift equivalent in 1H NMR spectra

5. Chemical Shift Equivalent and Nonequivalent Protons


5A. Homotopic and Heterotopic Atomsv If replacing the hydrogens by a

different atom gives the same compound, the hydrogens are said to be homotopic

v Homotopic hydrogens have identical environments and will have the same chemical shift. They are said to be chemical shift equivalent


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HC CHH

HH

H

Ethane

HC CHH

HH

Br

HC CHH

BrH

H

HC CHH

HBr

HHCC HH

HBr

H

HCC HH

BrH

H

HCC HH

HH

Br

v The six hydrogens of ethane are homotopic and are, therefore, chemical shift equivalent

v Ethane, consequently, gives only one signal in its 1H NMR spectrum

sam

e co

mpo

unds

same com

pounds


v If replacing hydrogens by a different atom gives different compounds, the hydrogens are said to be heterotopic

v Heterotopic atoms have different chemical shifts and are not chemical shift equivalent


HC CHH

HH

Br

HC CHH

HCl

BrBrCC HH

HCl

H

BrCC HH

HCl

H

BrCC HH

ClH

H

BrCC HH

HH

Cl

These 2 H’s are also homotopic to each other

different compoundsÞ heterotopic

same compoundsÞ these 3 H’s of the CH3 group are homotopicÞ the CH3group gives only one 1H NMR signal


HC CHH

HH

Br

v CH3CH2Br● two sets of hydrogens that are

heterotopic with respect to each other

● two 1H NMR signals


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v Other examples

(1) C CH

H

CH3

CH3Þ 2 1H NMR signals

(2) H

CH3H

H

H CH3

Þ 4 1H NMR signals


v Other examples

(3) H3CCH3

H H

H

H

H H

Þ 3 1H NMR signals


v Application to 13C NMR spectroscopy● Examples

(1) H3C CH3 Þ 1 13C NMR signal

(2)

CH3

CH3Þ 4 13C NMR signals


(3)

OHHO

Þ 5 13C NMR signals

(4)HO

HOÞ 4 13C NMR signals


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5B. Enantiotopic and Diastereotopic Hydrogen Atoms

v If replacement of each of two hydrogen atoms by the same group yields compounds that are enantiomers, the two hydrogen atoms are said to be enantiotopic


v Enantiotopic hydrogen atoms have the same chemical shift and give only one 1H NMR signal:

H3C Br

H H

H3C Br

H G

H3C Br

G H

enantiomers

enantiotopic



v If replacement of each of two hydrogen atoms by the same group yields compounds that are diastereomers, the two hydrogen atoms are said to be diastereotopic

CH3

H OHH3C

HaHb

dias

tere

omer

s

diastereotopic

CH3

H OHH3C

GHb

CH3

H OHH3C

HaG

chiralitycenter


v Except for accidental coincidence, diastereotopicprotons do not have the same chemical shift and give rise to different 1H NMR signals.(the difference in chemical shift may be small)

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Neuman projections help us see that diastereotopic experience different environments.

In every conformation Ha and Hb experience different environments.

They give rise to signals with different chemical shifts.

HbBrHa

H

dias

tere

omer

s

diastereotopic

HbBrG

H

GBr

Ha

H



6. Spin–Spin Coupling: More About Signal Splitting and Nonequivalent or Equivalent Protons


• Signal splitting arises due to spin-spin coupling.

• Spin-spin coupling effects are transferred primarily through bonding electrons and lead to spin-spin splitting.

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v Vicinal coupling is coupling between hydrogen atoms on adjacent carbons (vicinal hydrogens), where separation between the hydrogens is by three sbonds

Ha Hb3J or vicinal coupling

6A. Vicinal Couplingv Vicinal coupling between heterotopic

protons generally follows the n + 1 rule.

v Exceptions to the n + 1 rule can occur when diastereotopic hydrogens or conformationally restricted systems are involved



These hydrogens are more than 3 bonds away from any other hydrogens.

n = 0n + 1 = 1 (singlet)

These hydrogens are more than 3 bonds away from any other hydrogens.

n = 0n + 1 = 1 (singlet)


Recall…

These hydrogens are spin-spin coupled to two protons.

n = 2n + 1 = 3 (triplet)

These hydrogens are spin-spin coupled to five protons.

n = 5n + 1 = 6 (sextet)

These hydrogens are spin-spin coupled to two protons.

n = 2n + 1 = 3 (triplet)

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C C

H H

C C

H HA A

upfielddownfield

Bo

THE CHEMICAL SHIFT OF PROTON HA IS AFFECTED BY THE SPIN OF ITS NEIGHBORS

50 % ofmolecules

50 % ofmolecules

At any given time about half of the molecules in solution willhave spin +1/2 and the other half will have spin -1/2.

aligned with Bo opposed to Bo

neighbor aligned neighbor opposed

+1/2 -1/2

6B. Coupling Constants – Recognizing Splitting Patterns


• Protons that are coupled share a coupling constant (J).

• Coupling constants are determined by measuring the distance (in hertz) between each peak of a signal.

• A vicinal coupling constant is 6-8 hertz.

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X CHa

CHb

HbHbHa If we measure the separation of

peaks for two spin-spin coupled hydrogens, they have the same coupling constant (Jab).

This is called reciprocity of coupling constants.

6C. The Dependence of Coupling Constants on Dihedral Angle

v The magnitude of a coupling constant is related to the dihedral angle (f) between coupled protons.

H

H

f


v Karplus curve

v f ~0o or 180o

Þ Maximum value

v f ~90o

Þ ~0 Hz


v Karplus curve examples

Hb

Ha

Hb

Ha

f = 180ºJa,b = 10-14 Hz

(axial, axial)

Hb

Ha

Hb

Ha

f = 60ºJa,b = 4-5 Hz

(equatorial, equatorial)


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v Karplus curve examples

Hb

Ha

Hb

Ha

f = 60ºJa,b = 4-5 Hz

(equatorial, axial)


7. Proton NMR Spectra and Rate Processes


• Think of NMR like a camera with a slow shutter speed.

• Just as a photo will “blur” for an object moving rapidly, the NMR spectrum of a fast molecular process will be blurred.


Examples of rapid processes that “blur” NMR spectra

• Chemical Exchange of Protons

• Conformational Changes


Chemical Exchange Causes Spin Decoupling

We do not typically observe splitting between protons of the OH and CH2 group in ethanol.

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• Protons attached to electronegative atoms (such as O) with lone pairs of electrons can undergo rapid chemical exchange.

• These exchangable protons can be transferred rapidly from one molecule to another.

• The exchange is so rapid that the hydroxyl proton does not couple with C-H protons.



• Rapid exchange causes spin decoupling.• Spin decoupling is seen in the 1H NMR

spectra of alcohols, amines, and carboxylic acids.

• The signals of NH and OH protons are normally unsplit and broad.



• Protons that undergo rapid chemical exchange will exchange H with D from D2O.

• If placed in D2O, the 1H NMR signal from these protons will disappear.

• This is an easy method for identifying 1H NMR signals produced by exchangeable protons.

v Why don’t we see coupling with the O–H proton, e.g. –CH2–OH (triplet?)

● Because the acidic protons are exchangeable about 105 protons per second (residence time 10-5 sec), but the NMR experiment requires a time of 10-2 – 10-3 sec. to “take” a spectrum, usually we just see an average. Thus, OH protons are usually a broad singlet.


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+ 50 °C

- 30 °C

v Protons of alcohols (ROH) and amines (RNH2) may appear over a wide range from 0.5 – 5.0 ppm● Hydrogen-bonding is the reason for this

range

Proton NMR Spectra and Rate Processes



Conformational Changes

• At temperatures near room temperature, C-C single bonds can rotate rapidly.

• When we measure the 1H NMR spectra of compounds with single bonds that allow rotation, the spectra we obtain often reflects the hydrogen atoms in their “average” environment.



But, imagine for a moment that C-C bonds didn’t rotate…

We already said these protons are in the same environment…

If the C-C bond is frozen, these three hydrogens are notequivalent!

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However, the C-C bond rotates ~1 million times per second. Thus, they are all in the same “average” environment.


Conformational ChangesAt room temperature, cyclohexane chair conformations interconvert so fast that we only see one 1H NMR signal.


Conformational ChangesAt cold temperatures, cyclohexane chair conformations are slow.

The 1H NMR spectrum becomes very complicated due to complex spin-spin couplings between non-equivalent axial and equatorial protons.


Conformational ChangesAt – 100 oC, undecadeuteriocyclohexane gives only two 1H NMR signals of equal intensity.

Interconversions of the two conformations is slow at this temperature.

(The deuterons are invisible to the 1H NMR)

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v Unlike 1H with natural abundance ~99.98%, only 1.1% of carbon, namely 13C, is NMR active

8. Carbon-13 NMR Spectroscopy8A. Interpretation of 13C NMR

Spectra


8B. One Peak for Each Magnetically Distinct Carbon Atom


• Each distinct carbon produces one signal in a 13C spectrum.

• Splitting of 13C signals into multiple peaks is not observed in routine 13C spectra.

The odds of two 13C atoms (~1% abundance) being next to each other, to split each other, is very low!


• Although 13C-13C signal splitting does not occur in 13C NMR, 1H atoms attached can split 13C NMR signals

• To simplify the 13C NMR spectrum, such 1H-13C splitting is instrumentally eliminated.

• A spectrum with 1H-13C splitting eliminated is called broadband proton decoupled.

CH3 C CH2 CH3

H

OH

v Example:● 2-Butanol

Proton-coupled13C NMR spectrum


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CH3 C CH2 CH3

H

OH

v Example:● 2-Butanol

Proton-decoupled13C NMR spectrum


• In a boradband proton-decoupled 13C NMR spectrum, each carbon atom in a distinct environment gives a signal consisting of only one peak.

• Most 13C NMR spectra are obtained in the simplified broadband decoupled mode first and then in modes that provide information from the 1H-13C couplings.

8C. 13C Chemical Shiftsv Decreased electron density around an

atom deshields the atom from the magnetic field and causes its signal to occur further downfield (higher ppm, to the left) in the NMR spectrum

v Relatively higher electron density around an atom shields the atom from the magnetic field and causes the signal to occur upfield (lower ppm, to the right) in the NMR spectrum


Electronegative groups deshield the carbons to which they are attached.

The lower the electron density in the vicinity of a given carbon, the less the carbon will be shielded.

Cl CH2 CH CH3

OH

(a) (b) (c)

1-Chloro-2-propanol (a)(b)

(c)

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The CDCl3 solvent peaks at d 77 should be disregarded.

Cl CH2 CH CH3

OH

(a) (b) (c)

1-Chloro-2-propanol

solvent


BROMOCYCLOHEXANE

Cl

Cla

a

b

b

c

c

1,2-DICHLOROBENZENE

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8D. DEPT 13C Spectrav Distortionless Enhancement by

Polarization Transfer

v DEPT 13C NMR spectra indicate:• how many hydrogen atoms are bonded to

each carbon• the chemical shift information contained in a

broadband proton-decoupled 13C NMR spectrum.

v The carbon signals in a DEPT spectrum are classified as CH3, CH2, CH, or C accordingly


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Cl CH2 CH CH3

OH

(a) (b) (c)

1-Chloro-2-propanol

(a)(b) (c)


v The broadband proton-decoupled 13C NMR spectrum of methyl methacrylate


Cl CH2 CH CH3

OH

(a) (b) (c)

1-Chloro-2-propanol

(a)(b) (c)


v HCOSY● 1H–1H correlation spectroscopy

v HETCOR● Heteronuclear correlation

spectroscopy

9. Two-Dimensional (2D) NMR Techniques


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9A. 1H–1H COSY Cross-Peak Correlations

v HCOSY of 2-chloro-butane

H2

H1

H1

H3

H3

H4

H4

H2


9B. 1H–13C Heteronuclear Correlation Cross-Peak Correlations

v HETCOR of 2-chloro-butane

H1

H2

H3

H4

C1

C2

C3 C

4



Quiz 1


Quiz 2

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Quiz 3


Quiz 4

10. An Introduction to Mass Spectrometry


Mass spectrometry (MS) involves formation of ionsseparation of the ionsdetection of the ions

Based on mass and charge


Represents the formula weight of the detected ions. m/z is the mass (m) to charge (z) ratio.

Relative abundance of each detected ion

z usually = +1

base peak is the tallest peak

molecular ion

Small peaks with m/z +1 or +2 due to heavy isotopes

CH3CH2CH3

33

v In the mass spectrometer, a molecule in the gaseous phase under low pressure is bombarded with a beam of high-energy electrons (70 eV or ~ 1600 kcal/mol)

v This beam can dislodge an electron from a molecule to give a radical cation which is called the molecular ion, M+ or more accurately

11. Formation of Ions: Electron Impact Ionization

M 70 eV e-M


radical cationwith net +1 charge

EI

12. Depicting the Molecular Ion

CH3CH2 CH3

H3C OH H3C N CH3

CH3

H2C CHCH2CH3

Methanol Trimethylamine 1-Butene

Radical cations from ionizationof nonbonding on p electron


CH3CH2 CH3

H3C OH H3C N CH3

CH3

H2C CHCH2CH3

Methanol Trimethylamine 1-Butene

When a molecule contains O, N, or a p bond, we place the odd electron and charge there.

Sometimes the choice of “where” to locate the radical cation is arbitrary

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CompoundIonization

Potential (eV)CH3(CH2)3NH2 8.7C6H6 (benzene) 9.2C2H4 10.5CH3OH 10.8C2H6 11.5CH4 12.7

v Ionization potentials of selected molecules


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13A. Fragmentation by Cleavage at a Single Bond

v When a molecular ion fragments, it will yield a neutral radical (not detected) and a carbocation (detected) with an even number of electrons

v The fragmentation will be dictated to some extent by the stability of the carbocation generated:ArCH2

+ > CH2=CHCH2+ > 3o > 2o > 1o > CH3

+


v e.g.


More stable carbocation

Less stable carbocation

This fragmentation pathway is more predominant

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v As the carbon skeleton becomes more highly branched, the intensity of the molecular ion peak decreases

v Butane vs. isobutane

70eVe-

M+(58)

70eVe-

M+(58)

aCH3+

(43)a

b CH2CH3+(29)

b

CH3+(43)


Fragmentation


m/z = 29

m/z = 15

Smaller fragment peak

Larger fragment peak

Molecular ion peak

m/z = 44

13B. Fragmentation of Longer Chain and Branched Alkanes

v Octane vs. isooctane

M+(114)

(85)

(71)

(57)

(43)M+(114)

+

+

+

+

+(57)


37

v Masses are usually rounded off to whole numbers assuming:H = 1, C = 12, N = 14, O = 16, F = 19 etc.

Molecular ion (parent peak)

Daughterions[C8H18]

(M+, 114)[C6H13]

(85)

fragmentation

-CH3CH2 (29)

[C5H11](71)

-CH3CH2CH2 (29+14)


v Partial MS of octane (C8H18, M = 114)

11485

71

57M+

29 (CH3CH2)14 (CH2)



13C. Fragmentation to Form Resonance-Stabilized Cations


Alkenes ionize and undergo fragmentation to yield resonance-stabilized allylic carbocations.

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16C. Fragmentation to Form Resonance-Stabilized Cations

v Alkenes● Important fragmentation of terminal

alkenes t Allyl carbocation (m/e = 41)

R

(41)

R +


Ÿ Alkene Fragmentation– Fairly prominent M+

– Fragment ions of CnH2n+ and CnH2n-1+

– Terminal alkenes lose allyl cation if possible to form resonance-stabilized allylic cations

R - R

[CH2=CHCH2 CH2 CH3 ] CH2 =CHCH2+ + • CH2 CH3

+•


Carbon-carbon bonds next to an atom with a lone pair of electrons readily break because the resulting carbocation is resonance-stabilized.


Carbon-carbon bonds next to the carbonyl group of an aldehyde or ketone break to form a resonance-stabilized acylium ion.

acylium ion

39


Alkyl-substituted benzenes undergo loss of a hydrogen atom or methyl group to yield the tropylium ion

tropylium ionm/z = 91


Monosubstituted benzenes with other than alkyl groups lose their substituent to yield a phenyl cation.

phenyl ionm/z = 77Y = halogen, NO2, etc.

13D. Fragmentation by Cleavage of Two Bonds

v Alcohols frequently show a prominent peak at M - 18. This corresponds to the loss of a molecule of water

● May lose H2O by 1,2- or 1,4-elimination

+●


1,2-elimination: OH+ H2O

M (M - 18)

1,4-elimination:OH H OH

+ H2O

+ CH3CH2

M

(M - 18)Copyright © 2016 by John Wiley & Sons, Inc. All rights reserved.

40

v Alcohols● Most common fragmentation: loss of

alkyl groups

OH

M+(74)

a

OH OHCH3CH2 +b

(m/e = 45)

b


v Aldehydes● M+ peak usually observed but may

be fairly weak

● Common fragmentation patternt a-cleavage

RR H

OH C OR

C OH

+

+(m/e = 29)

acylium ion


hexanal

41

Mc Lafferty Rearrangement

v Ketones● a-cleavage

O a

a

b

b


42

OH H

OH OH

(m/e = 86)2º radicalobserved

i

i

OH

1º radical

OH

(m/e = 114)NOT observed

ii

ii


v Characteristics of McLafferty rearrangements1. No alkyl migrations to C=O, only H

migrates

OH

O

O

R

R

R

H

HX


v Characteristics of McLafferty rearrangements2. 2o is preferred over 1o

OH Hiii

OH

2º radical

OH

1º radical

not


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v Aromatic hydrocarbons● very intense M+ peaks ● characteristic fragmentation pattern

(when an alkyl group attached to the benzene ring): tropylium cation

CH3CH2

CH3

(m/e = 91)tropylium cation

rearrangement+

benzyl cation


14. Isotopes in Mass Spectra


v 13C and 12CAbout 1.1% of all carbon atoms are the 13C isotope

v About 98.9% of the methane molecules in the sample will contain 12C, and the other 1.1% will contain 13C


44

v Example● Consider 100 molecules of CH4

M : 16

H1

C12H1 H1

H1

H1

C13H1 H1

H1

H1

C12H1 H2

H1

M + 1 = 17

C12: 100 C13: 1.11

H1: 100 H2: 0.016Copyright © 2016 by John Wiley & Sons, Inc. All rights reserved.

M : 16

H1

C12H1 H1

H1

H1

C13H1 H1

H1

H1

C12H1 H2

H1

M + 1 = 17

1.11 molecules contain a 13C atom

4x0.016 = 0.064 molecules contain a 2H atom

Intensity of M + 1 peak:1.11+0.064=1.174% of the M peak

+●+●


≈

100

1.17m/z

rela

tive

ion

abun

danc

e

M

M +1

+●

+●


v Some elements that are common in organic molecules have isotopes that differ by two atomic mass units.

v These include 16O and 18O, 32S and 34S, 35Cl and 37Cl, and 79Br and 81Br.

v It is particularly easy to identify the presence of chlorine or bromine using mass spectrometry because multiple isotopes of chlorine and bromine are relatively abundant


45

• The natural abundance of 35Cl is 75.5% and that of 37Cl is 24.5%

• In the mass spectrum for a sample containing chlorine, we would expect to find peaks separated by two mass units, in an approximately 3 : 1 (75.5% : 24.5%) ratio for the molecular ion or any fragments that contain chlorine


• The natural abundance of 79Br is 51.5%, and that of 81Br is 49.5%

• In the mass spectrum for a sample containing bromine we would expect to find peaks separated by two mass units in an approximately 1 : 1 ratio (49.5% : 51.5%)


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14A. High-Resolution Mass Spectrometry


“Low-resolution” mass spectrometers measure m/z to the nearest whole number unit.

“High-resolution” mass spectrometers an measure m/z values to three or four decimal places.

Because the masses of the actual masses of individual atoms are not integers, a high-resolution mass spectrum can be used to determine molecular formulas.


v Example 1

O2, N2H4 and CH3OH all have M.W. of 32 (by MS), but accurate masses are different

t O2 = 2(15.9949) = 31.9898

t N2H4 = 2(14.0031) + 4(1.00783) = 32.0375

t CH4O = 12.00000 + 4(1.00783) + 15.9949 = 32.0262


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v Example 2

Both C3H8O and C2H4O2 have M.W. of 60 (by low-res MS), but their accurate masses are different.

t C3H8O = 60.05754

t C2H4O2 = 60.02112


Low Resolution Mass

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High Resolution Mass

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Identify which one of the following isomers of C6H14 has the C-13 NMR below.

A) CH3CH2CH2CH2CH2CH3 C) (CH3)2CHCH(CH3)2

B) CH3CH2CH2CH(CH3)2 D) CH3CH2C(CH3)3

A) CH3CH2CH2CH2CH2CH3 C) (CH3)2CHCH(CH3)2 B) CH3CH2CH2CH(CH3)2 D) CH3CH2C(CH3)3

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m/e = 72 ?

72

O

nuclear magnetic resonance and mass spectrometryhome.konkuk.ac.kr/~parkyong/classes/ch09.pdf ·...

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