Nuclear Magnetic Resonance(NMR) Spectroscopy -...

Chapter 2 - 89

Nuclear Magnetic Resonance(NMR) Spectroscopy

Because each proton in a substance is surrounded by electrons that have their own magnetic fields, the magnetic environment of a proton depends on the bonding, or chemical, environment in which the proton is involved

Chapter 2 - 90


a

b c

b

c

a

▶ The combination of IR and NMR data is often sufficient todetermine completely the structure of an unknown molecule.

Information from 1H-NMR

The environments and numbers of all the hydrogens in a molecule.

Intensity of signal

position (chemical shift), splitting, andnumber of signal

91

Integrals

Peak picking

ppm scale ppm

ppm

Chapter 2 - 91


1H-NMR chart

Chapter 2 - 92

Theory of NMR

In the absence of an external magnetic field these bar magnets are randomly orientated.

In an external magnetic field they orient either with(α spin state) or against(β spin state) the magnetic field.


The positively charged nuclei of certain elements (e.g., 13C and 1H) behave as tiny magnets: 1H and 13C have spin quantum number I =1/2

- Nuclei with I=0 (ex. 12C and 16O) do not have spin and do not respond to an external magnetic field

B0: magnetic flux density (T)

Chapter 2 - 93

Theory of NMR

The magnetic moment of the lower energy +1/2 state is aligned with the external field ( ↑ ), and that of the higher energy -1/2 spin state is opposed to the external field ( ↓ ).


Direction of an Externally Applied Magnetic Field (Bo)

Spin +1/2 (or α)Aligned

Spin -1/2 (or β)Opposed

Bo

EEnergy- 1/2 Opposed to Field

+ 1/2 Aligned with Field

The nuclei aligned with the magnetic field can be flipped to align against it if the right amount of energy is added (ΔE)

Chapter 2 - 94

Theory of NMR


Application of just the right radiofrequency causes the nucleus to “flip” to the higher energy spin state

The exact amount of energy required depends on the chemical identity(H, C, or other element) and the chemical environment of the particular nucleus.

Eabsorbed = E = (E-1/2 state - E+1/2 state) = h = hBo/2

The small energy difference between the two alignments of magnetic spin corresponds to the energy of radio waves according to Einstein’s equation E=h (energy of a photon)

The necessary frequency is : Bo/2

Spin state energy differenceE = hBo/2

When E = E, Spin flip occurs

Chapter 2 - 95

Theory of NMR


The amount of energy required depends on the strength of the external magnetic field (and therefore electron density around a nucleus).

The stronger the applied magnetic field (Bo), the greater the energy difference between the spin states, (the higher the radio frequency energy required to flip the nuclear spin)

Aligned against the applied field

Aligned withthe applied field

Chapter 2 - 96

Theory of NMR


Nuclei are surrounded by electrons. The strong applied magnetic field (Bo) induces the electrons to circulate around the nucleus.

The induced circulation of electrons sets up a secondary (induced) magnetic field (Bi) that opposes the applied field (Bo) at the nucleus. (*Lenz’s law)

Nuclei are shielded from the full applied magnetic field by the surrounding electrons because the secondary field diminishes the field at the nuclei

Actual magnetic field is less than Bo due to the induced magnetic field (Bi) which opposed the

Bo appliedBi induced

= B0 - Bi

I

Chapter 2 - 97

NMR Spectroscopy

Bo

++

Factors which lower e‐ density allow thenucleus to “see” more of the B0 beingapplied – resonance occurs at higher energy

Factors which raise e‐ density reduce theamount of B0 the nucleus “sees” –resonance condition occurs at lower energy

Less shielded (more deshielded) More shielded (less deshielded)

Less electron, smaller opposing Bi

More electron, larger opposing Bi

(higher frequency) (lower frequency)

Theory of NMR

The necessary frequency, Bo/2

Chapter 2 - 98

NMR Spectroscopy

Theory of NMR

The necessary frequency, Bo/2

= B0 - Bi

Chapter 2 - 99

Shielding / Deshielding


This represents the electron density of a C-H bond.

- If Z is an elelctronegative atom, the carbon becomes electron deficient and pulls some of the electron density away from the H.

- If Z is an EDG, more electron density ends up on the H.

→ How much electron density is on the proton depends on what else is attached to the carbon.

Chapter 2 - 100



• In acetic acid, the –CH3 protons arein an e- rich environment relative tothe –OH proton.

• Shielding of the electrons opposesB0 and therefore ΔE is lower thanthat observed for the –OH proton.Here ΔE is large as the full effect ofB0 is felt

Deshielded Shielded

← Increasing frequency ()

Bo/2

(B0 ↓ → ↓)

B0 ↓

↓

Chemical shift(ppm)(← increasing frequency, ν)

Chapter 2 - 101

3.442.55

SiCH3

CH3

CH3

CH3

More shielded

Less shielded

Extremely shielded!



Chapter 2 - 102



The electron density surrounding a given nucleus depends on the electronegativity of the attached atoms.

The more electronegative the attached atoms, the less the electron density around the nucleus in question: less shielded, or deshielded

Deshielding effects are generally additive. That is, two highly electronegative atoms (2 Cl atoms, for example) would cause more deshielding than only 1 Cl atom.

CH

HHH

CH

ClHH

CH

ClH

Cl

C and H are deshielded C and H are more deshielded

Chapter 2 - 103

Chemical Shift


Nuclei in different chemical environments in a molecule will absorb at slightly different frequencies

We define the relative position of absorption in the NMR spectrum the chemical shift. It is a unitless number but we assign ‘units’ of ppm or units.

For 1H, the usual scale of NMR spectra is 0 to 12 ppm (or ). (The usual 13C scale goes from 0 to about 220 ppm)

The zero point is defined as the position of absorption of a standard, tetramethylsilane (TMS):

This standard has only one type of C andonly one type of H. 1.90 2.55

SiCH3

CH3

CH3

CH3

Chapter 2 - 104

NMR Spectroscopy

Chemical Shift (δ)

- a reference compound is placed in the solution of the substance to bemeasured, and the resonance frequency of each proton in the sample ismeasured relative to the resonance frequency of the protons of thereference substance.

- The standard reference substance which is universally used istetramethylsilane (CH3)4Si, also called TMS: the protons of its methylgroups are more shielded than those of most other known compounds.

- Thus when another compound is measured the resonance of its protonsare reported in terms of how far they are shifted from those of TMS.

1.90

2.55 Electron crowding→ extremely shielded!

Chapter 2 - 105

Tetramethyl silane (TMS) is used as reference because it is - soluble in most organic solvents, - inert, - volatile, and - has 12 equivalent 1H (and 4 equivalent 13C)

→ one signal(peak) is producedTMS signal is set to 0 : most shielded (electron crowding)

TMS (tetramethyl silane)


Chapter 2 - 106


Chapter 2 - 107


Chemical Shift (δ)

The chemical shift for a given proton is in frequency units (Hz)

This value will change depending on the B0 of the particular spectrometer

By reporting the NMR absorption as a fraction of the NMR operating frequency, we get units, ppm, that are independent of the spectrometer

Chapter 2 - 108

- The Values of δ for a given proton will always be the same irrespective ofthe magnetic field strength of the instrument (whereas the absolutefrequency depends on field strength)

- ex. at 60 MHz the shift of the proton in CH3 Br is 162 Hz from TMS, while at300 MHz the shift is 810 Hz. However both of these correspond to thesame value of δ

162 Hz

60 MHz

810 Hz

300 MHz


δ =

= 2.7ppm

=

(a) Shielding effects : an electron shields the nucleus, the absorption shifts upfield.

(b) Deshielding effects : Decreased electron density deshields a nucleus, the absorption shifts downfield

Shielding and Chemical Shift

Chapter 2 - 109 Chapter 2 - 110

Chemical Shift


Proton Chemical Shift () vs. Electronegativity

-0.50

0.51

1.52

2.53

3.54

4.55

1.5 2 2.5 3 3.5 4 4.5

Electronegativity

H1

Che

mic

al S

hift

CH3 Si

CH3 C

CH3 N

CH3 O

CH3 F

1.90

2.553.04

3.44

3.98Down field

(deshielded)

up field(shielded)

Chapter 2 - 111

3.44

2.55

1H-NMR spectrum of Methyl t-butyl ether

- Alkene and aromatic ring hydrogens are deshielded by the circulation of π electronscf. A terminal alkyne hydrogen is shielded by the circulation of π electrons

Intensity = area : Relative # of H ex. Ha:Hb=1.5:1.0=3:2

Chapter 2 - 112

deshielding by the benzenering is even larger than foralkene double bonds

deshielding

Chapter 2 - 113

Chemical Shift


Applied Magnetic Field Strength – Bo is held constant

Shielding of Proton by Valence Electrons

Applied Radio Frequency - RF

Downfield upfield

Chapter 2 - 114

Magnetic induction in the π bond of a carbon-carbon double bond deshields vinylic hydrogens and shifts their signal higher frequency

Carbon-Carbon Double Bond Effect


ΔE becomes greater → deshielding effect

Diamagnetic anisotrpy

Chapter 2 - 115

The magnetic field induced by circulation of π electrons in an aromatic ring deshields the hydrogens on the ring and shifts their signal to higher frequency

Aromatic Effect


Chapter 2 - 116

Carbon-Carbon Triple Bond Effect

A carbon-carbon triple bond shieldsan acetylenic hydrogen and shifts its signal to lower frequency (to the right) to a smaller value


sp carbon is more electronegative than sp2 or sp3 : deshielding effect→ supposed to be found at higher frequency than double bond, but found at lower frequency than expected!

Chapter 2 - 117

Shielding cones for common π systems


(+) denotes shielding areas and (–) denotes deshielding areas.

Chapter 2 - 118

H

H H

H H

CH2

CCH3

O

7.2(5 protons)

3.6

2.1(3 protons)

(2 protons)

The nmr spectrum of phenyl acetone : a compound having three chemically distinct types of protons.

Chemical Equivalence


Chapter 2 - 119

Chemical Environment and Chemical shift

More shielded absorb at a higher field

Less shieldedabsorbs at alower field

More electronegative atoms deshield more and give larger shift values.

Effect decreases with distance

Additional electronegative atoms cause increase in chemical shift.


H CH

HH

H CH

ClH

H CCl

ClH

H CCl

ClCl

2.8 ppm

2.3 ppm

1.9 ppm

Chapter 2 - 120

- All of the protons in a molecule which are in chemically identicalenvironments will often exhibit the same chemical shift.

- Each compound gives rise to a single absorption peak in its nmr spectrum.The protons are said to be chemically equivalent.

H

H

H

H

H

H H2C

H2CH2C CH2

CH2

CH2

CH2

C

C

O

O

OCH3

OCH3

CH3

H

H

CH3

H

H

C

O

CH3CH3

Si

CH3

H3C CH3

CH3

CH3 C

O

OCH3 CH3 O OCH2Cl

Molecules giving rise to one NMR absorption peak all protons chemically equivalent

Molecules giving rise to two NMR absorption peaks two different sets of chemically equivalent protons

Chemical Equivalence


Chapter 2 - 121

H3C O CH3 H3CH2C Cl H3C O CH2CH3

All equivalent H’s signal

2 types of H’s 2 signals

3 types of H’s 3 signals

Chapter 2 - 122

Number of H type (signal) and intensity ratio?


Chapter 2 - 123

Schematic NMR Spectrometer


- Perpendicular to the RFoscillator coil is a detector coil.

- When no absorption of energyis taking place, the detectorcoil picks up none of theenergy given off by the RFoscillator coil.

- When the sample absorbs energy however the reorientation of the nuclearspins induces a radiofrequency signal in the plane of the detector coil, andthe instrument responds by recording this is a resonance signal or peak.

Chapter 2 - 124

Sample preparation

NMR Spectroscopy

- Use about 700 μL of deuterated solvent (for a 5-mm tube).

- If necessary, filter your sample using glass wool in the end of a pipette.

- The height of liquid should be about 4.5cm.- Label your tube (on the cap or topmost 2 cm of

the tube).- NMR tube caps must be on tightly to prevent

evaporation of solvent while waiting in the queue.- Typical concentrations are:

10-mg (H1)/50 mg (C13) for 300 MHz, 5 mg (H1) /20 mg (C13) for 400 MHz.

Solvent

Vortex mixer

Chapter 2 - 125

Choosing a solvent for NMR


- To avoid spectra dominated by the solvent signal, most 1H NMR spectra are recorded in a deuterated solvent. ex. CDCl3 (chloroform-d)

DMSO-d6 (dimethyl sulfoxide-d6)

- However, deuteration is not "100%", so signals for the residual protons are observed: For chloroform as a solvent (CDCl3), the residual signal is due to CHCl3, so a singlet signal is observed at 7.26 ppm. (DMSO-d6 = 2.50ppm)

- Most solvents used for NMR analysis are toxic.

Chapter 2 - 126

- The sample is dissolved in a solvent having no protons (usually CCl4or DMSO-d6) and a small amount of TMS is added to serve as aninternal reference.

- The sample cell is a small cylindrical glass tube which is suspendedin the gap between the faces of the pole pieces of the magnet.

- The sample is spun around its axis to ensure that all parts of thesolution experience a relatively uniform magnetic field.


Chapter 2 - 127

- The sample is positioned in the magnetic field and excited via pulsations in the radio frequency input circuit.

- The realigned magnetic fields induce a radio signal in the output circuit which is used to generate the output signal.

- Fourier analysis of the complex output produces the actual spectrum. - The pulse is repeated as many times as necessary to allow the signals to

be identified from the background noise.


The Fourier transform decomposes a function of time (a signal) into the frequencies that make it up

Chapter 2 - 128

Spin-Spin Splitting (N+1) rule

- Even in simple molecules one finds that each type of protonrarely gives a single resonance peak.

- For instance 1,1,2-trichloroethane there are two chemicallydistinct types of hydrogens:

C

Cl

Cl CH2

HCl


Chapter 2 - 129


- One would predict two resonance peaks in thenmr spectrum of 1,1,2-trichloroethane with anarea ratio of 2:1.

- In fact the high resolution nmr spectrum of thiscompound has five peaks: a group of threepeaks (called triplet) at 5.77δ and a group of twopeaks (called doublet) at 3.95δ.

- This phenomenon is called spin-spin splitting. Inan empirical fashion spin-spin splitting can beexplained by the n+1 rule.

C

Cl

Cl CH2

HCl


if a set of hydrogenshas n neighboring, non-equivalent hydrogens, it will be split into n + 1 subpeaks.

Chapter 2 - 130

Two neighbors giveA triplet (n+1=3)

One neighbors givesA doublet (n+1=2)



HbHa

Chapter 2 - 131



- Looking at the Ha signal: in addition to being shielded by nearby valence electrons, each of the Ha protons is also influenced by the small magnetic field generated by Hb next door (remember, each spinning proton is like a tiny magnet).

- The magnetic moment of Hb can be up or down (aligned with B0 or opposed to B0)

- In other words, in half of the molecules Ha is shielded by Hb(thus the NMR signal is shifted slightly upfield) and in the other half Ha is deshielded by Hb (and the NMR signal shifted slightly downfield).

- A single Ha peak is split into two sub-peaks (a doublet), one upfield and one downfield of the original signal.

Ha and Hb protons are spin-coupled to each other.

→upfieldshielded

←downfielddeshielded

Chapter 2 - 132



Combinations of Magnetic moment of two protons : 1 + 2 + 1

1 2 1

Chapter 2 - 133



Combinations of magnetic moment of three protons (Hc) : 1 + 3 + 3 + 1

Chapter 2 - 134



Pascal’s triangle

1 2 1

1 3 3 1

1

1 1

Chapter 2 - 135

135

C

H

H

H

C

H

I

H

a b

C

C

H

H

H

C

H

I

H

a b

CThree equivalent neighbors give a quartet (n+1=4) Two equivalent neighbors

gives A triplet (n+1=3)



Hb,c

Ha

Chapter 2 - 136

1) Protons that are equivalent by symmetry usually do not splitone another (equivalent protons)

2) Protons in the same group usually do not split one another

Spin-Spin Splitting (N+1) rule : Exceptions!


Chapter 2 - 137

Spin-Spin Splitting (N+1) rule : Exceptions 1) symmetry


Chapter 2 - 138

3) Protons The n+1 rule applies principally to protons in aliphaticchains or aliphatic rings,

but does not apply (in the simple way shown here) to protons ondouble bonds or benzene rings

Spin-Spin Splitting (N+1) rule : Exceptions!


Chapter 2 - 139



Chapter 2 - 140

Methylcyclopentane



Chapter 2 - 141

n=7

Chapter 2 - 142

Information derived from a 1H NMR spectrum




Chapter 2 - 147

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