NMR - NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

WHAT WE KNOW…

Tool for structure determination.

Used for both organic or inorganic compounds..?

Easily available..?

Expensive or cheap..? Affordable..?

Quantitative analysis

REFERENCES

INTERNET REFERENCES

http://www.youtube.com/watch?v=7aRKAXD4dAg

- Type ‘Magritek’ in youtube

http://www.youtube.com/watch?v=jRxgX-7FO8g

http://www.freelance-teacher.com/videos.htm

- Type ‘freelanceteach’ in youtube

• http://www.youtube.com/watch?v=uNM801B9Y84

VECTORS

Quantities which have both magnitude and direction.

Examples for vectors are Force, velocity, angular momentum, spin etc

When two vectors (P and Q) interact with each other, the resultant vector (R) has a different direction and magnitude, which is influenced by the properties of the interacting vectors (P and Q)

R = (P2+𝑸𝟐+ 𝟐𝑷𝑸𝒄𝒐𝒔θ)0.5

SPIN

Form of angular momentum. This is an intrinsic property shown by elementary particles and some atomic nuclei.

This is a vector quantity, has both magnitude and direction.

Particles which posses spin and have charge, interact with magnetic fields, since they generate their own magnetic field

Bose – Einstein Statistics

Fermi – Dirac Stats

• Indistinguishable particles

• One particle per energy level

• Particles have half integral spin

• Examples for Fermions are protons, electrons, NO, 2He3

• Indistinguishable particles

• Any number of particles can occupy a given energy level

• Particles have integral spin.

• Examples for Bosons are N2, H2, photons.

NUCLEAR SPIN

Nuclei consist of nucleons, each having spin quantum number of ‘Half’ (1/2)

When these nucleons combine to give the nucleus of an atom, the nucleus possesses ‘Spin’

For a nucleus, the number of allowed spin states is quantized and determined by it’s ‘Nuclear Spin Quantum Number’ (Symbol : I)

For each nucleus, there are (2I+1) allowed spin states, with integral differences ranging from +I to –I

+ I, (I-1), (I-2),…., (-I+1), (-I)

Exercise: Calculate the allowed spin states for a. Hydrogen (I= ½)b. Chlorine (I= 3/2).

You should get 2 spin states for Hydrogen i.e(+1/2 and -1/2)

4 spin states for Cl i.e. (-3/2, -1/2, +1/2, +3/2)

MORNING EXERCISE

NMR ACTIVE NUCLEI

All nuclei carry charge, they possess spin angular momentum

Only nuclei having spin quantum number I > 0 are NMR active.

Mass no Atomic no. Spin quantum no.

Examples NMR Activity

Odd Odd or Even Half Integral ½(H), 3/2 (13C), 5/2 (17O)

Active

Even Even Zero 0 (12C, 16O) Inactive

Even Odd Integral 1(2H, 14N), 3(10B) Active

OF ENERGY AND POPULATION

In the absence of an applied magnetic field, all spin states of a nucleus are degenerate (have equal energy)

In a collection of atoms, all spin states should be equally populated.

This situation changes, when an external magnetic field is applied.

Degeneracy is lost, the spin states have different energies.

Also, the population of atoms occupying each state changes.

ALIGN AND OPPOSE

For a Hydrogen nucleus, there are two spin states, (+1/2) and (-1/2)

Under an applied field, each nucleus can adopt only one of these states.

(+1/2) is of lower energy since it is aligned to the field. This is called α state

(-1/2) is of higher energy since it is opposed to the field. This is called the β state

NUCLEAR MAGNETIC RESONANCE

NMR occurs when nuclei aligned to applied field, absorb energy and oppose it.

This is a Quantized process , and energy absorbed should be equal to difference in energy between the two spin states.

Energy difference is also a function of applied magnetic field.

Eabs = E(1/2) – E(-1/2) = hν

Inaugural Derivation..!

Concentrate on the Board..!

VISUALISE

PRECESSION AND RESONANCE

COMPARE…

Source : Google Images

SPINNING TOPS AND NUCLEI

The nucleus behaves as a spinning top under the influence of an external field.

The nucleus undergoes precession around its own axis with an angular frequency ω.

Stronger the magnetic field, higher the ω.

Precession generates an oscillating electric field, of same frequency

If radio waves of same frequency are supplied externally, absorption occurs.

PROTON FLIPPING

Absorption occurs since the electric field generated by the nucleus and the electric field of the incoming radiation undergo coupling.

Energy can be transferred from the incoming radiation to the nucleus.

Spin change occurs from (+1/2) to (-1/2).

This condition is called Resonance.

ANATOMY

Source : Google Images

LABELLING THE PARTS

There are primarily 4 parts :

- The Magnet : With a controller to produce a specified magnetic Field

- Radio Frequency Source

- Detector : To detect the output signal

- Recorder : To record the output and plot it against the magnetic field.

FURTHER DETAILS

Sample is taken in a glass tube, placed between 2 poles of a magnet.

Sample is exposed to Radio Waves, frequency kept constant.

The tube is spun around at a steady rate, such that all parts of the sample get equal exposureto the radiation and the magnetic field.

Using electromagnets, the magnetic field strength is varied.

SOLVENTS

As the field strength is varies, the precessionalfrequency of the proton matches the frequency of incident radiation.

Small amount of the compound is taken in 0.5-1 ml of solvent.

Solvents used are Deuterated. Common ones are D2O, CDCl3, DMSO.

These solvents do not contain protons.

Deuterium though being NMR active, does not resonate along with a proton.

STRUCTURES

- Pure CDCl3 and DMSO do not show any peaks in 1H NMR.

- However, commercialsamples are not 100% pure and hence show peaks.E.g: DMSO – 2.5ppm,

CDCl3 – 7.32, D2O – 4.79*

Ref: Fulmer, Miller et al : NMR Chemical Shifts of Trace Impurities: CommonLaboratory Solvents, Organics, and Gases in Deuterated Solvents Relevant to the Organometallic Chemist, Organometallics 2010, 29, 2176–2179

Source : Wikipedia

FT NMR

The older NMR spectrometers used to excite nuclei of one type at a time.

Resonances of individual nuclei are recorded separately and individually.

Modern instruments use short (1-10 μs) bursts of energy (radiation) called Pulses.

A Pulse contains a range of frequencies, which is enough to excite all the types of protons at the same time.

Each type of proton, emits radiation of different frequencies. This emission is called a Free Induction Decay. (FID)

The intensity of this FID decreases over time as the nuclei lose their excitation.

The FID signal is a superimposed combination of all the frequencies emitted.

A mathematical method called ‘Fourier Transform’ is used to extract individual frequencies.

The FT breaks the FID into individual sine or cosine wave components.

From these components the individual frequencies can be determined.

The computer collects intensity vs time data and converts it into intensity vs frequency data.

The final spectrum obtained (I vs ν) is called the FT NMR Spectrum.

Left : Free Induction Decay SignalBelow: The FID Signal broken into a sine wave using FT.

ADVANTAGES OF FTNMR

Entire spectrum recorded, computerized and transformed in few seconds

400 spectra accumulated in approx. 13mins

Samples at very low conc. Can be measured.

Studies on nuclei with low natural abundance, and small magnetic moments ( 13C, 15N & 17O)

Improved spectra for sparingly soluble compounds.

Many spectra for same sample : Averaging

DEFLECTION

SHIELDING

Each proton is in a slightly different chemical environment within the molecule and hence not all protons resonate at the same frequency.

Electrons surrounding the protons shield (or protect) the protons from the applied magnetic field.

The electrons circulate and form a current.

This current generates a secondary magnetic field which opposes the applied field. This is called ‘Diamagnetic Sheilding’

QUANTITATIVE ASPECTS

Greater the electron density around the proton, greater the induced counter field.

Proton experiences a lower magnetic field, and hence the precessional frequency of the proton decreases.

Due to this the proton undergoes absorption at a lower frequency.

Since each proton is in a different environment, each has a different resonance frequency.

DESHIELDING

In some cases, like aromatic nuclei, closed loops of pi-electrons are found.

These electrons cause strong diamagnetic shielding around the aromatic nucleus, but the protons align themselves to the magnetic field and hence are deshielded.

In other words, when protons align themselves to the direction of the applied field, they are said to be deshielded.

MEASUREMENTS AND REFERENCES

The difference in absorption frequencies is very low (50-100 Hz in field of 60 MHz)

The exact resonance frequencies are not easily measurable, and hence a reference compound TMS (Tetramethylsilane) is used.

TMS protons are most shielded. TMS gives one sharp peak, which is taken as zero.

All other proton resonance frequencies are measured relative to TMS

NUMERICALS

Calculate the resonance frequency of a proton at a.) 1.41 Tesla.b.) 2.35 Tesla. c.) Find the ratio of the given field strengthsand the calculated resonance frequencies

From this we infer that, for a given proton, the shift in frequency from TMS is field dependent.

CHEMICAL SHIFT

Protons resonate at different frequencies depending on the strength of the applied field and the radiation used.

This can be confusing if chemists have spectrometers of different magnetic fields.

To avoid this confusion, a new parameter which is independent of the applied field is defined.

This parameter is called ‘Chemical Shift’

MATHEMATICALLY SPEAKING

• Values of δ for a given proton are same irrespective of the spectrometer frequency.

MORE NUMERICALS

For example : Bromomethane protons resonate at 162Hz in 60 MHz NMR, but in 100 MHz NMR they resonate at 270 Hz. Calculate δ for both.

Calculate the frequency and energy corresponding to 1 ppm shift from TMS in a.) 600 MHz NMR Spectrometerb.) 300 MHz NMR Spectrometer

SUMMARISING…

Chemical shift of frequency.

Shielding : Opposing the applied magnetic field, peaks at low δ values. E.g : Aliphatic, vinylic, benzylic protons etc

Deshielding : Alignment towards applied field. Peaks at higher δ values. E.g : Aromatic protons and protons of aldehydes and carboxylic acids.

So tell me where do ketone proton peaks appear..?

independent

FACTORS AFFECTING CHEMICAL SHIFT

Inductive or Electronegativity effects

Hybridization effects

Hydrogen Bonding

Solvent effects

Van der Waal’s deshielding

Geminal and Vicinal Coupling

Proton exchange

ELECTRONEGATIVITY EFFECTS

If electronegative elements are attached to the same carbon of interest, the absorption shifts downfield.

More electronegative the atom, more downfield is the absorption.

Multiple substituents have a stronger effect

Effect decreases over distance

This idea can be used to measure electronegativity of elements

ELECTRONEGATIVITY EFFECTS

HYBRIDIZATION EFFECTS

Higher the s character of the bond, closer the electrons are to the nucleus. Proton resonates downfield. Other effects can affect this order.

For sp3 hybridised C atoms

3o > 2o > 1o > strained ring

δ Values: 2 1 0

Sp2 and sp Hybrid Carbons

The s character is greater in protons attached to sp2 carbons. Appear at 4.5 – 7 ppm.

Aromatic Protons appear at 7-8 ppm, and aldehydic protons at 9-10 ppm. This high shift is caused by another effect called anisotropy.

For acetylic hydrogens, anisotropy results in shift at 2-3 ppm, though it has greater s character.

Protons attached to benzene rings are influenced by 3 magnetic fields.

i. External Magnetic fieldii. Magnetic field generated by circulating πelectrons.Iii. Magnetic field generated by valence electrons around the protons.

These fields interact with each other and caused anisotropy. (non uniformity).

Benzene protons lie in the deshielded region and hence they appear downfield.

MAGNETIC ANISOTROPY

MAGNETIC ANISOTROPY

All groups with π electrons cause anisotropy. In acetylene, the geometry of the generated field is such that the protons get shielded. Hence they appear upfield.

FINDING AROMATICITY

Hydrogen Bonding

In samples that can undergo intermolecular H bonds, the δ value is highly concentration dependent. Intramolecular H bonding is very slightly concentration dependent.

EXCHANGEABLE PROTONS

Protons which can undergo exchange show variable and broad absorptions.

VARIABLE ABSORPTIONS

Acids RCOOH 10.5-12.0 ppm

Phenols ArOH 4.0- 7.0 ppm

Alcohols ROH 0.5-5.0

Amines RNH2 0.5-5.0

Amides RCONH2 5.0- 8.0

Enols CH=CH-OH > 15

Functional Group Proton

Chemical Shift (ppm) Reason

Ethers R-O-CH 3.2-3.8 ppm Electronegativity of oxygens

Haloalkanes -CH-I 2.0-4.0 ppm

Electronegativity of halogen atom

-CH-Br 2.7-4.1

-CH-Cl 3.1-4.1

-CH-F 4.2-4.8

Nitriles -CH-C≡N 2.1-3.0 α- Protons deshieldedby C≡N group

Amines R-N-H

CH-N-

Ar-N-H

0.5-4.0

2.2-2.9

3.0-5.0

Protons undergo exchange

Resonance removes electron density from

nitrogen

Functional Group Proton ChemicalShift (ppm)

Reason

Ketones R-CH-C=O 2.1-2.4 Anisoptropy of C=Ogroup

Esters 2.1-2.5

3.5-4.8

Anisoptropy of C=Ogroup

Electronegativity of Oxygen

Carboxylic acids

R-COOH

-CH-COOH

11.0-12.0

2.1-2.5

Electronegativity of Oxygen

Signal usually broad.Exchangeable

Amides R(CO)-N-H-

-CH-CONH-

R(CO)-N-CH-

5.0-9.0

2.1-2.5

2.2-2.9

Variable absorption

Deshielded by carbonyl group

- Nitrogen’s electronegativity

CH 2

O

O

C H 2

Apart from finding the types of hydrogenspresent, NMR can also be used to find the relative number of hydrogens.

This can be done by calculating the area under each peak.

This is done by tracing a vertically rising line over each peak, this line is called the integral.

The integral rises by an amount proportional to the area under the peak.

INTEGRALS AND INTEGRATION

INTEGRALS AND INTEGRATION

CALCULATING THE HYDROGENS

CONCENTRATION ON THE BOARD..!

WORD FILE..!

ALCOHOLS

C-OH – 0.5-5.0 ppm

CH-OH – 3.2- 3.8 ppm

No spin-spin coupling if exchange is rapid.

Exchange promoted by increased temperature, small amounts of acidic impurities and also water content.

DEUTERIUM SHAKE

Used to identify –OH absorption.

Add D2O to the sample, and shake. Hydrogen gets replaced by D atom, and peak disappears.

A compound with molecular formula C8H10 has in its 1HNMR spectrum two signals δ 2.3(6H)

and 7.1(4H) deduce the structure

p-xylene

Deduce the structure of an organic compound(C7H8O having following spectral data

IR : 3300cm-1( br), 3050cm-1 (m) , 2950cm-1(str) NMR: δ7.1(m,5H) , δ4.4(s,2H) and 3.7(s,1H)

C6H5CH2-OHBenzyl alcohol

Deduce the structure of an organic compound (C8H9NO) having following spectral data IR :

3430cm-1, 1710cm-1 , NMR: δ7.3(m,5H) , δ2.2(s,3H) and 8.3(br,1H)

C6H5NH-CO-CH3acetanilide

A compound with molecular formula C8H8O shows following NMR peaks

Deduce the structure of the compound δ2.8 (2H, d): δ 7.28 (5H,m) δ 9.78(1H,t)

C6H5CH2CH=OPhenyl acetaldehyde

COUPLING CONSTANT (J)

Distance between peaks of a simple multiplets

Geminal Coupling 2J : Coupling between 2 protons on same carbon, this carbon can be C-13 .

ALKANES:

Three types of hydrogens :

- Methyl Hydrogens (-CH3): δ 0.7- 1.3 ppm

- Methylene hydrogens (-CH2-): δ 1.2 – 1.4 ppm

- Methine Hydrogens (-CH--): δ 1.4-1.7 ppm

Coupling Constant : 3J = 7-8 Hz

COUPLING CONSTANTS OF SOME GROUPS

ALKENES

Vinylic hydrogens : (C=C-H) δ 4.5 – 6.5 ppm

Allylic Hydrogens : (C=C-C-H) δ 1.6 – 2.6 ppm

3Jtrans = 11 – 18 Hz

3Jcis = 6 -15 Hz

2J = 0-3 Hz

4J = 0-3 Hz

H

H

a n g le = 1 0 9 o

2 J = 1 2 -1 8 H z

H

H

a n g le = 1 1 8 o

2 J = 5 H z

H

H

a n g le = 1 2 0 o

2 J = 0 -3 H z

H

H

C H 3

H 3 C

C H 3

O

a n g le = 1 0 7 o

2 J = 1 7 .5 H z

Karplus Relationship

Relates coupling constant with dihedral angle.

variation of Jvic is given by Karplus’s equation

3JH,H = A + B cosф + C cos2ф

Modified as :

If 0o˂ ф˂90o , Jvic = 8.5cos2 ф-o.28

If 90o˂ ф˂180o , Jvic = 9.5cos2 ф-o.28

Types of NMR spectrum

First order spectra:

Spectra shows multiplets obeying n+1 rule and intensities following Pascal’s triangle

n+1 rule is applied only if 1) ratio of difference in chemical shift (ΔνHz) of interacting nuclei to their coupling constant J(Hz) is more than 10, ΔνHz/JHz≥10

Values of J of all protons in neighboring groups with protons of interest must be identical

First Order Spectra:

J does not depend on field strength, chemical shift value depends on field strength

At field strength ΔνHz/JHz≥10 possible

Spectra becomes first order at higher field strength .

Second Order Spectra

If ratio ΔνHz/JHz ˂ 10 l 1H NMR becomes complicated

Splitting not obey n+1 rule

Intensities of peaks not belong Pascals tringle

Also known as non-first order spectrum

Limitations of the (n+1) rule

This is valid only when vicinal inter-proton coupling constants are exactly the same for every successive pair of carbons.

JAB = JBC = JCD = JDE ….

H 2 C

H 2C

CH 2

H 2C

CH 2

A

B D

CE

H

16

8 8

44 4 4

42 4 22 2

11333311

1 4 6 4 1

In the above molecule, count the number of 3J H-H couplings, for the 2nd carbon.

You will find 4 of them.Hence, 24 = 16

C

C

C

H

H

H

H

H

H

Rules are meant to be broken..?

Use 1,1,2-trichlorethane – predict spectrum. This obeys (n+1) rule.

Styrene oxide does not…

All protons of the side chainappear as quartets.

Reasons: Coupling constants not equal for vicinal protons.

Restricted rotation across the ring.

H

OH

H

HAHC

HC

CH CH2

J1 J2

J1

J2J2

CH2CH CH

OJ1 J2

J1

J2J2

CH2

Hindered Rotation

N,N- Dimethyl Formamide shows 2 distinct peaks for the methyl protons even though they are chemically equivalent.

Free rotation of the methyls around the nitrogen is possible.

This is due to resonance, the molecule is forced to adopt a planar geometry making the methyl magnetically non equivalent.

At higher temperatures, the rate of rotation increases and a single peak is equivalent.

Spin System Notations – Pople Notation

Each chemically different proton is given a capital letter A, B, C so on…

If there are more than one proton of the same type then numerical subscripts are used A2 , B2 etc

Protons with widely different chemical shift are denoted with letters far off in the alphabet. E.g AB protons are more closely related than AX or AY protons.

If three protons with different chemical shifts exists then notations like AMX, AMY etc are used.

For protons to give distinct peaks, the ratio of Δν/J (Both measured in Hertz) should be more than 10.

For AB protons this ratio is very low compared to that of AX protons.

Δν (Hz) is dependent on spectrometer frequency, but J (Hz) is independent of frequency.

Hence to get larger Δν/J spectrometer of higher frequencies have to be used.

In other words AB protons behave as AX protons in higher frequency spectrometer.

Which means that at higher frequencies, second order spectra become first order spectra.

Some Examples

1,1,2-trichloroethane coupling constant is 7Hz

Difference btw CH & CH2 protons is 1.8 δ (5.7-3.9 δ) corresponding to 108Hz (60MHz spectrum)

Hence ΔνHz/JHz ˃ 10

If Ha & Hb become more alike relative intensities of peaks do not follow Pascal’s triangle

Spectrum deviates from first order

During distortion of signals inner signal intensity increases outer signal intensity decreases

McConnell Equation