Nmr nuclear magnetic resonance spectroscopy
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Transcript of Nmr nuclear magnetic resonance spectroscopy
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