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900 MHz NMR

900 MHz NMR. Characteristics of Principal Spectrometric Methods. 1 H-NMR. 13 C-NMR. MS. IR. Scale. 0-15 ppm. 1-220 ppm. 50-4000 amu. 400-4000 cm -1. Sample. 1-2 mg. 10-20 mg. < 1 mg. < 1 mg. Molecular formula. Partial. Partial. Yes. No. Functional group. ~ yes. ~ yes.

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900 MHz NMR

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  1. 900 MHz NMR

  2. Characteristics of Principal Spectrometric Methods 1H-NMR 13C-NMR MS IR Scale 0-15 ppm 1-220 ppm 50-4000 amu 400-4000 cm-1 Sample 1-2 mg 10-20 mg < 1 mg < 1 mg Molecular formula Partial Partial Yes No Functional group ~ yes ~ yes Limited Yes yes yes yes Very limited Substructure yes yes No Very limited C-Connection Stereochem. & regiostereo-chemistry yes yes No Very limited

  3. ALL SPECTROMETERS HAVE SOME COMMON ESSENTIAL FEATURES • A source of electromagnetic radiations of the appropriate frequency range. • A sample holder to permit efficient irradiation of the sample. • A frequency analyzer which separates out all of the individual frequencies generated by the source. • A detector for measuring the intensity of radiations at each frequency, allowing the measurement of how much energy has been absorbed at each of these frequencies by the sample; and • A recorder – either a pen recorder or computerized data station, with a VDU for initial viewing of the spectrum, with the possibility of manipulation.

  4. NMR • NUCLEAR- study of nuclear spins • MAGNETIC- under the influence of applied magnetic field • RESONANCE- and to record the resulting resonance in nuclear spin through the absorption of Rf

  5. SPECTROSCOPIC TECHNIQUES IN ORGANIC CHEMISTRY AND THEIR USES

  6. The first published ‘high-resolution’ proton NMR spectrum (30 MHz) displaying the proton chemical shifts in ethanol

  7. Fundamental of NMR Spectroscopy?

  8. Angular momentum and magnetic moment µ = γJ Where µ is magnetic moment, J is angular momentum, and γ is the constant of proportionality called magnetogyric ratio Now: Angular momentum is directly proportional to nuclear spin quantum Number; so, J = h/2π * I Therefore, µ = h/2π * γI

  9. Nuclear Magnetic Resonance I is a property of the nucleus Nuclear spin m = g I * h/2π m -magnetic moment g - magnetogyric ratio I - spin quantum number h - Planck’s constant Mass # Atomic # I Odd Even or odd 1/2, 3/2, 5/2,… Even Even 0 Even Odd 1, 2, 3 m I is quantized; it has only certain values.

  10. Spin Quantum Numbers of Some Common Nuclei The most abundant isotopes of C and O do not have spin. Element1H 2H 12C 13C 14N 16O 17O 19F Nuclear Spin Quantum No1/2 1 0 1/2 1 0 5/21/2 ( I ) No. of Spin 2 3 0 2 3 0 6 2 States Elements with either odd mass or odd atomic number have the property of nuclear “spin”. The number of spin states is 2I + 1, whereI is the spin quantum number.

  11. Number of spin states or multiplicity: If we place an magnetically active nucleus in an external magnetic field, how many orientations it can adopt. Number of spin states is given by formula: m = 2I + 1 For example, for a nucleus with I = ½, m = 2 * ½ + 1 = 2 So it has two spin states (or, orientations, or multiplicities), They are +1/2 and -1/2. We can find out individual spin states by using the sequence: +I, +(I-1), ……-(I-1), -I

  12. NMR Instrumentation • An NMR machine is basically a big and expensive FM radio. • Magnet - Superconducting. Some electromagnets and permanent magnets still around. • Frequency generator - Creates the alternating current (at wo) that induces B1. Continuous wave or pulsed. • Detector - Subtracts the base frequency (a constant frequency very close to wo) to the output frequency. It is lower frequency and much easier to deal with. • Recorder - XY plotter, oscilloscope, computer, etc., etc. Bo N S Magnet B1 Recorder Frequency Generator Detector

  13. A Conventional 60 MHzNMR Spectrometer hn RF (60 MHz) Oscillator RF Detector absorption signal Recorder Transmitter Receiver MAGNET MAGNET ~ 1.41 Tesla (+/-) a few ppm N S Probe

  14. Nuclear Magnets in an External Magnetic Field (B0) Bo Bo > 0 Bo = 0 Oriented in a pattern Randomly oriented N S Each nucleus behaves like a bar magnet.

  15. Orientations of Nuclei with I = 1/2 Nuclei acquire orientation either Align with the external magnetic field, Or Align against the external magnetic field Bo Align with α state Align against β state Which state should be of lower energy and why?

  16. NUCLEAR SPIN STATES OF HYDROGEN NUCLEUS The spin of the positively charged nucleus generates a magnetic moment vector, m. m + + The two states are equivalent in energy in the absence of a magnetic or an electric field. m + 1/2 - 1/2 TWO SPIN STATES

  17. THE ENERGY SEPARATION DEPENDS ON Bo - 1/2 = kBo = hn DE degenerate at Bo = 0 + 1/2 Bo increasing magnetic field strength

  18. EFFECT OF A STRONG MAGNETIC FIELD WHAT HAPPEN WHEN A SPIN-ACTIVE HYDROGEN ATOM IS PLACED IN A STRONG MAGNETIC FIELD? ….. IT BEGINS TO PRECESS OPERATION OF AN NMR SPECTROMETER DEPENDS ON THIS RESULT

  19. Precessional Frequency When placed in an external magnetic field (B0), a magnetically active nucleus starts to undergo a particular motion, called precession. The frequency with which it precesses is called precession frequency. It is angular frequency (ω0). It can be converted to linear frequency (ν0). (ω0= 2πν0) This is also known as Larmor Frequency

  20. Larmor Frequency and B0 What is the relationship between Larmor frequency and applied magnetic field? Should Larmor frequency depend on B0? How?

  21. The Larmor Equation!!! (Angular) frequency of the incoming radiation that will cause a transition gyromagnetic ratio g o = gBo strength of the magnetic field g is a constant which is different for each atomic nucleus (H, C, N, etc) no = gB0/2π Here, no is linear frequency (Hz or MHz)

  22. Application of an External Magnetic Field (Putting your sample in the magnet) z w0 = g Bo = 2πν0 w0 - resonance frequency in radians per second, also called Larmor frequency n0 - resonance frequency in cycles per second, Hz g - gyromagnetic ratio Bo - external magnetic field (the magnet) w0 m Bo m Spin 1/2 nuclei will have two orientations in a magnetic field +1/2 and -1/2. w

  23. E and Bo If the difference of energy between β and α orientations is ∆E, then ∆E = hgB0/2π And since, ∆E = hν, So, ν = gB0/2π Using these equations we can calculate the frequency (or, energy) of the RF radiation which can be absorbed by magnetic nuclei placed in an applied magnetic field.

  24. WHAT IS “RESONANCE” ? ….Absorption of energy by the spinning nucleus

  25. N w Nuclei precess at frequency w when placed in a strong magnetic field. RADIOFREQUENCY 40 - 600 MHz hn NUCLEAR MAGNETIC RESONANCE If n = w then energy will be absorbed and the spin will invert. NMR S

  26. Nuclear Spin Energy Levels N -1/2 unaligned In a strong magnetic field (Bo) the two spin states differ in energy. +1/2 aligned Bo S

  27. Absorption of Energy quantized Opposed -1/2 -1/2 DE DE = hn Radiofrequency +1/2 +1/2 Applied Field Bo Aligned

  28. Resonance Frequencies of Selected Nuclei Isotope Abundance Bo (Tesla) Frequency (MHz) g (radians/Tesla) 1H 99.98% 1.00 42.6 267.53 1.41 60.0 2.35 100.0 7.05 300.0 2H 0.0156% 1.00 6.5 41.1 7.05 45.8 13C 1.108% 1.00 10.7 67.28 2.35 25.0 7.05 75.0 19F 100.0% 1.00 40.0 251.7

  29. POPULATION AND SIGNAL STRENGTH The strength of the NMR signal depends on the Population Difference of the two spin states Radiation induces both upward and downward transitions. induced emission resonance For a net positive signal there must be an excess of spins in the lower state. excess population Saturation = equal populations = no signal

  30. Boltzmann Excess Nβ/Nα = e-∆E/kT Or Nβ/Nα = 1+ (-∆E/kT) Or Nα/Nβ = 1 + ∆E/kT) We know that: ∆E = E – E Energy of radiation must match with this energy difference if resonance has to occur. The radiation having energy comparable to this energy demand falls in the region of radiofrequency(RF). This is the radiation of very low energy. Thus: ∆E = hν So, Nβ/Nα = 1+ (-hν/kT) or Nα/Nβ = 1 + ∆E/kT = 1 + hν/kT

  31. Calculation of Nα/Nβ What would be the value of DE for protons if Bo = 9.4 T. It is 4 x 10-5 Kcal / mol. What is the frequency, ν, ? (400MHz ?) Use this equation for energy: ∆E = hgB0/2π And this for frequency: ν = gB0/2π Then use this for Boltzmann Excess: Nα/Nβ = 1 + ∆E/kT Or Nα/Nβ = 1 + hν/kT The Nα/Nβ ratio is only 1.000064. In one million nuclear spins we have a difference of just 64: NMR is very insensitive when compared to UV or IR...

  32. MODERN INSTRUMENTATION PULSED FOURIER TRANSFORM TECHNOLOGY FT-NMR requires powerful computer

  33. Energy States for a Spin 1/2 System DE = g h Bo = h n Antiparallel -1/2 DE E +1/2 Parallel Bo > 0 Bo = 0 Therefore, the nuclei will absorb light with energy DE resulting in a change of the spin states.

  34. Effect of Static Magnetic Bo Field on Magnetic Moment

  35. Net Magnetic Moment z w m +1/2 Bo -1/2 m w

  36. The Net MagnetizationVector w z w one nucleus x Mo - net magnetization vector allows us to look at system as a whole many nuclei z w z y y x x

  37. Bloch Vector Model- Boltzmann Distribution Excess

  38. Effect of RF Pulse (B1) on Nuclear Spin and Resonance

  39. Relaxation • Application of RF pulse (lets say 90ox) bring the bulk magnetization vector to the xy plan. • The system re-establish the equilibrium by releasing energy through a process of relaxations. • >Spin-Lattice (Longitudinal Relaxation) T1 • >Spin-Spin (Transverse Relaxation) T2 • This relaxation or decay with time is detected as FID signals.

  40. A bit of a tickle and the Protons will Sing Because the atom is not static the magnets rotate around the external magnetic field like a gyroscope. Now that their voices are warmed up and ready to go, very little is needed to make the protons sing their song. We don’t use feathers to elicit a response - we use radio waves ! The spinning nuclei interact with the radio wave and are knocked out of their gyroscopic motion. We can detect the radio wave (energy) lost as the nuclei return back to their gyroscopic equilibrium. We do this by carefully placing a receiver coil to listen to the song. This is the NMR signal (in a time format).

  41. PROBLEM 1 • Explain behavior of spinning nuclei: • In the absence of magnetic field • Under the influence of magnetic field • When rf of appropriate energy is applied to the system

  42. Relaxation

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