Chapter 10 NMR Theory I.Identifying Molecules A.Physical and Chemical Tests 1)Known compounds are easy to identify by melting point, TLC, etc… 2)Literature values for chemical properties already exist 3)New compounds must be identified without this knowledge a)Elemental Analysis: C 63.24% H 14.12% O b)Chemical Tests: ROH will react with Na; ROR would not c)We need more informative tests B.Spectroscopy = analyzing molecules based on how they absorb radiation 1)Electromagnetic Radiation Spectrum

2)Molecules absorb “quanta” of light energy only when the light is of the correct energy—all or nothing 3)  E = h h = x erg.s = frequency 4)Absorption causes some electronic or mechanical motion called an Excitation to occur 5)Different types of excitations require different energies a)IR light causes excitation of vibrations b)UV light excites e- from filled to unfilled orbitals c)X-Rays excite e- from inner to outer shells d)Radio Waves cause alignment of nuclei with magnetic fields 6)A Spectrometer records the absorption as a visual plot of what energy of light and how much of that light is absorbed

II.Proton Nuclear Magnetic Resonance Spectroscopy (radio waves) A.Nuclear Spin 1)Hydrogen nuclei = 1 H = proton have a nuclear spin, often represented as 2)Because of its charge (+), the spinning creates a magnetic field 3)We can analyze the proton as a tiny bar magnet 4)Without a magnetic field, a collection of protons have all possible directions of spin 5)When we expose this collection to a magnetic field (H o ), the protons align with and against the magnetic field a)  spin state is aligned with the field and has lower energy b)  spin state is aligned against the field and has higher energy

6)Spectroscopy requires at least 2 energy states so that absorption can occur 7)We must put the sample in a magnetic field to do NMR 8)Resonance = when the frequency of the radio waves is just right to cause absorption and excitation from  to  states. 9)Relaxation = loss of energy by the proton to return to  state (heat)

 E between  and  depends on how strong the magnetic field is. a)For H o = 21,150 G,  E = 90 MHz (90,000,000 Hz) b)For H o = 42,300 G,  E = 180 MHz c)For H o = 70,500 G,  E = 300 MHz 11)  E (  ) = h = 9 x kcal/mol = very small amount of energy 12)Out of a million protons, only a few more will be  than  B.Other Nuclei and Resolution 1) 14 N, 35 Cl, 31 P, and 13 C all have nuclear spins and are NMR active 2)Different Nuclei absorb at different energies 3)High Resolution NMR looks at differences within a single nuclei type

C.NMR Instrumentation 1)Continuous Wave Instruments a)Scan through the frequency range and are slow b)Usually old, low field (90 MHz) instruments 2)Fourier Transform Instruments a)Pulse at all frequencies at once and are fast b)Requires computing to do math “Fourier Transform” for real spectra c)Usually high field ( MHz) supercooled electromagnets (4 K uses liquid He, surrounded by 77 K liquid N 2 ) III.Proton Chemical Shift and Structure Determination A.Position of an 1 H NMR peak 1)Electron density around the proton in question, determines how much of the magnetic field H o is “felt” by that proton 2)The frequency of radio wave needed to excite that proton (thus where that proton appears on the spectrum) depends on the H o field “felt” 3)Molecular Structure determines the electron density 4)Work backwards: spectrum to e- density to molecular structure

B.Electron Density effects on protons 1)Any bound 1 H is surrounded by orbitals and atoms 2)Electron donating and electron withdrawing ability (polarity) of neighbor atoms control how much electron density each 1 H is surrounded by 3)Electrons move to oppose H o, creating h o = local magnetic field 4)Consequently, the overall magnetic field felt at each proton is reduced or shielded due to the electrons of the molecule a)Adding electron density increases the shielding (need higher H o for resonance to occur) b)Removing electron density decreases shielding = deshielding (need a lower H o for resonance to occur) 5) 1 H NMR spectra have higher field moving to the right UpfieldDownfield

6)Every chemically different proton of a molecule has a different amount of shielding leading to a unique peak on the spectrum 7)2,2-dimethyl-1-propanol 1 H NMR Spectrum

C.Chemical Shift 1)The frequency of radio waves needed for resonance of a proton varies with the size of the magnetic field (can’t compare spectra from different instruments directly) a)90 MHz magnet would give proton resonance of f b)180 MHz magnet would give 2f as the resonance frequency c)Frequency is directly proportional to the size of the magnetic field 2)Pick a standard reference: Si(CH 3 ) 4 = tetramethylsilane (TMS) a)12 equivalent protons for a strong signal b)Resonance frequency is upfield (to the right) of almost any other proton type, so it doesn’t interfere in the spectrum 3)Use a unit that compares how far other proton resonances are from TMS a)Double the magnet size, double distance from TMS b)Now, spectra from different size magnets are directly comparable 4)Chemical Shift (  ) a)Unit ppm = parts per million (Hz/MHz) b)TMS assign as 0.00 ppm

5)For 2,2-dimethyl-1-propanol peaks are at 78Hz, 258Hz, 287Hz (90MHz) D.Functional Groups and Chemical Shift 1)Each functional group has a characteristic amount of shielding or deshielding due to the atom types that make it up 2)This leads to a specific chemical shift for protons near or on that group 3)Tables 10-2 and 10-3 in your book a)Alkanes or alkyl groups: lots of shielding, upfield resonances b)Any electronegative atoms: deshielding, downfield resonances 4)The closer a –CH 2 - group is to an electronegative atom, the larger the shift 5)Effect of electronegative atoms is cumulative 6)ROH, RNH 2, RSH chemical shifts are variable. They depend on how much hydrogen bonding occurs (solvent, temperature, concentration)