Chapter 14 NMR Spectroscopy Organic Chemistry 6 th








































































- Slides: 72

Chapter 14 NMR Spectroscopy Organic Chemistry 6 th Edition Paula Yurkanis Bruice 1 © 2011 Pearson Education, Inc.

Nuclear Magnetic Resonance (NMR) Spectroscopy Identify the carbon–hydrogen framework of an organic compound Certain nuclei, such as 1 H, 13 C, 15 N, 19 F, and 31 P, have non-zero value for their spin quantum number; this property allows them to be studied by NMR 2 © 2011 Pearson Education, Inc.

The spin state of a nucleus is affected by an applied magnetic field: 3 © 2011 Pearson Education, Inc.

The energy difference between the spin states increases with the strength of the applied magnetic field: 4 © 2011 Pearson Education, Inc.

absorb DE a-spin states b-spin states release DE Signals detected by NMR 5 © 2011 Pearson Education, Inc.

An NMR Spectrometer In pulsed Fourier transform (FT) spectrometers, the magnetic field is held constant, and a radio frequency (rf) pulse of short duration excites all the protons 6 simultaneously © 2011 Pearson Education, Inc.

The electrons surrounding a nucleus decrease the effective magnetic field sensed by the nucleus: Beffective = Bo – Blocal 7 © 2011 Pearson Education, Inc.

Chemically equivalent protons: protons in the same chemical environment Each set of chemically equivalent protons in a compound gives rise to a signal in an 1 H NMR spectrum of that compound: 8 © 2011 Pearson Education, Inc.

9 © 2011 Pearson Education, Inc.

The Chemical Shift The reference point of an NMR spectrum is defined by the position of TMS (zero ppm): The chemical shift is a measure of how far the signal is from the reference signal The common scale for chemical shifts = = distance downfield from TMS (Hz) operating frequency of the spectrometer (MHz) 10 © 2011 Pearson Education, Inc.

1 H NMR spectrum of 1 -bromo-2, 2 -dimethylpropane The greater the chemical shift, the higher the frequency The chemical shift is independent of the operating frequency of the spectrometer 11 © 2011 Pearson Education, Inc.

12 © 2011 Pearson Education, Inc.

Protons in electron-poor environments show signals at high frequencies Electron withdrawal causes NMR signals to appear at higher frequency (at larger values): 13 © 2011 Pearson Education, Inc.

Characteristic Values of Chemical Shifts 14 © 2011 Pearson Education, Inc.

15 © 2011 Pearson Education, Inc.

16 © 2011 Pearson Education, Inc.

Diamagnetic Anisotropy The unusual chemical shifts associated with hydrogens bonded to carbons that form bonds: The electrons are freer to move than the s electrons in response to a magnetic field 17 © 2011 Pearson Education, Inc.

The protons show signals at higher frequencies because they sense a larger effective magnetic field: benzene 18 © 2011 Pearson Education, Inc.

The alkene and aldehyde protons also show signals at higher frequencies: alkene aldehyde 19 © 2011 Pearson Education, Inc.

The alkyne proton shows a signal at a lower frequency than it would if the electrons did not induce a magnetic field: alkyne 20 © 2011 Pearson Education, Inc.

1 H NMR spectrum of 1 -bromo-2, 2 -dimethylpropane The area under each signal is proportional to the number of protons giving rise to the signal: 21 © 2011 Pearson Education, Inc.

Integration Line The area under each signal is proportional to the number of protons that give rise to that signal The height of each integration step is proportional to the area under a specific signal The integration tells us the relative number of protons that give rise to each signal, not the absolute number 22 © 2011 Pearson Education, Inc.

23 © 2011 Pearson Education, Inc.

Splitting of the Signals • An 1 H NMR signal is split into N + 1 peaks, where N is the number of equivalent protons bonded to adjacent carbons • Coupled protons split each other’s signal • The number of peaks in a signal is called the multiplicity of the signal • The splitting of signals, caused by spin–spin coupling, occurs when different kinds of protons are close to one another 24 © 2011 Pearson Education, Inc.

It is not the number of protons giving rise to a signal that determines the multiplicity of the signal It is the number of protons bonded to the immediately adjacent carbons that determines the multiplicity a: a triplet b: a quartet c: a singlet 25 © 2011 Pearson Education, Inc.

Equivalent protons do not split each other’s signal: 26 © 2011 Pearson Education, Inc.

27 © 2011 Pearson Education, Inc.

28 © 2011 Pearson Education, Inc.

The ways in which the magnetic fields of three protons can be aligned: 29 © 2011 Pearson Education, Inc.

30 © 2011 Pearson Education, Inc.

Splitting is observed if the protons are separated by no more than three s bonds: Long-range coupling occurs over systems, such as benzene 31 © 2011 Pearson Education, Inc.

More Examples of 1 H NMR Spectra Triplet: two neighboring protons Quintet: four neighboring protons 32 © 2011 Pearson Education, Inc.

Doublet: one neighboring proton Sextet: five neighboring protons Septet: six neighboring protons Triplets: two neighboring protons 33 © 2011 Pearson Education, Inc.

The three vinylic protons are at relatively high frequency because of diamagnetic anisotropy 34 © 2011 Pearson Education, Inc.

The signals for the Hc, Hd, and He protons overlap because the electronic effect of an ethyl substituent is similar to that of a hydrogen: 35 © 2011 Pearson Education, Inc.

The signals for the Ha, Hb, and Hc protons do not overlap because of the strong electron-withdrawing property of the nitro group: 36 © 2011 Pearson Education, Inc.

Coupling Constants The coupling constant (J) is the distance between two adjacent peaks of a split NMR signal in hertz: Coupled protons have the same coupling constant © 2011 Pearson Education, Inc. 37

38 © 2011 Pearson Education, Inc.

Summary 1. The number of chemical shifts specify the number of proton environments in the compound 2. The chemical shift values specify the nature of the chemical environment: alkyl, alkene, etc. 3. The integration values specify the relative number of protons 4. The splitting specifies the number of neighboring protons 5. The coupling constants specify the orientation of the coupled protons 39 © 2011 Pearson Education, Inc.

A Splitting Diagram for a Doublet of Doublets 40 © 2011 Pearson Education, Inc.

Complex Splitting JAC = JAB Triplet JAC > JAB Doublet of doublets 41 © 2011 Pearson Education, Inc.

The trans coupling constant is greater than the cis coupling constant: 42 © 2011 Pearson Education, Inc.

A Splitting Diagram for a Quartet of Triplets 43 © 2011 Pearson Education, Inc.

Why is the signal for Ha a quintet rather than a triplet of triplet? 44 © 2011 Pearson Education, Inc.

45 © 2011 Pearson Education, Inc.

The Difference between a Quartet and a Doublet of Doublets Methylene has three neighbors, appears as a quartet Doublet © 2011 Pearson Education, Inc. Doublet 46

When two different sets of protons split a signal, the multiplicity of the signal is determined by using the N + 1 rule separately for each set of the hydrogens, as long as the coupling constants for the two sets are different When the coupling constants are similar, the multiplicity of a signal can be determined by treating both sets of adjacent hydrogens as though they were equivalent 47 © 2011 Pearson Education, Inc.

Replacing one of the enantiotopic hydrogens by a deuterium or any other atom or group other than CH 3 or OH forms a chiral molecule: prochiral carbon Ha is the pro-R-hydrogen, whereas Hb is the pro-S-hydrogen; and they are chemically equivalent 48 © 2011 Pearson Education, Inc.

Diastereotopic hydrogens have different chemical shifts: 49 © 2011 Pearson Education, Inc.

Diastereotopic hydrogens are not chemically equivalent: 50 © 2011 Pearson Education, Inc.

The three methyl protons are chemically equivalent because of rotation about the C—C bond: We see one signal for the methyl group in the 1 H NMR spectrum 51 © 2011 Pearson Education, Inc.

1 H NMR spectra of cyclohexane-d 11 at various temperatures: axial equatorial The rate of chair–chair conversion is temperature dependent axial 52 © 2011 Pearson Education, Inc.

Protons Bonded to Oxygen and Nitrogen The greater the extent of the hydrogen bond, the greater the chemical shift These protons can undergo proton exchange They always appear as broad signals 53 © 2011 Pearson Education, Inc.

pure ethanol with acid 54 © 2011 Pearson Education, Inc.

A 60 -MHz 1 H NMR spectrum A 300 -MHz 1 H NMR spectrum 55 © 2011 Pearson Education, Inc.

To observe well-defined splitting patterns, the difference in the chemical shifts (in Hz) must be 10 times the coupling constant values 56 © 2011 Pearson Education, Inc.

13 C NMR Spectroscopy • The number of signals reflects the number of different kinds of carbons in a compound. • The overall intensity of a 13 C signal is about 6400 times less than the intensity of an 1 H signal. • The chemical shift ranges over 220 ppm. • The reference compound is TMS. 57 © 2011 Pearson Education, Inc.

58 © 2011 Pearson Education, Inc.

59 © 2011 Pearson Education, Inc.

Proton-Decoupled 13 C NMR of 2 -Butanol 60 © 2011 Pearson Education, Inc.

Proton-Coupled 13 C NMR of 2 -Butanol 61 © 2011 Pearson Education, Inc.

The intensity of a signal is somewhat related to the number of carbons giving rise to it Carbons that are not attached to hydrogens give very small signals 62 © 2011 Pearson Education, Inc.

DEPT 13 C NMR distinguishes CH 3, CH 2, and CH groups: 63 © 2011 Pearson Education, Inc.

The COSY spectrum identifies protons that are coupled: Cross peaks indicate pairs of protons that are coupled 64 © 2011 Pearson Education, Inc.

COSY Spectrum of 1 -Nitropropane 65 © 2011 Pearson Education, Inc.

The HETCOR spectrum of 2 -methyl-3 -pentanone indicates coupling between protons and the carbon to which they are attached: 66 © 2011 Pearson Education, Inc.

Unknown Identification Using Spectroscopy Example 1: 13 C-NMR of C 5 H 9 Br 67 © 2011 Pearson Education, Inc.

Example 1: 1 H-NMR of C 5 H 9 Br 68 © 2011 Pearson Education, Inc.

Example 1: IR of C 5 H 9 Br Answer: 69 © 2011 Pearson Education, Inc.

Example 2: 13 C-NMR of C 6 H 10 O 4 33. 4 24. 1 174. 4 Solvent: 70 © 2011 Pearson Education, Inc.

Example 2: 1 H-NMR of C 6 H 10 O 4 2. 21 1. 50 11. 97 71 © 2011 Pearson Education, Inc.

Example 2: IR of C 6 H 10 O 4 Answer: 72 © 2011 Pearson Education, Inc.