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Lecture

Topic 8

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Department
Chemistry
Course
Chemistry 2223B
Professor
Felix Lee
Semester
Fall

Description
Chemistry 2213a  Fall 2012  Western University Topic 8. Structure Elucidation: IR, C-NMR, H-NMR 13 1 A. Electromagnetic Radiation  Energy is transmitted through space in the form of electromagnetic radiation, which are oscillating waves of electric and magnetic fields, and this term used to describe all the component waves found in a continuous spectrum known as the electromagnetic spectrum. Structure Elucidation  2  Regardless of wavelength, all electromagnetic radiation propagates through space at the speed of light (c). It is characterized its wavelength ( in metres) and frequency (oscillations per second,  in s ).−1  The shorter the , the higher the  (more waves go by per second if the wavelength is shorter). B. Review of Infrared Spectroscopy −1  IR spectra are normally scanned from 4000 to 600 cm (wavenumbers, which are reciprocal of  in cm). Spectra can be divided into the functional group and fingerprint regions. Aspirin Structure Elucidation  3  The absorption of IR light causes vibrational excitation. Different functional groups absorb at different wavelengths, which in turn allows the identification of functional groups.  Absorption appears as peaks of lowered transmittance. Correlation tables list the characteristic absorptions, as well as their intensities, for various functional groups. Alkanes, alkenes, and alkynes Structure Elucidation  4  Sample spectra are shown below. Note that the alkene and alkyne signals near 3100 and 3300 cm , respectively, are only present if an alkenyl or alkynyl hydrogen is present. Structure Elucidation  5  Question: The IR shown is for which compound, H C=CH CH CH CH C2 or 2 2 2 2 3 HC≡CH CH CH CH CH 2 2 2 2 3 Structure Elucidation  6 Alcohols and ethers  Both contain a C–O bond, but only alcohols contain an O−H bond. Structure Elucidation  7 Benzene rings  Compounds with benzene rings give distinctive IR spectra. Structure Elucidation  8 Amines −1  If an N–H bond is present, the region from 3100 – 3500 cm will show two peaks in the case of 1°amines (two N–H bonds), but 2° amines will have just one peak. Structure Elucidation  9 C. Nuclear Magnetic Resonance Spectroscopy 1. Energy absorption by nuclei  An electron has a spin, and spinning charges create magnetic fields (spin quantum number, m = ±½s. A nucleus that has an odd mass number (total number of protons and neutrons) also has a spin. All nuclei with an odd mass number can have two spin states, +½ and −½ . o Examples include H, C, F, P 19 31  For nuclei with the two possible spin states, they can be viewed as spinning bar magnets. If such nuclei are placed in a magnetic field (B ), t0e two spin states assume two different energy levels. Structure Elucidation  10  Once the energy separation has been achieved, the nuclei can the promoted from the lower- to higher-energy state using radiofrequency radiation.  The absorption of radiation by nuclei in a magnetic field is called nuclear magnetic resonance, NMR.  For organic chemists, the two most useful types are: 13 o C NMR (“carbon NMR”) 1 o H NMR (“proton NMR”)  For all types of NMR, the stronger the magnetic field (B ),0the greater the energy separation between the two states. Structure Elucidation  11  Very strong magnetic fields are required to make the separation large enough 13 to be useful. For example, for C: o At B 0 7.5 T (Tesla), E = 0.0072 cal/mol o This energy separation corresponds to an electromagnetic frequency of 75 MHz and a wavelength of 1000 m (radiofrequency).  Most commercial NMR instruments consist of a fixed magnetic field, a radiofrequency generator to “radiate” the nuclei, a receiver to detect the absorption, and a recorder.  The sample is dissolved in a solvent that does not interfere with the magnetic field. The solvent chosen must be not contain nuclei with spin, so no hydrogens! A common solvent used is CDCl .3 Structure Elucidation  12 2. C-NMR spectroscopy  The most abundant isotope of carbon is C (98.9%), which has no nuclear 13 spin. Yet, all substances containing carbon will contaC isotope (1.1%), which does have spin and will give a signal in an NMR instrument.  The local magnetic fields of carbon atoms in a molecule are affected by all neighbouring atoms. This makes the actual magnetic field around a specific atom slightly different than the applied field.  Therefore, every different carbon environment in a molecule has a d13ferent energy separation and therefore absorbs a different radiofrequency. Thus, C- NMR spectra reveal the number of magnetically non-equivalent C atoms and their identity (chemical shifts).  We can determine the O 1 signal O OH citric acid number of diffe13nt C HO 4 signals atoms (and thus C signals) by looking at O HO OH molecular symmetry. 3 signals O O O Structure Elucidation  13  Instead of plotting signals using energy on the x-axis, they are plotted as chemical shifts () in units of ppm.  Furthermore, absorption 13gnals are plotted relative to the single C absorption signal of a tetramethylsilane (TMS) reference compound, TMS. So, the signal at 0 Si ppm, corresponds to that of TMS.  E13’s, and greater s-character in hybrid C orbitals, shift C signals to the left of TMS. This is termed downfield. Structure Elucidation  14 13  Shown below is a correlation table for C NMR. Structure Elucidation  15  Example: The C-NMR spectrum of 1-pentanol shows the five signals, with the ones furthest downfield closest to the OH.  Example: diethyl ether Structure Elucidation  16  Example: toluene  Question: Predict the number of signals for the following compounds. Do the spectra agree with their structure? Cl OH 1-pentene 4-chlorophenol Structure Elucidation  17  Important note: signal heights on C-NMR spectra do not correspond with the number of C atoms they represent (see 1-pentene). However, higher signals usually result from greater numbers of equivalent nuclei (see 4-chlorophenol). 13  The analysis of IR and C-NMR spectra, together with info about units of unsaturation, often gives enough information to elucidate molecular structure (see assigned problems). Structure Elucidation  18 3. H-NMR spectroscopy 13 1  Like C, H atoms have spins of ±½ and are NMR-active. Non-equivalent protons in a molecule will absorb a different energy of radiofrequency energy. Absorptions are plotted as chemical shifts that typically range from 0 – 13 ppm. 1  For example, methyl acetate contains two H environments.  In methyl acetate, the peak at 3.7 ppm is due to the threeO H protons of the methoxy group, while the 2.0 ppm peak is H due to the three protons of the ac3l CH group. H O H H  See textbook “How To 12.2” for more information on H how to determine the number of proton signals. Structure Elucidation  19  Chemical shifts are determined by the type, hybridization, and electronegativity 1 of the atom to which the H is bonded. H chemical shifts give us the same structural information for protons that C shifts give for carbon atoms. Structure Elucidation  20 1  However, H NMR has two extra features that make increase its usefulness for the elucidati
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