The resonance frequency of the H b protons is dependent upon the orientation of the H a protons with respect to the external magnetic field:
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1 Spin-Spin Splitting in Alkanes The signal arising from a proton or set of protons is split into (N+1) lines by the presence of N adjacent nuclei Example 1: Bromoethane The resonance frequency of the H b protons is dependent upon the orientation of the H a protons with respect to the external magnetic field: There are three possible orientations of the two H a protons Both aligned with the external magnetic field (Lowest energy state) Both aligned against the external magnetic field (Highest energy state) One aligned with and one aligned against the external field (Twice as likely to occur) 1
2 The result is a 1:2:1 triplet There are four possible orientations for the three Hb protons All aligned with the external magnetic field (Highest energy state) All aligned against the external magnetic field (Lowest energy state) Two aligned with and one against the external magnetic field (Three ways to obtain this possibility) Two aligned against and one with the external magnetic field (Three ways to obtain this possibility) The result is a quartet with intensities of 1:3:3:1 Note that the three H b protons are equivalent and do not split one another; rotations about the C-C bond cause all three protons to be equivalent In a similar fashion the two H a protons are equivalent and do not split one another 2
3 Example 2: 2-Butanone The 2-butanone molecule contains a methyl group as in the last example The H c protons are split by the two H b protons and appear as a triplet The H b protons are split by the three H c protons and appear as a quartet The H a protons are not split and appear as a singlet Splittings in alkanes typically operate over three bonds; the H b protons are four bonds away from the H a protons and hence these do not split one another Note that the H a and H b protons are shifted downfield due to their proximity to an electronegative atom (the carbonyl group) 3
4 Example 3: 1,1-dichloropropane The protons on the end carbons will be shifted downfield due to the chlorine atoms; could make assignments on this basis alone The molecule is symmetrical; all four Ha protons are equivalent Both Hb protons are equivalent The Ha protons are split into a triplet by the Hb protons The Hb proton is split into a pentet by the Ha protons 4
5 Example 4: 2-Bromobutane The H A protons are split only by the H B proton and should appear as a doublet The H D protons are split by the H C protons and should appear as triplet The H C protons will be split by the H D protons into a quartet and by the H B proton into a doublet; splitting will be complex The H B proton will be split by the H A protons into a quartet and by the H C protons into a triplet; splitting will be complex Those protons in close proximity to the bromine will be shifted downfield (towards higher ppm values) The H D protons are furthest from the bromine and should be shifted furthest downfield (towards lower ppm values) 5
6 Integration of Proton Spectra In a proton spectrum the area under each peak is proportional to the number of protons giving rise to the peak The process of determining the area under each peak is known as integration If splitting is equivalent, a peak arising from two protons will be larger than a peak arising from a single proton Splitting reduces peak height; a peak that appears as a doublet will be half the height it would be as a singlet Integration is useful in assigning the data from the preceding problem. 6
7 Example Problem 1: an unknown compound has the formula C 5 H 11 OH and gives the following proton NMR spectrum. Identify the unknown compound. Hint: the compound is an alcohol; such hydrogen atoms do not show splitting patterns with other hydrogen atoms, even if they are within three bonds. 7
8 Example Problem 2: A ketone with the molecular formula C 5 H 10 O gives the H-1 NMR spectrum shown below. Identify this compound. 8
9 Example Problem 3: A compound has the molecular formula C 4 H 7 ClO 2 ; identify the structure of the compound. (Hint: the compound is an ester; it will have one COO group). 9
10 Example Problem 4: The following is an H-1 spectrum of pentanone. Does this data support 2-pentanone or 3-pentanone? 10
11 Determination of Coupling Constants Coupling constants are constant, as their name implies and are usually measured in hertz To convert ppm into hertz, multiply by the transmitter frequency Example: the H C multiplet of 2-bromobutane (previous example) measured at 90 MHz (left) and 300 MHz (right) For the 90 MHz spectra: ( ) x 90 = 7.0 Hz ( ) x 90 = 7.3 Hz For the 300 MHz spectra: ( ) x 300 = 7.2 Hz ( ) x 300 = 7.5 Hz These values are typical for alkanes; typical values are 6-10 Hz The splitting (hertz) is very nearly the same; but the chemical shift difference is smaller in the 300 MHz spectra = ppm = ppm ppm / ppm = 3.22 = 300 MHz / 90 MHz At higher field the multiplets (in ppm) are tighter and less likely to be overlapped. 11
12 Comparison of Spectra at 90-Mhz and 300-MHz 90-MHz Proton Spectrum of 1-Bromobutane 300-MHz Proton Spectrum of 1-Bromobutane 12
13 13
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