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 Hb protons is dependent upon the orientation of the Ha protons with respect to the external magnetic field: There are three possible orientations of the two Ha 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)

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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 Hb 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 Ha protons are equivalent and do not split one another

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Example 2: 2-Butanone

The 2-butanone molecule contains a methyl group as in the last example

The Hc protons are split by the two Hb protons and appear as a triplet

The Hb protons are split by the three Hc protons and appear as a quartet

The Ha protons are not split and appear as a singlet

Splittings in alkanes typically operate over three bonds; the Hb protons are four bonds away from the Ha protons and hence these do not split one another

Note that the Ha and Hb protons are shifted downfield due to their proximity to an electronegative atom (the carbonyl group)

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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

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Example 4: 2-Bromobutane

The HA protons are split only by the HB proton and should appear as a doublet The HD protons are split by the HC protons and should appear as triplet The HC protons will be split by the HD protons into a quartet and by the HB proton into a doublet; splitting will be complex The HB proton will be split by the HA protons into a quartet and by the HC 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 HD protons are furthest from the bromine and should be shifted furthest downfield (towards lower ppm values)

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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.

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Example Problem 1: an unknown compound has the formula C5H11OH 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.

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Example Problem 2: A ketone with the molecular formula C5H10O gives the H-1 NMR spectrum shown below. Identify this compound.

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Example Problem 3: A compound has the molecular formula C4H7ClO2; identify the structure of the compound. (Hint: the compound is an ester; it will have one COO group).

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Example Problem 4: The following is an H-1 spectrum of pentanone. Does this data support 2-pentanone or 3-pentanone?

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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 HC multiplet of 2-bromobutane (previous example) measured at 90 MHz (left) and 300 MHz (right)

For the 90 MHz spectra: (1.020 – 0.943) x 90 = 7.0 Hz (0.943 – 0.862) x 90 = 7.3 Hz

For the 300 MHz spectra: (1.054 – 1.030) x 300 = 7.2 Hz (1.030 – 1.005) 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

1.020 – 0.862 = 0.158 ppm 1.054 – 1.005 = 0.049 ppm

0.158 ppm / 0.049 ppm = 3.22 = 300 MHz / 90 MHz

At higher field the multiplets (in ppm) are tighter and less likely to be overlapped.

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Comparison of Spectra at 90-Mhz and 300-MHz

90-MHz Proton Spectrum of 1-Bromobutane

300-MHz Proton Spectrum of 1-Bromobutane

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