NMR-2

Proton NMR

Proton NMR, generally referred to as 1H NMR, uses hydrogen nuclei to generate signals between 0 and 12ppm.

An 1H spectra of propanoyl chloride

Similar to how carbons in 13C NMR have different environments, the same hold true in 1H NMR. Different hydrogens can exist within the same environment. The peak area (listed in the diagram under each corresponding peak) is proportional to how many hydrogens exist within that environment. Within actual NMR spectra, the peak area is found by taking the integral l of the range the peak exists in.

The area under the peak, not the height of the peak, shows the density of the environment.

Exercise

match the above peaks to their corresponding hydrogens based on the peak density.

Hydrogen Equivalence

Despite both the third and fourth carbon having two hydrogen bonds, their hydrogens are not chemically equivilent.

Hydrogen are almost always equivalent to all other hydrogen attached to an atom (ie all four hydrogen in a methane molecule are chemically identical). However, not all similar chemical groups are equivalent

Not all hydrogens with the same chemical structure are chemically equivalent. On the above structure, the far left and far right CH3 groups are not chemically equivalent and therefor have separate peaks on a 1H NMR spectrum.

If Ha and Ha` are chemically equivalent, they have to be bonded to chemically equivalent atoms. The hydrogens on the opposing CH3 are chemically equivalent however the hydrogens on the CH2 are not.

Exercise

For each of these structures, identify chemically equivalent hydrogens

1H Splitting

Splitting in 1H NMR is determined differently than in 13C NMR. The amount of splits is determined by the by the amount of non-equivalent hydrogens within three bonds from the Hydrogen.

Fill in the table describing the relationship between peak multiplicity and the number of neighboring hydrogens within three bonds

chemical structure and 1H NMR spectrum of propane

The peak cluster at 0.9ppm for propane corresponds to Ha hydrogen atoms. The peak is split twice because within the 3 bond distance, are both of the Hb hydrogen atoms.

The peak cluster at 1.3ppm corresponds to the Hb hydrogen atoms. The peak is split three times despite having 6 hydrogens within the 3 bond distance. this is because both CH3 groups within its range are chemically equivalent and therefor, only one is counted for splitting the peaks.

Similarly to how in 13C NMR, the number of peaks is equal to N+1 where N is the number of bonded hydrogens, in 1H NMR, the number of peaks is equal to N+1 where N is the number of hydrogens within the three bond radius

Exercise

For each of the molecules above, determine whether the signal from the circled hydrogen is split by the squared hydrogen. (for the purposes of this exercise, treat any double or triple bonds the same as a single)

For each of the molecules above, label each hydrogen atom, giving the same label to hydrogens in the same environment, then determine for each hydrogen environment its integration value and its multiplet type (Remember: integration values are the area under the peak. They are proportional to the amount of hydrogens in the environment)

PPM Ranges for 1H NMR

below are the most common molecular structures and their corresponding chemical shift (ppm range)

Alcohols do not always follow the rules for predicting multiplicities.

Based on integration and chemical shift alone, assign each of the three peaks in the proton NMR spectrum of ethanol? Which of the peaks above does not follow the peak multiplicity rules we have learned so far?

Special Considerations

Exercise

Which of the above could serve as an NMR solvent?

Using what we know, assign each peak to its hydrogen environment and identify which peaks are caused by the solvent. Does the hydrogen in the alcohol group participate in splitting in this spectrum?

Assign each peak cluster and predict the integration value for each.

The proton NMR data for 1-bromopropane is as follows: H1: triplet (2) 3.2ppm; H2: multiplet (2) 1.81ppm; H3: triplet (3) 0.93ppm. (relative integration value in parentheses) * Draw the structure of 1-bromopropane * According to the splitting rule, does H1 split H3? * How many hydrogens split H2 * Upon closer inspection of of the NMR data, the H2 peak cluster is discovered to have 6 peaks; is this consistent with the answer to the previous question? * Why is any peak cluster with more than four peaks listed as a “multiplet”

Above is the 1H NMR spectrum for C4H8O, draw a possible structure for this molecule.

Sketch a possible NMR specturm for diethyl ether (CH3CH2OCH2CH3)

Cis and Trans Splitting

Splitting between hydrogens of trans of one another is larger than splitting between hydrogens that are cis of each other.

This can be used to determine whether the molecule is the trans (Z) or cis (E) version of its self

Conclusion