Spectral Analysis of Organic Compounds

Core Concepts

In this organic chemistry tutorial, we explore how to use techniques of IR and NMR spectral analysis to identify unknown compounds. We’ll briefly explore each technique before we walkthrough the analysis process using an example.

Topics Covered in Other Articles

Understanding Organic Structure

One of the most important skills a chemist can have is to identify an unknown substance. With an organic substance, any one analytical method usually only gives a few structural details about the unknown. As a result, an organic chemist often needs to run multiple analyses before they can make an identification. Saturation analysis combined with IR and NMR spectroscopy is one group of analytical techniques that allows for the identification of multiple common organic compounds.

Let’s take a brief look at how each of these three methods works. Then, let’s see how they work together in an example problem.

Pre-Spectral Analysis: Degrees of Unsaturation

To perform saturation analysis and calculate degrees of unsaturation, you need to know the molecular formula of your unknown. There are multiple ways you can find this out without knowing the full structure. Such methods include mass spectroscopy, combustion analysis if its combustible, byproduct analysis with stoichiometry, and a few others.

With the chemical formula, you can calculate the compound’s degrees of unsaturation using the proportion of each component. Specifically, you use the following formula:

Degrees of Unsaturation = (2(C) + 2 – (H + X – N))/2

C = Number of carbons in formula

H = Number of hydrogens in formula

X = Number of halogens in formula

N = Number of nitrogens in formula

To learn more about how the formula works, check out this article, which provides a deeper explanation. 

A compound’s degrees of unsaturation will always be an integer, which equals the sum of pi bonds and ring structures present. For instance, if you use the above formula with cyclopentadiene (C5H6), you will calculate 3 degrees of unsaturation. Accordingly, cyclopentadiene has 1 ring structure and 2 pi bonds.

Knowing our unknown’s degrees of unsaturation provides an important clue that helps inform our further analysis.

Spectral Analysis I: IR Spectroscopy

Next, we can use IR spectral analysis to further identify specific functional groups present in our unknown. Based on the geometry and chemistry of the unknown, its chemical bonds each vibrate at specific frequencies. IR theory states that a molecular structure absorbs infrared radiation only at frequencies that resonate with those of its bond vibrations. An IR spectrometer can pick up these absorbances, providing each resonance frequency associated with the molecule in the form of an IR spectrum. To learn more about the science behind IR spectroscopy, check out this article.

On an IR spectrum, certain “peaks” (or rather “valleys”) corresponds to an important resonance frequency of a bond found in your compound. 

ir spectral analysis of formic acid

The x-axis is displayed in units of inverse centimeters (cm-1), also called “wavenumbers”, which indicates the inverse of the resonance frequency. Generally, higher wavenumbers indicate a stronger bond. For instance, carbon-carbon double bonds (~1680cm-1) have higher wavenumbers than aromatic carbon-carbon bonds (~1500cm-1) but lower than carbon-carbon triple bonds (~2200cm-1). 

Importantly, the same bond can have different wavenumbers depending on the chemistry of the surrounding region in the molecule. A carbonyl group in an amide (~1690cm-1) usually has a different resonance than a carbonyl in a carboxylic acid (~1750cm-1). Conjugation can also affect wavenumbers, especially in carbonyls.

The y-axis indicates the intensity of the peak, which tends to increase with bonds with stronger dipole moments. For instance, carbonyl groups tend to have a way higher intensity than non-polar carbon-carbon double bonds.

Luckily, the IR resonance frequencies for most bonds can be found tabulated in many textbooks and online resources. A comprehensive list of common bond wavenumbers can be found here.

Spectral Analysis II: NMR Spectroscopy

Finally, once we know the most important functional groups, we can use NMR spectral analysis to better understand our molecule’s hydrocarbon skeleton. When an atom has nuclear spin, it expresses a magnetic dipole moment. This dipole moment consequently involves some energetic change from the atom’s ground state which NMR measures as a frequency. However, this only applies to atoms with nuclear spin, such as hydrogen-1 (1H) and carbon-13 (13C). If you’d like to learn more about the physics of NMR spectroscopy, check out this article.

In 1H NMR, the most common method, chemically different hydrogens give off measurably different frequencies. NMR spectroscopy quantifies these frequencies in ppm, with higher values indicating stronger dipole moments. Importantly, the magnetic fields of the other atoms near a hydrogen shift its nuclear spin, depending on the distribution of electrons in the molecule. Generally, more electron-poor or “de-shielded” hydrogens have higher ppm than electron-rich or “shielded” hydrogens. The NMR spectrogram reflects this “chemical shifting” of chemically unique hydrogens. A comprehensive list of common NMR chemical shifts can be found here.

nmr of 2-chlorobutane for spectral analysis

You may notice that peaks in the NMR often come grouped in twos, threes, fours, and so on. This comes from the “splitting” effect of neighboring hydrogens, which follows the “n + 1” rule; a group of four peaks indicates that there are three hydrogens on neighboring atoms, three peaks indicate two nearby hydrogens, and so on. 

Finally, you can integrate groups of peaks. The relative area under a group of peaks indicates the number of hydrogens at that frequency. Specifically, peaks representing three hydrogens have three times the area as those representing one hydrogen, half the area of those representing six hydrogens, and so on.

nmr of 2-chlorobutane analyzed spectrally

With all these intricacies, NMR is a powerful analytical tool as we’ll see in the next section.

Complete Spectral Analysis: An Example

To fully appreciate how each of these methods complements one another, let’s take a look at an example. Keep in mind: professors commonly use problems like this on college organic chemistry exams!

Let’s say you have an unknown compound produced serendipitously in the lab one day. You run a mass-spec on the sample and it has the following formula: C8H9ON. Next, you run IR and NMR spectroscopies on the sample, and the following spectra are produced:

IR for spectral analysis example
NMR for spectral analysis example

Unsaturation Analysis

To first understand our unknown, we need to calculate its degrees of unsaturation. With the formula, we find that the unknown has 5 degrees of unsaturation.

DoU = (2(C) + 2 – (H + X – N))/2 = (2(8) + 2 – (8 – 1))/2 = 5

IR Spectral Analysis

Second, we can notice that the IR frequencies at 3296cm-1 and 1662cm-1 probably have the most importance. The other IR peaks likely correspond to C-H and C-C bonds that don’t tell us much useful information. Upon looking at a table of IR frequencies, the peak at 3296cm-1 could correspond to an N-H bond or an O-H bond. The peak at 1662cm-1 could indicate a C=O or a C=C bond. Each of these bonds is possible given our formula.

ir spectrum annotated

NMR Spectral Analysis

Third, the NMR peak at 2.1ppm seems significant. Given its integration value of 27.9 and the smallest value on the chart of 9.2, this peak must correspond to three hydrogens. If we refer to a table of NMR frequencies, we see that 2.1ppm tends to correspond to methyl groups adjacent to a carbonyl. This makes sense since we previously noted that C=O was possible. Also, that would mean that there are no adjacent hydrogens to split the methyl hydrogens. Accordingly, the peak at 2.1ppm has no splitting whatsoever.

ketone identification from ir and nmr

Compound Identification

With the confirmation of the carbonyl, we can rule out O-H as the 3296cm-1 peak, since only one oxygen exists in the structure. Thus, we must have an N-H somewhere. Additionally, the carbonyl accounts for 1 of the 5 unsaturation units with 4 remaining. This likely means we need to add a phenyl group somewhere to absorb the rest of those unsaturation units.

Going back to the NMR, we can see a group of three split peaks between 7.0 and 7.5. The integration (19.4, 20.5, 10.7) indicates this group corresponds to five total hydrogens. When taking splitting into account, we see two hydrogens with one neighbor, two hydrogens with two neighbors, and one hydrogen with two neighbors. All of these data points indicate a monosubstituted phenyl group.

phenyl identification from nmr

This phenyl provides three pi bonds and one ring, accounting for the remaining unsaturation units. 

The final peak at 9.2ppm must then correspond to the hydrogen in the N-H bond. This peak has no splits, indicating no neighboring hydrogens.

amine identification with nmr and ir

For this to exist in our structure, we need to place the nitrogen between the phenyl and carbonyl, yielding our final structure.

spectral analysis final structure

This final structure thus has a secondary amide. For a final check, we can observe that the C=O bond in a secondary amide typically has a frequency of 1680cm-1. We can see that this is remarkably close to the 1669cm-1 wavenumbers in our IR spectrum.