# Introduction to spectroscopy and structure determination

### Introduction to spectroscopy and structure determination

#### Lessons

In this lesson, we will learn:

• To understand how chemists determine the structure of organic molecules.
• To recall the major types of structure determination.
• To understand the practical issues surrounding the methods of structure determination.
Notes:

• Now we understand what organic compound structures look like and how to communicate them, we can think about how we know the structures.
There are many ways we can find the structure of a molecule but most rely on studying how a molecule interacts with energy (radiation). This is called spectroscopy.
If we apply a magnetic field or fire some radiation at a molecular sample and put a detector behind it, what happens?
• If the molecules absorb the radiation, do they respond by moving in some way: vibrating or bending?
• If they scatter the radiation, did it scatter in some places but not in others?
• If they dont absorb the radiation or scatter it in any way, why not?

Spectroscopy answers these questions and builds understanding of molecules and functional groups so we can predict the results for new, unstudied compounds.

• An equation sometimes used on organic molecules to find rings and double bonds is the degree of unsaturation, or the index of hydrogen deficiency (IHD).
This doesnt involve any analytical instruments, but the molecular formula needs to be known:

• IHD = (C + 1) - $\large \frac{H\;+\;X\;-\;N}{2}$

Where:
• $C$ is the number of carbon atoms;
• $X$ is the number of halogens;
• $N$ is the number of nitrogen atoms.

The IHD value will return the number of rings and pi bonds combined. It does this because a cyclic ring (like cyclohexane) lacks two hydrogens compared to the chain version (e.g. hexane) in the same way as a pi bond leads to two less hydrogens on a molecule.
Using benzene (C6H6) as an example:

IHD (benzene C6H6) = (6 + 1) - $\large \frac{6\;+\;0\;-\;0}{2}$ = 4

This 4 includes the three pi bonds that make the aromatic system, and the cyclic hexane ring.

Another example, 4-chloro-1-pentene (C5H9Cl):

IHD (benzene C5H9Cl) = (5 + 1) - $\large \frac{9\;+\;1\;-\;0}{2}$ = 1

This value of 1 is for the C=C double (pi) bond that makes pentene an alkene.

• The most important types of spectroscopy to organic chemists are:
• NMR (nuclear magnetic resonance) spectroscopy:
• NMR applies a magnetic field that flips an atom's 'spinning' nucleus. After applying some radiation (radio waves) the nucleus drops back to its ordinary 'spin state' and emits energy that we can measure.
• We use the precise frequency of this energy to determine how atoms containing a spinning nucleus are connected in the larger chemical compound. NMR can be run on any nuclei that has non-zero spin and these include hydrogen and carbon atoms.
• Mass spectrometry:
• This works by ionizing and fragmenting molecular samples and then effectively recording the mass of the fragments.
• In this way, mass spectrometry is used to determine the molecular mass of the sample being analyzed. It can also identify specific elements due to the mass of protons in the nucleus.
• Infrared (IR) spectroscopy:
• This works by firing infrared radiation at a molecule. This radiation is absorbed at certain frequencies by particular bonds.
• In this way, IR spectroscopy is used to identify bonds and functional groups in an organic structure being analyzed,
• X-ray crystallography:
• X-ray crystallography fires X-ray radiation into a crystallized organic sample. The X-rays are scattered by atoms and the pattern of X-ray scattering is measured. The position of atoms in the crystal structure, and bond angles, is gathered from this.

• These techniques have varying usefulness and practical considerations:
• NMR spectroscopy is probably the most useful of these techniques:
• If you are running a reaction using a reactant you know the structure of, you expect hydrogen and/or carbon environments to change as your reactant becomes a chemically different 'unknown' product.
Comparing a reactant's NMR spectrum to this unknown product that has been made will show which environments have changed (signals present in reactants but not products) and which new environments have been created (signals present in products but not reactants).
• If starting from a totally unknown structure, NMR can give you the background hydrocarbon skeleton 'parts'.
From here, using infrared spectroscopy to find the functional groups will tell you the particular bonds that connect the skeleton up!
• IR spectroscopy gives important information but it needs supporting evidence:
• In large, complicated molecules there will often be more than one of a particular bond or functional group.
• Detecting individual bonds does not relay much about the molecule's larger structure or where they are attached in a molecule. You'll need NMR for that!
• For simple molecules with few functional groups, a lab spot test could be more appropriate to 'test' for functional groups.
• Mass spectrometry gives chemists a 'ball-park' figure for how large an unknown molecule is by giving the molecular mass, so it usually only plays an early or supporting role in identifying a structure.
• The other methods (NMR, IR) can come in to give you a hydrocarbon skeleton and the functional groups that connect the skeleton. You use these to suggest a structure with the molecular mass you already know.
• X-ray crystallography is not always available because it requires a crystalline solid sample, and structures are often hard and time-consuming to resolve. When it is done successfully though, it is normally considered a final, decisive piece of evidence – a 'slam dunk' - to show a molecular structure.

• WORKED EXAMPLE:
A typical process of identifying an unknown molecule might go like this:
An unknown organic compound is found as a side product of a reaction and isolated.
• The first stage in structure determination would usually be to collect a mass spectrum of the molecule.
• For example, the mass spectrum shows a molecular ion peak of 138. This means the structure has a molecular mass of 138 g mol-1. It also reveals the formula to be C7H6O3.
With this alone, if we had absolutely no idea of the structure, we can only speculate:
C7H6 is a nearly 1:1 carbon:hydrogen ratio, which strongly suggests there is a phenyl (-C6H5) ring of some sort in the structure. This leaves the remaining formula with one carbon and three oxygen atoms.
Any ring attachment would replace a hydrogen, so other parts of the structure could be a functional group made of one carbon atom, three oxygen atoms and possibly one to three hydrogens, depending on how many functional groups are attached to the "phenyl ring".

Summary: We have a few ideas of what the core skeleton is, but the mass spectrum alone doesnt confirm anything. The larger the molecular mass and formula gets, the less certainty we have. We need to go at least one method further.
• Using 13C and 1H NMR spectroscopy, we can more or less confirm the hydrocarbon skeleton using the process of elimination through the molecular formula.
• 13C NMR shows six carbon atoms in an aromatic ring, and another signal that is consistent with a carbon in an ester or carboxylic acid group It would also reveal a carbon singly bonded to an oxygen atom, such as in an alkoxy or hydroxyl group.
• 1H NMR shows four aromatic protons and a –COOH proton in a carboxylic acid.
-NMR spectra is a huge step here to confirm what this structure is – an aromatic ring with two attachments! It will also reveal the relative positions of these two attachments on the aromatic ring (we'll learn this later).
- From our formula C7H6O3, the carbon NMR spectra shows six carbons in an aromatic ring, and the hydrogen NMR spectra shows four aromatic hydrogens. Four aromatic signals means two have been replaced – there are two attachments on this ring.
- Subtracting the aromatic carbon and hydrogen atoms, we're left with CH2O3. We have NMR that suggests a carboxylic acid, and this signal is very distinct in NMR spectra. This must be one of the two attachments to the ring.
- Subtract the carboxylic acid (-COOH) from this and we are left with HO, or -OH, a hydroxyl group.
Summary: We have conclusive evidence for the majority of the structure, but arrived at the OH attachment partly by the process of elimination, and NMR for –OH groups is sometimes inconclusive. If we look at IR spectra, we should be able to confirm functional groups.
• IR spectra here would be used to confirm the functional groups we believe we have. O-H bond stretches are quite distinct, in both alcohols and carboxylic acids.
• Introduction
How do we know about molecular structure?
a)
What is spectroscopy?

b)
Finding the degree of unsaturation (IHD)

c)
How different analysis methods work.