Traditional Methods of Analysis - Empirical and Molecular Formulae

19 Modern Analytical Techniques

Combustion Analysis

eg. A combustion analysis of a 0.44g sample of an unknown compound yields 0.88 g CO2 and 0.36 g H2O. If the sample a molar mass of 132 g/mol, what is the molecular formula of the sample?

Once we have the molecular formula, we would need to use other methods to determine the structure of the compound. Which? Text book pp226-227

Mass spectrometry (this is not absorption!)

Revision: Mass spectrometers are calibrated so as to give Mr for the fragments produced, however, we are really plotting abundance of an ion on the y-axis against m/z on the x-axis (mass/charge).

Note: the workings of a spectrometer

Basically, electrons bombard a sample of the unknown compound, and knock off electrons, usually one of them...

M+ is called the “molecular ion” because it is the original molecule with one less electron and no fragmentation. This tells us the Mr for the compound and is represented as M/z not m/z.

M + e- M+ + 2e-

Fragmentation occurs for two reasons:

1. Thermodynamic instability of the molecular ion compared to possible smaller molecules.

2. The energy absorbed in the collision causes vibrations which overcome the energy required to fragment the molecule.

In general:

• bigger molecules fragment easier.

• the fragmentation of a particular molecule produces characteristic fragments which can also break down.

• after fragmentation, only one molecule is charged and the other is a stable molecule or a radical (how this happens depends on the relative stabilities like with carbonium ions)

• heteroatoms (eg. Cl, O;..) can lose electrons easily (from lone pairs) and disturb the local bonds making fragmentation more likely (eg on either side of a C=O or C-Cl)

• Useless low level noise is eliminated from spectra.

• The most intense peak is called the base peak (given the value 100)

Benzene:

• forms stable fragments as it is hard to break up, and usually is involved in the base peak. A stable fragment that may be formed is C6H5

+ (m=77)

• Fragments can correspond to the formula 13n but the mass destruction of the ring causes chaos.

Isotopes

Abundance is the key factor here. Why?

About 1% of C is 13C so there is a small peak called “M+1” . This may be significant with longer chain HCs (see p89) but is usually ignored with smaller molecules. O also exists as 18O but this is a tiny proportion.

Important: The molecular ion (M+) is assigned to the most abundant isotope although we can speak of multiple “parent ion peaks”.

This effect becomes important with the…

HalogensEg. 81Br and 79Br (50%)

37Cl and 35Cl (25% and 75%)

These isotopes lead to the formation of doublets and triplets.

Consider the molecular ion: CH3CH2Br+

Sketch the mass spectrum…

Consider the molecular ion: CH3CH2Cl+

Now, consider the ions

CH2BrCH2Br+ and CH2ClCH2Cl+

Text book pp228-229

Extra pointers to interpret these mass spectra:

• If 2 fragments of the spectrum have m/z > 1/2 M/z, then they overlap in the original molecule.

• Every fragment (mass m) will have a corresponding fragment (M-m) although this may be a radical or a molecule which does not show on the spectra.

• A difference of 14 between peaks indicates the loss of CH2

• Methyl fragments are unstable –CH3 (m=15)

• The molecular ion may not be formed in some cases eg. Alcohols easily lose the H atom so M-1 is more significant.

• Big molecules are more likely and easier to fragment.

Distinguish between the mass spectra of pentan-2-one and pentan-3-one…

Identify the peaks

Simple ChromatographyA method used to separate and identify the components of in a mixture. It’s a good way to make sure that a product in a synthesis does not contain impurities

What is the stationary phase and the mobile phase here?

What will determine how far the component travels?

What is the Rf?

Column Chromatography Thin Layer Chromatography

Text book pp224-225

Gas Chromatography (GC)

Notes: - The column can vary from a

metre to several metres long

- The internal walls of the column are coated with a solid or liquid which constitutes the stationary phase.

- The mobile phase is an inert gas (He or N2)

- A detector and computer gives us a graph as shown

- The area under the peak gives the concentration of the components

HPLC - High Performance Liquid ChromatographyNotes: - The solvent is forced through the

tube under a pressure of 400-500bar

- The stationary phase consists of compacted silica powder which leads to better separation of the components

- The detector uses UV absorption

- Retention factor is calculated, but they depend on: the solvent, the pressure and the temperature of the column

- The results look like this…

Text book pp230-231GC and HPLC are both examples of column chromatography

HPLC-MSRead pp232-233 of the text book and think about the following…

Limitations of LC and HPLC on their own

Advantages of coupling a HPLC column with a MS

Applications of HPLC-MS

Problems encountered in Drugs testing

Questions pp230-231

Nuclear Magnetic Resonance - NMRMoving charges produce a magnetic field, including spinning atomic nuclei. Although the nuclei of 16O and 12C do not spin, those of 1H 13C do!

This leads to 2 energy levels if an external magnetic field is applied, and the proton will jump up when it absorbs electromagnetic radiation (Radio Waves!)

This is called NMR, but more specifically we will need to deal with 1H NMR (or proton magnetic resonance) and 13C NMR.

19 Modern Analytical Techniques

Important points to remember:

• The environment of the 1H or the 13C will affect the magnetic field (and the energy levels). Eg. The electrons in the local bonds reduce the effect of the magnetic field applied.

• The AREA under each of the peaks (not the height as sometimes stated) gives the number of atoms in that particular environment

• The standard for NMR is tetramethylsilane (CH3)4Si. Why?

This is mixed with the sample (CCl4 solvent). There are 12H atoms in the same environment in TMS giving a strong absorption peak which can be used as a standard reference.

• The closer these bond electrons to the proton, the greater this shielding. An electronegative atom at the other end of the bond will de-shield the proton and this resonates earlier.

Let’s look at 13C NMR first

Each peak identifies a carbon atom in a different environment within the molecule. In this case there are two peaks because there are two different environments.

Consider the following spectrum:

Ethanol has a peak at about 60 because of the CH2OH group.

It also has a peak due to the RCH3 group. The "R" group this time is CH2OH. The electron pulling effect of the oxygen atom increases the chemical shift slightly from the one shown in the table to a value of about 18

C-13 NMR spectrum for 1-methylethyl propanoate. Draw the displayed formula…

NB. The text book refers to the height of the peaks as indicating the number of carbon (or hydrogen) atoms in a particular environment, but technically it should be the area under the peak, which must be given in any question

Another more complex example….

Read pp236-239 and carefully look through the worked examples

Try to make sense of the spectrum

Try the following problem…

Remember:

• The easier it is for resonance to occur, the more “downfield shift”.

• De-shielding causes shift downfield.

• Methyl protons are closer to the TMS due to little de-shielding

• The area of the peaks gives the relative number of protons in a particular environment (this information will be provided)

• H-bonding shifts the peak downfield.

This is the spectrum for Ethanal. What does it tell us about the structure?

Now let’s look at 1H NMR…..

Low resolution NMRHere we are concerned with:

• the number of peaks (ie. the number of different environments) • their relative area (the number of protons in each environment) • chemical shift - this indicates the type of environment (how de-shielded a proton is)

Text Book pp240-242

Point: The OH group can vary widelyRead pp243-245 and carefully look through the worked examples

Identify the peaks of this proton NMR