CHEM COMPLEMENT
TABLE 1.8.4 SUMMARY OF SPLITTING PATTERNS Splitting
Splitting patterns can be complex, but one useful feature is seen in the spectrum in fi gure 1.8.3. The peak at 3.8 ppm, due to the CH2 protons, is split into four, indicating that there are three neighbouring protons (the CH3 protons). Similarly, the peak at 1.8 ppm is a triplet. These protons have two neighbouring protons (the CH2 group). This illustrates the rule that a proton that has n equivalent neighbouring protons gives a signal peak that is split into a multiple of n + 1 peaks.
TABLE 1.8.4 SUMMARY OF SPLITTING PATTERNS Splitting
pattern
Looks like this Gives the following information
Splitting pattern
Looks like this Gives the following information Singlet
singlet peak
The neighbouring functional group is most likely a has two hydrogen atoms bonded to it.
Doublet
doublet peak
The neighbouring carbon atom has one hydrogen atom bonded to it.
Quartet
Analysing 1H spectra
A.9.2
Analyse 1H spectra. © IBO 2007
Figure 1.8.3 The NMR spectrum of chloroethane. The peak at 1.8 ppm is a triplet and that at 3.8 ppm is a quartet.
increasing signal intensity
Worked example 1
Consider the following NMR spectrum of a compound with molecular formula C3H8O. Use table 1.8.1 (p. 57) to help determine the structure of the molecule.
Figure 1.8.4 The NMR spectrum of C3H8O.
absorbance
6
chemical shift (ppm)
5 4 3 2 1 0
Solution
• There are four sets of peaks in the spectrum, indicating four different environments for the 1H nuclei.
• Note that the formula is C3H8O, so if there are four different environments, there must be a hydroxyl, –OH, group to provide the fourth position for the hydrogen atoms to be bonded.
• The triplet peak centred at δ = 3.8 ppm could correspond to the hydroxyl proton. Since it is a triplet, the adjacent carbon must have two 1H atoms bonded to it.
• The triplet peak at δ = 0.9 ppm will most likely correspond to methyl protons (chemical shift = 0.9–1.0 ppm). Since it is a triplet, the adjacent carbon must have two 1H atoms bonded to it.
• The quartet peak at δ = 1.4 corresponds to CH2 protons that must be adjacent to a methyl group, CH3, since it is a quartet and the chemical shift corresponds to that of a CH2 group (chemical shift = 1.3–1.4 ppm).
• Finally, the singlet at δ = 2.1 should correspond to the CH2 group that is closest to the hydroxyl group. The high electronegativity of the oxygen atom will increase the chemical shift of the CH2 which would normally be in the range 1.3–1.4 ppm.
This spectrum does not have an integration trace;
however, we have enough information to determine the structure. The structure of the molecule is shown in fi gure 1.8.5.
C C
C H H
H O
H
H H
H H
Figure 1.8.5 The compound with molecular formula C3H8O is propan-1-ol.
CHAPTER 1 MODERN ANALYTICAL CHEMISTRY
Worked example 2
A compound has the molecular formula C3H6O. Consider the following NMR spectrum and use table 1.8.1 (p. 57) to help determine the structure of the molecule.
10 8 6 4 2 0
chemical shift (ppm)
Figure 1.8.6 The NMR spectrum of C3H6O.
Solution
• There are three sets of peaks in the spectrum, indicating three different environments for the 1H nuclei.
• Note that the formula is C3H6O, so the three different environments means that hydrogen atoms are bonded only to carbon atoms. There is no hydroxyl group, so the single O will be either an aldehyde or a ketone. With only three carbons and three different environments, a ketone is not possible (would have only one environment for 1H).
• The triplet peak centred at δ = 9.7 ppm will correspond to the proton in the aldehyde functional group. The very large chemical shift is due to the adjacent carbonyl group, which deshields the proton. Since it is a triplet, the adjacent carbon must have two 1H atoms bonded to it.
• The multiplet peak at δ = 2.4 must belong to hydrogens in the middle of a chain since the peak has split into more peaks than an adjacent CH3 group would cause (a quartet) and the chemical shift corresponds to that of a CH2 group (chemical shift = 1.3–1.4 ppm) that has been deshielded by the adjacent carbonyl group (increasing the chemical shift).
• The triplet peak at just over δ = 1.0 ppm will most likely correspond to methyl protons (chemical shift = 0.9 – 1.0 ppm). As it is a triplet, the adjacent carbon must have two 1H atoms bonded to it.
• This spectrum does not have an integration trace; however, we have enough information to determine the structure.
• The structure of the molecule is shown in Figure 1.8.7.
C C
C H H
H H
H H
O
Figure 1.8.7 The compound with molecular formula C3H6O is propanal.
1 a State the full name and draw the structure of the compound represented by the letters TMS.
b Explain why TMS is used as a reference standard in 1H NMR spectroscopy.
2 a Explain what is meant by the term chemical shift.
b Describe how chemical shift is used as a reference standard to identify the environment of the 1H nuclei that correspond to a given peak.
3 Examine the following NMR spectrum and answer the following questions.
absorbance
5 4 3 2 1 0
TMS
D (ppm)
a Copy and complete the following table, using table 1.8.1 to help you.
Chemical shift (δ) Type of 1H nucleus Splitting pattern Neighbouring carbon
3.8 –CH2– Quartet
2.6 Singlet
1.3 Triplet Has 2 Hs
b Using the information you have gathered in the table, draw the structure of this compound, which has the molecular formula C2H6O.
4 Explain how NMR spectra would help to distinguish between two isomers such as H3C–O–CH3 and H3C–CH2OH.
5 The NMR spectra of the following compounds all have one feature in common. Describe the feature: C6H6, (CH3)4C, CH2Cl2, HClC=CHCl 6 The 1H NMR spectrum for C4H8 is given below. Examine the spectrum and
answer the following questions.
absorbance Peak set A
(relative area = 1)
Peak set B (relative area = 3)
TMS
Section 1.8 Exercises
CHAPTER 1 MODERN ANALYTICAL CHEMISTRY a Copy and complete the following table using table 1.8.1, or the IB Data
Booklet © IBO 2007, to help you.
Chemical shift (δ) Type of 1H nucleus Splitting pattern Neighbouring carbon
5.4 1.5
b Using the information you have gathered in the table, draw the structure of this compound.
7 Consider this NMR spectrum of a compound with molecular formula C4H8O2.
a State how many different 1H environments there are in this molecule.
b Comment on the comparison between the number of 1H environments and the number of carbon atoms in the molecular formula.
c Peak A is a quartet. Explain the impact of this observation on your determination of the structure of this molecule.
d Assuming that there is not an –OH group in this compound, explain why peak B is a singlet.
e Draw the structure of the molecule that matches this NMR spectrum.
8 The following two NMR spectra belong to two structural isomers, each with the molecular formula C3H6O2. One of these spectra corresponds to ethyl methanoate and the other to propanoic acid.
X Y TMS
Z Peak area ratio 1:2:3
10 9 8 7 6 5 4 3 2 1 0
absorbance
D (ppm)
I
a Draw the structural formula of ethyl methanoate and of propanoic acid.
b Describe the effect that being bonded to the O–C=O functional group will have on the chemical shift of hydrogen atoms.
c Describe one feature that is identical in the two NMR spectra.
d Identify which molecule corresponds to each of the two NMR spectra.
absorbance
10 9 8 7 6 5
A
B
C
TMS
4 3 2 1 0
Peak area ratio 2:3:3
D (ppm)
TMS X
Y Z Peak area ratio 1:2:3
10 11
12 9 8 7 6 5 4 3 2 1 0
absorbance
D (ppm)
II
Two types of column chromatography—high performance liquid
chromatography (HPLC) and gas–liquid chromatography (GLC)—are used routinely in analytical laboratories. Both use the same principles as the simple column chromatography described in section 1.6, but both involve the use of sophisticated (and expensive) instruments.
A schematic diagram of a gas chromatograph is shown in fi gure 1.9.1. The stationary phase is packed in a long column, (often several metres long) which is then coiled into a helix. Typically, the column is several millimetres in diameter, although some columns used are even thinner, of capillary size.
The column is fi lled with fi ne granules of an inert solid. These granules serve as a support for a high boiling point, viscous liquid. In this technique, known as gas–liquid chromatography (GLC), separation of a mixture depends on the interaction of the components of the mixture with the stationary and gas phases. In GLC, the mobile phase is a gas, called a carrier gas, which is inert and is usually nitrogen, helium or argon.
Figure 1.9.1 Schematic diagram of a gas–liquid chromatograph.
column filled with high boiling point liquid
(stationary phase) detector oven
carrier gas, usually nitrogen (mobile phase)
sample injection port
waste
recorder
The sample to be separated must be vaporized. This limits the use of gas chromatography to substances that do not decompose when heated, often substances with molar masses of less than 300 g mol–1. The sample is injected, using a syringe, into a steady gas fl ow at the start of the column. Very small samples are required, around several microlitres (µL). The components of the sample adsorb to the stationary phase and partition occurs as they desorb and are carried along in the gas stream. The rate of fl ow of the gas is controlled by the temperature, and the entire column is housed in an oven.
Components are detected as they elute from the column, either by their effect on the thermal conductivity of the gas or, more commonly, by their effect on the
1.9 CHROMATOGRAPHY
HLA.10.1
Describe the techniques of gas–liquid chromatography (GLC) and high-performance liquid chromatography (HPLC).
© IBO 2007
Gas chromatography
WORKSHEET 1.4 GLC
CHAPTER 1 MODERN ANALYTICAL CHEMISTRY Results are presented as a graph. Identifi cation of each component is based on
the retention times, Rt. The retention time of a component is compared to that of standard samples run through the column under the same conditions. An identical retention time indicates identical components. Quantifi cation is based on the area under each peak, which is calculated automatically by an attached computer. A calibration curve is prepared by measuring peak areas for samples of known concentration.
Gas–liquid chromatography is extremely sensitive, and can detect specifi c components in masses as small as 10–12 g. It can be used to separate complex mixtures such as the hydrocarbons, that make up crude oil. Gas–liquid chromatography is frequently used in combination with mass spectrometry and infrared spectroscopy, and together these techniques make an incredibly powerful tool for identifying and quantifying mixtures. The mixtures are separated using GLC. As the separated components elute from the column, they are passed directly into a mass spectrometer, which is sensitive enough to detect and measure the concentration of minute quantities. A computer connected to the mass spectrometer matches the spectra to a database of spectra of known substances. This technique is used to detect narcotics, anabolic steroids, analgesics and other banned substances in the urine samples of athletes, and to detect toxins in food samples.
Figure 1.9.2 A gas chromatograph (left) connected to a mass spectrometer (right) provides a powerful analytical tool for forensic chemists.
Other uses of GLC in identifying compounds that can vaporize without decomposing include:
• analysing blood alcohol levels
• monitoring pollutants in air and other gas samples, such as gases from underground mines
• analysing foods—for example, checking the fatty acid content of vegetable oils and determining the presence of certain food colourings
• identifying the source of oil spills by fi ngerprinting crude oil found at different locations.
A.10.2
Deduce which chromatographic technique is most appropriate for separating the components in a particular mixture.
© IBO 2007
CHEM COMPLEMENT
Name that cockroach
Gas chromatography is being used to aid in the identification of cockroach species! Cockroaches pose a major problem in many parts of the world, where they consume large quantities of grain and other food crops. Different species of cockroach present different problems and so require different methods of control, but first the species concerned needs to be identified. Biologists traditionally identify species by comparing features with those listed in guides for identification.
This can be time-consuming and, for closely related species, quite difficult. Enter the chemist armed with a gas chromatograph! It has been found that the outer, waxy layer of a cockroach is distinctive for each species. By dissolving this waxy coating and analysing it using gas chromatography, the species is found with some certainty in about 30 minutes.
Figure 1.9.3 Which species of cockroach is this? GLC may be used to efficiently and accurately identify it.
CHEM COMPLEMENT
Over the limit?
Motorists are often stopped for random breath testing.
An initial screening test consists of blowing into a white tube connected to a black box. This test, based on an electrochemical method, gives an estimate of the driver’s blood alcohol content (BAC). If this estimate indicates a BAC over the legal limit (ranging from 0 in countries such as Pakistan, Saudi Arabia and Hungary, through 0.05% in many countries of the world, to 0.08% in countries such as the US, Malta and Ireland), the driver must be tested using a breathalyser.
A breathalyser is essentially a type of colorimetric analysis.
A measured amount of the driver’s breath sample is bubbled through an acidified potassium dichromate (K2Cr2O7) solution.
Any alcohol present reacts according to the equation:
3CH3CH2OH(aq) + 2Cr2O72–(aq) + 16H+(aq)
→ 3CH3COOH(aq) + 4Cr3+(aq) + 11H2O(l) The extent of the colour change (from chromium(VI), orange, to chromium(III), green) caused by the breath sample is used to determine the amount of ethanol present. This colour is compared with that of a test sample containing only the acidified potassium dichromate solution.
The BAC is most reliably tested by GLC of a blood sample.
Nitrogen gas is blown through a measured sample of the driver’s blood. The nitrogen removes any alcohol from the blood and carries it, as a vapour, through the GLC column.
Any alcohol eluted from the column is detected and measured electronically. By comparing the peak area with those of alcohol samples of known concentration, the BAC can be calculated.
A schematic diagram of a high performance liquid chromatograph is shown in fi gure 1.9.4. The stationary phase is a solid packed in a column that is much shorter than those used in GLC (usually 10–30 cm in length). The nature of this solid varies with the particular analysis being conducted. The solid particles are small and tightly packed, so the mobile phase (the solvent) is pumped through the column to ensure an effi cient fl ow rate. The solvent under pressure gives HPLC its alternative name: high pressure liquid chromatography.
Components separate as they pass through the column as a result of their different degrees of adsorption and desorption.
As the components elute from the column they are detected in the solvent stream by measuring a variety of physical properties of the eluted components, including refractive index, conductivity, fl uorescence and, most commonly, absorption of UV–visible light. The recorder produces a printout, known as a chromatogram.
Figure 1.9.5 shows a chromatogram for a mixture of three hydrocarbons. The horizontal axis shows the retention time.
The most strongly adsorbed component (octane, C8H18) has the longest retention time. The least strongly adsorbed component (butane, C4H10) moved quickly through the column and so has a shorter retention time. The order of retention times can be explained in terms of the increasing strength of van der Waals’ forces with increasing molecular size. The larger molecules, with the stronger van der Waals’
forces, adsorb more strongly to the column and so take longer to be eluted. The relative amounts of each hydrocarbon are calculated by reference to the vertical axis. The area under each peak provides a measure of the relative amounts of each hydrocarbon. Here, there is
approximately twice the amount of octane as hexane.
High performance liquid chromatography
liquid (mobile phase)
solid-filled column (stationary phase)
detector
recorder pump
syringe for injecting sample
waste
Figure 1.9.4 Schematic diagram of a high performance liquid chromatograph.
CHAPTER 1 MODERN ANALYTICAL CHEMISTRY Identifi cation of components on the chromatogram is
made by comparing retention times with those of known substances obtained under the same conditions. The concentration of each component is determined by comparing peak areas with those of the same substance at known concentrations. The peak areas for samples of known concentration are determined and plotted against concentration. In this way, a calibration curve is
constructed to determine unknown concentrations.
Calibration is conducted with each test, as results depend on factors such as column type and length, and the fl ow rate.
peak areapeak area
2% v/v
3% v/v
4% v/v
?% v/v
ETHANOL STANDARDS A standard curve is plotted.
UNKNOWN
ethanol concentration (% v/v)
0 1 2 3
Figure1.9.6 Determining concentration using standard solutions in HPLC.
HPLC is used for non-volatile molecular substances (with relative molar mass of approximately 300 g mol–1 or greater). It is well suited to temperature-sensitive materials that would decompose if used in a gas–liquid chromatograph.
It requires only a small sample (approximately 10 µL) and is effi cient and relatively fast. Advances made in shortening the column length have meant that the procedure is now faster and less expensive than previously, as shorter column lengths means less solvent is needed and the retention times are shorter.
Figure 1.9.5 A chromatogram for a mixture of three saturated hydrocarbons, obtained using HPLC.
HPLC has a wide range of uses in testing and analysing temperature-sensitive compounds.
These include:
• testing pharmaceutical products
• quality control of insecticides and herbicides
• analysing foods and beverages, such as analysing the additives (antioxidants) in margarine, sugars in fruit juice and vitamins in other foods
• analysis of alcoholic beverages
• polymer analysis
• analysis of oil
• biochemical and biotechnology research
• detecting drugs, such as barbiturates, in blood samples.
1 In all forms of column chromatography, the material packed into the column is usually fi nely divided to produce a large surface area of solid.
Explain why this is important.
2 For each of the following techniques describe:
i the mobile phase ii the stationary phase.
a Gas–liquid chromatography
b High performance liquid chromatography
3 By considering the types of samples that can be analysed using the two methods, describe an advantage of HPLC over GLC.
4 List two types of samples that could be analysed by HPLC, but not by GLC, and explain why they cannot be analysed by GLC.
5 a Describe the general features of a sample that can be analysed using GLC.
b Describe the general features of a sample that can be analysed using HPLC.
c List three uses of GLC.
d List three uses of HPLC.
6 When answering the following questions, select from the following list of terms:
peak heights, peak areas, peak widths, retention time (Rt) State which feature of a chromatogram obtained during an HPLC analysis is:
a used to identify the components of a mixture
b used to determine the concentration of each component of the mixture c least affected by a change in concentration of the sample components.
Figure 1.9.7 An HPLC being used in a chemistry laboratory.
Section 1.9 Exercises
CHAPTER 1 MODERN ANALYTICAL CHEMISTRY 7 A gas chromatogram of a mixture containing the
alkanes pentane (C5H12), butane (C4H10), octane (C8H18) and decane (C10H22) is shown to the right.
Identify the four peaks, A to D, on the chromatogram. Explain your choice.
8 An analytical laboratory used HPLC to analyse samples of soft drink containing the additives vitamin C (A), saccharin (B) and caffeine (C).
The chromatogram obtained is shown below.
retention time (min) 0
A B
C
2 4 6 8
Explain how this chromatogram can be used to provide information about the soft drink:
a qualitatively b quantitatively.
9 Compare the design and function of an HPLC and a GLC.
a State three similarities between them.
b State two differences between them.
10 Gas–liquid chromatography was used to test the alcohol content of a new white wine blend. A set of ethanol standards was prepared and, each standard was injected into the GLC column. The peak area for each standard was determined and used to generate the calibration curve shown below. A wine sample was diluted by a factor of 3 and then injected into the column. The peak area for the diluted sample was found to be 12.
peak area (units)
ethanol concentration (% v/v)
0 2 8
5 10 15 20 25 30
0
4 6 10
a Determine the volume of ethanol in 100 cm3 of the wine.
b Suggest an alternative method for the determination of the ethanol
b Suggest an alternative method for the determination of the ethanol