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RAFFLES INSTITUTION (Junior College) H3 CHEMISTRY 2009/2010

Lecture Notes – Analytical Techniques (I) (Basic Principles of Spectroscopy & UV/Vis Spectroscopy)

Prepared by: Ms Ng Xinyi

________________________________________________________________________ A Lecture Outline

1. Analytical techniques in synthesis and drug development 2. Basic principles of spectroscopy 2.1 Electromagnetic spectrum 2.2 Energy and energy levels 2.3 Types of spectroscopy 3. UV/Vis Spectroscopy 3.1 Instrumentation

3.2 Molecular orbital theory 3.3 Analysis of UV/Vis spectra

3.3.1 Nature of electronic transitions 3.3.2 Chromophores and conjugation 3.3.3 Quantitative analysis using UV/vis spectroscopy: Beer-Lambert’s Law

B Learning outcomes

Candidates should be able to: a) recognize the importance of analytical techniques in drug development, for example, in

isolation of drugs, and in the determination of drug structure and purity b) understand qualitatively the range of wavelengths (frequencies, energies) with different

types of radiation used in spectroscopy c) understand the concept of quantized energy levels [quantal formulae and selection rules

are not required] d) explain that ultraviolet/visible absorption in organic molecules requires electronic

transitions between energy levels in chromophores which contain a double or triple bond, a delocalized system or a lone pair of electrons [detailed theory of why chromophores have absorptions of appropriate energy is not required, nor is any consideration of molecular orbitals involved]

e) predict whether a given organic molecule will absorb in the ultraviolet/visible region f) explain qualitatively how increasing conjugation in an organic molecule decreases the

gap between energy levels and hence shifts the absorption towards longer wavelength g) use Beer’s Law, absorbance = lg(I0/I) = εcl, where ε is taken merely as a constant

characteristic of the substance concerned, to calculate the concentration of a given species in solution

h) apply uv/visible spectroscopy to quantitative analysis of a given species in solution C References & Websites

1. Introduction to Spectroscopy by Pavia, Lampman and Kinz 2. Thinking Straight about Pharmaceutical Chemistry by Wee et. al. 3. http://www.chemistryexplained.com/Ny-Pi/Pharmaceutical-

Chemistry.html#ixzz0VJLCyCYW- What is Pharmaceutical chemistry about? 4. http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/UV-Vis/spectrum.htm#uv1 – A

concise guide to the various aspects of UV/vis spectroscopy

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1. Analytical techniques in synthesis and drug development

Candidates should be able to recognize the importance of analytical techniques in drug development, for example, in

isolation of drugs, and in the determination of drug structure and purity

Prior to the nineteenth century, schools of pharmacy trained pharmacists and physicians how to prepare medicinal remedies from natural organic products or inorganic materials. Herbal medications and folk remedies dating back to ancient Egyptian, Greek, Roman, and Asian societies were administered without any knowledge of their biological mechanism of action. It was not until the early 1800s that scientists began extracting chemicals from plants with purported therapeutic properties to isolate the active components and identify them. By discovering and structurally characterizing compounds with medicinal activity, chemists are able to design new drugs with enhanced potency and decreased adverse side effects. Drug development and synthesis is the core of pharmaceutical chemistry. Pharmaceutical chemists are involved in the development and assessment of therapeutic compounds.

So how do we know if a compound that has been synthesized is the correct compound? How do we ensure that there are no impurities present in the compound synthesized? How can we tell the structure of an unknown compound that has poisoned many people?

We can answer these questions by making use of various analytical techniques to elucidate the structure of the compound by using nuclear magnetic resonance (NMR) spectroscopy or check if there is a change in the functional group by using infra-red spectroscopy. Impurities can also be detected using thin layer chromatography. To fully identify the structure of an unknown compound, purification is needed and a combination of data by analyzing the purified compound using different spectroscopic methods is required. No single spectroscopic method can determine exactly the structure of a large compound by itself. Hence, a study of the various spectroscopic and separation techniques would be necessary in drug development to determine drug structure and purity. 2. Basic Principles of Spectroscopy

Candidates should be able to understand qualitatively the range of wavelengths (frequencies, energies) with different types

of radiation used in spectroscopy understand the concept of quantized energy levels [quantal formulae and selection rules are

not required] What is spectroscopy? Spectroscopy is the study of the _____________________________________________________. The result of this interaction is an exchange or transfer of energy between the two. This transfer of energy may involve either an _______________ or ______________ of energy in the form of EM radiation. Certain structural features/properties of compounds will undergo an absorption or emission of energy in the form of EM radiation. This absorption/emission of energy therefore enables us to detect these structural features. By piecing together various features that different probes reveal, we can deduce the structure of the unknown compound.

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2.1 Electromagnetic spectrum

The spectrum of electromagnetic radiation is classified by wavelength into radio, microwave, infrared, the visible region we perceive as light, ultraviolet, X-rays and gamma rays. The behavior of electromagnetic (EM) radiation depends on its wavelength. Higher frequencies have

_________________ wavelengths, and lower frequencies have ___________________ wavelengths.

All electromagnetic radiation travels at a constant speed, the speed of light (3 x 108m s-1), and is composed of photons. Photons are particles with considerable wave-like properties, that have no mass, but have associated with them a quantity of energy due to wave particle duality. They can thus be considered as packets of energy, and the amount of energy associated with each photon is proportional to its frequency. Relationship between photon energy and its frequency This simple relationship was first put forward by Max Planck, and the constant of proportionality, h, is called Planck’s constant.

Since E α f, [1] where E = energy in Joules; f = frequency in Hertz; h = 6.62 x 10-34 J The frequency and the wavelength of a particular radiation are related by an equation involving the speed of light, c. [2] Combining [1] & [2] gives us the equation: [3]

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Thus we can see that the higher the frequency or the shorter the wavelength of the photon, the more energy is associated with it. Energy-per-mole-of-photons

Area of spectrum Average frequency,

f/Hz

Average wavelength,

λ/m

Average energy per

mole X-ray 1 x 1017 3 x 10-9 20,000 kJ mol-1 UV 1 x 1015 3 x 10-7 200 kJ mol-1 IR 1 x 1013 3 x 10-5 2 kJ mol-1

microwave 1 x 1010 3 x 10-2 2 J mol-1 radio waves 1 x 108 3 0.02 J mol-1

2.2 Energy and energy Levels What happens when EM energy is absorbed? All energy is _________________, i.e. a particle such as a molecule or atom can only associate

with a ___________________ amount and this energy is determined by quantum rules.

In order for a molecule to acquire more of any of these forms of energy, it has to absorb a packet of energy of a particular size, which is determined by the size of the gap between the energy levels concerned.

The particle that absorbed the EM radiation is promoted to a higher energy state(excited state). Transitions between energy states are therefore detected by spectroscopy.

Energy difference between the ground state and 1st excited state is E1. The promotion of electron requires a photon energy hf1 (electronic transition).

Energy difference between the ground state and 1st excited state is E1, requiring photon energy hf1 so that E1=hf1

If a photon of lower energy(and frequency) hf2 irradiates the particle, the energy is insufficient to bring about a transition, so no absorption takes place

Worked Example 1 (i) Calculate the energy of a photon of yellow light with a wavelength of 600 nm.

E = hc/λ = (6.62 x 10-34 x 3 x 108)/(600 x 10-9) ∴ E = 3.31 x 10-19 J

(ii) Calculate the energy per mole of photons of yellow light.

To convert energy per photon into energy per mole of photons, we multiply this value by the Avogadro constant.

E = 3.31 x 10-19 x 6.02 x 1023 = 200 kJ mol-1

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What are the various types of energy a molecule can have? Restricting ourselves to molecules for the moment, we can identify various types of energy associated with particles.

1. electronic energy. This is associated with the movement of electrons between electronic energy

levels (or orbitals) in the molecule. 2. vibrational energy . This is associated with the regular periodic stretching-and-contracting

and the bending of bonds within the molecule. 3. rotational energy. This is associated with the speed of rotation of the molecule around various

axes. 4. nuclear magnetic energy. This is associated with the energy of interaction of a magnetic

field with those nuclei that have magnetic moments. The following diagram illustrates the different energy levels a molecule can have.

Energy level Energy gap electronic energy levels

Several hundreds of kJ mol-1

Molecule interacts with UV/vis light (λ=100nm - 700nm) an electron is likely to be changing from one electronic

energy level to another. vibrational energy levels

About 22 kJ mol-1 Molecule interacts with photons in the infrared region (λ ≈ 5 x 10-6m or 5000nm)

molecule changing from one vibrational energy level to another.

rotational energy levels

Less than 1 kJ mol-1

Molecule interacts with photons in the far-infrared to microwave region (λ= 0.1mm - 10mm)

molecule changing from one rotational energy level to another.

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2.3 Types of spectroscopy For our syllabus, we will only be looking at 3 different areas of EM spectrum absorbed by molecules and what knowledge of structures of molecules we can gain from the frequencies of absorption. Absorption spectroscopy involves the measurement of how much radiation is absorbed at each frequency, as we “sweep” a range of frequencies. 3 types of spectroscopy

1. Ultraviolet/visible (UV/vis) spectroscopy • Absorbance against wavelength (λ) spectra

2. Infra-red (IR) spectroscopy • Transmittance against frequency (or wavenumber) spectra

3. Nuclear Magnetic Resonance (NMR) spectroscopy • Absorbance against change of frequency in parts per million (δ)

3. UV/Vis spectroscopy UV-visible spectroscopy is the study of the ________________ of ultraviolet and visible radiation by molecules. 3.1 Instrumentation

A diagram of the components of a typical spectrometer is shown in the above diagram. As a beam of radiation is absorbed in its passage through a sample, the intensity of radiation decreases. The __________________ is defined as the ratio of the intensities of the emerging beam and the incident beam, expressed as a percentage.

I = intensity of emerging beam Io = intensity of incident beam T = 100 x (I/Io) T = transmittance

The absorbance of a sample is defined as the negative logarithm of (I/Io) ratio

The relationship between absorbance and transmittance is illustrated in the following table.

Over a short period of time, the spectrometer automatically scans all the component wavelengths in the manner described. The ultraviolet (UV) region scanned is normally from 200 to 400 nm, and the visible portion is from 400 to 800 nm.

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3.2 Molecular orbital theory (Not in syllabus) There is a second major theory of chemical bonding whose basic ideas are distinct from those employed in valence bond theory. This alternative approach to the study of the electronic structure of molecules is called molecular orbital theory. The theory applies the orbital concept, which was found to provide the key to the understanding of the electronic structure of atoms, to molecular systems. The Schrödinger wave equation is a mathematical function to study the wave-like behavior of an electron in a molecule. This function reflects the probability of finding an electron in any specific region. Molecular orbitals are obtained by combining the atomic orbitals on the atoms in the molecule. The total number of molecular orbitals is always equal to the total number of atomic orbitals contributed by the atoms.

Example: H2 molecule One of the molecular orbitals in this molecule is constructed by adding the mathematical functions for the two 1s atomic orbitals that come together to form this molecule. Another orbital is formed by subtracting one of these functions from the other, as shown in the figure below.

One of these orbitals is called a bonding molecular orbital because electrons in this orbital spend most of their time in the region directly between the two nuclei. It is called a sigma ( ) molecular orbital. Electrons placed in the other orbital spend most of their time away from the region between the two nuclei. This orbital is therefore an antibonding, or sigma star ( *), molecular orbital. Electrons are added to molecular orbitals, one at a time, in accordance with the Pauli Exclusion Principle, starting with the lowest energy molecular orbital. The two electrons associated with a pair of hydrogen atoms are placed in the lowest energy, or bonding, molecular orbital, as shown in the figure above. This diagram suggests that the energy of an H2 molecule is lower than that of a pair of isolated atoms. As a result, the H2 molecule is more stable than a pair of isolated atoms. Example: O2 molecule

Figure : Molecular orbital of O2 molecule

Similarly, if we have a molecular that is involved in π bonding, a π and π* orbital is created.

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3.3 Analysis of UV/vis spectra Candidates should be able to

explain that ultraviolet/visible absorption in organic molecules requires electronic transitions between energy levels in chromophores which contain a double or triple bond, a delocalized system or a lone pair of electrons [detailed theory of why chromophores have absorptions of appropriate energy is not required, nor is any consideration of molecular orbitals involved]

explain qualitatively how increasing conjugation in an organic molecule decreases the gap between energy levels and hence shifts the absorption towards longer wavelength

predict whether a given organic molecule will absorb in the ultraviolet/visible region 3.3.1 Nature of electronic transitions UV/vis spectroscopy involves the excitation of a particle in an electronic ground state to an electronic excited state. Specifically, an electron is excited from an occupied orbital of a lower energy (LUMO) to an unoccupied orbital of a high energy (HOMO). The highest occupied molecular orbital and lowest unoccupied molecular orbital are often referred to as the HOMO and LUMO, respectively. When a photon is absorbed by a molecule an electron gains the photon’s energy and is promoted to a higher energy orbital. The molecular orbitals in molecules are of three types, bonding (σ or π), non-bonding (n) and anti-bonding (σ* or π*). The electrons are normally resident in the bonding orbitals, or in the non-bonding orbitals if they are lone pairs on atoms such as nitrogen or oxygen. The higher energy orbital to which they are promoted is almost invariably an anti-bonding orbital.

In the above electronic transitions between energy levels, the energy involved in each transition decreases from left to right. Recall that a lower energy gap requires lower frequency/longer wavelength of EM radiation. Hence, transitions of π → π*, n → π* and n → σ* normally produce absorption in the UV/visible region. i.e. only molecules with n or π electrons give characteristic UV/vis spectra. In methanol (formaldehyde) CH2=O, the two electronic transitions absorbing UV radiation are shown below:

σ∗

π∗

n

π

σ

λ = 300 nm

λ = 190 nm

Electronic energy levels and transitions Possible transitions in some organic molecule

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3.3.2 Chromophores and conjugation A system containing the electrons responsible for the absorption of radiation is called a _______________________.

A typical UV spectrum is shown in on the left. The three absorbances are due to three different electronic transitions (n → π* or π → π*), but it is not possible to allocate a specific transition to each wavelength.

Different chromophores absorb strongly at different wavelengths. Even with a mixture of compounds, the UV spectrometer could selectively detect the presence of a particular compound, if its optimum UV absorption wavelength is sufficiently different from that of the other compounds.

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The following table summarises the wavelengths at which some chromophores absorb radiation.

R

CR

O

R CH CH CH CH R

R CH CH CH O

O

O

N

HN NH2

O

N

N NH

N

NH2

(benzene)

(cytosine)

(adenine)

190 nm (>10,000),

Absorption wavelengths

230 nm (15,000)

300 nm (30)

230 nm (12,000)

184 nm (60,000), 203 nm (7400), 254 nm (204)

300 nm (5000)

267 nm (6130)

260 nm (13,500)

O O (quinone) 242 nm (24,000), 281 nm (400), 434 nm (20)

values in brackets after the wavelengths are the molar extinction coefficients (see text) As can be seen from the above table, many different chromophores absorb at nearly the same wavelength. The position of the UV/VIS absorption is therefore inadequate in the identification of compounds.

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This chromophores system is either an isolated bond, or more usually a conjugated system of double bonds. Conjugation of π bonds reduces the energy gap between the π and the π* orbitals, and hence shifts the absorption towards longer wavelength:

Effect of increasing conjugation on the energy gap between π to π*

More examples of conjugated systems:

1. Benzene absorbs far UV light very strongly at 180nm, weaker absorption at 200 nm and a group of much weaker bands at 254 nm region due to various π to π* transitions

2. β-carotene has a highly conjugated π electron system absorbs visible blue light at 480nm

(lower frequency than UV, so lower energy) Since blue light is absorbed, we observe beta carotene to have a red-orange colour

Effect of lone pair on absorbance The incorporation of an atom with a lone pair of electrons can cause a major shift of absorbance to ____________________ wavelengths, since the n → π * energy gap is smaller than the π → π * gap.

Worked example 2 Diphenylmethanone, shown below, absorbs in the UV/vis region. Predict and explain the wavelength, relative to propanone, with which diphenylmethanone will absorb energy.

Absorption λ/ nm

296nm (colourless)

400nm (orange, since violet absorbed)

π → π* energy gap/kJ mol-1 λ/nm R-CH=CH-R 690 190 RCH=CH-CH=CH-R 520 230 R-CH=CH-CH=CH-CH=CH-R 235 275

N N

CH

CH

Note that with increased conjugation, there is more than one possible π π* transition, hence multiple absorption bands will be observed for the UV spectrum of a conjugated system. Since conjugated π systems will reduce energy gap and absorb at longer wavelengths, highly conjugated π systems will appear coloured.

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Worked example 3 Study each of the molecules drawn below. Circle the molecule(s) which will show more than one absorption peak in the UV/vis region.

H3C

H

CH2 3.3.3 Quantitative analysis using UV/vis spectroscopy: Beer-Lambert’s Law

Candidates should be able to use Beer’s Law, absorbance = lg(I0/I) = εcl, where ε is taken merely as a constant

characteristic of the substance concerned, to calculate the concentration of a given species in solution

apply uv/visible spectroscopy to quantitative analysis of a given species in solution Beyond detection of the presence of a particular compound, the main use of UV/VIS spectroscopy is to analyse quantitatively how much of a particular compound is present in a solution. This makes use of Beer’s Law: A = log10(Io/I) = εcl A = absorbance Io = intensity of incident radiation I = intensity of emerging radiation ε = the molar extinction coefficient c = concentration of compound (in mol dm-3) l = path length of the absorbing solution (in cm) (the cells used in spectrophotometry are usually 1 cm2 in cross section, so l = 1.0 cm) The molar extinction coefficient (units: _____________________) is how much light of a particular wavelength is absorbed by 1.0 cm of a 1 mol dm–3 solution of a particular chemical i.e. how well a compound absorbs at that wavelength. It is important to note that molar absorptivity is wavelength and sample specific. Hence, 2 different samples of the same concentration have different absorbance at a given wavelength (provided both compounds absorb at this wavelength).

CH3 CH2

OH

CH3

CH

O Cl

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How do we carry out a quantitative analysis using Beer’s Law?

If the molar extinction coefficient,ε, is known, the concentration can be calculated directly from the absorbance using the Beer’s Law equation. Worked Example 4 A solution of quinone (ε242nm = 24 000) had an absorbance of 0.574 in a 1.0 cm cell at 242 nm. What is the concentration of the solution?

** Do note that the deviations from Beer’s law can arise and the application of Beer’s law to a mixture containing 2 independent components is further elaborated in Appendix.

Step 1. Run the UV/VIS spectrum of the compound in question, and of other compounds that may be present in the mixture.

Step 2. Choose a wavelength where there is a large absorption of the compound in question but only a small absorption of the other compounds present.

Step 3. Prepare standard solutions of the compound in question, to cover the range of concentrations expected. eg. standard solutions 1 – 4 each contain a different known conc. of the compound of interest

Step 4. Using a UV/VIS spectrophotometer (capable of measuring absorbance highly accurately at a particular wavelength), run these prepared solutions at the same wavelength to record their absorbances.

Step 5. Plot a calibration curve of absorbance against concentration. Step 6. Place the sample of unknown concentration in the

spectrophotometer, and measure the absorbance. Step 7. Read the concentration of the compound in question from

the calibration curve.

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Appendix

More on Beer’s Law…

Deviations from Beer’s law may occur under the following circumstances:

1. chemical factors such as dissociation, association, complex formation or polymerization of species resulting in the conversion of one chemical species to another causes change in measured absorbances

2. analyte concentration is so high such that it causes a change in the refractive index of the medium and affects the ability of molecules to absorb radiation

3. detected radiation is not monochromatic but polychromatic radiation instead 4. stray light striking the detector decreases measured absorbances

For a more complicated example, consider a mixture in solution containing two components (X & Y) at concentrations cX and cY. These 2 components behave independently of one another and obey Beer’s Law. To make quantitative determination of the individual constituents of the mixture, we make use of the following standard addition method; where absorbance at any wavelength, λ, is, for same unit path length l, given by

A(λ) = εX(λ)cXl + εY(λ)cYl

Therefore, measurements at two wavelengths yield two equations in two unknowns and will suffice to determine the concentrations cX and cY as long as the molar absorbances of the two components, εX and εY are known at both wavelengths.

Worked Example When measured with a 1.00cm cell, 8.50 x 10-5 mol dm-3 solution of species X gave absorbances of 0.129 and 0.794 at 475 and 700nm respectively. Under the same circumstances, 4.65 x 10-5 mol dm-3 solution of species Y gave absorbances of 0.567 and 0.0863 at 475 and 700nm respectively. Calculate the concentration of X and Y in the solutions that yielded the following absorbance data in a 1.25cm cell: (a) 0.502 at 475nm and (b) 0.912 at 700nm. We need to calculate ε475,X ε700,X ε475,Y ε700,Y first. For species X, At λ = 475nm, 0.129 = ε475,X (8.50 x 10-5)(1.00) ε475,X = 1520 mol dm-3 cm-1 At λ = 700nm, 0.764 = ε700,X (8.50 x 10-5)(1.00) ε700,X = 8990 mol dm-3 cm-1

….. For species Y, ε475,Y = 12200 mol dm-3 cm-1; ε700,Y = 1780 mol dm-3 cm-1

At λ = 475nm, 0.502 = (1520)(1.25)cX + (12200)(1.25) cY – [1] At λ = 700nm, 0.912 = (8990)(1.25)cX + (1780)(1.25) cY – [2] Solving by simultaneous equations, cX = 2.34 x 10-5 mol dm-3 and cX = 7.64 x 10-5 mol dm-3

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RAFFLES INSTITUTION (Junior College) H3 CHEMISTRY 2009/2010

Tutorial 1: UV/vis Spectroscopy

Prepared by: Ms Ng Xinyi ________________________________________________________________________ 1) In UV spectroscopy, what experimental variables must be controlled to assure reproducible

absorbance data? 2) A solution of guanine of concentration 2 x 10-4 mol dm-3 in a 1.00 cm path length cell has an

absorbance of 1.56 at 275 nm. What is the molar extinction coefficient of guanine at this wavelength?

3) The molar absorptivity for the complex formed between bismuth(III) and thiourea is 9.32 x 103 L

cm─1 mol─1 at 470 nm. Calculate the range of permissible concentration for the complex if the absorbance is to be no less than 0.15 nor greater than 0.80 when the measurements are made in 1.00-cm cells.

4) The logarithm of the molar absorptivity of acetone in ethanol is 2.75 at 366nm. Calculate the

range of acetone concentrations that can be used if the transmittance is to be greater than 10.0% and less than 90.0% with a 1.50-cm cell.

5) A standard solution was put through appropriate dilutions to give the concentrations of iron

shown below. The iron(II)-1,10-phenanthroline complex was then developed in 25.0 mL aliquots of these solutions, following which each was diluted to 50.0 mL. The following absorbances (1.00-cm cells) were recorded at 510 nm:

Original Fe(II)

conc, ppm A510

4.00 0.160 10.0 0.390 16.0 0.630 24.0 0.950 32.0 1.260 40.0 1.580

i. Sketch a calibration curve from these data. ii. A 1.0 g sample of a food product was crushed and extracted with 25.0 mL of water, after

which iron(II)-1,10-phenanthroline complex was developed and the solution diluted to 50.0 mL. The absorbance of this solution at 510nm in 1.00-cm cell was found to be 1.470. Determine the percentage mass of iron in the food product.

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6) The UV spectra of paracetamol and caffeine are shown in figures 2.30 and 2.31.

a) Suggest suitable wavelengths to use for the estimation of the concentrations of each of

these two drugs in solution.

b) Given the following molar extinction coefficient for relevant peaks, calculate the concentration of each drug in the solution used for its spectrum.

c) *(Refer to appendix) Paracetamol pills sometimes have small quantities of caffeine added in order to speed up the rate of absorption of the drug through the gut wall. A solution made up of such a pill had measured absorbances of 0.85 at 245nm and 0.45 at 275nm. Calculate the ratio of the molar amounts of paracetamol and caffeine in the pill.

Peak Molar Extinction coefficient (ε)

245nm in paracetamol spectrum 15 100

275nm in caffeine spectrum 18 200