PL1001 Pharmaceutical Chemistry Organic Spectroscopy PDF

PL1001 Pharmaceutical Chemistry Organic Spectroscopy; an introduction Dr Lynda Storey N548 RSE [email protected]  Areas Covered Electromagnetic spectrum Ultraviolet and Visible Spectroscopy Infrared Spectroscopy  Learning Objectives Understa...

PL1001 Pharmaceutical Chemistry Organic Spectroscopy; an introduction Dr Lynda Storey N548 RSE [email protected]  Areas Covered Electromagnetic spectrum Ultraviolet and Visible Spectroscopy Infrared Spectroscopy  Learning Objectives Understand the origin of electromagnetic radiation Relate energy, frequency and wavelength of EM radiation With regard to UV-Vis Spectroscopy – Understand the types of molecules that absorb in UV-Vis spectroscopy – Understand the concepts of conjugation and chromophores – Explain how UV spectroscopy can indicate the presence of various structural features in a molecule  The Electromagnetic Spectrum Organic spectroscopy involves the interaction between parts of a molecule and The Electromagnetic Spectrum Such interactions give us very important information about the structure of the molecule. Electromagnetic radiation is energy that displays both particle & wave properties.  The Electromagnetic Spectrum A particle of electromagnetic radiation is called a photon. The relationship between the energy (E) of a photon and the frequency (n) of the photon is: E = hn – Where energy is measured in joules (J) – Frequency is measured in Herts (Hz) or cm-1 – H is a constant called Planck’s constant = (6.626x10-34 J s)  The Electromagnetic Spectrum Electromagnetic radiation consists of waves of 2 vectors at right angles to each other: E = Electric vector B = Magnetic vector  The Electromagnetic Spectrum A wave like this may be described in terms of its wavelength (λ) or frequency (n) , i.e. waves per second (Hz) 1λ – Wavelength, λ, (distance between equivalent points on two successive waves, in metres) and – Frequency, n (the Greek letter nu) (the number of cycles per second, s-1, or Hz)  The Electromagnetic Spectrum The velocity (c) of light (all regions) in a vacuum is constant c = 2.998 x 108 m sec-1 Speed of light in air is approximately equal to that in a vacuum so c can be used for all calculations This c is a product of wavelength and frequency: c = λn The Energy of electromagnetic radiation is directly proportional to frequency and inversely proportional to wavelength E = hn = hc/l Another common unit used in infrared spectroscopy is wavenumber (cm-1) σ = 1/ λ It’s used in preference to wavelength as it is directly proportional to energy.  Example Calculations Calculate the frequency of a wavelength of 525 nm c = nl n=c/λ = 2.998 x 108 m sec-1 / 525 x 10-9 m = 5.71x1014 sec -1 What is it’s energy? E = hn = 6.626x10-34 J s X 5.71x1014 sec-1 = 3.78x10-19 J  The Electromagnetic Spectrum Majority of light sources are continuum sources – Polychromatic light = light which contains radiation of more than one wavelength Sunlight contains wide range of wavelengths – Peak output is in the visible region 800 to 400 nm  Spectral Output of the Sun X-ray UV Visible Infrared Radio Relative Spectral Irradiance 10-2 10-6 10-10 10-14 10-18 10 nm 100 nm 1000 nm 10 μm 100 μm 1 mm 1 cm 10 cm Wavelength  Strength of Radiation Energy of a monochromatic radiation depends only on the frequency or wavelength of its waves – Monochromatic = single wavelength Does not depend on the intensity of its beam Photons in the UV region carry more energy than those in the IR The greater the energy of the photon = the more powerful the radiation The Electromagnetic Spectrum  Ultraviolet (UV) and Visible (VIS) Spectroscopy This is when E.M. radiation in the 180-400 nm (UV) and 400-780 nm (VIS) interacts with molecules possessing conjugated double bonds. UV-Vis light have just the right energy to cause an electronic transition. The normal electronic configuration of a molecule is known as its ground state – electrons are in orbitals with the lowest energy. Absorption of UV-Vis light at an appropriate wavelength LUMO Excited states causes an excitation of an electron from the ground state to the excited state. Lone pairs of electrons HOMO Ground states  Ultraviolet (UV) and Visible (VIS) Spectroscopy An atom does not absorb photons unless the photons have enough energy to bring about transition of electrons from one energy level to another Only photons which carry the exact amount of energy required for transitions of electrons to the next permitted level can bring about transitions UV-Vis light have only enough energy to cause the two electronic transitions shown in red.  n → π* (n to pi star) electronic transitions The promotion of a nonbonding electron (n) (i.e. a lone pair of electrons) into a π antibonding molecular orbital. ΔE is small for this transition so longer wavelengths are required. π → π* (pi t o pi star) electronic transitions The promotion of an electron from a π bonding molecular orbital into a π antibonding molecular orbital. ΔE is larger for this transition so shorter wavelengths are required.  An example Acetone has both π electrons and nonbonding electrons, so there are two absorption bands. Under normal circumstances bands below about 210nm are not detected because O2 (in the air) absorbs in this region and covers up any other bands. So to see the one at 195 nm the spectrophotometer would be purged with N2.  An example O O acetone Methyl vinyl ketone n → π* λmax = 274 nm λmax = 324 nm π → π* λmax = 195 nm λmax = 219 nm The λmax for methyl vinyl ketone are both higher than those of acetone because this is a conjugated system. Conjugation raises the energy of the HOMO and lowers the energy of the LUMO, so less energy is required for an electron transition. LUMO π* LUMO π* HOMO π HOMO π  Chromophores A chromophore is that part of a molecule that is responsible for a UV or visible absorption spectrum. The carbonyl group is the chromophore of acetone. Only molecules that contain conjugated π bonds absorb in the UV – chromo = colour spectrum λmax (nm) 217 256 290  Chromophores Other types of common chromophores include conjugated carbonyl compounds and aromatics: C C C O Extra double bonds that extend the conjugation in these chromophores lead to a higher λ of absorption: C C Absorbs at a longer λ than benzene Phenylethene (styrene) Only the chromophore absorbs the chromophore UV light O  Visible Absorption Some compounds contain so many conjugated π bonds (very long chromphore) they absorb in the visible region and are coloured. UV/Vis absorption spectrum of β-carotene 2.5 2.0 Absorbance 1.5 1.0 0.5 0.0 190 270 350 430 510 590 670 750 Wavelength / nm β – carotene (orange)  Absorption of Light by 3.0 Dyes Visible region 2.0 Absorbance Yellow dye 1.0 Red dye Blue dye 0.0 190 270 350 430 510 590 670 750 830 Violet Blue Yellow Red Wavelength / nm Indigo Green Orange UV Spectrum To record a spectrum: – a solution of the sample is made – solution is then scanned with the full range of wavelengths required If any peaks are recorded then a chromophore must be present in the molecule Able to get information on which possible chromophores are present – But not able to determine structure (see next lecture)  Quantitative analysis of drugs UV/Vis absorption very important in drug analysis – Quantification of drugs in formulations (without excipient interferences) – Determine pKa of some drugs – Determine partition coefficients and solubilities of drugs – Determine release of drugs from formulations with time (dissolution testing) – Monitor reaction kinetics of drug degradation – UV spectrum used as one of the pharmacopoeial identity checks  Summary Covered the electromagnetic spectrum – Made up of radiation of different wavelengths Able to relate energy to wavelength and frequency – E = hn = hc/λ – c = λn – For a transition to occur, E must be exactly equal to difference in energy levels Only chromophores will absorb radiation – Double bonds, lone pairs of electrons, conjugated systems  Points to consider from today’s lecture What type of bonds are required to produce an absorption spectrum? Which bonds cause a longer wavelength of absorption? Do you understand the relationship between energy and wavelength from the EM spectrum? Do you understand how to relate energy, frequency and wavelength in calculations?