1. 1 Instrumental Analysis Tutorial 2

2. 2 Objectives By the end of this session the student should be able to: 1.Describe the grating principle of work. 2.Describe the advantage of the photodiode array detector. 3.State the advantage of decreasing slit width. 4.Identify the absorbing species in organic molecules. 5.Specify the effect of conjugation on absorbance. 6.Calculate the wavelength obtained at a certain angle of reflection. 7.Calculate the concentration of a substance given absorbance data.

3. 10 6 photons 500 nm 0.7x10 6 photons Spectrophotometer

4. 4 Wavelength selector – Filters Filters permit certain bands of wavelength (bandwidth of ~ 50 nm) to pass through. The simplest kind of filter is colored glass, in which the coloring species absorbs a broad portion of the spectrum (complementary color) and reflects other portions (its color) Disadvantage: They are not very good wavelength selectors and can’t be used in instruments utilized in research. This is because they allow the passage of a broad bandwidth which gives a chance for deviations from Beer’s law.

5. 55 Schematic diagram of monochroamator

6. 6 The grating One peak corresponds to one valley destructive waves, they vanish. Now, what would constructive waves look like where double the intensity results?

8. 8 Light source Grating Rotating the grating changes the wavelength going through the sample slits Sample The components of a single beam spectrophotometer Separates white light into various colors Detector

9. 9 Wavelength selector – Grating Example: An echellette grating that contains 1450 grooves/mm was irradiated with a polychromatic beam at an incident angle 48 deg to the grating normal. Calculate the wavelengths of radiation that would appear at an angle of reflection of 20 and 10 deg. Solution:  To obtain d (constant spacing between grating’s grooves):  For reflection angle 20 deg: Grating equation: n = d (sin  i + sin  r ) Wavelength (nm) n =1  r (deg) 748 632 20 10 ii rr d Calculation of the first-order reflections are shown in the table

10. 10 Detector Photodiode Array Spectrophotometers Ordered arrangement of photodiode Multiple wavelengths detected simultaneously –no need to select single  at a time for drawing of the spectrum. Sample Source Grating Polychromator Photodiode Array white light

11. 11 Advantages unique to array detectors 1- Fast spectral acquisition (in a fraction of a second). Dispersive instruments require several minutes. 2- No wavelength selection needed, all wavelength are recorded simultaneously. In a dispersive instrument only one narrow band reaches the detector at any time. 3- No source intensity lost because we do not use slits as in conventional spectrophotometers.

12. 12 Electronic Spectra and Molecular Structure The electronic transition that takes place in the UV-Vis regions of spectrum is due to the absorption of radiation by: Specific types of functional groups or bonds (called CHROMOPHORES ) that contain valence electrons with relatively low excitation energy.

13. 13 Absorption of UV-Vis Radiation by Organic Compounds Types of Absorbing Electrons The electrons contributing to absorption of UV-Vis radiations by organic compounds are: 1. Electrons that participate directly in bond formation between atoms (  and  electrons) 2. Nonbonding or unshared outer electrons (n electrons) that are localized on atoms such as oxygen, nitrogen, sulfur and the halogens.

14. 14 Electronic molecular Energy levels Electronic transitions between the allowed energy levels can be brought about by absorption of UV-Vis radiations Four types of transitions are possible: Energy needed increases and hence the wavelength decreases   * > n  * >   * > n  * Not allowed

15. 15 ChromophoreExampleExcitation λ max (nm) ε C=CAlkenesπ π*17115,000 C≡CAlkynesπ π*18010,000 C=OKetones n π* π π* 290 180 15 10,000 N=O Nitro- compounds n π* π π* 275 200 17 5,000 C-X X=Br X=I Alkyle halidesn σ* 205 255 200 360 Absorption characteristics of some common chromophores  The energy required for   * and n  * (185-290 nm) allows the use of these transitions to identify organic compounds The high value of ε for   * indicates that these transition are the most easily occurring and thus the most probable transition occurs between these levels.  Absorption spectrum due to   * and n  * transitions are generally not observed in the ordinary accessible UV region (occur at < 180 nm because they require high energy)  Note that all the above transitions are observed in the ultraviolet region. That is the reason why most simple organic compounds are colorless

16. 16  -electrons are considered to be further delocalized by conjugation. The effect of this delocalization is to decrease the energy gap between  and  *. Absorption maxima are thus shifted to longer wavelengths (Bathochromic shift). The higher the extent of conjugation, the longer the wavelength of transition. Also, the molar absorptivity (ε) roughly doubles with each new conjugated double bond (Hyperchromic effect). Some highly conjugated organic compounds such as carotene and caffeine absorb radiations at longer wavelength enough to be observed in the visible region i.e., they are colored Effect of conjugation  11 ** 22  2 *  1 * CH 2 =CH 2 CH 2 =CH-CH=CH 2 conjugated system

17. 17  -CAROTENE LYCOPENE PARACETAMOL HIGHLY ‘CONJUGATED’ MOLECULES ARE COLOURED

18. 18 Example A 1.43 x 10 -4 M solution of the pure substance B (M. Wt. = 185) has A = 0.572, whereas a solution D (0.1358 g/L of a pharmaceutical preparation containing substance B) has a transmittance of 36.2% at 284 nm in a 1-cm cell. Calculate the percent of B in the pharmaceutical preparation. Solution:  Absorbance of substance B in the pharmaceutical preparation (solution D): A = – log T = – log (36.2/100) = 0.44  Concentration of B in pharmaceutical preparation D = 1.1 x 10 -4 M  Mass of B dissolved in 1 L of the pharmaceutical preparation: % B = (0.02/0.1358) x 100 = 14.98%

19. 19 ِِ 1.Draw a schematic diagram of a double beam scanning spectrophotometer describing briefly the role of each component in the spectrophotometer. 2.Would you use a tungsten lamp or a deuterium lamp as a source of 300 nm radiation? 3.A solution with A = 0.76 (3.0 cm cell) is doubled in concentration and measured in a 2 cm cell; calculate the absorbance of the measured solution. 4.A solution of Cu(II) ion prepared by dissolving 0.4 g of CuSO 4.5 H 2 O in 100 ml distilled water has an absorbance of 0.576 at 790 nm in a 3 cm cell. Calculate the molar absorptivity of [Cu(H 2 O)] 2+ complex. (Atomic masses: Cu = 63.5; S = 32; O = 18; H = 1). Exercise