Option A: Modern Analytical Chemistry Review

1. Option A: Modern Analytical Chemistry Review

2. Analytical Techniques • State the reasons for using analytical techniques • The main use of analytical analysis is to determine chemical structure and composition, and to determine purity of substances. Structure is found through instrumentation. • State that the structure of a compound can be determined by using information from a variety of analytical techniques singularly or in combination • Information from only one technique is usually insufficient to determine or confirm a structure.

3. Principles of Spectroscopy • Describe the electromagnetic spectrum • Distinguish between absorption and emission spectra and how each is produced. • Emission Spectroscopy: the analysis of light energy emitted by excited atoms, ions, or molecules as they transition back to their ground state. • Absorption Spectroscopy: when radiation is passed through a sample, some of the energy is absorbed by the sample to excite an atom, ion, or molecule to the excited state. A spectrometer analyzes the transmitted energy relative to the incident energy. This energy is quantized. • Describe the atomic and molecular processes in which absorption of energy takes place • Absorption in the UV and visible region and is due to transitions in which electrons excite to high levels Molecular vibrations. (Typically used to identify metals in a sample i.e. Iron in blood, Lead or Calcium in water, Magnesium in vitamins. • Absorption due stretching and bending, occur in the Infrared (IR) region resulting in changes in polarity of the molecules (Identify types of bonds/functional groups) • Molecular rotations occur in the microwave region For every bond type, there is a unique bond strength, and there is a different amount of energy necessary to vibrate, rotate, etc.

4. Infrared (IR) Spectroscopy • Describe the operating principles of a double-beam IR spectrometer. • The spectrometer has a source of infrared light (hot coil of nichrome wire) that emits radiation over the whole frequency range of the detector. • The beam is then split into two beams of equal intensity. • One beam is passed through the sample and one through the reference chamber. • The rotating disk alternates the two light beams passing through. The intensity of the two beams are compared and the wavelength over which the comparison is made is dispersed on the detector via the prism. Generally, this detector converts the reading into electrical signals for storage. • Describe how information from an IR spectrum can be used to identify bonds. • IR helps to confirm or eliminate functional groups present in a molecule when an organic compound is exposed to electromagnetic radiation, it absorbs energy of a certain wavelength and transmits energy of other wavelengths. Based on this information about energy released or absorbed, information about functional groups can be deduced Functional groups have characteristic IR absorptions that do not change from one compound to another. • Explain what occurs at a molecular level during the absorption of IR radiation by molecules • When IR radiation strikes a molecule, it causes the molecule to stretch, bend, or vibrate, thus changing the polarity of the molecule The energy needed to stretch, bend, or vibrate is quantized A molecule can vibrate at specific frequencies relative to specific energy levels • Analyze IR spectra of organic compounds (Practice using in Class handout)

5. Mass Spectrometry • Determine the molecular mass of a compound from the molecular ion peak. • Look at the farthest right peak to determine the mass of the entire molecule. This may vary by one or two grams per charge due to isotopes. • Analyze fragmentation patterns in a mass spectrum to find the structure of a compound. (Use in class practice handout) • When a sample is introduced into the mass spectrum and ionized, some of the molecular bonds are broken and produce fragments. These fragments are also deflected by the magnetic field and show up on the detector. The fragments can help determine the molecular structure of the molecule. Examples: • Mr -15)+ is a loss of CH3. • Mr -17)+ is a loss of OH. • Mr -29)+ is a loss of C2H5 or CHO. • Mr -31)+ is a loss of CH3O. • Mr -45)+ is a loss of COOH.

6. Nuclear Magnetic Resonance (NMR) Spectroscopy • Deduce the structure of a compound given information from its H NMR spectrum. (Practice Using in class handout) • The number of hydrogen environments is shown by the number of peaks on the chart. The relative height of the peaks is equivalent to the number of members of each hydrogen environment. NMR Theory The hydrogen nucleus acts like a “spinning top.” The spinning nuclei act like a spinning magnet which can interact with an external magnetic field. A spinning Hydrogen nucleus will orient itself so that it is parallel or anti-parallel to an external magnetic field. Parallel orientations are more stable than anti-parallel, therefore they are lower in energy. When a nuclei is radiated with energy of the correct frequency, the lower energy (parallel) orientations flip to the higher energy (anti-parallel) state. When a flip occurs, it is now in a resonance state and exhibits nuclear magnetic resonance. Absorption frequency is not the same for all hydrogen nuclei. Local magnetic fields from surrounding electron clouds will shield the external magnetic field. Each cluster of groups creates a different amount of shielding. • Outline how NMR is used in body scanners. • Protons in water, lipids, carbohydrates, etc. each give different signals depending on their hydrogen environments. An image of the whole body can be built up by placing the patient inside a magnet of a MRI machine. MRI is especially useful for diagnosing tumors, cancer, multiple sclerosis, hydrocephalus, and torn ligaments. The radiation associated with Nuclear Magnetic Radiation are low-energy radio waves, therefore they are not harmful to the human body. • The MRI scan can detect heart defects and also changes in the thickness of the muscles around the heart. MRI gives detailed images and can be used to find tumour in brain and examine soft parts such as liver, kidney, spleen. • MRI is dangerous as it has small risk to the foetus in the first 12 weeks. Its disadvantage is that the scan may augment claustrophobia.

7. Atomic absorption (AA) spectroscopy • State the uses of AA spectroscopy. • Used to determine concentrations of an element or ion to one part per million. AA can also be used to determine the concentrations of metals in water, blood, soil, oil, and food. • Describe the principles of atomic absorption. • Atomic Emission spectra are obtained by giving atoms of an element sufficient electrical or heat energy and recording the light emitted as the electron transitions back to the ground state. AA is the reverse of emission spectroscopy. In AA, it is the energy absorbed by the electrons as they are promoted from a lower state to a higher state that is measured. By the amount of light absorbed, the concentration of a particular element can be quantized. AA utilized a double-beam principle outlined in IR spectrometer. There is a different light source necessary for each element being analyzed. The AA utilizes light of a specific frequency passing through the sample (analate). • Describe the use of each of the following components of the AA spectrophotometer: fuel, atomizer, monochromatic light source, monochromatic detector and readout.  • AA uses light of a specific frequency passing through the sample. • The sample is first put into a fine mist/aerosol in the nebulizer. • The mist is mixed first with the fuel and the oxidizer, then burned. • Monochromatic light from the light source is passed through the vapor sample. • Amount of light absorbed by the sample is detected by converting it into an electrical signal • Double-beam principle compares the amount of light from the source to the amount of light which passed through the flame. The difference between the beams is detected and converted into an electrical signal by the photomultiplier. This is the amount absorbed by the photomultiplier. • A calibration curve must first be determined by comparing its absorption to the control • Determine the concentration of a solution from a calibration curve. • The concentration of each element can be determined because there is a direct relationship between the concentration and the absorption. If the same sample and the same path length are maintained, the Absorption remains proportional to the concentration. A plot of Absorption vs. Concentration should produce a straight line. A=-logTf A = log (lo/la)

8. Chromatography • State the reasons for using chromatography. • Chromatography can be used to separate mixtures containing very small amounts of the individual components. Coupled with other techniques, it can be used to separate and identify complex mixtures quantitatively and qualitatively. It can also be used to determine how pure a substance is. • Explain that all chromatographic techniques involve absorption on a stationary phase and partition between a stationary phase and a mobile phase. • In each type of chromatography, there are two phases: a stationary phase which stays fixed and a mobile phase which moves. Chromatography relies on the fact that in a mixture, the components have different tendencies to absorb onto a surface or dissolve in a solvent. This makes it possible to separate them. • Generally classified by their main separation process: • Adsorption chromatography • Stationary phase = solid • Mobile phase = liquid or gas • Mixture of components are adsorbed on the surface of the stationary phase • Example: column chromatography, thin layer chromatography (TLC) • Partition chromatography • Stationary phase = non-volatile liquid film supported on an inert solid • Mobile phase = liquid or gas • Mixture of components are partitioned between the liquid film and the mobile phase • Interpretation: high travel = Rf closer to 1 = the component dissolves more readily in the mobile phase and the component moves rapidly along the stationary phase. • Interpretation: low travel = Rf closer to 0 = the component (X) remain largely adsorbed on the stationary phase and the components moves slowly along the stationary phase. • Example: paper chromatography, gas-liquid chromatography (GLC)

9. Outline the use of chromatography • Paper Chromatography: • A small amount of the mixture is placed onto the paper approximately 1 cm from the edge. Then, the paper is suspended in a small quantity of the solvent (eluent) in a closed container. • The closed container allows the atmosphere to be saturated, and prevent evaporation of the solvent from the paper to give better and faster separation. • As the solvent rises up the paper, the components in the mixture divide between the two phases depending on their relative solubility. As the solvent nears the top, a mark is made to record the level, then the paper is removed and dried. • Some components are colored and can be seen with the naked eye, however, some necessitate staining (maybe with iodine) or irradiation with an ultraviolet lamp. • Solutes have individual Retention factors (Rf) for each eluent. This factor is found by measuring the distance from the original spot both to the center of the particular component and to the solvent front. • Rf=(distance moved by solute)/(distance moved by solvent (eluent)) = x/y • If substances have similar Rf values, the paper can be turned 90⁰ and treated with a different solvent. This is known as two-way chromatography • Thin-Layer Chromatography (TLC): • Similar to paper chromatography, but instead of paper, a thin layer of solid (such as alumina or silica), or an inert support (such as glass) are used. The separated components can be recovered pure by scraping off the section containing the component and dissolving it in a solvent. Pregnancy tests may use TLC • Column Chromatography: • Separates the components of a mixture for further use rather than identification • Stationary Phase: alumina or silica gel Column is set up by packing the dry stationary phase on top of a piece of glass wool in a long column with a tap at the end and then saturating it with the eluting solvent. The sample is added at the top and as it moves down the column, more eluent is added After some of the components have been eluted, it is possible to elute more tightly help components

10. Thin-Layer Chromatography (TLC):

11. Column Chromatography:

12. Visible and ultraviolet (UV-Vis) spectroscopy • Describe the effect of different ligands on the splitting of the d orbitals in transition metal complexes. • Coloration has to do with an unoccupied d level; hence, Zn, Cd, and Hg are colorless. The different ligands are actually not all on the same level—the dx2-y2 and dz2 are slightly higher than the dxy, dxz, and dyz ligands. • Describe the factors that affect the colour of transition metal complexes. • Electron transitions within the d sublevel absorb energy in the visible light spectrum. When white light falls on a solution containing the complex ion of a transition metal, some of the light is absorbed. The solution transmits the remaining light. The light is the complimentary color of the absorbed light. Factors that affect d-d splitting and hence color include the nature of the transition element, the oxidation state, the type of ligand, and the shape of the complex ion (tetrahedral, square planar, octahedral) • State that organic molecules containing a double bond absorb UV radiation. • Molecules that contain unsaturated double bonds can absorb UV radiation. UV light tells us about the presence of a double-bond in a molecule. Molecules that have double bonds alternating with single bonds are called conjugated dienes. Compounds absorbing UV radiation include alkenes, aromatic hydrocarbons/arenes, and chlorophyll. • Describe the effect of the conjugation of double bonds in organic molecules on the wavelength of the absorbed light. • As energy is gained, the double bond is elevated to an anti-bonding state. With UV radiation, the energy absorbed corresponds to the amount necessary to raise the energy level of a pi electron in an unsaturated solution to a higher empty energy level. • Predict whether or not a particular molecule will absorb UV or visible radiation. • When a wavelength of UV with the correct quanta of energy needed to excite a pi electron to a higher energy level is absorbed, energy is absorbed. This absorbed energy is detected and recorded. This requires a large amount of energy. Most compounds absorb in the UV region and therefore appear colorless. Conjugated dienes require less energy to promote pi electrons to a higher energy level. Only conjugated dienes show UV absorption above 200nm because of the delocalizes electrons and corresponding resonance structures. • Determine the concentration of a solution from a calibration curve using the Beer-Lambert law. • Beers law states that the concentration of each element can be determined because there is a direct relationship between the concentration and the absorption. A = abc where A=absorbance, a=absorbicity constant, b=path length, and c=concentration Since the path length and the absorbicity are held constant, absorbance is directly proportional to the concentration. A plot of Absorbance v. concentration should produce a straight line.