Colour by design: 6.8, 7.3. 13.6

Descripción

a level chemistry Mapa Mental sobre Colour by design: 6.8, 7.3. 13.6, creado por Clelia Serra el 21/05/2015.
Clelia Serra
Mapa Mental por Clelia Serra, actualizado hace más de 1 año
Clelia Serra
Creado por Clelia Serra hace más de 9 años
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Resumen del Recurso

Colour by design: 6.8, 7.3. 13.6
  1. UV and visible spec

    Nota:

    • Explain why there are many shades of blue. The shade depends on what other colours are absorbed besides red
    • Describe the absorption spectrum of grey pigment. Partially abssorbs all frequencies of light.
    1. Many unsaturated molecules and those with conjugated systems absorb UV and visible light
      1. The delocalised electrons in these systems require less energy to become excited compared with electrons in single bonds
        1. Conjugated system has C=C bonds, benzene rings or/and N=N
        2. Absorb UV, eyes do not detect UV, appears colourless
          1. Spectrometer measures the quantity of light absorbed by the solution at each wavelength
            1. Source of radiation: UV and visible
              1. Causes electronic transitions
              2. Light split into two beams: one through sample, one through pure solvent
                1. Intensity of the spectrum depends on concentration, and the distance the light travels in solution
                2. X - axis wavelength nm
                  1. Y- axis intensity of absorption
                  2. Identify the colour of the solution, find the wavelength at which absorption is greatest.
                    1. Absorption is most intense and the wavelengh of lambda max increases for molecules with large delocalised systems
                      1. For organic molecules with delocalised systems; the longer the conjugated chain, the more intense the absoption, longer the wavelength of lambda max
                  3. Reflectance spectrum - UV and visible light shone onto the surface of the sample and any reflected light is collected and analysed
                    1. Analyse pigments on old paintings
                      1. Opposite of absorption spec
                      2. Shape and width tells you shade + purity
                      3. Gas-liquid chromatography

                        Nota:

                        • Different types of chromatography: paper, thin layer, gas-liquid, high-pressure liquid.
                        1. Method of seperating and identifying the components of a mixture
                          1. All types depend on equilibrium set up and the components a mixture distribute themselves between the stationary phase and the mobile phase.
                            1. Each component has a different affinity for stationary phase.
                              1. A higher temperature will tend to excite the molecules into gas phase. Because they evaporate more readily or because they are so energetic that the attractions of the liquid no longer hold them
                                1. At the beginning, compounds which spend most of their time in the gas phase will pass quickly through the column and be detected. Increasing the temperature a bit will encourage the slightly "stickier" compounds through. Increasing the temperature still more will force the very "sticky" molecules off the stationary phase and through the column.
                              2. Outlet can be directly connected to a mass spectrometer - this will give the relative molecular mass
                                1. Methyl esters of fatty acids are used to run g.l.c traces rather than the acid themselves.
                                  1. Because methyl esters are more volatile, than fatty acids, have less of an affinity for the column. Chromatograms can be produced more quickly
                                2. Mobile phase - unreactive carrier gas, coated on the surface of finely divided solid particles
                                  1. Stationary phase - small amount of high boiling point liquid
                                    1. 3 things can happen to the molecule; condense at stationary phase, dissolve in the liquid of stationary phase, remain in gas phase
                                      1. Dissolve in the liquid, so remains in stationary phase for longer.
                                  2. Main parts: syringe containing sample, inert gas carrier, column, thermostatically controlled oven, recorder, detector, outlet tube
                                    1. As each component emerges from the column, a peak is recorded on a chromatogram.
                                      1. The area under each peak is proportional to the amount of that component in the mixture
                                        1. Use this to work out the relative amount of each component
                                        2. The time that a component takes to emerge is the retention time
                                          1. Factors that affect retention time: length and packing of the column, nature and flow of carrier gas, temperature of the column.
                                            1. Need to calibrate with an instruments with known compounds, and keep conditions constant
                                        3. Oils and fats

                                          Nota:

                                          • How does the shape of a saturated trimester differ from that of an unsaturated trimester? more linear
                                          • Why are saturated fats solids? Intermolecular bonds are stronger because they act over a shorter distance due to packing
                                          • The degree of unsaturation can be determined by measuring the mass of iodine which react with 100g of oil or fat. Explain why Each C=C reacts with an iodine molecule, more iodine used, the more unsaturated the oil or fat is.
                                          1. Naturally occurring triesters of propane-1,2,3-triol and long chain carboxylic acids
                                            1. Mixed triester has different fatty acids attached
                                              1. 3 alcohol groups in each molecule, 3 carboxylic acid can attach to each alcohol group
                                              2. Natural oils and fats can be broken down into the sodium salt of the fatty acid and glycerol by heating with dilute NaOH solution
                                                1. If the acid, rather than the carboxylate ion is required, can treat the sodium salt with dilute HCl
                                                2. Hydrogenation
                                                  1. Addition of hydrogen to unsaturated oils
                                                    1. Use a Nickel catalyst, high pressure and high temperature
                                                      1. Heterogeneous catalyst, different physical state
                                                      2. Produces a more solid fat - used in the manufacture of margarine
                                                        1. Not all C=C bonds are hydrogenated, this gives a more spreadable fat
                                                      3. Saturated can pack closer together. Unsaturated have Z double bonds, causing molecules to kink
                                                        1. Attractive bonds between molecules will be weaker
                                                      4. Oils have a higher proportion of unsaturated triglycerides
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