SCH 410: Techniques in Organic Chemistry Course Lecturer: Dr. Jenipher Odak Contact: odakj68@gmail.com or jodak@maseno.ac.ke, +254751612967 & +254722562460 . Course content Modern techniques in organic synthetic and analytical chemistry; micropreparative scale of single and multistep synthesis of known organic molecules; methods of monitoring course of reactions; TLC, GLC, UV, IR, HPLC, MS and NMR. • Techniques in Organic Chemistry involves the theory and procedures on the main organic laboratory techniques which include distillation, extraction, crystallization, chromatography, spectroscopy e.t.c • Organic synthesis is a branch of organic chemistry involving construction of organic compounds from smaller subunits that have functional groups via organic reactions. • Organic synthesis involves the conversion of a substrate to the desired product molecule. Scales of reactions in the lab 1. Preparative scale: running experiments on the largest scale compatible with the equipment available. >10 g. 2. Analytical scale: mini scale, lab scale 1 g-10 g. 3. Micro scale: these are reactions that can be carried out conveniently on small scale i.e., 100 mg or less, and sometimes on as little as a milligram. Organic synthesis can be divided into three steps 1. Reaction: Transformation of a substrate/s into the desired product/s using appropriate reactants, reagents and catalysts under suitable conditions (temp, press, light, etc) In most cases one or more of the reagents is in liquid phase. 2. Purification: separation of desired product from by-products and remaining reaction components. This involves the use of i. wet chemistry techniques such as extraction, filtration, decantation, sublimation, evaporation, distillation, centrifugation and/or crystallization for crude separation. ii. Chromatography (separation of organic compounds) 3. Identification/analysis i. Purity determination: mpt apparatus, spectroscopy and/or chromatography ii. Structure determination: spectroscopy, spectrometry and/or crystallography iii. quantitative analysis SEPARATION TECHNIQUES ➢ Separation depends on the chemical and physical properties of compounds eg solubility, pH, polarity, melting point, boiling point etc. 1. Decantation • It is a process of separating a liquid from a few large solid particles by carefully pouring away the liquid above the particles after settling. • If the sample contains a large number of solid particles or the particles are fine, or suspended particles, then decantation will not be appropriate but filtration centrifugation would be a better separation method. Disadvantages - A small amount of liquid is always left in the solid residue - Very fine solid particles take time to settle out and any disturbance can mix them with the liquid being poured off. - Cannot be used to separate suspended particles. 2. Filtration • Filtration is used for the physical separation of solids and liquids. Uses of filtration in the organic chem laboratory: -To separate a solid product from a reaction mixture or recrystallization solution -To remove solid impurities from a solution -To separate a product solution from a drying agent after an extraction -To separate a solvent from a drying agent. Types of filtration (a)Gravity filtration • It is the most common filtration method which is used to separate insoluble solid from a solid/liquid mixture by passing through a filter paper in a filter funnel by gravity. • Gravity filtrations are used in the organic laboratory for several purposes: - to remove a drying agent from an organic solution, during a recrystallization where the desired product is completely dissolved in a hot solution but insoluble impurities remain, - when coloured impurities are present in a hot recrystallization solution the mixture is treated with activated charcoal and then gravity-filtered to remove the charcoal. when one of the products of a reaction pptates and the precipitate can be separated from the solution by filtration. NB: If the compound being recrystallized is colourless and the recrystallization solution is deeply colored after the compound dissolves, treatment with activated charcoal removes the colored impurity. Activated charcoal has a large surface area and a strong affinity for highly conjugated colored compounds, allowing it to readily adsorb these impurities from the recrystallization solution. Using too much charcoal, however, may cause some of the compound you are purifying to be adsorbed hence reducing your yield. (b) Vacuum filtration • Vacuum filtration is used to rapidly and completely separate a solid from the liquid with which it is mixed. • The recovery of the crystallized product from a recrystallization procedure is a common application of vacuum filtration in the organic chemistry lab • Vacuum filtration is also employed when it is necessary to use a filter aid, such as celite and activated charcoal, to remove very finely divided insoluble solids from a solution when the solution but not the solid is the desired product. The vacuum source for a filtration can be either a water aspirator or a compressor-driven vacuum system. NB: Heavy-walled tubes must be used in vacuum filtration so that it does not collapse from the effect of atmospheric pressure when the vacuum is applied. The solution to be filtered is drawn through the filter paper by application of a vacuum to a filter flask with a side arm adaptor referred to as a Buchner flask. Differences between gravity and vacuum filtration Gravity Vacuum Ideal for collection of the filtrate Ideal for collection of solids since it is faster Larger volume of solution required Small volumes of solution required Large amount of solid to be collected Small amount of solid to be collected Solids with large particle size Solids with small (fine) particle size Normal filter funnel/conical flask used Buchner/hirsch funnel and flasks used NB: Micropore filters remove particles as small as 0.5 μm diameter from solutions for instrumental analysis by NMR spectroscopy, polarimetry, or high-pressure liquid chromatography to remove very fine particles that would interfere with obtaining a correct measurement. 3. Centrifuging • Centrifuges are devices or apparatus that can be used to separate insoluble materials (usually a solid) from a liquid, where normal filtration does not work well e.g. a suspension of very fine (tiny) solid particles. • The centrifuge separates particles from a solution based on their: size, shape, density, viscosity of the medium and rotor speed of the centrifuge. A centrifuge basically consists of a motor which spins a rotor containing the experimental samples. The differences between centrifuges are in the speeds at which the samples are centrifuged and the volumes of samples. The centrifuge consists of carriage or glass tube holder, mounted on an electrically motor driven vertical axle. The carriage holds the balanced sample tubes of equal amounts of the solid-liquid mixture. All tubes are initially in a vertical position before the motor is switched on. The tube carriage is rotated at high speed in a fully enclosed container. On rapid rotation of the carriage, the tubes whirl round horizontally and the centrifugal force causes the more dense insoluble material particles to move outwards, separating from the liquid. • When rotation ceases the solid particles end up at the 'bottom' of the sample tubes with the liquid above. • After centrifuging the liquid can be decanted off and the solid is left at the bottom of the sample tube. • When operating a centrifuge, the sample tube must be counterbalanced by another centrifuge tube filled with an equal volume of water or sample. • A centrifuge containing unbalanced tubes vibrates excessively, is noisy and may move around on the bench top. • A balanced centrifuge makes a steady uniform noise at full speed Types of centrifuges - High speed - Table top - Clinical - Microfuges 4. Simple Distillation Distillation is used to separate miscible liquids or separate a liquid from a solution. Fill the distilling flask with sample 1/3-1/2 full. Add a few boiling stones or stir bar to the flask. Position the thermometer bulb just below the arm of the three-way adapter, where vapors turn toward the condenser. Be sure all of the connections are secure (especially between the distilling flask and 3way adapter: potential of fire!). Simple distillation is often applied for separation when: - Purification of a liquid which is relatively pure (less than 10% liquid contaminants) E.g., removal of water from organic solvents such as toluene, diethylether, dimethyl sulfoxide, etc. - Separation of a pure volatile material from a non-volatile solvent - separation of a liquid from a soluble solid contaminant - Separation of miscible liquids whose Bpt differs by at least 70°C eg Octane (Bpt 125 0C) and pentane (Bpt: 36 0C) NB: Cannot be used to separate a mixture of miscible liquids especially if the boiling points are relatively close. 5. Fractional Distillation • Continuous repetition of redistillation process in fractional distillation gives good separation of the volatile liquid components. • The more the theoretical plates (more surfaces or beads) added to the column, the more the time used for separation and more energy is used to keep reevaporating the liquid in the fractionating column. • Fractional distillation is used to separate complex mixtures of miscible liquids whose bpts are relatively close together. It can be used for separation alcohol/water Advantages over simple distillation • Gives much better separation between liquids • Can separate liquids whose Bpt are close (<700C) • Can purify complex mixtures of liquids Disadvantages over simple distillation • Set up is more complicated takes longer for liquids to distill • Consumes more energy Fractional Distillation Cont… During distillation, the components of the mixture are controlled by Raoult’s law. Raoult’s law states that in an ideal solution the partial pressure (PA) of component A at a given temperature is equal to the vapor pressure PoA of pure A multiplied by the mole fraction of A (XA) in solution. Consider an ideal solution of A and B: XA = nA/(nA + nB) , XB = nB/(nA + nB); nA and nB represent the number of moles of components A and B respectively. XA + X B = 1 PA = XAPoA PT (total vapor pressure) = PA + PB PB = XBPoB Calculate the mole fraction of each compound in a mixture containing 95.0 g ethanol, and 5.0 water. n(ethanol) = 95.0 /46 = 2.07 moles ethanol n(water) = 5.0 /18 = 0.28 mole water n(total) = n(ethanol) + n(water) = 2.07 + 0.28 = 2.35 moles X (mole fraction of ethanol) = Xeth = neth/ nt = 2.07 /2.35 = 0.88 X (mole fraction of water) = Xw = nw / nt = 0.28 /2.35 = 0.12 An ideal binary solution of miscible liquids A and B has the following parameters: mole fraction of A is 0.40 and the vapour pressure of a pure liquid A (PoA) is 1710 mmHg, if the vapour pressure of a pure liquid B (PoB) is 127 mmHg. Calculate the composition in mole percentage of the vapor from distilling the solution at 150 oC and 760 mmHg. (also called "rotavap”) Ptotal = PA + PB = XAPAo + XBPBo = (0.40 x 1710) + (0.60 x127) = 760.2 mmHg % A = (0.40x1710) x100/ 760.2 = 90% % B = (0.60 x 127) x100 /760.2 = 10% 6. Evaporation Conversion of a liquid to a gas or vapour. It can be used for removing the liquid from a solution, leaving behind a solid (solute). It can be done quickly with gentle heating or left out to 'dry up' slowly. 7. Crystallization It is often done from a hot concentrated solution, because most substances are more soluble in the hotter liquid. Solid crystals will 'grow' out of the solution because the solution is too concentrated for the entire solid to remain dissolved at that temperature. On cooling a hot concentrated solution, crystals form since the solubility of substance decreases with decrease in temp. Crystallization usually results in a greater loss of material than distillation but the degree of purity is significantly higher than that of distillation. Recrystallization technique is most commonly used in organic chemistry and involves dissolving the solute in a suitable hot solvent, cooling the solution to room temperature or below, and collecting the crystals that deposit on standing. Sublimation Can be used for purification of solid with high vapor pressure. It involves warming the solid to a temperature below its melting point and condensing the vapors on a cold surface. This technique is useful only if impurities associated with the component being sublimed have a substantially different vapor pressure at the sublimation temperature. It is often used as a final purification step in the preparation of an analytical sample. Examples of substances that undergo sublimation bezophenone naphthalene benzoic acid Camphor 7. Extraction The separation is based on the selective partitioning of the analyte between two immiscible phases in which one phase containing a solute S, is brought into contact with a second phase and the solute partitions itself between the two phases. The equilibrium constant is reached the distribution constant, or partition coefficient. If KD is sufficiently large, then the solute will move from phase 1 to phase 2, if the partition coefficient is sufficiently small the solute will remain in phase 1. Example for separation of a mixture: A mixture containing aspirin (KD = 100), ibuprufen (KD = 20) and benzocaine (KD = 1000) in water can be separated by addition of dichloromethane to the mixture in separating funnel (the KD values were obtained for the substances when the mixture in water is partitioned with acetone). KD KD = [ibuprufenAcetone] [ibuprufenwater] = 20 = [Benzocaine Acetone] [Benzocaine water] = 1000 Benzocaine will be separated since it has a highest KD value which enables it to move from the water into the DCM phase. Aspirin and ibuprofen having low KD values tend to remain in the water phase hence will not be separated. NB: Small amounts of ibuprofen and aspirin will also be extracted in the DCM layer. PROBLEM: A fourth-year student used Partion Coefficient (KD) to separate organic compound W. One gram of W in 100cm3 of water was shaken with 5cm3 of ether. Determine the mass of W that was extracted in ether. Given that the KD of compound W between ether and water is 40. SOLUTION: Let x be the mas of organic compound in ether. Concentration in ether = x/5 g/cm3. Mass in water = (1-x) grams Concentration in water = (1-x) grams/100 g/cm3 KD=Concentration of W in ether/ Concentration of W in water. 40 = x/5/(1-x)/100 40 = x/5/(1-x)/100 40 (1-x)/100 = x/5 200-200x = 100x 300x = 200 X = 0.67g example: Given compound A, K (ether:water) = 4.0, how much of A can be extracted from a solution of 10.0 g of A in 100 mL of water with a single portion of 100 mL of ether? X / 100 mL ether Kc = 4.0 = ------------------------------(10.0 - X) / 100 mL water X = 8.0 grams of A extracted into the ether -same as above, but extract two times with 50 mL of ether each time. X / 50 mL ether first extraction: Kc = 4.0 = ------------------------(10.0 - X) / 100 mL water X = 6.67 grams of A extracted Y / 50 mL ether second extraction: Kc = 4 0 = ----------------------------(3.33 - Y) / 100 mL water Y = 2.22 grams of A extracted total extracted = X + Y = 6.67 + 2.22 = 8.89 grams ==> multiple extractions with smaller amounts of solvent are more efficient than a single extraction with the same total amount of solvent. Solve the following problems Q1. The vapor pressure of benzene is 95.1 torr and that of toluene is 28.4 torr at 25.0 °C Calculate the vapor pressure of the ideal binary solution that contains 74.0 g of benzene in 48.8 g of toluene at the same temperature. (C=12, H=1). Q2. Given that the partition coefficient between ether and water for compound A is 8.5. What mass of A would be extracted from a solution of 10g of compound A in 100g ml of water by: (i) a single extraction using 100ml of ether (ii) two extractions of 50 ml of ether (iii) 4 extractions of 25ml of ether.
