Study Guide-Chem 233
Study Guide-Chem 233 233
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Chemistry 233: Exam 1 IR and primer and Labs 16 Multiple Choice: 20 points: Prelab Quizzes Short answer: 80 points: Homework (general theory, as well as mathematical problems). Questions about the procedure (general procedure and why you did certain steps at certain times) Study Question, Prelab slides, Textbook to review the chemical theory of IR and labs 16) Theory analysis and procedure Lab 1: Characterization of the organic solids and liquids by elemental analysis, melting point, Boiling point, and Infrared Spectroscopy (identification of unknown organic compounds by MP, BP, and infrared spectroscopy) Procedure: Unknown organic liquid/solid 1. Molecular formula: Determining Molecular formula from elemental analysis and Molecular Formula. 2. Index of Hydrogen Deficiency: Determine the number of possible combinations of double bonds, triple bonds, and rings) (note: IHD>4 suggest benzene ring, Degrees of unsaturation, Saturation: Maximum number of hydrogens IHD: 2Carbon+2 –HydrogenNitrogen+ Halogens 2 Ignore oxygen!! 1 Halogens: F, Cl, Br, 3. Infrared Spectroscopy: Identification of functional groups present based on vibrational frequencies Is affected by… Hybridization Resonance Signal intensity: Strong or weak, polarity) Shape (broad or narrow, hydrogen bonding) Amines (two peaks) PEAKS: • Double bonds: 1600–1850 cm1 Sp3:2900 ` • Triple bonds: 2100–2300 cm 1 Sp2:3100 • XH bonds: 2700–4000 cm 1 Sp: 3300 In IR spectroscopy, the absorption of radiation corresponds to transitions among different molecular vibrationalrotational levels within the same electronic state, normally the ground state, of the molecule; these excitations require energies of 1–36 kcal/mol. The data from IR spectroscopy are most useful for determining the presence or absence of functional groups in a molecule. For instance, examining the appropriate regions of an IR spectrum will show whether or not carbon carbon multiple bonds, aromatic rings, carbonyl groups, or hydroxyl groups are present. This technique does not give quantitative information regarding the elemental composition of a compound, nor does it allow assignment of an exact structure to an unknown compound unless the IR spectrum of the unknown is shown to be identical to that of a known compound. 4. Boiling Point (liquids): When the equilibrium vapor pressure of the liquid=the atmospheric pressure (P =P atm Miniscale procedure: An accurate boiling point may be determined with as little as 0.5–1.0 mL of liquid using the method. Working at a hood, place the liquid in a long, narrow Pyrex test tube, and add a small, black carborundum boiling stone; do not use a white marble chip, as bumping is more likely. Clamp the test tube and position the thermometer about 2 cm above the level of the liquid using a second clamp. Bring the liquid rapidly to a vigorous boil using a suitable heating device suggested by your instructor. You should see a reflux ring move up the test tube, and drops of the liquid should condense on the walls of the test tube. Control the amount of heating so the liquid does not boil out of the test tube. Be sure the bulb of the thermometer is fully immersed in the vapor of the boiling liquid long enough to allow the equilibration required for a good temperature reading to be obtained. Record temperatures of the vapors. Cool to below the boiling point, and repeat two more times to get an accurate result. 2 B.P are the least accurate of the technique, so performing multiple measurement will increase your likelihood of getting accurate result Melting point (solids): Many organic compounds undergo a change in crystalline structure just before melting, perhaps as a consequence of release of the solvent of crystallization. The solid takes on a softer, “wet” appearance, which may also be accompanied by shrinkage of the sample in the capillary tube. These changes in the sample should not be interpreted as the beginning of the melting process. Wait for the first tiny drop of liquid to appear. Melting usually occurs over a range of a degree, perhaps slightly more. Accordingly, a meltingpoint range of a compound is typically reported with the lower temperature being that at which the first tiny drop of liquid appears and the higher temperature is that at which the solid has completely melted. Thus, for our purposes, the start of melting is defined as the temperature at which the first tiny droplet of liquid can be detected The presence of an impurity generally decreases the melting point of a pure solid. If recrystallizing a sample changes an originally broad melting range to a narrow one, the reasonable conclusion is that the recrystallization was successful in purifying the solid. Should the meltingpoint range remain broad after recrystallization, the sample may be contaminated with solvent and additional dry Heating the apparatus too quickly also leads to depress readings o In order to maximize time, take one quick measurement by heating the instrumo t ~12 C until you determine the approximate m.p. Then cool the instrument to ~1015 C below that point and take a slower measurement by heating the instrument ~12 C until you get an exact reading To be lower than the correct melting point: Any of the following: heating the sample too quickly (lag between real temperature and what thermometer reads), using an uncalibrated thermometer, measuring a sample with an impurity (usually, see b for exceptions), measuring a wet sample. To be higher than the correct melting point: Any of the following: using too large a sample size, packing a sample too loosely in the capillary tube, using an uncalibrated thermometer, measuring a sample with significant quantities of an impurity with much higher melting point than the sample itself. To be broad in range: Any of the following: measuring an impure sample, measuring a wet sample, heating the sample too quickly, using too large a sample size, measuring a sample made up of large crystals. Lab 2: Chromatographic Methods: Separation of Dyes and Spinach Pigments by Column and TLC. Column Chromatography: is used to separate of two or more compounds. It takes advantages of the different in polarity (separation of charge or partial charge) between these substances. Compound that are more polar will have a greater affinity (be more attracted) to polar stationary 3 phases such the silica gel. What is needed. Mobile Phase: Liquid (solvent) Stationary phase: Packed solid (silica gel) (absorbent) Solid support: none Procedure “Dry packing” (cotton, silica gel, sand….) 24 drops of methylene bluemethyl orange mixture on the silica baseline (Stationary Phase, SiO 2 O2 Add solvent of choice (note each dielectric constant) Partition Coefficient (K )c K c Solute in stationary phase Solute in Mobile Phase Larger K c solute has greater affinity to stationary phase and smaller affinity for mobile phase. Solutes with larger values move more slowly because they spend more time with the stationary phase, which are nor moving compare with smaller K c Solute with larger difference in K easier to separate. c Smaller K =csolute has smaller affinity for stationary phase and greater affinity for mobile phase. Polarity and K vac es Polarity is the major factor affecting Kc values in column and thin layer chromatographys Separating of charge, as in a molecular dipole. The polarity of solvents is often measured by their dielectric constants. Higher dielectric constant: more polar “polar attracts polar”, “nonpolar attracts polar” Polar molecule: (larger molecular dipole) Nonpolar: (small molecule dipole) Relative polarities of functional Groups Carboxylic acid Alcohol Amine 4 Ketone Decreasing polarity Ester Ether Alkane halide Decreasing retention time on a column chromatography (shorter time because of the solvent is more nonpolar and attract more to the mobile phase than the polar stationary phase) Less polar: Smaller K c Shorter Retention Time More polar: Larger K :cLonger Retention Time Increasing distanced traveled by Solute Increasing Retention Factor (R )fon TLC plate (Nonpolar attracted to mobile phase) since stationary is polar Mobile phase: Liquid (acetyl acetate and hexane) Stationary Phase: Bound solid (silica gel) Solid support: thin aluminum or glass sheet R:f Distance traveled by substances Distance traveled by solvent If the substance was polar it will be attracted more to the stationary phase and will not move far. If the substance was nonpolar it will be attracted more to the mobile phase and traveled farther up the TLC plate The mobile phase rises up the TLC plate by capillary action Decreasing affinity for the stationary phase phase (decreasing K ) c Chromatography is the process of separating mixtures of two or more compounds based on their relative affinities to two immiscible phases: Stationary and mobile phases Elute power of some liquid*Mobile phase Eluent: The sample or mobile phase Eluting power: the ability of a solvent to move through a stationary phase of a chromatography device Increasing eluting power with polar stationary phase (water, methanol, ethanol…) Increasing eluting power with nonpolar phase (hexane, ether, toluene...) Silica Gel Porous solid 5 Slightly acidic Polar stationary phase Can be made nonpolar replacing H Use as a desiccant and food preservative Thinlayer Chromatography: is based on mechanism identical in column chromatography described previously. Stationary phase is bound as a thin layer solid support of aluminum or glass. The amount of absorbent is significantly less. Mostly use to analyze the progress of a reaction by comparing a reactant spot with a spot from the reaction mixture. Procedure: Extraction Grind two fresh spinach leaves Add 5 mL of methylene chloride to the mixture and mix gently Filter the mixture using vacuum filtration Pour the filtrate into a separatory funnel Add 5 ml of water to the solution in the separatory funnel and swirl gently. Vigoursly swirling or shaking will form an emulsion. Carefully drain the organic layer (bottom layer) from the separatory funnel into a clean flask. Discard the remaining aqueous layer and then transfer the organic layer back into the separatory funnel. Wash the remaining organic layer with 2 times 5 ml of water by repeating drainage of the remaining organic material Add the organic layer to flask Add small quantities of Na 2O 4o the solution until the solid no longer clump together. Efficient drying of the solution is paramount to obtaining accurate TLC results. Decant solution into a clean test tube Concentrate the by evaporating approximately half of the methylene chloride in the fume hood TLC Obtain TLC plate. Draw a dot approximately 1cm above the bottom. Above the level of the mobile phase in the beaker Immersed the open end of a melting point capillary into the extract. With a quick motion apply the extract to the TLC plate on the spot drawn and allow the solvent to evaporate from the plate. Spot needs to be concentrate. (the smaller the spot size applied the better the separation Place the TLC into the beaker cover the beaker with watch glass. (This will prevent the 6 mobile phase from changing composition during elution and also assure that the atmosphere surrounding TLC plate is saturated with solvent vapors, which is essential for proper elution). Remove the TLC from the beaker when the eluent is 0.51.0 cm from the plate Repeat each step for each of the mobile phase Pigments isolate from spinach leaves fall into three main structural categories: Chlorophyll: Color appearance green (A, B) B: additional aldehyde group (makes it slightly more polar) Pheophytins; are conjugate acids of chlorophylls. Appearance dull gray (A, B) Slightly less polar Readily formed in acidic solution and presence of enzymes Process of dematalation dull bright green vegetables when cooked Pheophytin a, is form by chlorophyll a, since grinding spinach leaves in a mortar allows the chlorophylls to mix with cell enzymes that would otherwise be separated. Water present in the spinach leaves act as a sufficient proton source. Chlorophyll b, is less demetalated; therefore, pheophytin b may not be observed B is slightly more polar because of the aldehyde present Carotenoids: yellowtoorange appearance Eyesight Carrots, photosynthesis Animal cannot synthesize carotenoids, alkene conjugation, efficient free radical’s scavengers, radicals are delocalized, and have been implicated as cancerreducing agents. Carotenes: Nonpolar and water insoluble, highly aliphatic Lycopene Beta carotene Xanthophyll’s: Oxygenated forms of carotene and are significantly more polar Lutein Zeaxanthin Both have bright yellow appearance Thinlayer chromatography Column Chromatography 7 Lab 3: Separation of Liquids by simple distillation and analysis by gas chromatography Simple distillation allows separation of distillates from lessvolatile substances that remain as pot residue at the completion of the distillation. In the ideal case, only a single component of the mixture will be volatile, so the distillate will be a pure com pound. Procedure simple distillation allows isolation of the various components of the mixture in acceptable purity if the difference between the boiling points of each pure substance is greater than 40–50 °C. Build the apparatus Thermometer should be slightly below the entrance to the condenser to ensure the end is immersed in the vapor Distill 30ml of a 1:1 mixture of EtOAc/BuOAc at a rate of approximately 1 drop per second Record temperature Collect Three fraction: Keep them cover with a wash glass after collecting to prevent the composition to change due to evaporation. Fraction 1: The still head temperature remains close to the boiling point of ethyl acetate Fraction 2: The stull head temperature begins to increase. This may be rapid or gradual depending on your rate of heating. Fraction 3: The still head temperature stabilizes at or near the boiling point of butyl 8 acetate Finally, to check the effectiveness of the separation using GC analysis, determine the molar percentage of each of the fraction May be affected by impurities Equilibrium vapor is directly proportional to temperature: Particles to escape from liquid or solid phase in a close system B.P = when equilibrium vapor pressure of liquid= the atmospheric pressure B.P= rapid/spontaneous vaporization throughout the liquid (bubbles) Compounds with higher vapor pressures have lower Boiling points Compounds with lower vapor pressures have higher Boiling points More volatile liquids have lower boiling points Less volatile has higher boiling points Volatile: readily vaporizes at low temperatures: high relatively equilibrium vapor pressure at relatively low temperature Boiling point of Mixtures: Atmospheric: 760 torr or 1 atm Bp= P =tot atm Daltons law: total pressure above a liquid (P )=sum of partial pressure (P ) of tot x each component in the mixture P tot a+ P b+ P c.. Raoults Law: partial pressure (P ) xs the mole fraction (N ) xultiplied by the equilibrium vapor pressure P = N *P o x x Mole fraction: N x= N /x N +x ) y As T increases, the P aA P inb eases until P eqtot the atmospheric pressureat which point the mixture boils Intermolecular forces affect boiling points Exist between neighboring atoms They are nonbonding forces compared to intra Weakest attraction force Influence physical properties Primarily influence by bond dipole and molecular dipoles Types of intermolecular forces Iondipole (positive or negative) Hydrogen Bonding (OH(F, N,O) 9 Dipoledipole Ioninduced dipole DipoleInduced dipole Induced dipoleinduce dipole (London dispersion) Ethyl acetate and Butyl acetate study in this lab have different boiling points due to London Dispersion Forces London dispersion: More atoms More electrons More induced dipole More attractive forces Lower equilibrium vapor pressure = higher boiling point Temperature Composition diagrams During condensation, Vapor condenses, no change in composition (solid arrows), the composition of the liquid is the same as the original vapor. From gas to liquid. Filter into the cylinder During Vaporization, the composition of the more volatile (higher vapor pressure, lower boiling point) component increases: form liquid to gas. Likewise, the composition of the less volatile (lower vapor pressure and higher boiling temperature) component decrease. Condensing and vaporization continues, the vapor becomes more more and more concentrated in the more volatile component. Each condensation step represents one theoretical plate of separation. More theoretical plates the better. Theoretical plates depend on the surface area available for vaporization/condensation and on the height of the column Fractional distillation provided more theoretical plates than simple distillation because a hempel column filled with Rashings to increase the surface area and distance through which the vaporization/condensation steps can occur 10 Gas Chromatography: Kc: (A)s m(A) Mobile phase: Helium nonpolar Stationary phase: Carbowax Solid support: Typically, diatomaceous earth (celite: diatoms) Packed column example 11 Comparison Column and thinlayer chromatography: K c Depends primarily on polarity of solute, mobile phase and stationary phase Gasliquid Chromatography: K cdepends on polarity of solutes and stationary phase Mobile phase is always helium or H 2 K calso depends on equilibrium vapor pressure More volatile solute= lower b.p=Higher vapor pressure= larger (A) m =smaller K cshorter retention time (ethyl acetate is more attracted to the mobile phase since its less polar than butyl acetate , first to distillate. Analysis of GLC: If polarity are very similar than they have similar dielectric constants. Therefore, polarity cannot be the determining factor Greater molecular weight means greater London dispersion = lower vapor pressure= higher boiling point (less volatile) Lower vapor pressure = lower (A) =lmrger K =movc more slowly through the column longer retention time Higher vapor pressure= lower boiling point = smaller K valc s = shorter retention time Reduce retention time Reduce the amount of the stationary phase Increase the carrier gas flowrate Reduce the polarity of the stationary phase Increase the column temperatur 12 Thermal Conductivity: the ability of a substance to conduct/transfer heat Electric current is pass through the filament Solute passes through the cell with the mobile phase Because of the solutes thermal conductivity, the temperature of the filaments decrease A change in filament temperature causes a change in resistance of the filament The resistance change is measure electronically Increasing concentration of solute= larger change in resistance = larger detector response (area) Determining Mole % from GLC plate: Correction factor because slightly different thermal conductivity Lab 4: Separation of liquids by fractional distillation and analysis by gas chromatography Fraction distillation similar to simple but has a smaller temperature range. Materials may differ Hempel column packed with Raschig rings, The most important requirements for performing a successful fractional distillation are (a) intimate and extensive contact between the liquid and the vapor phases in the column (b) maintenance of the proper temperature gradient along the column 13 (c) sufficient length of the column (d) sufficient difference in the boiling points of the components of the liquid mixture. Each of these factors is considered here. Pippete bulbs: seal air inside, provided insolation to keep temperature gradient Raschigs rings: increase the surface area inside of the column, which increase the number of theoretical plates, allowing better separation Copper wire may be use to prevent rashing rings from following into still plot Use boiling points if necessary Still pot no bigger than 100 ml Distillation of a typical binary mixture. The head temperature should first rise to the normal boiling point of the more volatile component and remain there until that component is mostly removed. The head temperature may then drop somewhat, indicating that the more volatile component has largely been removed. As additional heat is provided to the still pot, the less volatile component will begin to distill, and the head temperature will rise to the boiling point of the second component. If the separation is efficient, the volume of this fraction, which contains a mixture of the two components, will be small. The head temperature should then remain constant at the normal boiling point of the less volatile component until most of it has distilled Temperature Gradient Easily establish as the column is insulated Many vaporization/condensation cycles take place inside the hempel column. The vapor phase becomes more and more concentrated in the more volatile component with each cycle, since the more volatile has o greater vapor pressure and since the Nx increases with each cycle (Raoult law: P= P N x Theoretical plates: Factors affecting, as mention again Temperature gradient Surface area Column length Rate of gas flow HEPT: Height to a theoretical plate Measure of the efficiency of a fractionating column or GC column Lower HEPT: less height needed for 1 theoretical plate = more efficient column HETP: Column height or length (L) Theoretical plate (N) 14 Simple distillation and Fractional Distillation Graphs Lab 5: Steam Distillation of (S) (+) –carvone from caraway seeds and (R) () carvone form spearmint leaves. Steam distillation is a mild method for separating and purifying volatile liquid or solid organic compounds that are immiscible or insoluble in water. This technique is not applicable to substances that react with water, decompose on prolonged contact with steam or hot water, or have a vapor pressure of less than about 5 torr at 100 °C. Advantages and Requirement of steam distillation o Distill liquids with high boiling points at temperature < 100 C Avoid high temp simple /fractional distillation, which could cause decomposition. Nonreactive with H 2 Immiscible with H 2 Does not decompose at 100 Co o Vapor pressure is greater than or equal to 5 torr at 100 C Procedure Steam distillation Weigh 5g of (caraway seeds or spearmint leaves) Transfer it to a flask Assemble the steam distillation apparatus 15 Steam is generated by heating in situ Add 150 ml of hot water in your flask sample Also add water you separatory funnel (dropping funnel) Heat the flask using a heating mantle and place a graduated cylinder below the vacuum adapter in which to collect the distillate Adjust the heating mantle to produce distillation at the maximum rate possible without splashing or foaming the condenser When the rate becomes steady, open the stopcock of the separatory funnel to admit replacement of water at the rate it distills out the flask (so that the volume of water in the flask remains nearly constant) Collect 75 ml. Record the temperature at every 10 ml interval. Distillate will be cloudy because carvone and water are immiscible. Extraction: Using the separatory funnel, extract the distillate 3 times with 7 ml of methylene chloride. Carry the extraction by inverting the stopped separatory funnel 10 times. Do not shake vigorously a stable emulsion will form which layers separate very slowly. Since methylene chloride and water are immiscible you will notice two layers The less dense layer is on top (H2O) Like dissolves like, there carvone (organic) is soluble in C2 C2 (organic): Bottom layer Combine the methylene Chloride extract in a flask Add anhydrous sodium sulfate (Na SO )2and4swirl to remove water. The exact amount added depends on how much water is present Remove the drying agent by gravity filtration in a beaker Spot your TLC plate using the dried solution of carvone in methylene chloride Remove most of your solvent using the heat mantle set up in the hood. Leave only 0.5 ml of solvent. The exact amount of methylene chloride left will depend on how much carvone you isolated. (It has an oily appearance) Characterization: The resulting mixture should be a clear sample of the essential oil dissolved in a small sample of methylene chloride (dichloromethane) Obtain the IR spectrum Prepare a TLC plate containing the isolated oil and authentic sample of (R)carvone. Do a cospot. Put the TLC plate in a solvent system containing ethyl acetate and hexane. You may use the UV lamps to analyze your TLC plate. Calculate the Rfvalue of the oil and compare it to the sample of pure carvone. What the major compound is in your isolated oil? Carvone Baeyer Test 16 Final TLC plate, circle all the circles seen in the UV light . Then exposed the plate to the KMnO solutio4 provided Next heat the plate with a heat gun Record the results Principle: Immiscible component (X) in a heterogeneous mixture with H O 2 Raoults law revised P x= P xo The partial pressure P ofi ach component i of a mixture of immiscible, volatile substances at a given temperature is equal to the vapor pressure P of the pure i compound at the same temperature and does not depend on the mole fraction of the compound in the mixture. In other words, each component of the mixture vaporizes independently of the others. Since x is not soluble in water, P dox not depend on its mole fraction in the mixture. This relationship applies to each component in mixture, including water. Dalotn law stil applies: P : P +P o t x H2O P t P x+ P H20 B.P (mixture): P =TP atm The total vapor pressure is always higher than the most volatile component (Higher vapor pressure, lowest boiling point) which is always water in steam distillation. The B.P of the mixture is always lower than that of the lowest boiling point component againalways water (100 C) in steam distillation (The presence of the second liquid causes both to boil at a temperature lower than the regular boiling points of either liquid) Because the two are immiscible, they will steam distill at a temperature around 90°C, less than 100°C (the boiling point of water) and well below 218 °C. In all steam distillations, the distillate collected will be a mixture of both liquids, but since they are immiscible, they can generally be separated easily via extraction. Analysis of the Isolated oil Thinlayer chromatography Compared R vafues of isolated oil with authentic sample of carvone Find a good solvent (ethyl acetate and hexane) Use a cospot 17 Are the R falues of the isolated and authentic sample of carvone the same or different. What does the result Imply? Indeed, it is carvone being isolated from the seeds or spearmint leave Is the oil mainly one component or a mixture of the component? (mainly a one component) Infrared Spectroscopy Obtain IR of isolated oil. Compare the IR spectra with the one of carvone If identify (Alkene and Ketone) Smell: Does your oil smell like caraway (s)carvone or spearmint (R)carvone. Receptor in the nose can’t tell the difference, IR and TLC cannot tell you which enantiomer of carvone you isolated only your nose Our noses are chiral (receptors for smell) , they can detect the stereochemical difference in the odors, therefore interact with each enantiomer differently Optical rotation can differentiate (can plane polarizelight) between enantiomers. IR of Carvone Enantiomer Enantiomer have the same physical properties including IR vibrational frequencies Melting point Boiling point Retention factor Thinlayer Gaschromatography Color Viscosity Densities These are compounds have an asymmetric carbon or chiral center and are nonsuperimposable mirror images Determine chiral center use Cahningold Prelog based on atomic number (hydrogen should be in the back and is given #4 (R)clockwise (S)counterclockwise Be sure you can draw enantiomers Properties of enantiomer: 18 How they interact with plane polarized light Rotate the planepolarize light Chiral molecules are optically active Enantiomer will rotate the plane of the light the same magnitude but in opposite direction. Differ in biological properties, such as smell and pharmaceutical action. For pharmaceutical, slight difference in 3D spatial arrangement can make the difference between targeted treatment and undesired side effects Our enzyme and receptor are stereospecific (Chiral) Stereoisomers of carvone: Two monoterpenes (10 carbon) found in spearmint leaves and caraway seed often used as flavors. 2 isoprene carbon fragments found in carvone Lab 6: Base Extraction of benzoic acid from acetalide recrystallization of products Melting point Benzoic acid: 121123 Acetanilide: 111115 19 Recrystallization of solids is a valuable technique to master, because it is one of the most common methods used to purify solids General Principle: Solvent must me properly selected Dissolve a moderate quantity of the substance to be purified at an elevated temperature, but only a small quantity at low temperatures Does not react with the substance to be purified Dissolve impurities readily at low temperature or not dissolve them at all Be easily separated from the purified product, low boiling point and evaporate readily Extraction Procedure: 1 g of binary mixture acetanilide (neutral) and benzoic acid (acidic) Transfer to a beaker dissolve the sample in 10 mL of methylene chloride Next, transfer solution to a separatory funnel Then, add 3 M NaOH (each 3 mL) creates two layer Do this 2 times for 2 portions, Do it one at a time Each time extract the Organic layer that is located in the bottom of the flask Next, combine all portions for the aqueous layer in separated beaker and the organic layers as well Then, for the aqueous extract add 2 X 10 mL methylene chloride to ensure removal of acetanilide suspended or dissolved in water. Again do it one at a time At this point your two compounds are separated (Acetanilide in the organic layer and sodium benzoate in the aqueous layer) CRYSTILLIZATION Acetanilide (Methylene chloride): Crude Acetanilide combine all methylene chloride extracts and remove residual water by drying with Anhydrous sodium sulfate. Then gravity filter into a clean flask Then gently boil the methylene chloride until it is all gone. You will be left with crude acetanilide Do not heat vigorously it will cause you product and turn brown Sodium Benzoate (conjugate base of benzoic acid): Crude benzoic acid 20 Place the aqueous solution in a beaker Cool it with ice bath Then, neutralize it with 3M HCl Be sure all the benzoic acid precipitates out of the solution Vacuum filter the solid using a Buchner funnel Then wash the collected sample with a few milliliters of cold water. Recrystallization Benzoic acid Weigh the benzoic acid Mix the sample with 10 mL of boiling water per gram of sample and bring to gentle boil (0.6 g = 6 mL) Continue to add hot water to the boiling solution until the solid dissolves. 1 g of benzoic o acid in about 15 ml of water at 95 C. o Allow the filtrate to cool to room temperature and then to 0 C in an ice bath. Collect the crystals of benzoic acid by suction filtration. Allow air to aspirate through the crystals for 5 minutes or more to ensure they are dry. Weigh and obtain Melting point and IR spectrum. Calculate the percent yield obtained Acetanilide Weight the sample Mix sample with boiling water using the same procedure as with benzoic acid. 1 g of acetanilide will dissolve in about 20 mL of water at 95 C. Adjust the water base on sample Same procedure as above… Collect the crystals of acetanilide by suction filtration. Allow air to aspirate through the crystal fro 5 minutes or more to ensure they are dry Weigh and obtain melting point and IR spectrum. Calculate percent yield 21 Three Readily ionizable functional groups are separated by acid or base extraction Hydroxide act as a base: depronates Hydronium act as an acid: Pronates 22 Lewis Theory: The Lewis theory of acids and bases states that acids act as electron pair acceptors and bases act as electron pair donors. PKA Values PKA values: The compound with the lowest value is more acidic PKA values can be use to compare basicity: Pronate each of the bases and then look at the PKA values and compare The stronger the acid the weaker base. Effects stability of the conjugate base The equilibrium will always favor formation of the weaker acid (higher pa value) Atom. Which atom bears the charge? (How do the atoms compare in terms of electronegativity and size? Remember the difference between comparing atoms in the same row vs. atoms in the same column.) Resonance. Are there any resonance effects that make one conjugate base more stable than the other? Induction. Are there any inductive effects that stabilize one of the conjugate bases? Orbital. In what orbital do we find the negative charge for each conjugate base? Solubility: Like dissolves like The purpose of adding an acid or a base is to change the solubility of one compound to that it can be separated from the other 23 Choosing an effective extracting acid or base: Liquidliquid extraction is one of the most common methods for removing an organic compound from a mixture. Filtration is the primary technique used to separate solids from liquids. It is important to perform filtrations properly to avoid loss or contamination of your product, regardless of whether it is a solid or liquid. Gravity filtration is the filtering technique most commonly used to remove solids such as impurities, decolorizing carbon (Sec. 2.18), or drying agents (Sec. 2.24) from liquids prior to crystallization, evaporation, or distillation Vacuum filtration is a technique for collecting crystalline solids from solvents after recrystallization or precipitation. A typical apparatus is shown in Figure 2.54. Either a Büchner funnel or a smaller Hirsch funnel is used, with the latter being better suited for isolating quantities of solid ranging from 100 to 500 mg For many experiments, it is necessary to remove the excess solvents to recover the product Recrystallization of solids is a valuable technique to master, because it is one of the most common methods used to purify solid 24 The process of recrystallization involves dissolving the solid in an appropriate solvent at an elevated temperature and allowing the crystals to reform on cooling, so that any impurities remain in solution. The choice of solvent is perhaps the most critical step in the process of recrystallization, since the correct solvent must be selected to form a product of high purity and in good recovery or yield. The solvents commonly used in recrystallizations range widely in polarity, a property measured by the dielectric constants , listed in Table 3.1. Those solvents with dielectric constants in the range of 2–3 are considered nonpolar and those with constants above 10 as polar. Occasionally a mixture of solvents is required for satisfactory recrystallization of a solute. The mixture comprises only two solvents; one of these dissolves the solute even when cold and the other one does not. The logic of this will become clear when you read about this technique under “Dissolution.” Several small extractions are more effective than one large extraction (same volume): 25 Requirement of an extraction solvent:( ACID OR BASE) Does not react irreversibly with the solute extracted Immiscible with original solvent Selectively removes the desire compound (larger K>1 for compound to be extracted and small K for the rest) Easily separated from the solute crystallization/precipitation then filtration; or distillation REQUIREMENT FOR Recrystallization SOLVENT: 26 27
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