CHE 102 Organic Chemistry Lecture Notes PDF

GENERAL CHEMISTRY II CHE 102: SYNTHESIS OF ORGANIC COMPOUNDS Originally, all organic compounds were isolated from natural sources. However, as chemists developed an understanding of the chemical behavior of organic compounds, they began to devise methods of synthesizing compounds from simple starti...

GENERAL CHEMISTRY II CHE 102: SYNTHESIS OF ORGANIC COMPOUNDS Originally, all organic compounds were isolated from natural sources. However, as chemists developed an understanding of the chemical behavior of organic compounds, they began to devise methods of synthesizing compounds from simple starting materials. Moreover, some of the compounds synthesized by the newly developed methods had never been observed to occur naturally. In organic synthesis, chemists attempt to transform simple, readily available compounds into more complex molecules with desirable physical and chemical properties. Some syntheses are designed to make biologically active compounds that are otherwise available only from natural sources and in only small quantities at high cost. Other synthetic approaches are designed to make new compounds, similar to naturally occurring ones but even more powerful in biological activity, such as medications to fight disease. The approach that a synthetic organic chemist takes is to apply knowledge of a wide variety of reaction types and reaction mechanisms to devise a synthetic scheme for assembling simple molecules into more complex structures. Running an organic reaction is usually the easiest part of a synthesis. The challenge lies in isolating and purifying the product from the reaction because organic reactions seldom give quantitative yields of one pure substance. To isolate the pure organic compound obtained either from synthesis or from natural sources, the organic chemist must employ one or a combination of purification methods. PURIFICATION TECHNIQUES FOR ORGANIC COMPOUNDS INTRODUCTION: Organic compounds are isolated either from natural sources or from reactions mixtures. These compounds are seldom pure and are usually contaminated with small amounts of other similar compounds, which are found to exist together or formed during the reaction. In order to characterize them, it is important to purify them. PURIFICATION TECHNIQUES FOR SOLID There are four important purification techniques for solid organic compounds: 1. Crystallization 2. Fractional Crystallization 3. Sublimation and 4. Solvent Extraction 1. Crystallization: - When isolated from organic reactions, solid organic compounds are commonly contaminated with impurities - Purification can be effected by crystallization by using a suitable solvent or mixture of solvents The Purification of solids by crystallization is based on two facts: 1  a. Most solids are more soluble in hot solvents than in cold solvents b. The solubilities of different compounds in a given solvent are different The process of the crystallization technique consists of seven discrete steps; 1. Choosing the solvent and or solvent pairs (or mixture of solvents) 2. Dissolving the solute (or the impure or crude organic sample) in the chosen solvent or solvent mixtures at or near the boiling point 3. Decolorizing the solution by simply boiling the solution with activated charcoal 4. Filtering the hot solution from suspended particles such as insoluble materials and dust 5. Crystallizing the solute by allowing the hot solution to cool to room temperature or below, causing the dissolved organic substance to crystallize out 6. Collecting and washing the crystals – that is, separating the crystals from the supernatant solution (or “mother liquor”) by filtration usually under suction 7. Drying the product and testing for purity by using a melting point determination The most important step in the crystallization procedure is the choice of a suitable solvent or use of a suitable solvent mixture. The most desirable characteristics of a solvent for recrystallization are a. It should have a high solvent power for the substance to be purified at elevated temperatures and a comparatively low solvent power at room temperature or below b. It should dissolve the impurities readily or not at all c. It should yield well-formed crystals of the purified compound d. It should have a relatively low boiling point so that it can be easily driven off from the crystals, after the crystals are separated from the mother liquor e. It should not react chemically with the substance to be purified Dielectric constant is a general index for the polarity of a solvent. Since like dissolves like, polar solvent will dissolve polar and ionic compounds; whereas, non-polar solvents will dissolve non- polar solutes. Some common solvents for crystallization are given below in order of decreasing polarity: # Solvent Boiling point oC Dielectric constant at 25oC 1 Water 100 79 2 Acetonitrile 80 38 3 Methanol 65 33 4 Ethanol 78 24 5 Acetone 56 21 6 Methylene chloride 41 9 7 Acetic acid 118 6 8 Chloroform 61 5 9 Diethyl ether 35 4 10 Benzene 80 2.3 11 Dioxane 101 2.2 2 12 Carbon tetrachloride 77 2.2 13 Petroleum ether 30-60 2 60-90 2 90-120 2 14 Cyclohexane 81 2.0 15 Pentane 36 1.8 2. Fractional Crystallization This technique is used when it is clearly known that a solid is a mixture of two substances and a separation of the two can be carried out by fractional crystallization from a solvent in which both are soluble but to different degrees. By using the chosen solvent obtained by trial, a process of dissolution and crystallization is repeated a number of times and in later stages, the mother liquor of the previous filtrationcan be used as the solvent to prevent loss. Since one substance is more soluble than the other, the more soluble ones is separated into solution and the less soluble obtained by filtration. The relative solubilities of the two substances and the proportions in which they are present, are the deciding factors as to which of the two will separate first. 3. Sublimation This technique is used when an organic solid has a high vapor pressure particularly at a temperature below its melting point. The solid is warmed at a temperature below its melting point and then condensing the vapors on the cold surface. It is used to separate volatile solids, which pass directly into vapour state on heating from a non-volatile solid. It can be used for a mixture of solid substances, such as camphor, benzoic acid, ammonium chloride, iodine etc., containing non-volatile substances, when heated, and the volatile solid change directly into vapour without passing through the liquid state. 4. Solvent Extraction This technique is used to separate the components of organic mixtures based on the distribution of a solute between two immiscible phases or solvents. It is an equilibrium process which is governed by Nernst Distribution Law. If we assume that the solute A, distributes itself between an aqueous and organic phase, the resulting equilibrium will be written as Aaq ↔Aorg Where the subscripts aq = aqueous phase and org = organic phase According to the Nernst distribution law, the ratio of the concentrations of A in the organic and aqueous phases is constant at a constant temperature and is known as the Distribution Coefficient, K: 3 K = [Aorg] / [Aaq] ------------------------ equation 1 To establish the experimental conditions required to extract the organic solute from one solvent to another, we can apply the distribution law as follows: Consider Vaq mL of an aqueous solution containing a0millimoles of A; then we can extract this with Vorg mL of an immiscible organic solvent. At equilibrium, a1millimoles of A will remain in the aqueous layer. Since, C = n/V (L), then, a1 = [Aaq]1* [Vaq]; that is, [Aaq]1 = a1 / [Vaq] Therefore, [Aorg]1 = (a0 – a1) / Vorg Substitute these values in equation 1 as follows: K = {(a0 – a1) / Vorg}* {Vaq / a1}; and this can be rearranged to obtain a1: a1 = {Vaq/(Vorg.K + Vaq)} * a0 ---------------------------------------equation 2 We can compute similarly the number of millimoles, a2, remaining after the second extraction with an identical volume of solvent to be: a2 = {Vaq/(Vorg.K + Vaq) * a1 ---------------------------equation 3 Since we know a1 from equation 2, then we can substitute in equation 3 to obtain a2 = {Vaq / (Vorg.K + Vaq)}2 * a0 After n extractions, the number of millimoles of A remaining in the aqueous solution is: an = {Vaq / (Vorg.K + Vaq)}n * a0 ---------equation 4 Since an =[Aaq]n* Vaq; and a0 =[Aaq]0 * Vaq Where [Aaq]n is the concentration of A in the aqueous phase after n extractions, then equation 4 can be rewritten as [Aaq]n = {Vaq / (Vorg.K + Vaq)}n * [Aaq]0 --------------------------equation 5 So, in a solvent extraction, an efficient extraction is achieved when several small lots of the solvent is used rather than a single large one. Example: 4  The distribution coefficient of I2 between CCl4 and H2O is 85. Calculate the concentration of I2 remaining after extracting 50.0 mL of an aqueous 1.00 x 10-3 M solution of I2 with (a) 50.0 mL CCl4 (b) Two 25.0 mL portions of CCl4, and (c) Five 10.0 mL portions. Answer: Use equation 5, [I2aq]n = {Vaq / (Vorg.K + Vaq)}n * [I2aq]0 (a) [I2aq]1 = {50.0 / (50*85 + 50)}1 * 1.00 * 10-3 = 1.16 * 10-5 M (b) [I2aq]2 = {50.0 / (25*85 + 50)}2 * 1.00 * 10-3 = 5.28 * 10-7 M (c) [I2aq]5 = {50.0 / (10*85 + 50)}5 * 1.00 * 10-3 = 5.29 * 10-10 M These results show that efficient extraction is achieved with several small volumes of solvent rather than a single large one. In the isolation of organic compounds from aqueous solutions use is made of the fact that the solubility of many organic substances in water is considerably decreased by the presence of dissolved inorganic salts. This is called the salting out effect. PURIFICATION TECHNIQUES FOR LIQUIDS Distillation is the most common and most important method for the purification of liquids. We can use distillation to separate a liquid organic from non-volatile impurities. Also, we can use distillation to separate one volatile liquid from another with due modifications of the technique as follows; DISTILLATION is a purification technique used to separate a solution of a solid in a liquid and for separating a solution of two liquids whose boiling points are different. The principle involves the conversion of a liquid into its vapors upon heating and then cooling the vapors back into the liquid. Depending on the difference in boiling points of liquids, we can have the following types of distillation techniques: Simple Distillation; Fractional Distillation; Distillation Under Reduced Pressure or Vacuum Distillation and; Steam Distillation. SIMPLE DISTILLATION: It is used for separating liquids having boiling points differing by 10-20 degrees. The liquid having the lower boiling point distills over first, and the other liquid component is left behind. In this process, vaporization and condensation occur side by side. Example: Simple distillation of a Cyclohexane- Toluene mixtures FRACTIONAL DISTILLATION: It is used for separating two liquids in any mixture, which have boiling points within a narrow range of temperatures. In such cases, simple distillation does not give complete separation and a modified version called fractional distillation is employed. 5 DISTILLATION UNDER REDUCED PRESSURE OR VACUUM DISTILLATION: The lowering of pressure on the surface of a liquid lowers its boiling point. As a result of this, a liquid can be boiled and distilled, without any decomposition, at temperature much below its normal boiling point. The decrease in boiling point with reduction in pressure can be predicted approximately by making use of two rules: (a) The Troutan Rule, which states that, for normal unassociated liquids, the ratio of the molar heat of vaporization in calories to the normal boiling point on the absolute scale is approximately constant and has a value of 21 cal deg-1 mol-1. (b) And (b) Clausius-Clapeyron Equation – if P1 and P2 are the two pressures at which the boiling points are respectively T1 and T2, then, ln (P2/P1) = H/R (1/T1 – 1/T2) where H is the molar heat of vaporization and R is the gas constant Based on these rules, we can obtain the reduced boiling point TR at the reduced pressure P (in mm) from its normal boiling point TN by using the following equation: TR = 10.5 TN/(17.133 – lnP) Example: The normal boiling point of chlorobenzene is 132oC (132 + 273 = 405K). What is the boiling point at a pressure of 25 mm? Answer: TR = 10.5 TN/(17.133 – lnP) TR = 10.5 * 405 / (17.133 – ln 25) TR = 4252.5/13.91 = 306 K or 33oC. A useful general rule is that when the pressure is reduced from 760 mm to 25 mm, the boiling point of high boiling liquid is reduced by about 100oC. STEAM DISTILLATION: This technique is used for separating/purifying liquids, which are immiscible with water, volatile in steam, & have high vapor pressure at the boiling temperature of water. This technique can be used to purify many high boiling liquids which may decompose at or below their normal boiling points. Consider a mixture of two immiscible liquids A and B in equilibrium with vapor at a specified temperature. The total vapor pressure is the sum of the vapor pressures PA and PB of pure A and B respectively: 6 PT = PAo + PBo The mole fraction χA and χB of A and B in the vapor will then be χA = PAo/ PT and χB = PBo/ PT Since: χA = nA / (nA + nB) and χB = nB/ (nA + nB) Where nA and nB are the number of moles of A and B; so, nA / nB = χA / χB = (PAo/ PT) / (PBo/ PT) = PAo / PBo Also; nA = WA/MA and nB = WB/MB; where WA and WB are the weights or masses of A and B while MA and MB are the Molecular weights or masses of A and B. Hence we can have; (WA * MB) / (MA * WB) = PAo / PBo On rearranging the above equation, we have; WA / WB = (MA * PAo) / ((MB * PBo) This equation relates the relative masses of the two substances in the vapor to their molar masses and vapor pressures. If the vapor is condensed, the ratio WA/WB would express the relative masses of A and B in the condensate. Example; Consider the distillation of a mixture of water and bromobenzene, which is almost completely immiscible with water. At 95.3oC, the vapor pressure of water is 641 mm and that of bromobenzene is 119 mm. What is the relative mass ratio of bromobenzene to water in the distillate? The molecular mass of bromobenzene is 157 g/mol and that of water is 18 g/mol. Answer: If we represent bromobenzene as BB and water as H2O, then; Use the equation: WA / WB = (MA * PAo) / ((MB * PBo) WBB / WH2O = (119 * 157) / (641 * 18) = 1.6 Despite the much lower vapor pressure of bromobenzene, on a weight basis it will distill 1.6 times as fast as water – all because of its much greater molecular weight. 7 OTHER SEPARATION AND PURIFICATION METHODS CHEMICAL METHOD: The Chemical method can be adopted if there is a substantial difference in acidity and or basicity between the organic substance to be isolated and the contaminants with which it is associated. Under these circumstances, a rapid and efficient separation can usually be achieved through simple Ether Extraction. A general outline for separating a mixture of ACIDIC (HA), BASIC (B) and NEUTRAL (N) components are given in the scheme below: Separation Scheme of Acidic (HA), Basic (B) and Neutral (N) Substances by an Extraction Procedure. HA + B + N Distribute between ether and excess dilute HCl Ether Aqueous HA + N BH+Cl- Add excess aq. NaOH Add excess aq. NaOH and ether Ether Aqueous N Na+A- Ether Aqueous Add excess dilHCl B Na+Cl- And ether Ether Aqueous HA Na+Cl- The fact that organic salts are distributed almost entirely in the aqueous layer and all other organic substances almost entirely in the ether layer forms the basis for this scheme. The acidic, basic, and neutral components separated in this way can be purified by adopting the usual techniques. CHROMATOGRAPHY: To identify, purify and/or separate constituents of a mixture that are present in very small amounts. The principle behind this technique is the differential adsorption of the various components of a mixture between two different phases that are as follows: Fixed or stationary phase and Mobile or Moving phase. 8 The stationary phase may either be a solid or liquid and the moving phase also called the mobile phase may either be a liquid or gas. There are four categories of chromatography: Type Mobile Stationary Name Method of fixing the stationary phase Phase Phase Liquid Solid Column Held in a tubular column Liquid Solid Ion exchange Finely divided ion-exchange resin held in a tubular column Liquid Liquid Paper Held in the pores of a thick paper Liquid Solid or Thin Layer Finely divided solid held on a glass or Liquid Chromatography plastic plate (TLC) III Gas Solid Gas-solid (GSC) Held in a tubular column IV Gas Liquid Gas-liquid Adsorbed on a porous solid held in a (GLC) tube or adsorbed on the inner surface of a capillary tube We are going to discussed type I and type II techniques as follows: COMMONLY EMPLOYED CHROMATOGRAPHIC TECHNIQUES: Column Chromatography; Paper Chromatography: Ascending and Descending Paper Chromatography; Radial Paper Chromatography; Thin Layer Chromatography. Column chromatography: In column chromatography, the stationary phase is a solid like alumina (Al2 O3), that is used as the adsorbent and the mobile phase is a liquid or the solvent. The separation is based on the fact that a given adsorbent will adsorb to varying extents the components of a given mixture present in solution. If such a solution is poured down a column filled with a powdered adsorbent, the substances that are most readily absorbed are retained in the upper layers of the column. The substances which are less readily absorbed are gradually absorbed as they pass further down the column so that partial separation of the constituents of the mixture has been achieved. If, at this stage, the pure solvent is allowed to flow through the column, the less easily adsorbed substances present in the upper parts pass into solution and are re-adsorbed at lower levels. The result is the formation of a series of almost distinct layers containing individual components in the order of decreasing ease of adsorption down the column. This is known as development of the chromatogram. As more and more solvent is allowed to flow the substances get eluted one after another such that the most strongly adsorbed component is carried down last. The right choice of adsorbent and the solvent is critical for good separation. The main application of column chromatography is the separation of mixtures into pure individual components. It can also be used to a. Purify compounds by the removal of small amounts of contaminants b. Determine the homogeneity of chemical substances c. Compare compounds considered to be identical 9  d. Concentrate materials from dilute solutions such as those obtained when natural products are extracted from plants or trees by extraction with large volumes of organic solvent. ION EXCHANGE CHROMATOGRAPHY: In this chromatography, the stationary phase is a solid or ion exchange resin which consists of highly polymerized cross-linked organic material containing a large number of acidic or basic groups. Although the resins are insoluble in water, the active groups are hydrophilic and have varying degrees of affinity for ionic solutes. There are four types, namely; a. strongly acidic (sulphonatedpolysterene resin) for cation exchange; (b) weakly acidic (carboxylic polymethacrylate resin), also for cation exchange; (c) Strongly basic (quarternary ammonium polystyrene resin for anion exchange; and (d) Weakly basic (polyamine polystyrene or phenol formaldehyde resin) for anion exchange. The mobile phase is a liquid such as water or buffered aqueous solution of a suitable pH.The basis of separation is chemical exchange rather than adsorption. PAPER CHROMATOGRAPHY: In this chromatography, the stationary phase is a liquid (adsorbed water that is always present in filterpaper that is supported by the ~22% cellulose molecules of the papers) and the mobile phase is a liquid solution that is a mixture of water and one or more organic solvents.The basis of separation is the partition of the various components between two liquid phases.In Ascending Paper Chromatography, the mechanism of the upward motion of the solvent is the capillary action on the pores of the paper. In the Descending Paper Chromatography, the paper is anchored in a solvent through the top of the chamber, and the solvent moves downwards. Here the mechanism is gravitational motion assisted by capillary action. For identification purposes, solutes from the original mixtures will have migrated with the paper at different rates forming a series of separated spots – when colored can be determined directly; otherwise, it will be sprayed with suitable locating reagents or indicated by fluorescence in ultraviolet light. Spots are characterized by their retardation factor, Rf, which is defined as Rf = (distance moved by substance)/(distance moved by the solvent front). The Rf value is characteristic of a particular species and can be used for the qualitative identification of the unknown species. Identical Rf values for a known and unknown compound using different solvent systems provides good evidence that the two are identical especially if they are run side by side along the same strip of paper.Spots can also be quantitatively determined by cutting out and washing out by suitable solvents followed by a suitable quantitative technique. Paper chromatography is important in biochemistry with small and complex samples and also in protein chemistry where mixtures of amino acids have to be separated and identified. THIN LAYER CHROMATOGRAPHY: In this chromatography, the basis can be either adsorption or partition depending upon the mode of selection of a stationary phase which can be a solid or liquid (commonly, the stationary phase is a thin layer of an adsorbent coated on Flat glass strip). The solvent (mobile phase) moves up the layer due to the capillary action and thus 10 causes the separation of constituents of the mixture. The constituents are identified by measuring their RF values. Example Separation of vitamins by thin layer chromatography QUALITATIVE ANALYSIS: The systematic qualitative analysis of organic compounds includes the following general scheme of analysis: General Scheme of Analysis A. Preliminary Test Note physical characteristics: solid, liquid, color, and odor. Compounds that are yellow to red in color are often highly conjugated. Amines often have a fish-like odor, while esters usually have a pleasant fruity or floral smell. Acids have a sharp, biting odor. Odors can illicit information about your unknown; it is wise to sniff them with caution. Some compounds can have corrosive vapors or make you feel nauseous. B. Physical Constants Determine the boiling point or melting point. Distillation is recommended in case of liquids. It serves the dual purpose of determining the boiling point as well as purifying the liquid for subsequent tests. C. Solubility Tests The solubility of the unknown in the following reagents provides very useful information. In general, about 1 mL of the solvent is used with approximately 0.1 g or 0.2 mL (2-3 drops) of the unknown compound. Procedure for Determining Solubility of Organic Compounds The amounts of material to use for a solubility test are somewhat flexible. Use 2-3 drops of a liquid or approximately 10 mg of a solid. Unless the solid is already a fine powder, crush a small amount of the solid on a watch glass with the back of a spatula. Do not weigh the solid; simply use enough to cover the tip of a small spatula. 1) Water Solubility Add approximately 6 drops of water to the test tube containing your unknown. Shake the tube and/or stir with a glass stirring rod. A soluble unknown will form a homogeneous solution with water, while an insoluble liquid will remain as a separate phase. A liquid which is soluble in water may be either a low molecular weight polar compound of up to 5 carbon atoms or less. Check the pH of the water to determine if your unknown is partially or completely soluble in water and whether your compound has changed the pH of the water.  pH paper turns red: water soluble acidic compound  pH paper turns blue: water soluble basic compound  pH paper does not change color: water soluble neutral compound or insoluble compound 11 An organic compound which is soluble in water is typically a low molecular weight polar compound of up to 5-6 carbon atoms or less. 2) 5% NaOH Solubility Add approximately 1 mL of 5% NaOH in small portions of about 6 drops each to the test tube containing your unknown. Shake test tube vigorously after the addition of each portion of solvent. Solubility will be indicated by the formation of a homogeneous solution, a color change, or the evolution of gas or heat. If soluble, then your unknown is behaving as an organic acid. The most common organic acids are carboxylic acids and phenols. Carboxylic acids are usually considered stronger acids than phenols, but both of these acids will react with NaOH (a strong base). 3) 5% NaHCO3 Solubility Add approximately 1 mL of 5% NaHCO3 in small portions of about 6 drops each to the test tube containing your unknown. Shake test tube vigorously after the addition of each portion of solvent. Solubility will be indicated by the formation of a homogeneous solution, a color change, or the evolution of gas or heat. If soluble, then it is a strong organic acid. If not, then it is a weak organic acid, if it dissolves in NaOH. The most common weak organic acid are phenols. Typically, only a carboxylic acid will react with NaHCO3. 4) 5% HCl Solubility Add approximately 1 mL of 5% HCl; in small portions of about 6 drops each to the test tube containing your unknown. Shake test tube vigorously after the addition of each portion of solvent. Solubility will be indicated by the formation of a homogeneous solution, a color change, or the evolution of gas or heat. If your compound is HCl-soluble, then it is an organic base. Amines are the most common organic base. If insoluble in all solutions, then your unknown is not an acidic or basic organic compound. D. Group Classification Tests (a) Tests for Aldehydes and Ketones (i) 2,4-DNP Test for Aldehydes and Ketones 12 Aldehyde or Ketone 2,4-dinitrophenylhydrazine reagent. Positive test Formation of a precipitate is a positive test. (ii) Tollen’s Test for Aldehydes Aldehyde Tollens reagent. Positive Test Formation of silver mirror or a black precipitate is a positive test. (iii) Jones (Chromic Acid) Oxidation Test for Aldehydes Aldehydes 13 Jones reagent (chronic acid in sulfuric acid). Positive Test A positive test for aldehydes and primary or secondary alcohols consists in the production of an opaque suspension with a green to blue color. Tertiary alcohols give no visible reaction within 2 seconds, the solution remaining orange in color. Disregard any changes after 15 seconds. (iv) Iodoform Test for Methyl Ketones Ketone Positive Test Formation of solid iodoform (yellow) is a positive test. (Iodoform can be recognized by its odor and yellow color and, more securely, from the melting point 119o-123oC). (b) Tests for Alcohols (i) Jones Oxidation for Primary and Secondary Alcohols Alcohol Jones reagent (chronic acid in sulfuric acid). Positive Test A positive test for aldehydes and primary or secondary alcohols consists in the production of an opaque suspension with a green to blue color. Tertiary alcohols give no visible reaction within 2 seconds, the solution remaining orange in color. Disregard any changes after 15 seconds. (ii) Lucas Test for Secondary and Tertiary Alcohols 14 Alcohols Positive test Appearance of a cloudy second layer or emulsion  3o alcohols: immediate to 2-3 minutes  2o alcohols: 5 -10 minutes  1o alcohols: no reaction (c) Tests for Halides (i) Silver Nitrate in Ethanol Test 1% ethanolic silver nitrate solution (ii) Sodium Iodide in Acetone Test 15% solution of sodium iodide in acetone, Positive Test The formation of a white precipitate indicates the presence of halides. 15  (iii) Beilstein Test Positive Test A green flash is indicative of chlorine, bromine, and iodine, but NOT fluorine. (d) Tests for Unsaturation (i) Bromine (in Methylene Chloride) Test for Multiple Bonds Alkene Alkyne Positive Test Discharging of the bromine color without the evolution of hydrogen bromide gas is a positive test. (ii) Baeyer Test for Multiple Bonds (Potassium Permanganate Solution) Alkene 16 Alkyne Positive Test The disappearance of the KMnO4's purple color and the appearance of a brown suspension of MnO2 is a positive test. (iii) Ignition Test for High Degrees of Unsaturation Aromatic compounds often burn with a smoky flame. Positive Test A sooty yellow flame is an indication of an aromatic ring or other centers of unsaturation. (e) Tests for Carboxylic Acids (i) pH of an Aqueous Solution for Carboxylic Acids See procedure for solubility tests with water. (ii) Sodium Bicarbonate Test for Carboxylic Acids Carboxylic Acid Positive Test Evolution of a carbon dioxide gas is a positive test for the presence of the carboxylic acid and certain phenols listed in the Complications section. (f) Tests for Phenols (i) Iron (III) Chloride Test for Water-Soluble Phenols 17 Phenol 1% aqueous iron (III) chloride solution. Positive Test A red, blue, green, or purple color is a positive test. (ii) Iron(III) Chloride - Pyridine Test for Water-Insoluble Phenols Phenol 1% solution ferric chloride in methylene chloride. Add a drop of pyridine and stir. Positive Test (b) Addition of pyridine and stirring will produce a color if phenols or enols are present. (g) Tests for Nitro Groups (i) Iron (II) Hydroxide Test for Nitro Groups Nitro Compounds Positive Test A positive test is the formation of the red-brown precipitate of iron(III) hydroxide. METHOD OF DETERMINING PURITY After a substance has been isolated and purified, it is necessary to check whether it is really pure. There are several criteria of purity. 18 MELTING POINT DETERMINATION: For solids, the most important criterion is its melting point. A pure solid melts sharply at its melting point to give a clear liquid while an impure solid melts over a range of temperatures to give a semi-solid mass. If a solid melts sharply and the melting point corresponds to the value reported already for the solid, it can generally be assumed that the solid is pure. Sometimes when a substance is suspected to be identical with a known organic compound, it may be further confirmed by determining the so called MIXED MELTING POINT. This depends on the principle that impurities generally lowers the melting point of a compound. To confirm whether a substance suspected to be a particular compound is indeed that particular compound, the substance and the authentic sample of the compound are mixed in about equal quantities and the melting point of the mixture determined. If the melting point is not lowered, then the two substances are taken to be identical. BOILING POINT DETERMINATION: For liquids, the most commonly used criterion is its boiling point. A pure liquid boils at a constant temperature and the entire liquid is converted into its vapor at this temperature which is the boiling point of the liquid. An impure liquid will have more than one boiling point corresponding to more than one compound. REFRACTOMETRY Refractometry is a technique that measures how light is refracted when it passes through a given substance, in this case, an unknown compound. The amount by which the light is refracted determines the refractive index. Refractive index can be used to identify an unknown liquid compound, or it can be used as a means of measuring the purity of a liquid compound by comparing it to literature values. The closer the refractive index is to the literature values, the purer the sample. Refractive index is defined as the ratio of the velocity of light in air to the velocity of light in the medium being measured: nD = [V air]/(V liquid) Refractive index is temperature dependent. To adjust for temperatures greater than 20oC, please use the following formula: nD20 = nDT + 0.00045(T - 20oC) POLARIMETRY Polarimetry measures the extent to which a substance interacts with plane polarized light (light which consists of waves that vibrate only in one plane); whether it rotates plane polarized light to the left, to the right, or not at all. If the substance rotates plane polarized light to the left or to the right, it is called optically active. To be optically active, a compound must have a chiral center. A chiral center is a carbon that has 4 different groups attached to it. Depending on the orientation of these four different groups about the chiral carbon, the compound may rotate plane polarized light to the left or to the right. If a compound does not have a chiral center, it will not rotate light at all. The number of degrees and the direction of rotation are measured to give the observed rotation. The observed rotation must be corrected for the length of the cell used and the solution concentration. Comparing the corrected observed rotation to literature values can aid in the 19 identification of an unknown compound. Use the following equation to correct the observed rotation: [a]lit = [(a)obs]/(l x c) where: a = specific rotation (degrees) (literature value), l = path length (dm), and c = concentration (g/ml). DETERMINING THE STRUCTURE OF AN ORGANIC COMPOUND The analysis of the outcome of a reaction requires that we know the full structure of the products as well as the reactants In the 19th and early 20th centuries, structures were determined by synthesis and chemical degradation that related compounds to each other Physical methods now permit structures to be determined directly. We will examine briefly the use of some of these instrumental methods: MASS SPECTROSCOPY A mass spectroscope measures the exact mass of ions. An organic sample can be introduced into a mass spectroscope and ionized. This also breaks some molecules into smaller fragments. The resulting mass spectrum shows: 1) The heaviest ion is simply the ionised molecule itself. We can simply record its mass. 2) Other ions are fragments of the molecule and give information about its structure. Common fragments are: species formula mass methyl CH3+ 15 + ethyl C2H5 29 + phenyl C6H5 77 INFRARED SPECTROSCOPY. Absorbing infrared radiation makes covalent bonds vibrate. Different types of bond absorb different wavelengths of infrared: Instead of wavelength, infrared spectroscopists record the wavenumber; the number of waves that fit into 1 cm. (This is easily converted to the energy of the wave.) 20 For some reason the spectra are recorded backwards (from 4000 to 500 cm-1 is typical), often with a different scale below 1000 cm-1 (to see the fingerprint region more clearly) and upside- down (% radiation transmitted is recorded instead of the absorbance of radiation). The wavenumbers of the absorbed IR radiation are characteristic of many bonds, so IR spectroscopy can determine which functional groups are contained in the sample. For example, the carbonyl (C=O) bond will absorb at 1650-1760cm-1. Summary of absorptions of bonds in organic molecules Infrared Spectroscopy Correlation Table Minimum Maximum Bond Functional group (and other notes) wavenumber (cm-1) wavenumber (cm-1) C-O 1000 1300 Alcohols and esters N-H 1580 1650 Amine or amide C=C 1610 1680 Alkenes C=O 1650 1760 Aldehydes, ketones, acids, esters, amides O-H 2500 3300 Carboxylic acids (very broad band) C-H 2850 3000 Alkane Alkene (Compare intensity to alkane for rough C-H 3050 3150 idea of relative number of H atoms involved.) O-H 3230 3550 H-bonded in alcohols N-H 3300 3500 Amine or amide Free –OH in alcohols (only in samples diluted O-H 3580 3670 with non-polar solvent) Absorptions listed in cm-1. 21 NMR SPECTROSCOPY Nuclear Magnetic Resonance (NMR) Spectroscopy is one of the most useful analytical techniques for determining the structure of an organic compound. There are two main types of NMR, 1H-NMR (Proton NMR) and 13C-NMR (Carbon NMR). At A-level we only need to know about 1H-NMR. NMR is based on the fact that the nuclei of atoms have a quantised property called spin. When a magnetic field is applied to a 1H nucleus, the nucleus can align either with (spin +1/2) or against (spin -1/2) the applied magnetic field. These two states have different potential energies and the energy difference depends on the strength of the magnetic field. The strength of the magnetic field about a nucleus, however, depends on the chemical environment around the nucleus. For example, the negatively charged electrons around and near the nucleus can shield the nucleus from the magnetic field, lowering the strength of the effective magnetic field felt by the nucleus. This, in turn, will lower the energy needed to transition between the +1/2 and -1/2 states. Therefore, the transition energy will be lower for nuclei attached to electron donating groups (such as alkyl groups) and higher for nuclei attached to electron withdrawing groups (such as a hydroxyl group). In an NMR machine, the compound being analysed is placed in a strong magnetic field and irradiated with radio waves to cause all the 1H nuclei to occupy the higher energy -1/2 state. As the nuclei relax back to the +1/2 state, they release radio waves corresponding to the energy of the difference between the two spin states. The radio waves are recorded and analysed by computer to give an intensity versus frequency plot of the sample. This information can then be used to determine the structure of the compound. ULTRAVIOLET AND VISIBLE SPECTROSCOPY While interaction with infrared light causes molecules to undergo vibrational transitions, the shorter wavelength, higher energy radiation in the UV (200-400 nm) and visible (400-700 nm) range of the electromagnetic spectrum causes many organic molecules to undergo electronic transitions. What this means is that when the energy from UV or visible light is absorbed by a molecule, one of its electrons jumps from a lower energy to a higher energy molecular orbital. Let’s take as our first example the simple case of molecular hydrogen, H2. As you may recall from section 2.1A, the molecular orbital picture for the hydrogen molecule consists of one bonding σ MO, and a higher energy antibonding σ* MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ* orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO). 22 If the molecule is exposed to light of a wavelength with energy equal to ΔE, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the σ to the σ* orbital. This is referred to as aσ - σ* transition. ΔE for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm. When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a π - π* transition. Because π- π* energy gaps are narrower than σ - σ* gaps, ethene absorbs light at 165 nm - a longer wavelength than molecular hydrogen. The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated pi systems. In these groups, the energy gap for π -π* transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores. STEREOISOMERISM Isomerism is the occurrence of different compounds having the same molecular formula. There are two kinds of isomerism namely: constitutional isomerism and stereoisomerism. 23  Isomers Constitutional Isomers Stereoisomers (connectivity differences) (configurational differences) Enantiomers Diastereomers Geometric Optically active Meso Constitutional Isomerism or structural isomerism is the occurrence of different compounds with the same molecular formula but different connectivity; that is the order in which atoms are attached to each other. There are three types of connectivity differences, any one of which gives rise to constitutional isomers: a. Different carbon skeletons as in butane and 2-methylpropane: CH3-CH2-CH2-CH3 CH3-CH-CH3 CH3 Butane 2-Methylpropane b. Different placements of the functional group(s) on the same carbon skeleton, as in 1- propanol and 2-propanol: OH CH3-CH2-CH2-OH CH3-CH-CH3 1-propanol 2-propanol c. Different functional groups together with different carbon skeletons, as in ethanol and dimethyl ether CH3-CH2-OH CH3-O-CH3 Ethanol Dimethyl ether STEREOISOMERS: These are different compounds having the same connectivity but differ in their configurations. CONFIGURATION describes the relative orientations in space of the atoms of a stereoisomer, independent of changes that occur by rotation about single bonds. Configuration is different from conformation. Conformation describes the relative orientations of the atoms of a compound caused by rotation about single bonds. There are two classes of stereoisomers: 1. ENANTIOMERS: these are stereoisomers that are nonsuperimposable mirror images of each other 2. DIASTEREOMERS are stereoisomers that are not enantiomers; that is, they are not mirror images of each other. 24 SUMMARY CONCEPTS OF STEREOISOMERISM: Stereoisomers - Compounds that have the same molecular formula and the same connectivity, but different arrangement of the atoms in 3-dimensional space.Stereoisomers cannot be converted into each other without breaking bonds. Enantiomers - Nonsuperposable mirror images, or chiral molecules which are mirror images. Chiral, or asymmetric carbon - A tetrahedral carbon atom bearing four different substituents. Chirality centers, or stereocenters- Asymmetrically substituted atoms in a molecular structure. The most common type encountered in this course will be the chiral carbon described above. Diastereomers- Stereoisomers which are not enantiomers (or mirror images). Meso compounds, or meso forms - Symmetric, or achiral molecules that contain stereocenters. Meso compounds and their mirror images are not stereoisomers, since they are identical. Optical activity - The ability of chiral substances to rotate the plane of polarized light by a specificangle. Dextrorotatory - Ability of chiral substances to rotate the plane of polarized light to the right. Levorotatory - Ability of chiral substances to rotate the plane of polarized light to the left. Specific rotation - The measured angle of rotation of polarized light by a pure chiral sample underspecified standard conditions (refer to textbook for a description of these). Racemic mixture, racemic modification, or racemate- A mixture consisting of equal amounts of enantiomers. A racemic mixture exhibits no optical activity because the activities of the individual enantiomers are equal and opposite in value, thereby canceling each other out. Optical purity - The difference in percent between two enantiomers present in a mixture in unequalamounts. For example, if a mixture contains 75% of one enantiomer and 25% of the other,the optical purity is 75-25 = 50%. Absolute configuration - A description of the precise 3-dimensional topography of the molecule. Relative configuration - A description of the 3-dimensional topography of the molecule relative to an arbitrary standard. Absolute and relative configurations may or may not coincide. 25