Experimental Objectives Determination of an unknown alcohol, carboxylic acid and ester Experiment 2,3 and 4 CHEM 203 Fall 2017 Experimental Objectives The purpose of this experiment is to determine in three experiments the identity of an unknown carboxylic acid, alcohol and the ester formed by the condensation of the two. You will be introduced to the laboratory operations of extraction, distillation and heating under reflux, as well as the analytic techniques of GC, IR and refractive index.

Overview Week 1 Purify a carboxylic acid by liquid-liquid extraction Recrystallize the material, dry and obtain a melting point Week 2 Purify and unknown alcohol by distillation – determine boiling point Obtain an IR spectrum of the alcohol and the refractive index Week 3 Condense the alcohol and carboxylic acid by the Fisher Esterification Analyze the product ester by odor, IR, refractive index or melting point This will be a full lab report – with introduction, experimental, data, discussion and conclusion sections

Part 1 - Extractions: Liquid-liquid Introduction A. Solubility of organic materials in water Organic compounds that contain highly polar groups up to about 4 carbons are highly soluble/miscible in water Once the “4-carbon” threshold has been reached, mono-functional molecules become increasingly insoluble in water The exception is multifunctional molecules: If the ratio of polar groups to carbons in the molecules skeleton is better than 1:4 the molecule will be highly soluble

Extractions: Liquid-liquid Introduction B. Measure of Solvent Polarity – Dielectric Constant The ability of a solvent to insulate opposite charges from one another is known as the dielectric constant. It is a measure of the solvents ability to dissolve polar substances The power of solvation in chemistry should not be underestimated! For example, to separate the Na+ and Cl- atoms within sodium chloride you could either heat the crystal until it melts at 800 °C or simply dissolve it in H2O at room temperature! Remember the rule of “like dissolves like”, polar solvents will solvate polar substances and non-polar solvents will solvate non-polar substances

Extractions: Liquid-liquid Introduction C. Common Organic Solvents (compared to water) Solvent Dielectric Constant, e Density Miscible with H2O? Protic Solvents Water 79 1.00 Yes Methanol 33 Ethanol 25 Acetic Acid 6 Aprotic Solvents Dimethyl Sulfoxide 47 Acetonitrile 38 Dimethyl Formamide 37 Acetone 21 Methylene Chloride 9 1.34 NO Chloroform 7 1.48 Ethyl Acetate 0.90 NO (sparingly) Diethyl Ether (ether) 4.3 0.71 Benzene 2.3 0.86 Hexane 1.9 0.66

Extractions: Liquid-liquid II. Distribution Coefficients A. Distribution of a solute in two immiscible phases If a substance is solvated in either water or an organic solvent (that is not miscible with water) and the two phases are in contact with one another, the substance will distribute across the two layers according to the relative solubilities of the compound in each layer This distribution is an equilibrium known as the partition coefficient, Kp: Aqueous Concentration of Substance in organic phase Grams/mL organic Kp = = Concentration of Substance in aqueous phase Grams/mL aqueous Organic Assuming an organic solvent more dense than water

Extractions: Liquid-liquid II. Distribution Coefficients Example: Suppose you have 100 mg of isobutyric acid dissolved in 30 mL of H2O and add 30 mL of CH2Cl2 and mix thoroughly. How much of the isobutyric acid can you expect to find in each layer, given that the Kp for isobutyric acid for water : CH2Cl2 is 3.0? First, assume that the amount in the organic layer = X Therefore whatever portion goes into the water is 100 – X Using the equation for Kp: Water CH2Cl2 Grams/mL organic X/30 mL Kp = = = 3.0 Grams/mL aqueous (100-X)/30 mL Solving for X we find: X = 75 mg in the CH2Cl2 layer and 100-X = 25 mg in the water

Extractions: Liquid-liquid II. Distribution Coefficients Given that most organics of interest (over 4 carbons) have negligible solubilities in water (high Kp), this method is best used to quickly separate water soluble by-products of a reaction from the organic soluble products For example, consider the following Grignard reaction: Water The reaction is performed in ether, and treated with water to protonate the product alcohol A separatory funnel would be used to separate the CH2Cl2 layer (which would contain almost all of the alcohol) and the aqueous layer would contain all of the magnesium salts Evaporation of the CH2Cl2 affords the product CH2Cl2

Extractions: Liquid-liquid II. Distribution Coefficients This method would be very poor for separating one organic from another, however. The most important exception to this is the separation of organic acids and bases, which we will explore in this experiment

Extractions: Liquid-liquid II. Distribution Coefficients B. Modification of the Distribution Coefficient through pH adjustment Consider benzoic acid: CH2Cl2 soluble The Ka (acid dissociation constant) for benzoic acid is ~5, which implies that only 1 out of 100,000 (105) molecules of benzoic acid are ionized at any point in time (an example of a weak acid) Benzoic acid possesses more than 4 carbons per polar functional group. This, coupled with the Ka for benzoic acid would explain the fact that the partition coefficient for benzoic acid for the CH2Cl2 -water system is over 6 (i.e. for 100 mg of benzoic acid distributed over equal volumes of CH2Cl2 /water, 86% of benzoic acid is in the CH2Cl2 layer).

Extractions: Liquid-liquid II. Distribution Coefficients B. Modification of the Distribution Coefficient through pH adjustment However, if instead of water being used, we use a 3M NaOH solution: CH2Cl2 soluble Water Soluble The equilibrium is driven completely forward in the reaction of a strong base with a weak acid. Sodium benzoate is entirely soluble in water, and not soluble in CH2Cl2 You have now fully moved benzoic acid from the CH2Cl2 layer to the aqueous layer!

Extractions: Liquid-liquid II. Distribution Coefficients B. Modification of the Distribution Coefficient through pH adjustment If we take the water layer that contains the separated sodium benzoate and adjust the pH with conc. HCl, the acid-base reaction is reversed in favor of the acid. Once again if this mixture was treated with CH2Cl2, benzoic acid would return to the CH2Cl2 layer Water soluble CH2Cl2 Soluble

Extractions: Liquid-liquid II. Distribution Coefficients B. Modification of the Distribution Coefficient through pH adjustment Likewise, we can move an organic base from an organic layer to the aqueous layer and back, using the opposite procedure. Consider aniline, a weak organic base. Unlike benzoic acid, it does not have an ionization reaction under aprotic conditions. In contact with water (a protic solvent) the following equilibrium is established: Ether soluble Again, the equilibrium for this equation is strongly to the left, and aniline remains in the CH2Cl2 layer in a two-phase CH2Cl2 -water system

Extractions: Liquid-liquid II. Distribution Coefficients B. Modification of the Distribution Coefficient through pH adjustment If we treat the CH2Cl2 layer containing aniline with a strong acid, say 3M HCl, a ammonium salt is formed which is more water soluble and the aniline is extracted into the water layer: CH2Cl2 soluble Water Soluble Note, this is the opposite behavior from benzoic acid, where treatment with strong base moves it to the aqueous layer

Extractions: Liquid-liquid II. Distribution Coefficients B. Modification of the Distribution Coefficient through pH adjustment If the water layer containing the ammonium salt is treated with base, the reaction is reversed. Aniline could then be extracted with the addition of CH2Cl2 : Water soluble CH2Cl2 soluble

Extractions: Liquid-liquid II. Distribution Coefficients B. Modification of the Distribution Coefficient through pH adjustment The distribution coefficients of neutral organics (those that do not contain acidic or basic groups such as –COOH, -SO3H, -PO3H2 or –NH2, -NHR or –NR2) are not affected by pH changes This provides the basis for the separation of organic acids and bases from other organics

Extractions: Liquid-liquid II. Distribution Coefficients C. This Experiment: You will purify an unknown carboxylic acid by extraction First, dissolve entire sample in CH2Cl2

Extractions: Liquid-liquid II. Distribution Coefficients C. This Experiment: You will purify an unknown carboxylic acid by extraction Second, add aqueous 10% NaOH and mix

Extractions: Liquid-liquid II. Distribution Coefficients C. This Experiment: You will purify an unknown carboxylic acid by extraction The carboxylic acid will separate from the impurities:

Extractions: Liquid-liquid II. Distribution Coefficients C. This Experiment: You will purify an unknown carboxylic acid by extraction Drain the CH2Cl2 and set aside – you can wash this with more NaOH to “pull” more carboxylic acid out! Drain the former “upper” layer into a separate beaker. If you reacidify the collected upper solutions, the carboxylic acid is no longer ionic, no longer soluble and will precipitate!

Extractions: Liquid-liquid II. Distribution Coefficients C. This Experiment: Isolate the precipitated carboxylic acid by vacuum filtration Recrystallize the product – this will help dry the H2O Allow to dry – determine melting point