University of Illinois at ChicagoUICCHEM 233: Organic Laboratory I Prelab LectureLab 6:Base Extraction of Benzoic Acid from Acetanilide followed by Recrystallization and mp Determination.CH2Cl2HNOOONaOHH2OOHO+ OHOO+H2O(pKa = 4.20) (pKa = 15.7)KeqOHOcrude solid: not pure/clean depressed mp possibly discoloredHN CH3Opure solid: white crystals sharp mp narrow mp rangeH2Oamine(conjugate base)ammonium cation(conjugate acid)+ +R NH2HOH2protonationR NH3UICUniversity of Illinois at Chicago 2009, Dr. Chad L. LandrieCHEM 233: Organic Chemistry Laboratory 1SlidePrelab Lecture: Lab 6Todays Goals21.Separate ~1.0 g of a 1:1 w/w mixture of benzoic acid and acetanilide by base-extraction.2.Purify each isolated solid by recrystallization from boiling water.3.Measure melting points and compare to reported values.4. Determine % yield obtained for each solid based on mass of starting mixture.OOHbenzoic acid(m.p. = 121-123 C)HN CH3Oacetanilide(m.p. = 111-115 C)UICUniversity of Illinois at Chicago 2009, Dr. Chad L. LandrieCHEM 233: Organic Chemistry Laboratory 1SlidePrelab Lecture: Lab 6General Scheme for:Base Extraction of 1:1 Benzoic Acid/Acetanilide3OHOHNOCH2Cl2 CH2Cl2HNOOONaOHAdd 3 MNaOH (aq)H2OseparatelayersHNOCH2Cl2H2OOONa1.Neutralize/ Protonate with 3 M HCl 2.Vacuum filter precipitated benzoic acidevaporateCH2Cl2OHOcrude solid: not pure/clean depressed mp possibly discoloredHNcrude solid: not pure/clean depressed mp possibly discoloredCH3O1. recrystallizefrom boilingwater

2. vacuum filterOHOpure solid: white crystals sharp mp narrow mp rangeHN CH3Opure solid: white crystals sharp mp narrow mp range1. recrystallizefrom boilingwater

2. vacuum filterCH2Cl2 is denser than water = bottom layerUICUniversity of Illinois at Chicago 2009, Dr. Chad L. LandrieCHEM 233: Organic Chemistry Laboratory 1SlidePrelab Lecture: Lab 6Three Readily Ionizable Functional Groups That are Separable by Acid or Base Extractions4ROOHROOOH+ +H2Ocarboxylic acid(conjugate acid)carboxylate anion(conjugate base)OOH H2Ophenol(conjugate acid)phenoxide anion(conjugate base)O+ +H2Oamine(conjugate base)ammonium cation(conjugate acid)+ +R NH2HHOH2deprotonationdeprotonationprotonationR NH3R R*Hydroxide and hydronium were chosen as general examples; other specic acids and bases may be used.The counterion on the left of the equilibria will depend upon the specic acid or base on the left.UICUniversity of Illinois at Chicago 2009, Dr. Chad L. LandrieCHEM 233: Organic Chemistry Laboratory 1SlidePrelab Lecture: Lab 6Solubility--Like Dissolves Like5ROOOR RNH3RNH2OHRROO Hwater soluble ionic and very polar small organic ionwater insoluble; organic solvent soluble nonionic, but polar functional groups i.e. organic solvent = CH2Cl2Water Polarity VSEPR = water is bent; bond angles of 104.5 strong molecular dipole inorganic and very polar (!=80) well suited to dissolve ionic compounds, including small organic anions and cationsHOHUICUniversity of Illinois at Chicago 2009, Dr. Chad L. LandrieCHEM 233: Organic Chemistry Laboratory 1SlidePrelab Lecture: Lab 6Choosing an Effective Extracting Acid or Base6 For an Effective Acid or Base Extraction, Keq must be > 1.0Base ExtractionpKeq = pKa (acid left) - pKa (acid right)Keq = 10-[pKa (acid left) - pKa (acid right)] Acid ExtractionHB HA B AKeqorganic baseextractingacid (left)extractingconjugatebaseorganicconjugateacid (right)BH HA BAKeqorganic acid (left)extractingbaseextractingconjugateacid (right)organicconjugatebase remember:p = -log10 For derivation of this equation, see text page 154.Example: Example:pKeq = 4.20 - 15.7Keq = 10-(-11.5) = 3.16 x 1011Since Keq is large (>>1), aqueous hydroxide is a good extracting solvent for benzoic acid.pKeq = -8.0 - 5.20Keq = 10-(-13.2) = 1.58 x 1013Since Keq is large (>>1), aqueous HCl is a good extracting solvent for 4-methoxyaniline.OHO+ OHOO+H2O(pKa = 4.20) (pKa = 15.7)KeqNH3COHHHCl +(pKa = -8.0)N+Cl(pKa = 5.20)H3COHHHKeqUICUniversity of Illinois at Chicago 2009, Dr. Chad L. LandrieCHEM 233: Organic Chemistry Laboratory 1SlidePrelab Lecture: Lab 6Several Smaller Extractions are More Effective than One Large Extraction with Same Volume7FA =VoKVx + VoFA = fraction of solute remaining in original solventK= partition coefcient = ([Ax]/[Ao]) " larger K = more efcient extracting solventVo = volume of original solventVx = volume of extracting solvent (per extraction)n = number of extractions( )nFA =202*5 + 20( )3FA =202*15 + 20( )1FA = 0.30 FA = 0.40 Example one:3 extractions x 5 mL each (15 mL total) from 20 mL of Vo.K = 2.Example two:1 extractions x 15 mL each (15 mL total) from 20 mL of Vo.K = 2.**30% of solute remains in the original solvent.** **40% of solute remains in the original solvent.**UICUniversity of Illinois at Chicago 2009, Dr. Chad L. LandrieCHEM 233: Organic Chemistry Laboratory 1SlidePrelab Lecture: Lab 6General Requirements for Extracting Solvent81.Does not react irreversibly with the solute.2.Immiscible with original solvent.(i.e. H2O and CH2Cl2)3.Selectively removes desired component.(i.e. large K for component to be extracted and small K for the rest.)4.Easly separated from the solute.(i.e. cyrstallizatio/precipitation then ltration,distillation)UICUniversity of Illinois at Chicago 2009, Dr. Chad L. LandrieCHEM 233: Organic Chemistry Laboratory 1SlidePrelab Lecture: Lab 6Recrystallization9Desired Product(yellow/large circles)+Impurities(small/blue circles)Dissolve product and impurities in a minimum amount of solvent at high temperature.Cool solution.Product crystallizes out of solution.Impurities remain dissolved.Vacuum lter solution through a Buchner funnel.Mother liquor (ltrate) contains impurities.ltrateRequirements for Recrystallization Solvent1.Not reactive with desired product(s).2. Desired product is completely soluble at elevated temperature, but only slightly solubleor not soluble at allat room temperature.3. Undesired components (impurities) are highly soluble at all temperatures (small temperature coefcient).4.Easily removed from crystalline product by ltration and evaporation.