Expt 2 Report Crutchfield PDF

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Lajae Crutchfield Chm 205 Section TX Vaitheesh 10-10-2020 Expt 2: Extraction

Objective: The objective of this lab is to identify solvent layers of two immiscible solvents; partition a compound between two immiscible solvents and determine KD; conduct liquid-liquid extraction with aqueous acids/bases and organic solvents; use drying agents, and use microscale techniques with centrifuge tubes. In this experiment, a mixture containing a carboxylic acid, phenol, and a The neutral compound was extracted and identified Introduction:  Liquid-liquid extraction is used to separate compounds within a mixture using two immiscible solvents. Separating a weak organic acid from a weak organic base. Immiscible solvents are solvents that do not form a homogeneous mixture when they are mixed together. The solvents form two distinct layers when allowed to separate and the less dense solvent forms the layer on top and the denser layer on the bottom. When the compound is mixed with the solvents and the mixture separates into layers, the solubility of the compound will determine which layer the compounds will reside in. If the compound is soluble in the less dense layer, it will be in the upper layer (organic layer.) Likewise, if the compound is more soluble in the denser layer, it will reside in the bottom layer (aqueous layer.) The partition coefficient (KD) is “the ratio of the concentration of a solute in the organic phase to its concentration in the water phase.”Extraction is dependent on the solubility of the compounds in the solvents.

Extraction can be enhanced by the incorporation of acids and bases. Change in pH can change the solubility of a compound with the addition of an acid or base. Distribution Coefficient: KD= C2/C1 Equation: RCO2H + NaOH ⇌RCO2-Na+ +H2O (organic soluble)

(water-soluble)

ArOH + NaOH⇌ArO-Na+ + H2O (organic soluble)

(water-soluble)

RCO2H+NaHCO3(aq)⇌RCO2-Na+ CO2(g) + H2O ArOH + NaHCO3(aq)⇌no reaction RNH2 + HCl(aq) ⇌ RNH3+Cl-(aq) (organic soluble)

(water soluble)

Hazards: ● Diethyl ether is widely used to extract solvent, but it is extremely flammable due to high volatility. In this lab we used methyl tert-butyl ether as a substitute. ● Combing a carboxylic acid with aqueous sodium bicarbonate leads to gas evolution Procedures part 1: Determination of Partition Coefficient of Benzoic Acid 1. Weigh 100mg benzoic acid into a centrifuge tube 2. Add 2mL of MtBE and dissolve the compound 3. Add 3mL of DI water into the centrifuge tube and cap it 4. Shake contents for 30 seconds 5. Remove cap and allow layers to separate 6. Remove the bottom layer using Pasteur pipet

7. Transfer the organic layer to the conical vial and add 50mg of anhydrous sodium sulfate (drying agent) 8. Recap vial and let sodium sulfate dry the organic phase for 5min 9. Transfer dried organic phase via a Pasteur pipet to a tared and dry conical vial containing a boiling chip 10. Rinse solid with an additional 500 microliters of MtBE 11. Evaporate organic solvent in a fume hood using a warm sand bath 12. Reheat vial to remove traces of solvent 13. Dry vial Procedure part 2: Separation of a Sample Mixture Containing Three Components An unknown mixture that contains: A carboxylic acid, a phenol, and a neutral compound 1. 100 mg of unknown 2. Dissolve 100mg of the unknown in 3mL of MtBE in a centrifuge tube 3. Add 1.5mL of 5% NaHCO3, cap and shake gently 4. Once layers have separated draw off the bottom layer and transfer it to a clean and labeled centrifuge tube 5. Repeat with two addition 1.5mL portions of 5% NaCHO3 6. Back extract the aqueous portions by adding 1mL of MtBE to the tub, shaking gently, and drawing off the MtBE layer 7. Combined NaHCO3 extractions can now be treated with 10% HCL solution until acidic 8. Precipitate should form (carboxylic acid) and is isolated by vacuum filtration with a Hirsch funnel

9. Extract organic solution three times with 1.5mL of 5% NaOH combining extractions into a centrifuge tube 10. Back extract with MtBE and carefully acidify aqueous extract with 10% HCl solution 11. You should get a solid precipitate 12. Wash the organic layer with 1mL of DI water and dry over sodium sulfate 13. Remove a layer from the drying agent and evaporate the MtBE solvent with gentle heating Data: Part A: Determination of Partition Coefficient for Benzoic Acid Initial weight of benzoic acid= 115 mg Weight of conical vial= 19.95 g Weight of anhydrous sodium sulfate =50 mg Weight of conical and benzoic sulfate= 50 mg Final Weight of conical and benzoic acid =20.060 g X= final weight- initial weight Weight of benzoic acid = 20.060g- 19.95g= 0.11 g= 110 mg Mo+Ma= 115 Ma=115-Mo= Ma=115-110= 5mg Corg= 110/2= 55 mg/ml C-water= 5/3 =1.66mg/ ml KD= Corg/Cwater( aq)= 55/1.66= 33.13 Part B: Separation of a sample mixture containing three components by liquid- liquid extraction; and identification by melting point. Unknown #16 Weight of unknown= 0.154g Neutral Weight of conical vial without neutral component= 19.972 g Weight of conical vial with neutral component= 20.013g Weight of neutral=( final mass after evaporation in sand bath- initial mass)= Weight of neutral= 0.041g Melting point= 82-83 C % yield=(pure sample / crude sample) *100 =( 0.041g /0.154g)*100= 26.6 % Neutral= 9-fluorenone Phenol Weight of weighing paper= 0.396g Weight of weighing paper and phenol= 0.421g

Weight of phenol = 0.421g-0.396g= 0.025 g Melting point= 92.6-98.6 C % yield= (pure sample / crude sample) *100 =( 0.025 g /0.154g)*100= 16.2 % Phenol= 4-t-butylphenol Carboxylic acid Weight of weighing paper= 0.405 Weight of weighing paper and carboxylic acid= 0.446g Weight of carboxylic acid= 0.041g Melting point= 134-135 C % yield=(pure sample / crude sample) *100 = (0.041g /0.154g)*100= 26.6 % Carboxylic acid= trans-cinnamic acid Discussion - I expected to discover the names of the unknowns during this experiment. In this experiment I did exactly that. I was able to obtain the melting points of the unknown and compare them with table 2.1 (attached to lab report). The weight of the neutral was 0.041g,the melting point was from 82 to 83 C, and the percent yield was 26.6%. Neutral 9- fluorenone is the closest to the values obtained in a melting point range of 80-83C. Phenol’s weight was 0.025g with a percent of 16.2% and a melting point of 92.6 to around 98.6C. The values obtained are close to the 4-t-butylphenol values from 96 to 99. Lastly, the carboxylic acid is believed to be trans-cinnamic acid since it has a melting point of 132 to 135 C and the barcoding to the data that was collected it had a melting point of 134135 C, a mass of 0.041, and percent yield of 26.9%. Conclusion -

Liquid- liquid extraction, solvent extraction, is a way to separate compounds made up of mixtures while using immiscible solution. In this experiment we identified the solvent layers with two immiscible solutions, liquid- liquid extraction with aqueous acid/ base and organic solvent with the use of a drying agent. I learned the techniques of microscale with centrifuge tubes. In part A of the

experiment the KD value is 33.13 and in part B I discovered the neutral was 9fluorenone, the phenol was 4-t-butylphenol, and the Carboxylic acid was trans-cinnamic acid. The differences between my actual results and and and my expectations did not vary much because it was an unknown solution and I had no idea what to expect from it. I learned that minor errors can be made in the slightest ways with weighting and calculating melting points. You have to be very precise and pay close attention to details. My results could have been more efficient in calculating the melting points. The liquid- liquid extraction technique can be used to purify compounds or to separate mixtures of compounds. Post-lab 1.Traumatic acid has a KD of 5.8 between MtBE and H2O. If 75 mg of traumatic acid is added to a tube containing 3mL of water and 2mL of MtBE how much traumatic acid would be in each layer after thorough mixing? -

The equation used to determine KD =C1/C2. KD= 5.8 = ((75 mg-x)/2 mL)/(x mg/3mL)

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Solving for X { x =15.41 mg}.

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15.41 mg would be found in the aqueous layer while 59.59 mg would be found in the organic layer.

2. Given a solution of 10 mg of traumatic acid dissolved in 100 mL of water, and using the partition coefficient given for traumatic acid in question 1, show that extracting the 100 mL of an aqueous solution with two 10 mL portions of MtBE would recover more of the traumatic acid from the aqueous solution than one extraction using 20 mL of MtBE. -

The first extraction of the two MtBE 10 mL portions: 5.8 = ((10mg-x)/10mL)/(xmg/100mL) x=6, so there would be 4mg dissolved in the organic layer.

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The second extraction: 5.8 = ((6mg -x)/10mL)/(xmg/100mL) x=4

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2mg dissolved in the organic layer for a total of 6mg on traumatic acid in the organic layer.

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using 20 mL of MtBE: 5.8 = (10mg -x)/20mL)/(xmg/100mL) x=5, so there would be 5mg of traumatic acid in the organic layer.

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You get a higher yield of traumatic acid by performing two extractions, each with 20mL of MtBE.

3. Give an equation for each of the organic salts below, showing how you would recover the parent compounds? a. Sodium 3-chlorobenzoate sodium 3-chlorobenzoate(aq) + HCl(aq) → 3-chlorobenzoic acid + NaCl(aq) 3-Cl-C6H4-COONa + HCl → 3-Cl-C6H4-COOH + NaCl The acid can then be extracted with an organic solvent b. sodium 3-methylphenolate sodium 3-methylphenolate(aq) + HCl(aq) → 3-methylphenol + NaCl(aq) 3-CH3-C6H4-ONa + HCl → 3-CH3-C6H4-OH + NaCl The phenol can then be extracted with an organic solvent c. ethylmethylpropylammonium (ion) (CH3)(CH3CH2)(CH3CH2CH2)NH+(aq) + NaOH(aq) → (CH3)(CH3CH2)(CH3CH2CH2)N + H2O + Na+(aq) The amine can then be extracted with an organic solvent 4. Suppose that just before adding a drying agent to an organic solvent that was used to extract an aqueous solution, you noticed that there are still tiny droplets in the organic layer. Can you proceed with adding a drying agent? What should you do? -

No, if there are still water droplets in your organic layer, those water droplets must be removed before the addition of the drying agent because the drying agent can only

remove minute amounts of water. You must use a pipet to remove the water droplets if they are visible in the organic layer. 5. Devise a general extraction scheme for separating the following pairs a) an organic base with an organic neutral compound (B+N) -

To separate an organic base mixed with a neutral compound, add aqueous HCl to the solution. The addition of HCl converts the base to an ammonium chloride salt which is water-soluble. Tthe neutral compound remains in the organic layer and the base is now in the aqueous layer.

b) an organic acid (HA) and a phenol (ArOH) -

To separate an organic acid from a phenol a weak base such as NaHCO3 must be used. The addition of the weak base will convert the organic acid to a salt which is water-soluble and will be found in the aqueous layer. The phenol will remain in the organic layer and the acid will be in the aqueous layer.

6. When two immiscible solvents are mixed, two layers form. Liquid-Liquid extractions commonly use organic solvents that are less dense than aqueous solutions, and therefore for the top layer. An important exception to this rule is chlorinated solvents, which are often more dense than aqueous solutions and form the bottom layer. Suppose you were unsure which layer was during your extraction. What simple method do you think you could use to determine which of the two layers was the aqueous layer? -

The most simple is to add water to the solution and observe which layer it dissolves in. Whichever layer this is will be the aqueous layer. Another simple method is to add a simple salt such as NaCl to the solution. Whichever layer the salt dissolves into is the aqueous layer.

7. Potassium carbonate is an excellent drying agent, but it should not be used with some classes of organic compounds. Would it be a better choice to use in drying and ether solution containing an acid (RCOOH) or a base (RNH2)? Why?

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Knowing that potassium carbonate is basic, it would perform much better as a drying agent for a basic compound. If the potassium carbonate were added to the acid, an acid-base reaction would occur, producing water instead of dehydrating the sample

Table 2.1...