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williamson ether synthesis reaction lab report by Hanna Thomson...

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Hanna Thomson Lab 10 Erica Tuesday 8am

Williamson Ether Synthesis: Preparation of Phenacetin from Acetaminophen Methods and Background The purpose of this experiment was to prepare phenacetin by Williamson ether synthesis using acetaminophen and iodoethane as our starting materials, by deprotonation of the alcohol with a strong base potassium carbonate, and a workup with sodium hydroxide. We then purify our product by recrystallization and verify our reaction went to completion by preforming TLC, Infrared spectroscopy, Nuclear Magnetic Resonance spectroscopy, and analyzing the melting point. Williamson ether synthesis is an organic reaction in which an ether is formed from either an alcohol, or an organohalide. In our reaction we preformed in this lab specifically, we transformed an alcohol group on acetaminophen to an ether using a strong base and an alkyl halide. The first step in the reaction is the deprotonation of the alcohol by the use of a base, in this case we used potassium carbonate to deprotonate the alcohol considering the protons of alcohol groups are not highly acidic with a Pka of around 16-18 meaning it will be considerably harder to deprotonate. After deprotonation the reaction proceeds through an Sn2 substitution reaction with the addition of an alkyl halide, in this experiment 1-iodoethane, where the deprotonated alcohol acts as the nucleophile that attacks the electrophilic carbon of the 1iodoehtane, and the iodine leaves during the process. However, since the second step of this reaction involves an Sn2 reaction there is a possibility of a competing E2 reaction happening as well. To eliminate the possibility of competing E2 reactions happening, we used a primary alkyl halide instead of a secondary or tertiary alkyl halide. The mechanism for step one of the reaction is shown below in figure 1.

figure 1: Step 1: deprotonation. Image from Williamson Ether Synthesis lab 10 in Organic Chemistry Lab Manual and Course Materials by Landrie, McQuade, and Yermolina

Hanna Thomson Lab 10 Erica Tuesday 8am

The mechanism for step two, the substitution portion of this reaction is shown below in figure 2.

Figure 2: Sn2 reaction, from Williamson Ether Synthesis lab 10 in Organic Chemistry Lab Manual and Course Materials by Landrie, McQuade, and Yermolina.

Phenacetin is a drug used for pain relief and a fever reducing agent, however it was banned by the FDA in the 1980s because of its link to kidney disease and cancer, it was also found that phenacetin is metabolized in to acetaminophen in our bodies after consumption. Acetaminophen is the active ingredient in Tylenol and acts as a pain reliever and fever reducer as well. Williamson ether synthesis is productive in a lot of pharmaceutical compounds such as melatonin, fluoxetine, tamoxifen, and vitamin E which all contain ethers. Our starting material acetaminophen has an alcohol group present, which means in the IR spectra we would expect to see an O-H stretch around 3200-3550(1/cm), however after preforming the reaction we would expect to see the absence of this stretch in our spectra because of the replacement with an ether. Instead, we would expect to observe N-H stretches around 3400(1/cm), C=O around 1600-1800(1/cm) and a new C-O stretch around 10001300(1/cm) with the formation of an ether. For our NMR results, we would expect to see an additional quartet around 1.8 and 4ppm from the formation of the CH2CH3 attached to the ether. While running TLC the first time after refluxing the reaction, we would expect our reaction mixture to have a higher Rf value than that of the sample of pure acetaminophen, because of the ether being less polar than an alcohol group. After the reaction is complete, we would expect to see the same thing with a higher Rf value than the pure sample of acetaminophen because of the difference in polarity. The melting point of pure phenacetin is 134 degrees Celsius, so if our reaction went to completion we would expect to see a melting point around 134 degrees Celsius. Our results concluded that our reaction went to completion after running the initial TLC plate and seeing that our reaction mixture had a higher Rf value than that of pure

Hanna Thomson Lab 10 Erica Tuesday 8am

acetaminophen (0.97 vs 0.6) meaning our product was less polar than acetaminophen, which is what we were expecting with the loss of the OH group and the addition of an ether. After the complete reaction and post workup, we took another TLC in which we saw the same results. After running IR spectroscopy, we observed no O-H stretch around 3200-3500(1/cm) which means we did not have an alcohol group present in our product vs. the alcohol group we had in the starting materials. We observed an ether stretch at 1042.06-1115.96(1/cm) that wouldn’t have been present before the reaction as well. As for NMR, we observed two quartets at 1.321.81ppm and 3.86-4.18ppm, which corresponded to the CH2 and the CH3 of the iodoethane that was substituted and formed an ether at the end of our reaction. We also observed the melting point range to be between 132.5 and 133 degrees Celsius, compared to the 134 degrees Celsius of the actual melting point. All of these factors concluded that our reaction went to completion and an ether was formed.

Procedure We began by measuring out 1.3 grams of acetaminophen (already in powdered form) and transferred it to a 50mL round bottom flask. To the 50mL round bottom flask containing acetaminophen, we added 2.5 grams of potassium carbonate and 15mL of 2-butanone along with one boiling stone, and then under the fume hood because iodoethane is toxic, we added 1mL of iodoethane to the mixture. We then allowed the reaction to reflux for 1 hour using the reflux apparatus shown below in figure 3.

Figure 3, reflux apparatus. Image from Williamson Ether Synthesis lab 10 in Organic Chemistry Lab Manual and Course Materials by Landrie, McQuade, and Yermolina

After refluxing for 1 hour,

disassembled the reflux apparatus and allowed the reaction to cool below its boiling point and vacuum filter the solids out by washing it with 5mL of ethyl acetate twice. We took a TLC of our

Hanna Thomson Lab 10 Erica Tuesday 8am

reaction mixture with three spots, one spot containing pure acetaminophen obtained from the fume hood, a co-spot containing both a sample of our reaction mixture and pure acetaminophen, and then a spot with only our reaction mixture, and eluted with a mixture of 4:1 ethyl acetate/methylene chloride and observed the spots under a UV light and calculated the Rf values. After running TLC, we transferred the remaining liquid to a separatory funnel and extracted with 20mL of 5% sodium hydroxide and separated the organic layer that formed, added it back in to the separatory funnel and extracted with 20mL of water, separating the remaining organic layer again. We transferred the remaining organic layer to an Erlenmeyer flask and dried the solution with sodium sulfate until the solid no longer clumped together and the solution became clear. We then transferred the liquid to a 50mL round bottom flask and removed the solvent using the rotary evaporator as shown below.

After there was only solid remaining in our 50mL receiving flask and all the solvent has been evaporated using the rotary evaporator, we recrystallized using the minimum amount of hot ethanol to dissolve the solid, and then cooled the solution to room temperature and then placed it in an ice bath until crystals formed. We removed the crystals from the flask and vacuum filtered the solids drying our final product, and weighed our final product to calculate percent yield. We then preformed a final TLC using our final product dissolved in ethyl acetate, a sample of pure acetaminophen, and a co-spot containing one spot of both mixtures. We ran NMR with our 100mg of our product dissolved in about 0.5mL of chloroform, and we also ran IR

Hanna Thomson Lab 10 Erica Tuesday 8am

spectroscopy and analyzed the melting point using the melting point apparatus to verify our results and confirm our reaction went to completion. Data and calculations Reaction Table (table 1) Mmol

Equivalence

-

Reaction weight or volume (g or mL) 1.3g

8.5996

1.0

138.21

-

2.5g

18.0884

2.103

155.97 72.06 179.216

1.94(g/mL) 0.805(g/mL) -

1.0mL 15mL 1.54g

12.4383 167.569 8.5996

1.447 19.5 1.0

Compound

Molecular weight (g/mol)

Density (g/mL) or (mmol/mL)

Acetaminophe n Potassium carbonate Iodoethane 2-butanone phenacetin

151.17

Sample calculation for calculating theoretical yield of phenacetin from starting material: same equivalence as acetaminophen (1.0), meaning same mmol 8.5996, divide mmol by 1000 to get mols 0.0085996mols of phenacetin, multiply by molar mass (0.0085996mols X 179.216g/mol) to get the amount of theoretical grams =1.54g of phenacetin theoretically. Sample calculation for finding mmol and equivalence for potassium carbonate: given 2.5g of potassium carbonate, divide by the molar mass (2.5g/138.21g/mol) to find mols = 0.0180884 and multiply by 1000 (0.0180884 X 1000)= 18.0884mmols of potassium carbonate. To find equivalence, divide the mmol by the smallest number of mmol observed in reaction table (divide by 8.5996 mmol of acetaminophen)= 2.103. Rf values and TLC observations TLC before workup (figure …)

Solvent line traveled 3.0cm Left spot=pure acetaminophen Middle spot=co-spot with acetaminophen and our reaction mixture Right spot=reaction mixture

Hanna Thomson Lab 10 Erica Tuesday 8am

Rf calculations= (distance spot traveled/distance solvent traveled)=Rf value Rf value(s) of pure acetaminophen (2.0cm/3.0cm)=0.67 Rf value(s) of co-spot: (spot closest to starting point)=(2.2cm/3.0cm)=0.73, (spot closest to solvent line)=(2.8cm/3.0cm)=0.93 Rf value(s) of reaction mixture (spot closest to starting point)=(2.1cm/3.0cm)=0.7, (spot closest to solvent line)=(2.9cm/3.0cm)=0.97. Rf Values Table 2 Spot Identity

Rf Values

Pure acetaminophen

0.67

Co-spot

Bottom= 0.73, top=0.93

Reaction mixture

Bottom=0.70, top=0.97

TLC after workup (figure…)

Solvent line traveled 3.5cm Left spot=pure acetaminophen Middle spot=co-spot with pure acetaminophen and our final product Right spot=our final product Rf calculations= Rf value(s) of pure acetaminophen (1.9cm/3.5cm)=0.54 Rf value(s) of co-spot (spot closest to starting point)=(2.1cm/3.5cm)=0.6, (spot closest to the solvent line)=(3.0cm/3.5cm)=0.86 Rf value(s) of reaction mixture (spot closest to starting point)=(2.1cm/3.3cm)=0.6, (spot closest to solvent line)=(3.1cm/3.5cm)=0.89. Rf values table 3 Spot identity Pure acetaminophen

Rf values 0.54

Hanna Thomson Lab 10 Erica Tuesday 8am

Co-spot Our final product

Bottom=0.6, top=0.86 Bottom=0.6, top=0.89

Melting Point Analysis (Table 4) Trial 1 2 3

Melting Point 132.5 C 132 C 133 C

IR spectroscopy (table 5) Functional Group

Wavelength (1/cm)

C-O N-H

1042.06-1115.96 About 3400

C=O

1505.08-1656.67

NMR spectroscopy (Table 6) Chemical Shift Integration Multiplicity Proton Arrangement 6.8-7.00 2.12 Doublet Aromatic protons 7.36-7.57 2.26 Doublet Aromatic protons 3.86-4.18 2.12 Quartet CH2 bonded to ether 2.22 3.0 Singlet CH3 on carbonyl 1.32-1.81 3.15 quartet CH3 bonded to CH2 Note: there was supposed to be a singlet integrated to 1 for the N-H proton but it is not present in our NMR graph. Percent yield Calculations Formula: (actual yield (g)/theoretical yield (g))x100=percent yield Our actual yield measured: 0.31g of phenacetin Our theoretical yield from table 1 (reaction table): 1.54g of phenacetin ((0.31g)/(1.54g))x100=20.13% yield of final product. Conclusion The purpose of this experiment was to synthesize phenacetin by Williamson ether synthesis with acetaminophen as the starting material, deprotonating with potassium carbonate, and a substitution reaction with iodoethane forming an ether and our final product,

Hanna Thomson Lab 10 Erica Tuesday 8am

phenacetin. We verified our results and that our reaction went to completion by performing a final TLC, IR and NMR spectroscopy, and analyzing the melting point of our final compound. After preforming the initial TLC, we found that the Rf of our reaction mixture was higher than that of acetaminophen, which means our reaction proceeded as expected since acetaminophen is more polar with the presence of an alcohol group, resulting in a lower Rf value. We found the same results in the TLC we performed after the workup and the complete reaction, meaning our product was less polar because of the presence of an ether instead of an alcohol group. The melting point of our product was observed to be around 132.5-133 degrees Celsius when the actual melting point of phenacetin is 134 degrees Celsius. The melting point of acetaminophen is near 137 degrees Celsius so we can conclude we were observing the melting point of phenacetin and our reaction went to completion. After performing IR and NMR we observed peaks in IR around 1042.06-1115.96 (1/cm) corresponding to the ether formed, along with peaks around 3400(1/cm) corresponding to the N-H stretch, and a C=O stretch at 1505.08-1656.67(1/cm). For NMR, we observed two quartets at 1.32-1.81ppm as well as 3.86-4.18ppm integrated to 3 and 2, corresponding to the CH2-CH3 group attached to the ester formed from the substitution part of the reaction, however we were supposed to observe an N-H singlet that was not present in our graph. The presence of the ester stretch in IR and the CH2-CH3 peaks in NMR conclude that our reaction went to completion. Sources

Gilbert, J.C., and Martin, S.M., Experimental Organic Chemistry, 6th edition, Cengage Learning, Boston, MA, 2015 Landrie, C.L., McQuade, L.E., Yermolina, M.V., Organic Chemistry: Lab Manual and Course Materials, 8th edition, Hayden McNeil, LLC, Plymouth, MI, 2018...