Williamson Ether Synthesis Preparation of Phenacetin from Acetaminophen PDF

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Williamson Ether Synthesis: Preparation of Phenacetin from Acetaminophen Partners: Victor, Zahra April 12, 2016 Methods and Background The goal of this lab is to use the Williamson ether synthesis and prepare a phenacetin. Phenacetin is prepared by taking Acetaminophen (which is in Tylenol), and mix it with iodoethane in the presence of a base. The hydrogen that is phenolic then proceeds to deprotonate due to the fact that it is acidic hydrogen. The potassium carbonate will be the solvent that deprotonates the phenolic hydrogen in the reaction; in addition the reaction will reflux for an hour, then after filter the product and recrystallizing it. In the end, we will use IR spectroscopy to label the functional groups of our final product, phenacetin and take the melting point. The percent yield is then calculated of the final product.

Figure 1: A Williamson synthesis reaction that will be done. The Williamson ether synthesis is an old reaction; specifically it is a Sn2 reaction between a deprotonated alcohol "alkoxide" and an alkyl halide that forms ether. Note that we’re forming and breaking a bond on carbon bonded to the OH group. The synthesis reaction usually involves two steps. The first step is the deprotonating of an alcohol by a base to form an alkoxide ion. Then the second step is where the Sn2 reaction takes place, where the newly formed alkoxide ion acts as the new nucleophile and the alkyl halide will act as the electrophile. Since it is an Sn2 reaction, the methyl and primary alkyl halides react more quickly than the secondary or tertiary alkyl halides, won’t react as quickly. Therefore the Williamson ether synthesis is often insufficient with secondary halides; the yield won’t be as good as if it were a primary halide (Figure 2, below, is an common example of the synthesis)

Factors that could help increase the Sn2 are determining a good electrophile and nucleophile. The rate is carbon dependent, of which oxygen bond is connected. Which will tell us the strength of the nucleophile; hence the nucleophile will always be an alkoxide. Also, the reactions will produce more rapidly is the kinds of solvents used in the reaction synthesis. The reaction will produce more rapidly when it’s polar. Aprotic solvents are used because the solvent and hydrogen bond will solvate the nucleophile. So, the commonly used solvents such as dimethyl sulfide (DMSO), and dimethyl formamide (DMF), and acetonitrile(CAN) are commons solvents used in the Williamson ether synthesis reactions. Experimental Procedure Obtain and crush 4 tablets of 325mg strength Tylenol using a mortar and pestle, this should add up to 1.3g acetaminophen. Make sure to weight it on the balance. Add the resulting powder to a 50 mL round-bottom flask. Add 2.5 g of potassium carbonate, 15ml of 2butanone, and one boiling stone to the acetaminophen. In the fume hood, add 1 mL of iodoethane to the reaction mixture, and then set up your apparatus as seen in (figure 3, shown below). Reflux the reaction mixture for 1 hour. Cool the reaction mixture to below its boiling point and vacuum filer the solids. Wash the bottle 2x with 5 mL of ethyl acetate. Then take a TLC of the reaction mixture before workup. The TLC plate should contain three lanes: 1) pure acetaminophen; 2) co-spot (reaction mixture + pure acetaminophen); 3) reaction mixture. Examine the TLC plate under UV light and circle the spots you observe. Now, transfer the filtrate to a separatory funnel and extract the solution (organic layer, will be on top), separate with 20ml of 5% NaOH(aq) and 20ml of water. Transfer the cloudy organic layer to a clean Erlenmeyer flask and dry using sodium sulfate (add Na2SO4 until the solid no longer clumps). Swirl the flask as you add the Na2SO4- the phenacetin solution should become clear. Decant the dried phenacetin to a 50 mL round bottom flask, and remove the solvent using the rotary evaporator. The “rotovaps” will increase the surface area and create a solid. Recrystallize the resulting solid using hot ethanol. Remember that for recrystallization you want to use the minimum amount of hot ethanol possible to just dissolve the solid, and that you should keep both ethanol and the beaker with your solid + ethanol hot at all times during the dissolution process. Once all the phenacetin is dissolved, remove the solution from heat and let it cool first to room temperature and then in an ice bath. Scratch to induce recrystallization if none occurs in the ice bath. Filter the final product and dry attached to vacuum. Weight your final product and calculate percent yield. Make sure you do this before performing any tests or characterization. Characterize the product by TLC, melting point analysis, and IR spectroscopy. For the TLC, prepare a tiny amount of solution of your product in ethyl acetate and use the same solvent system as you did the reaction mixture. Calculate the Rf values of all spots on all TLC plates.

Figure 3

Data Acquisition/Calculations: I. TLC Analysis Rf values where taking using the formula Rf = (distance travelled by solute / distance travelled by solvent) Table 1: Reaction TLC Spot Name

Rf Value

1. Pure acetaminophen

1.5 cm / 3.5 cm = 0.43

2. Co-spot (acetaminophen & reaction product)

.43 and .63

3. Reaction product

2.2 cm / 3.5 cm = 0.63

Table 2: Final Product TLC Spot Name

Rf Value

1. Pure acetaminophen

1.3 cm / 3.3 cm = 0.39

2. Co-spot (acetaminophen & final product)

.39 and .73

3. Final product

2.4 cm / 3.3 cm = 0.73

II. IR Spectroscopy Table 3: Final Product IR Spectroscopy Wavenumber (cm-1)

Functional Group

3280.75 cm-1

N-H stretch

2927.27 cm-1

Alkene or Arene

1657.20 cm-1

C-O stretch or Vinyl Ether

III. Melting point

Table 4: Final Product Melting Point Analysis Compound

Melting Point (Celsius)

Phenacetin

134.2 °C

IV. Percent Yields and Theoretical Yields Table 5: Summary of Reaction Table Compound

Molecular Weight

Rxn Weight or Volume (g or mL) 1.3 g

mmol

Equivalents

151.17

D (g/mL) or M (mmol/mL) n/a

Acetaminophe n Potassium Carbonate

8.60

1.0

138.21

n/a

2.5 g

18.1

2.10

Iodoethane 2-Butanone

155.97 72.06

1.94 g/mL 0.805 g/mL 3.00 mol/L

1.0 mL 15 mL

12.4 167.6

1.4 19.5

Theoretical yield = the theoretical yield is given by: Theoretical Yield = (Starting Volume) x (Density) x (Molecular Weight) x (Reaction Stoichiometry) x (Molecular Weight) (1.3 g) x (1 mol / 151.7g) x (1 mol Acetaminophen / 1 mol Phenacetin) x (179.2 g / 1 mol) = 1.53 grams of phenacetin (Theoretical yield)

Percent yield = (actual yield / theoretical yield) x 100 Actual yield = .56 grams Phenacetin [ 37.51g – 36.95g = 0.56g ] Theoretical yield = 1.53 grams Phenacetin [(.56g) / (1.53)] x 100 =

36.6% Yield of Phenacetin

Conclusion The goal of this experiment was to further our understanding of the Williamson Ether Synthesis through the reaction of Acetaminophen to prepare Phenacetin, in addition with follow-up analysis utilizing TLC, IR, and melting point tests.

Our TLC test produced the results with Acetaminophen holding an Rf value of .43 and Phenacetin containing an Rf value of .63. A possible reason for their different Rf values is due to their varying functional groups. Acetaminophen contains an alcohol functional group, where Phenacetin contains an ether functional group. Therefore, the alcohol is more polar than ethers and the stationary phase is polar Acetaminophen will be more attracted to the stationary phase and not move up the TLC plate as far as Phenacetin. Due to having two differences in polarity, Phenacetin will contain a larger Rf value since it is less polar. When we gathered our data from our IR analysis, we determined the presence of three functional groups, an N-H stretch at 3280.75 cm-1, an arene at 2927.27 cm-1, and the presence of vinyl ether from 1657.20 cm-1. The N-H stretch and arene functional groups do not provide significant insight into the reaction, because they are in both are reactants and products. The absence of an O-H stretch at approximately 3300 cm-1 tells us that we have had a successful reaction. Our melting point analysis experimentally determined our final product’s melting point to be 134.2 °C, which is approximate to the determined melting point of Phenacetin at 134-136 ° C. These similar melting points suggests these two products have alike intermolecular attractions, our group then disclosed that our final desired product was phenacetin. Our percent yield was 36.6% of phenacetin, this is not a good because we started with 1.3 grams of reactants and we only received 0.56 g back. Some possible errors that might have occurred in the experiment might have been during recrystallization. Not scraping the entire solid from the flask might have affected the final weight hence giving an inaccurate percent yield. Also, evaporation could have occurred form not covering the beaker, in the beginning, causing in loss of substance. Another possible errors is not allowing the substance to cool at its boiling point may have affected the final yield. Our goal for our experiment was achieved, even though our percent yield wasn’t what we expected it to be. However, based on melting point analysis and IR spectroscopy results, our experiment was successful for producing Phenacetin from Acetaminophen utilizing the Williamson Ether Synthesis.

Reference Gilbert, John C., and Stephen F. Martin. Experimental Organic Chemistry. Cengage Learning, Massachusetts, 2011, 5th Ed, pp.81-83, 153-161, 461-465, 92, 93-101....