Lab 10 William Ether synthesis PDF

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William Ether synthesis...

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Lab 10: William Ether synthesis: preparation of phenacetin from acetaminophen. Methods and background: The purpose of this lab is to prepare phenacetin using william ether synthesis. The process of william ether synthesis is started by taking acetaminophen and mixing it with iodoethane. In this case, potassium carbonate is the solvent that causes deprotonation of the phenolic hydrogen because it is an acidic hydrogen. The william ether synthesis is a Sn2 reaction between the deprotonated alkoxide and alkyl halide that forms an ether. It is a two step process. The first step is the alcohol deprotonation which is done by using the alcohol to form the alkoxide ion. The second step is the Sn2 substitution which happens when the alkyl halide functions as the electrophile and the alkoxide ion function as the nucleophile. There are limiting factors with both the first and second steps that should be taken into consideration when using the synthesis. In the first step, a strong base needs to be used in order to deprotonate the oxygen due to the fact that alcohols are very acidic. If the strong base is not present, the alcohol cannot function as a nucleophile so it cannot move onto the second step of the synthesis. The limitations on the second step are due to the competition between elimination and substitution reactions and the alkyl halide used. Sn2 reactions occur best with a sterically unhindered alkyl halide. A fast reaction occurs when methyl and primary alkyl halides are favored. William ether synthesis favors the reaction that uses a primary alkyl halide and the alkoxide ion. The rate of reaction in the second step depends on the strength of the nucleophile. The reaction rate can also depend on the leaving group because if there is a more suitable leaving group present like alkyl iodide, it will be an easy break of the carbon-hydrogen bond. The solvent that pushes the reaction to occur quickly is the polar aprotic solvent.

We used a reflux apparatus to prevent solvents and reactants used from being evaporated. The flameless heat used in the reflux apparatus allows for the return of the volatile material back into the flask. The reflux apparatus assures less time for the completion of the reaction because of the increased temperatures. After using the reflux apparatus, we used thin-layer chromatography to examine the progress of the reaction. The comparison is between the pure sample of acetaminophen and phenacetin.

On the TLC plate, three spots were created, the first was a pure spot of acetaminophen, the second spot was a co-spot of the acetaminophen and the mixture created, and the third spot was the mixture. After TLC was completed, a UV light was used to view and circle the spots in order to calculate the Rf values. After the TLC process was completed, the extraction needed to be done. The extraction was a select removal process of a specific component of the mixture. The layers separated due to their different densities and the immiscible phases that come into contact with one another. Then a rotary evaporator was used to remove any solids that formed at room temperature. The solids were removed by reducing the pressure of the solvents. The solvents we needed to evaporate were 2-butanone and ethyl acetate. Experimental procedure: Before beginning the lab, all safety attire must be worn. To begin the experiment, crushed acetaminophen should be added to a round bottom flask and dissolved with 2-butanone. Potassium carbonate and a

boiling stone was added to the flask. Using a syringe, iodoethane was added to the flask and the flask was swirled intermittently to mix the contents. The flask was clamped to a heating mantle and a condenser was added on top. This is the reflux apparatus. Once assembled, the heating mantle was raised, the power was turned on, and the water turned on where it was then refluxed for an hour. After an hour, the heating mantle was lowered and the flask was cooled to room temperature. Once cooled, the excess potassium carbonate was filtered. Ethyl acetate is then added to the flask and then poured onto the funnel to ensure all the product is collected in the filtrate. Washing with ethyl acetate is repeated once more. Next a TLC is taken of the filtrate. The three spots are acetaminophen, co-spot, and the reaction mixture. The TLC plate is then placed in the eluding chamber and the solvent is allowed to elute up the TLC plate. Once near the top, the tlc plate is taken out and the solvent line is marked. The excess solvent was dried and the TLC plate was seen under UV light. The filtrate was then transferred to a separatory funnel. The solution was then washed with 5% sodium hydroxide to remove any excess acetaminophen. The cap was placed on the funnel and shaken with intermittent venting. The layers were allowed to settle and then separated. The organic layer on top was transferred back into the separatory funnel. The organic layer was then washed with water. The separatory funnel was capped and shaken with intermittent venting. The layers were allowed to settle and separate. The organic layer which is on top was transferred to a flask. The organic layer was then dried with sodium sulfate. The solution is then decanted into a round bottom flask. The flask was then attached to a rotary evaporator. The flask was clamped with a blue keg clip to the bunk trap. The vacuum pump was turned on and closed to create suction. The flask was spun and the water bath was turned on. The atmospheric pressure lowered and the water bath raised the vapor pressure. The spinning allows for more surface area so the solvent evaporates faster. Once the solvent is removed, the vacuum pump is turned off and the air is allowed back into the system. The spinning is stopped and the flask is removed from the rotovap. The phenacetin is transferred to a conical flask. The crude phenacetin is recrystallized in ethanol. A small amount of hot ethanol is added and brought to a boil. Additional ethanol is added until the phenacetin has completely dissolved in the minimum amount of hot ethanol. Once dissolved the flask is allowed to cool to room temperature and the put on ice to induce

recrystallization. The crystals were then vacuum filtered and dried to produce pure phenacetin. The recrystallized phenacetin was then characterized by IR. Make sure to clean the machine after you are done. The phenacetin was also analyzed by melting point to compare to literature values. A small amount of sample was transferred to a melting point capillary and placed inside the apparatus. The melting point was recorded when it began to melt and when it fully melted.

Data Analysis

Compound

MW

D (g/mL) or M

Rxn weight

mmol

equivalents

acetaminophen

151.17

-

1.3 g

8.6

1

Ethyl iodide

155.97

1.94 g/mL

1.58 g

10.1

1.2

K2CO3

138.21

-

2.50 g

18.1

2.1

2-butanone

72.06

0.805 g/mL

0.80 g

11.1

1.3

Calculations: Mmol acetaminophen: 1.3 g/ 151.17 g/mol x 1000 mmol/ 1 mol= 8.6 mmol Mmol ethyl iodide: 1.58 g/155.97 g/mol x 1000 mmol/1mol=10.1 Mmol K2CO3: 2.50 g/ 138.21 g/mol x 1000 mmol/ 1mol= 18.1

Mmol 2-butanone: 0.80 g/ 72.06 g/mol x 1000mmol/ 1 mol= 11.1

Compound

MW

d or M

Rxn Weight

mmol

equivalents

acetaminophen

151.17

-

6.73 g

43.4

1

Ethyl iodide

155.97

1.94 g/mL

4.19 mL

52.1

1.2

K2CO3

138.21

-

12.6 g

91.1

2.1

2-Butanone

72.06

0.805 g/mL

5.04 mL

56.3

1.3

Calculations: Rxn weight acetaminophen: UIN/ 10^8= 673261106/ 10^8= 6.73 Mmol acetaminophen: 6.73 g/ 155.17 x 1000 mmol/ 1 mol= 43.4 Mmol ethyl iodide= mmol acetaminophen * equivalent ethyl iodide =43.4 * 1.2= 52.1 mmol volume=mmol * 10^-3 * (MW/d) =52.2*10^-3*(155.97/1.94)= 4.19 mL Mmol K2CO3= mmol acetaminophen * equivalent of K2CO3 =43.4 *2.1=91.1 mmol Rxn weight= (mmol/1000) * MW =(91.1/1000) * 138.21= 12.6 g Mmol 2-Butanone=mmol acetaminophen * equivalent 2-butanone =43.3 * 1.3= 56.3 mmol volume= mmol*10^-3 *(MW/d) =56.3*10^-3*(72.06/0.805)=5.04 mL Calculation percent yield: Theoretical yield= 8.60 mmol acetaminophen * (1 mmol phenacetin/ 1 mmol acetaminophen)*(1 mol/ 10^-3) * (179.22 g/ 1mol)= 1.54 g phenacetin Actual yield= %yield * theoretical yield = 0.94*1.54=1.45 g Table 2 calculation Theoretical yield=43.4 mmol acetaminophen * (1 mmol phenacetin/ 1 mmol acetaminophen)*(1 mol/ 10^-3) * (179.22 g/ 1mol)= 7.77 g phenacetin Actual yield= 0.94*7.77=7.30 g Table 1: Reaction TLC

Spot name

Rf value

Pure acetaminophen

1.6 cm/ 4 cm=0.4 cm

Co-spot (acetaminophen & mixture)

0.45 and 0.64

Mixture

2.3 cm/ 3.6 cm= 0.64

IR Acetaminophen Wavenumber (cm^-1)

Functional group

3280.50 cm

N-H stretch

2927.46 cm

Alkene

3073.05-3250.65

aromatic

HNMR Acetaminophen

2.1= N-H 6.3= aromatic ring and O-H bond 7.3= aromatic ring and O-H bond 9.2= aldehyde

IR Phenacetin

1230= CH2-O (ether) 1490= aromatic ring 1630= C=C 2940= Sp3C 3300= N-H

HNMR Phenacetin

1.2= Sp3C and CH3 2.0= Amide 4.2= R-CH2-O-R 6.5= aromatic ring 6.9=aromatic ring 7.8= N-H ring Melting point:

Measured value= 132-134 degrees celsius Literature value= 134 degrees celsius

Conclusion: The purpose of this lab was to create phenacetin using william ether synthesis. We used a rotary evaporator to remove excess solvent and the remaining solid was purified by ethanol. After the rotovap was used, tests like TLC, IR, NMR, and melting point were performed and then the percent yield was calculated. The TLC gave us results with Acetaminophen having an Rf value of 0.4 and Phenacetin having an Rf value of 0.64. Their different Rf values are most likely due to the different functional groups present. Acetaminophen has an alcohol functional group while Phenacetin has an ether functional group. The alcohol functional group is more polar than the ether and the acetaminophen will be more attracted to the stationary phase and not travel as far up the TLC plate. Phenacetin is less polar so it will have a larger Rf value. The IR spectroscopy analysis allowed us to determine the presence of several functional groups in the phenacetin. There is an Ether at 1230, aromatic ring at 1490, C=C at 1630, Sp3C at 2940, and N-H at 3300. The lack of O-H stretch at around 3300 means that it is a successful reaction. The N-H stretch does not mean much because it is in both the product and reactants. After doing the melting point analysis we determined that the melting point of the final product was between 132-134 degrees celsius and the literature value melting point is 134 degrees celsius. The melting point similarities means that the products have similar intermolecular attractions. This also means there are not many impurities present in the product. In conclusion, the process of this experiment gave us insight to using the rotary evaporation device. It also allowed us to successfully identify Phenacetin using TLC plate analysis, melting point analysis, and IR and NMR spectroscopy. Post Lab questions 1. The first limitation is that the alkyl halide must be primary because if it isnt then elimination would compete over substitution and we would end up with alkene as a major product. This is

why we used ethyl iodide as a primary alkyl halide to avoid this. The second limitation is that strong alkoxide ions are used which are high substituted. And lastly, secondary halides give mixtures of substituted and eliminated products. We can use strong hindered bases and a primary halide to avoid the formation of an elimination product. 2. The purpose of adding OH is to dissolve acetaminophen. Salt is soluble in an aqueous medium. 3. Acetaminophen has a phenol functional group which can be deprotonated by a strong base. So when the sodium hydroxide is added the acetaminophen turns into an ionic compound that dissolves in the aqueous layer. But it does not have a functional group that is deprotonated by the sodium hydroxide so the NaOH has no effect on it and it remains as the organic layer. 4. No the reaction would not proceed as expected because cyclohexanol is not acidic so if we wanted the reaction to proceed as expected, we would need a stronger base to make sure that the first step of the reaction occurs. 5.

6. Acid catalyzed ether formation is the best way to get a hindered ether formation. You cannot do william ether synthesis in this case.

Citations: Gilbert, J.C., and Martin, S.M., Experimental Organic Chemistry, 5th Edition, Cengage Learning, Boston, MA, 2011 Landrie, C.L., McQuade, L.E., Yermolina, M.V., Organic Chemistry: Lab Manual and Course Materials, 8th edition, Hayden McNeil, LLC, Plymouth, MI, 2018 http://what-when-how.com/organic-chemistry-laboratory-survival-manual/reflux-laboratory-manual/...