Synthesis of phenacetin lab report

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Chemistry

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Dec 6, 2023

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Table 1: Yield results Yield Theoretical 2.078 g Percent Yield 61.6% Actual 1.28 Recovery Percent Recrystallized 75.92% Figure 1: Equation and mechanism Javier Rios Exp.11 Synthesis of Phenacetin 1
Figure 2: Infrared Spectroscopy of acetaminophen Javier Rios Exp.11 Synthesis of Phenacetin 1
Figure 3: Infrared Spectroscopy of Phenacetin Javier Rios Exp.11 Synthesis of Phenacetin 1
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Figure 4: HNMR of Acetaminophen Javier Rios Exp.11 Synthesis of Phenacetin 1
Figure 5: HNMR of Phenacetin Table 2: Reagents table Compound MW (g/mol) mP ( ) bP ( ) Density (g/mL) Aspects P-aminophenol 109.13 187.5 284 1.29 Orange powder Phenacetin 179.22 137.5 242.245 1.24 White powder(recrystallized ) Clear crystals (crude) Acetaminophe n 151.16 168 420 1.293 Golden Acetic 102.9 -73 139.5 1.08 Drying agent Javier Rios Exp.11 Synthesis of Phenacetin 1
Anhydride Phosphoric Acid 97.995 42.4 407 1.87 Ethyl Iodide 155.97 -108 72 1.936 Sn2 Reaction Potassium Carbonate 138.205 899 Decompose s 2.29 5% Sodium hydroxide 39.997 323 1388 1.05 Cleaned solution in organic state Methyl/ Ethyl/ Ketone 72.11 -86.67 79.59 0.805 Source: Pubchem.gov Table 3: TLC Results Name R f 1 R f 2 Crude Phenacetin 0.18 0.308 Known Phenacetin 0.308 N/A Recrystallized Phenacetin 0.308 0.323 Solvent= 20% Dichloromethane, 80% Ethyl acetate R f = Solute Distance Solvent Distance = 6.5 cm Table 4: Infrared Spectroscopy analysis of Acetaminophen (From Attached Spectra) Frequency Type of bond 3323.16 N-H stretch 3160.04 O-H stretch 3108.16 C-H aromatic ring 1650.9 C=O Secondary amide 1561.27 C=C Stretch aromatic ring 835.87 Para disubstituted benzene ring Table 5: Infrared Spectroscopy analysis of Phenacetin (From Attached Spectra) Frequency Type of Bond 3281.07 N-H 1043.01 C-O 1506.49, 1549.08 Aromatic 1644,1656.6 Secondary amide Javier Rios Exp.11 Synthesis of Phenacetin 1
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2927.05 CH sp3 alkyl 825.51 Di-substituted ring Table 6: H NMR of Acetaminophen (From Attached Spectra) Chemical shift Splitting pattern Integration Protons 9.62 Singlet 1 A 9.12 Singlet 0.8 B 7.35, 7.32 Duplet 2.05 C 6.68,6.65 Duplet 2.07 D 2.08 singlet 1.12 E 1.97 singlet 1.96 F Table 7: H NMR of Phenacetin (From Attached Spectra) Chemical shift Splitting Pattern Integration Protons 7.38 Duplet 1.82 A 7.17 Singlet 0.79 B 6.85 Duplet 2 C 4.04 Quartet 2 D 2.14 Singlet 2.72 E 1.42 Triplet 2.94 F Discussion: For this experiment the synthesis of phenacetin was produced from p-aminophenol mixed with acetic anhydride, phosphoric acid, potassium carbonate and ethyl iodide. The percent yield as provided in Table 1 was 61.6% proving this experiment had a fair/good success since the phenacetin recrystallized was greater than 50%. However, there could have been a human error when recrystallizing and doing extraction since some amount was lost throughout the process making the weight less than the theoretical. The recrystallized recovery percentage was 75.2% which was calculated from wet weight and dry weight by subtracting the solvent lost which was 27.8%. The melting point of the crude acetaminophen was 167.3 -168.8 and recrystallized acetaminophen 167-169. The melting point of the crude phenacetin was 135.5-136.8 Javier Rios Exp.11 Synthesis of Phenacetin 1
and recrystallized phenacetin 137.2-137.6. The observed melting points also support this experiment and had a good success rate since the crude melting points are a little below the expected value, but the recrystallized melting points are very close to the expected value proving this synthesis pure. The TLC and its R f ' s values from Table 3 show 2 spots on the crude line, one was a small circle at 2.1cm, but there was a big loop at 1.2 cm. For the recrystallized there was one clear value at 2 cm (0.308 R f ) but there was barely noticeable circle at 2.1 cm (0.323 R f ), this shows phenacetin is a polar compound since it traveled a small scale and the solvent contained two polar compounds. The HNMR and Infrared Spectroscopy are analyzed on Tables 4-7, it shows the most noticeable signals for two different compounds acetaminophen and phenacetin. The groups to distinguish are the alcohol group in acetaminophen and the secondary amide on phenacetin. It can also be distinguished from the amount of protons around the molecules and their position. Questions: In the acetylation step, why does the amino group of p-aminophenol acetylate rather than the OH of the phenol? Because the amino group is a better nucleophile compared to the OH group since Nitrogen is less electronegative making it to have a higher tendency to donate lone pairs compared to Oxygen. What is the name of the reaction for the second step that produced phenacetin? Discuss any limitation of using this name reaction for ether synthesis? The name is Williamson ether synthesis and It is an Sn2 reaction between an organohalide and a deprotonated alcohol. Some limitations are deciding if this step would go through Sn2 or elimination, and this depends on the base used and what type of alkyl halide it is. In this case is a primary carbon and we are using the nucleophile alkoxide (produced from K 2 CO 3 ) which would favor Sn2 since it is less hindered. If it was a secondary carbon then it would prefer elimination. Javier Rios Exp.11 Synthesis of Phenacetin 1
Why does the phenol group of acetaminophen alkylate rather than the amide nitrogen? Because the hydrogen attached to the oxygen is more acidic compared to the hydrogen attached to the amide nitrogen. Also, after deprotonation from potassium carbonate the oxygen is stabilized by resonance with the help of the ring. In the second step, why is potassium carbonate used? Please explain by showing its reaction and address whether potassium carbonate can be used to form an ether for the following reaction. It is used because it deprotonates the alcohol making putting a negative charge to act as a nucleophile. The potassium carbonate would not be able to form an ether because the alkoxide ion is not stabilized by resonance and it could only lead to elimination of the alcohol. In the purification of phenacetin, why does one wash the organic layer with 5% sodium hydroxide? Javier Rios Exp.11 Synthesis of Phenacetin 1
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Because it can help to purify the solution by removing the acidic compounds in the mixture. Since its organic and it was previously reacting with dichloromethane NaOH can get rid of the phenoxide amide ion floating on this organic layer. Javier Rios Exp.11 Synthesis of Phenacetin 1