CHM 206 Expt 17 Report
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Experiment 17: Synthesis of Acetophenetidin
Objective:
The purpose of this experiment is to synthesize acetophenetidin using two methods: one involving the formation of an ether linkage and the other involving the formation of the amide bond. The goals of this experiment are to understand the identity of the products and find the advantages of each synthetic pathway.
Table of Reagents:
Physical Constants
Molar Mass (
g
mol
)
Boiling Point (°C)
Safety Precautions
Acetic Anhydride
102.09
139.5
Can cause burns.
Hydrochloric Acid 36.458
100 (aqueous solution)
Corrosive; avoid skin
contact or breathing of fumes.
p
-Phenetidine
137.182
254
Toxic and harmful on
skin contact.
p
-Acetamidophenol
151.16
Harmful if swallowed.
Sodium Methoxide 54.03
92
Causes sever skin irritation and burns, poisonous.
Bromoethane
108.97
38
Flammable, ensure adequate ventilation, do not get in eyes, skin or on clothing.
Ligroin
86.178
Flammable liquid, carcinogenic, aspiration hazard if swallowed.
Ethanol
46.07
78.37
Highly flammable, causes skin and eye irritation.
Reaction Equations/Experimental Equipment and Apparatus:
Part A:
Part B:
Procedure:
Part A: Amide Synthesis of Acetophenetidin (done on second day of the lab)
1.
Collect a vial that contains 2.0 graphs of p
-phenetidine from the fume hood.
2.
Transfer the p
-phenetidine sample to a beaker containing 38 mL of dilute HCl.
3.
Use a pipet to take some of the
p
-phenetidine/HCl solutions and rinse the remaining traces from the vial.
4.
Add this rinse to the same beaker.
5.
Heat the solution to just below its boiling point on a hot plate (~ 80 °C) and carefully add a pea-sized amount of decolorizing carbon to the hot solution using a spatula.
6.
Stir the hot solution with a glass rod for 2-3 minutes.
7.
Gravity filter the hot solution through a piece of fluted filter paper in a plastic funnel into a 250 mL beaker.
8.
Keep the reaction warm.
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9.
Obtain 9 mL of the sodium acetate solution and 1.8 mL of acetic anhydride and add sodium acetate solution to the decolorized solution of p
-phenetidine in HCl, swirl once or
twice.
10. Then immediately add the acetic anhydride.
11. Periodically swirl or stir the solution for 5 minutes while warmed gently on the hot plate.
12. Cool the crystals in an ice bath and collect crystals by vacuum filtration using the Buchner funnel.
13. Resist the urge to stir the crystals in the funnel – will make the filtration less efficient and
will not adequately remove the solvent from the crystals.
14. Allow air to pass through for 5 minutes and allow the product to dry on a watch glass.
15. Determine the mass of the crude product, crude % yield, melting point.
16. Recrystallize the product (instructions in 17b, with the solvent of water).
17. Then determine the mass, % yield, and IR of the recrystallized product.
Part B: Willamson Ether Synthesis of Acetophenetidin (done on first day of the lab)
1.
Place 2.5 mL of 30% sodium methoxide in methanol, 4 mL of 100% ethanol, and 10 mmol (1.51 g) of p
-acetamidophenol into a 25 mL round-bottom flask.
2.
Add a boiling chip.
3.
Assemble the apparatus for reflux with a water-cooled condenser.
4.
Lift the condenser away from the flask and use a syringe to add 15 mmol (1.15 mL) of bromoethane to the flask.
5.
Immediately replace the condenser, carefully turn on the water flow to the condenser to a slow trickle and heat the flask with a heating mantle – to a moderate reflux.
6.
Keep the temperature at 70 – 80 °C and the height of the vapors between 1
4
- 1
2
of the way up the inner tube.
7.
Reflux the mixture for 45 minutes.
8.
Once the reflux time is over, remove the heating mantle and slowly pour 12 mL of deionized water through the top of the condenser and into the hot reaction mixture. 9.
Allow the reaction mixture to cool to RT, then turn off the water and remove the condenser.
10. Transfer the reaction mixture into a 100 mL beaker and use 3 mL of water to rinse the round-bottom flask.
11. Add this rinse to the solution in the beaker. 12. Cool the beaker in an ice bath to crystallize the product.
13. Collect the crystals using vacuum filtration.
14. Allow the product to dry for 5 minutes in the funnel and then add to a watch glass.
15. Determine the mass of the dry rude product, crude % yield, and melting point.
16. The next steps are for recrystallizing the crude product (for both part a and b).
17. Determine an appropriate recrystallization solvent system, either in water, 95% ethanol, and ligroin.
18. Take about ~20 mg of acetophenetidin and place this amount in 3 separate test tubes.
19. Add about 2-3 mL of solvent to each of the tubes, and check the solubility at the boiling point, RT, and when cooled in ice.
20. Once the appropriate recrystallization solvent is determined, conduct a recrystallization on 0.5 g of crude material by placing 0.5 g of crude into a 125 mL Erlenmeyer flask.
21. To help with gentle boiling, add a wooden applicator stick and dissolve the sample using the minimum amount of hot (boiling) solvent.
22. Keep the flask on the hot plate and at the solution’s boiling point.
23. Once all the material has been dissolved in boiling solvent, remove the flask from the heat and allow the saturated solution to cool slowly to RT.
24. Then cool in an ice bath until ample crystals have formed.
25. Collect the purified crystals using the Buchner funnel.
26. Obtain the mass, melting point, and IR spectra of the purified crystals.
Observations:
Part A:
The starting sample turned from a dark reddish-brown to yellow-brown when placed in water.
The decolorizing carbon turned the solution black which then turned clear after hot gravity filtration.
The crude product was a white-pink powder which when recrystallized turned into pure white crystals.
Water was the best recrystallization agent as it only partially dissolved the crystals.
Part B:
The recrystallized product came out as white crystals.
Water was the best recrystallization agent, only partially dissolving the crystals.
Data and Results:
Part A:
Theoretical Yield of Crude Product:
2.012
g p
−
phenetidine
∗
1
mol
137.18
g
∗
1
mol p
−
phenetidine
a
1
molacetophenetidin
∗
179.22
gacetophenetidin
1
molacetophenetidin
=
2.629
gacetophenetidin
Experimental Yield of Crude Product: 2.885 g
Percent Yield of Crude Product: 2.885
2.629
∗
100
=
109.74 %
Melting Point of Crude Product: 130.1 – 133.5 °C
Theoretical Yield of Recrystallized Product: 0.505 g
Experimental Yield of Recrystallized Product: 0.266 g
Percent Yield of Recrystallized Product: .233
.505
∗
100
=
46.14%
Melting Point of Recrystallized Product: 132.3-134.7 °C
IR Spectra of Crude Product:
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IR Spectra of Recrystallized Product:
Part B:
Theoretical Yield:
1.511
g p
−
Acetamidophenol
∗
1
mol
151.17
g
∗
1
mol p
−
Acetamidophenol
1
molacetophenetidin
∗
179.22
g
1
molacetophenetidin
=
1.791
g acetophenetidin
Experimental Yield of Crude Product: 1.137 g
Percent Yield of Crude Product: 1.137
1.791
∗
100
=
63.48%
Melting Point of Crude Product: 129.1 – 131.3 °C
Theoretical Yield of Recrystallized Product: 0.498 g
Experimental Yield of Recrystallized Product: 0.234 g
Percent Yield of Recrystallized Product: 0.234
0.498
∗
100
=
46.99%
Melting Point of Recrystallized Product: 132.5 - 133.1 °C
IR Spectra of Crude Product: TA said not to find it
IR Spectra if Recrystallized Product:
Discussion and Conclusion:
In this lab we used two methods to produce acetophenetidin: Amide Synthesis and Willamson Ether Synthesis. Amide synthesis consists of the formation of the amide group while Willamson Ether Synthesis consists of the formation of the ether linkage.
For Part A of the lab, we somehow got over 100% yield of our crude product (109.74%). This could have been due to the fact that some impurities were still in the crude product, the product did not dry out fully when doing the vacuum filtration. For the recrystallized product we only got a 46.14% yield. The IR showed that there were some impurities in the crude product since the IR Spectra of the recrystallized product had much more defined peaks. The peaks were at 3280.83. 2981.85, 1657.79, 1604.24, and 1043.03 cm
-1
, showing the N-H bond in the amine, C-H bonds in an aromatic ring, carbonyl functional group, aromatic ring, C-O bond in ether respectively. The melting point also proves that there were impurities as the melting point of the crude product was 130.1-133.5 °C compared to the melting point of the recrystallized product which was 132.3-134.7 °C.
For Part B of the lab, we got a percent yield of the crude product of 63.48%. Moderating the reflux temperature better could have increased the percent yield, making this a possible source of error. The percent yield for the recrystallized product was 46.99% which could mean
that the crude crystals had some impurities that were removed. This can be proved by the increase of the melting temperature: MP of the crude product was 139.1-131.3 °C while the MP of the recrystallized product was 132.5-136 °C. The recrystallized IR for part b was very similar to the IR of the recrystallized product of part a. There were peaks at 3282.67, 2982.43, 1658.17, 1604.32 and 1042.34 cm
-1
representing the N-H bond in the amine, C-H bonds in the aromatic ring, carbonyl functional group, aromatic ring, and C-O bond in the ether respectively. The results of the two parts came out well with the recrystallized product from part A being the purest due to the IR and melting point.
Post Lab Questions:
1.
Everyone had relatively similar melting points. Comparing the NMR results would be the
best way to determine if the samples are identical from the two synthesis reactions.
2.
The reagent would dissolve quicker since HCl is more acidic and polar and would protonate the amine group.
3.
Sodium acetate will work as a pH buffer which would allow a purer and higher yield product.
4.
Willamson Ether Synthesis is less green than amide synthesis since one of the byproducts
of the Williamson Ether Synthesis is CO
2
.
5.
Ligroin is a volatile petroleum fraction and is nonpolar.
1.8
mLacetic
anhydride
∗
1.08
g
mL
∗
1
mol
102.09
g
=
0.019
molacetic acid
6.
p
−
Acetamidophenol
:10
mmol
∗
1
mol
1000
mmol
∗
151.16
g
1
mol
=
1.51
g p
−
acetamidophenol
Bromoethane
:15
mmol
∗
1
mol
1000
mmol
∗
108.97
g
1
mol
=
1.12
gbromoethane
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