Preparation of Acetaminophen | Experiment 2, Lab Reports of Organic Chemistry

Download Preparation of Acetaminophen | Experiment 2 and more Lab Reports Organic Chemistry in PDF only on Docsity! Experiment 2 Preparation of Acetaminophen p-aminophenol material used: 2.1 g acetic anhydride: MW =102 material used: 2.0 mL density: 1.10 g/mL NH2 OH OH NHCOCH3 + CH3 O CCH3 O O + CH3CO2H Note: The quality of p-aminophenol is sufficient that the Norit treatment should not be necessary. 1. The NH3+ group in not reactive in this form 2. Therefore, CH3CO2-Na+ is added NH3 OH + Cl- NH3 OH OH NH2 + CH3CO2H + Cl- + CH3CO2- Na+ The solubility is increased while maintaining a small amount of the free reactive amino group in equilibrium with the ammonium salt Why does the acetic anhydride react with the -NH2 and not the -OH group as in salicyclic acid? NH2 OH It is a matter of kinetics; the –OH of the phenol is about as reactive as the -OH in H2O; therefore since water is present in a much higher concentration, some acetic anhydride reacts with water The –NH2 group is more reactive 1. Heat your reaction mixture on a hot water bath 2. Cool your reaction mixture and scratch with a glass rod, if necessary, to achieve crystallization (wait about 1 hr, use this time to work on the aspirin experiment) 3. Vacuum filter your crystals and recrystallize your acetaminophen from hot (boiling ) water using a hot plate. 4. Identify the product by its melting point and determine the yield (next week). Recrystallization of Solids 1. Dissolve the solid in a minimum amount of hot solvent. 2. If necessary, filter the hot solution if everything doesn’t dissolve, by keeping everything hot. 3. Allow the hot solution to remain undisturbed; cool in ice water after the solution has reached room temperature. 4. Vacuum filter the solid and wash with a small amount of cold solvent. 5. Vacuum filtration is used to quantitatively remove the mother liquor which contains soluble impurities. lec 1 Temperature Solubility g/100mL Solvent selection 1. Polar solvents dissolve polar solutes 2. Non-polar solvents are usually best for non-polar solutes 3. Boiling temperature of solvent is a consideration The Digimelt Apparatus 1. Press the START TEMP button and use ▲and ▼arrows to set the start temperature. 2. Repeat step 1 to set the ramp rate (° C/min) and STOP TEMP (end temperature). 3. Press START/STOP to preheat oven to desired temperature. {PREHEAT is lit}. 4. When the ready light is lit, insert filled capillary into oven. Press START/STOP to begin temperature ramp. {MELT} indicator is lit. 5. After the sample is melted press START/STOP to end the experiment. {COOLING} is lit. Experiment 3 Isolation of caffeine from tea N N CH3 CH3 N N CH3 O O Caffeine is a natural product produced by a variety of botanicals that constitutes about 5% of the dry weight of tea leaves. It is a stimulant and used in a variety of over the counter drugs. In this experiment we will isolate caffeine from a host of other compounds present in tea by taking advantage of the solubility of caffeine in water. In addition to the caffeine that is water soluble, some tannins, chlorophyll and various flavonoids and other compounds are also somewhat soluble in water giving the brewed tea its characteristic color and beneficial effects. Calcium carbonate is added to help precipitate some of the tannins. In this experiment you will be asked to determine the % caffeine in dry tea leaves. You will use 14 g of tea or 6 tea bags (13.6g) and 200 mL of water. After brewing and filtering, you will measure the amount of water recovered. You will use the fraction of liquid recovered in your final calculation to determine how much caffeine was originally present in the tea leaves. Molecular structure of the flavone backbone (2- phenyl-1,4-benzopyrone) O O Extraction: Suppose you allowed two immiscible liquids to come in contact with each other and and one of the liquids had a solute dissolved in it. What would happen to the solute? How is an organic compound soluble in water recovered from an aqueous solution? solute N N CH3 CH3 N N CH3 O O N N CH3 CH3 N N CH3 O O caffeine Suppose caffeine has a solubility in water of 1 g /100 mL and a solubility in CH2Cl2 of 10 g/100 mL. How would the caffeine partition itself if we place 11 g of caffeine in contact with 100 mL of water and 100 mL of CH2Cl2 ? solubility of caffeine in water/methylene chloride = 1/10 1g/100mLH2O]/[10g/100mLCH2Cl2] = 1/10 1g/100mLH2O]/[10g/100mLCH2Cl2] = 1/10 How would caffeine partition itself if we only dissolved 1g in the same quantity of water and CH2Cl2 ? x = amount in H2O; 1-x =amount in CH2Cl2 xH2O/[1-x]CH2Cl2 = 0.1 x = 0.091 g in water 1- x = 0.909 in CH2Cl2 One extraction with 100 mL CH2Cl2 yields 0.909 g caffeine Two extractions with 50 mL CH2Cl2 yields 0.972 g caffeine Separatory funnel Proper Use of a separatory funnel 1. Use only cold or cool materials 3. Place stopper, invert and quickly vent via the stopcock any pressure generated 2. If any gas is evolved, make sure it has completely evolved 4. Shake carefully to avoid emulsions while venting periodically Suppose for the sake of argument, you recover only 150 mL of tea from the 200 mL of water you started with from 14 g of tea leaves. Assume that you didn’t lose any water to evaporation during the brewing process, this means you recovered 75 % of the original liquid. The remaining liquid was lost in the tea leaves and in the process of filtering. If you recovered 0.4 g of caffeine from the 150 mL of tea following extraction, this means that the total amount of caffeine in the 14 g of tea leaves is 0.4/0.75 or 0.53 g of caffeine in 14 g of tea leaves. 0.53/14 *100 = 3.8% by weight of caffeine in tea