Data collected: 1.4573g Aluminum Chloride 0.5174g Anhydrous Magnesium Sulfate 46
ID: 1022337 • Letter: D
Question
Data collected: 1.4573g Aluminum Chloride 0.5174g Anhydrous Magnesium Sulfate 46.4838g Pre-weighed 14/10 round bottom flask 46.7288g Product + round bottom flask
Chem 2520 Experiment 3 Friedel-Crafts Acylation: Synthesis of 4-Methoxyacetophenone Objectives Synthesize 4-methoxyacetophenone by the Friedel-Crafts acylation of anisole with acetic anhydride. Characterization with IR and NMR Chemicals Quantity Molar mass Melting point Boiling point Density (g/mL) 1.082 Substance (g/mol) 102.1 133.3 rc) 138- gh Acetic anhydride 0.6 mL 1.45 g 140 Aluminum chloride, anhydrous Anisole dichloromethane Product 4-methoxy acetophenone magnesium sulfate anhydrous Petroleum ether Sodium chloride, saturated Sodium hydroxide (3M) 0.55 984.9 154 40 0.995 1.325 108.1 7 mL 50.2 38-39 0.5 g 120.4 1.0 mL 1.0 mL 1.0 mL Chemical Hazards Acetic anhydride- corrosive and lachrymator 4-methoxyacetophenone-irritant and light sensitive 4-methoxyacetophenone- irritant Petroleum ether- flammable and toxic Anisole - irritant and hygroscopic Aluminum chloride-corrosive and moisture sensitive Dichloromethane -toxic and irritant Sodium Hydroxide -corrosive and toxic
Explanation / Answer
Q.1: moles of anisole taken = mass / molar mass = 0.55 g / 108.1 g/mol = 0.005088 mol
mass of acetic anhydride taken = volume x density = 0.6 mL x 1.082 g/mL = 0.6492 g
Hence moles of acetic anhydride taken = mass / molar mass = 0.6492 g / 102.1 g/mol = 0.006358 mol
From the balanced chemical reaction it is clear that 1 mol of anisole reacts with 1 mol of acetic anhydride to form 1 mol of the product.
Hence anisole is the limiting reactant that is exhausted completely and decides the amount of product formed.
Hence 0.005088 mol of the product(4-methoxy acetophenone) is formed.
Hence theoritical mass of product(4-methoxy acetophenone) formed = 0.005088 mol x 150.2 g/mol = 0.7642 g
Actual mass of product(4-methoxy acetophenone) formed = 46.7288 g - 46.4838g = 0.245 g (answer)
Hence percentage yield = (0.245 g / 0.7642 g) x 100 = 32.1 % (answer)
Q.2: The product 4-methoxy acetophenone has a sharp peak at 1710 cm-1 (due to -C=O of ketone) which is absent in anisole.
In acetic anhydride there is no stretch beyond 1800 cm-1 due to absence of benzene ring and -OH/-OR etc. groups. Peaks at around 1730 cm-1 are found due to -C=O groups.
Q.3: There are only two different types of H in the reactant anisole which are benzene H (5 Hs, 6 - 8.5 ppm), and -OCH3 hydrogen (3 Hs, 3.3 - 4.0 ppm).
However in the product, 4-methoxy acetophenone, there are 4 benzene hydrogens ( 2H, 2H, two doublets). Hence the two doublets in the product are due to two different types of 4 benzene Hs on each side of para-substituent. There are also -OCH3 hydrogen (3 Hs, 3.3 - 4.0 ppm) and -COCH3 hydrogen (3 Hs, 2 - 2.7 ppm) in the product.
Q.5: Nitration of aromatic ring (benzene) forms nitrobenzene.
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