The glassware needs to be dry. Please find the 25 ml. round-bottom flasks and co

Question

The glassware needs to be dry. Please find the 25 ml. round-bottom flasks and condensers in the oven and return them clean to the oven at the end of the experiment. As always, all glassware needs to be washed first with water, then acetone. Aluminum chloride is quite moisture sensitive. Be sure to keep the vials of anhydrous Aluminum Chloride closed between weightings to avoid the adsorption of moisture. Cool the glassware to room temperature. Assemble the apparatus as shown below, by placing a magnetic stir bar in the 25mL flask, then adding the condenser and drying tube on top. The drying tube should have fresh granular CaCI; diving agent. Add 3.75 mmol anhydrous Aids and 6 ml. methylene chloride to the flask. Next, add 1.5 mmol powdered biphenyl, and then add 0.15 mL acetyl chloride (CH_3COCI) by syringe. Stir the mixture a few minutes at room temperature, then heat to gentle reflux foe 10 minutes using flask sand bath. Note any color changes. Cool to room temperature, then remove the condenser and drying tube, and place the flask in an ice hath. Add 3 mL of 3 M HCl slow Is to the flask Remove the stir bar. cap the flask with a stopper, and shake, with venting in between shaking. Any remaining aluminum salts should dissolve in the aqueous layer. Now transfer the entire contents to a test tube (TTI). You can rinse the flask with a little methylene chloride to ensure everything is transferred over. Draw off the bottom layer (organic! with a Pasteur pipet and place it into

Explanation / Answer

I will complete data according to every compound:

1. Biphenyl:

m = moles * MW = 154.2 * 0.0015 = 0.2313 g * 1000 mg/g = 231.3 mg

melting point (according to literature) = 69.2 °C

Equivalence = 0.0015 eq.

2. Acetyl chloride:

density reported (in literature) = 1.1 g/mL

m = d * V = 1.1 * 0.15 = 0.165 g

moles = 0.165 / 78.5 = 2.11x10-3 moles * 1000 = 2.11 mmoles

Equivalence = 2.11x10-3 eq.

3. AlCl3

m = 133.3 * 0.00375 = 0.49988 g * 1000 = 499.88 g

Equivalence = EW = MW/valence of aluminium = 133.3 / 3 = 44.43 g/eq

n° eq = 0.49988 / 44.43 = 0.01125 eq.

4. Product.

moles: as it's a 1:1 relation the moles of the product would be the same as the main reactant, in this case biphenil so the moles are 0.0015 moles or 1.5 mmoles.

mass = 0.0015 * 196.2 = 0.2943 g * 1000 = 294.3 mg

Hope this helps

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