Question

How do I find the theoretical masses of the elimination product and the addition product? The molecular weights are provided in table

Aldol Addition and Condensation Alm. You will perform two reactions, one of which is an aldol addition and one provides the aldol condensation product. Both procedures will be performed ALDOL ADDITION conditions ALDOL CONDENSATION MSDS - ME Reactants Table Reactant(s) 2-acetylpyridine 4-nitrobenzaldehyde methanol Sodium carbonate water MW Density BPMP D 11.12 12.04 USD ROISS 0.180 MW BP MP Density MSDS Products Table Product(s) 3-hydroxy-3-(4-nitrophenyl)-1- (pyridin-2-yl)propan-1-one 27.834 (E) 3-(4-nitrophenyl)-1-(pyridin- 2-yl)prop-2-en-1-one. 254,74 Procedure: You and your lab partner should work together to set the following two reactions up to run simultaneously. Reaction Conditions 1 1) In a 50-ml. round-bottomed flask, with a spin bar, add 10mL of H2O solution which contains 1.0 mmol of 2-acetylpyridine. Stir the mixture vigorously. 2) Place 1.0 mmol of 4-nitrobenzaldehyde in a separate vial and dissolve using a minimum amount of warm MeOH. While still warm, add the solution of 4-nitrobenzaldehyde to the stirring mixture in the round-bottomed flask. 3) To this mixture add 0.25 mmol of Na2CO3 dissolved in 5 mL of H20. 4) After 1 hr of vigorous stirring, collect the solid product using vacuum filtration and wash with H2O (3 x 10mL) and a minimal amount of cold MeOH. 5) Allow the product to dry on the vacuum, determine the yield, melting point and record the IR spectra of your product. Reaction Conditions 2 1) In a 50-ml round-bottomed flask, with a spin bar, add 10mL of H2O solution which contains 1.0 mmol of 2-acetylpyridine. Stir the mixture vigorously. 2) Place 1.0 mmol of 4-nitrobenzaldehyde in a separate vial and dissolve using a minimum amount of warm MeOH. While still warm, add the solution of 4-nitrobenzaldehyde to the stirring mixture in the round-bottomed flask.

Answer 1

I am assuming you want to know the theoretical yield of the reaction: ( theoretical yield and theoreitcal mass are same)

so according to the reaction 1 mole of 2-acetylpyridine reacts with 1 mole 4-nitrobenzaldehyde to give out 1 mole of

a) in first case 3-hydroxy -3-(4-nitrophenyl)-1-(pyridin-2-yl)propane-1-one

b) in second case (E)-3-(4-nitrophenyl)-1-(pyridin-2-yl)prop-2-en-1-one

solving for first case,

1 milli mole = 1/ 1000 moles

no. of moles = weight of the reactant taken / molecular weight of the reactant

solving with respect to 2- acetylpyridine

wt. of acetylpyridine taken = no. of moles x mol. wt.

= 1/1000 x 121.14

therefore wt. of 2- acetylpyridine = 0.12114 g

now,

1 mol 2-acetylpyridine = 1 mole of addition product

121.14 g of 2-acetylpyridine = 272.256 of addition product (i.e the first product in your product(s) table)

1g of 2- acetylpyridine = 272.256 / 121.14 g of addition product

therefore , 0.12114 g of 2- acetylpyridine = 272.256/121.14 x 0.12114 g of addition product

= 0.2723 g of addition product

this is the theoretical yield of a) i.e addition product

now similarly for b) i.e. elimination product

1 mole 2- acetylpyridine = 1 mole elimination product

121.14 g of 2 acetylpyridine = 254.24 g of elimination product

1g of acetylpyridine = 254.24 / 121.14 g of elimination product

therefore , 0.12114 g of 2- acetylpyridine = 254.24 / 121.14 x 0.12114 g of elimination product

= 0.2542 g of elimination product (i.e the second product in your product(s) table)

This is the theoretical yield of b) i.e. elimination product.

you can also solve this w.r.t 4-nitrobenzaldehyde but the result won't change