Question

Calculate the percent yield of the aldol condensation-dehydration reaction.

I did the following

Put 0.8 mL aldehyde, 0.2 mL ketone, 4 mL ethanol, 3 mL of 2M sodium hydroxide in a flask. Then swirled it for 15 min. Then I added 6 mL ethanol and 4 mL of 4% acetic acid. I put the solution on ice and crystals formed. I ended up with 0.305 g of product. Please show me how to calcualte my percent yield for my product.

Answer 1

Only the cross aldol product will result as aldehyde has no alpha H and ketones are less reactive in self aldol condensation.

Let us calculate masses of aldehyde and ketone from given volumes and densities

Mass = Density x volume

For aldehyde, density = 1.019 g/mL and volume = 0.8 mL

So Mass of an aldehyde = 0.8 x 1.019 = 0.8152 g

For ketone, density = 0.791 g/mL and volume = 0.2 mL

So. Mass of ketone = 0.791 x 0.2 = 0.1582 g

Total mass of aldehyde+ ketone = 0.8152+0.1582 = 0.9734 g of reactants.

From mechanism it is clear that 1 mol aldehyde + 1 mol ketone = 1 mole enone product

i.e. 120 g aldehyde + 58 g ketone = 178 g reactants gives 160 g enone product

so we can write, 178 g reatants = 160 g enone product

Then, 0.9734 g reactant = say ; Mcal' g enone product

so, Mcal = (160 x 0.9734) / 178

Mcal = 0.875 g of enone product

But practically observed weight of enone product is (say) Mobs = 0.305 g

Then % yield calculated as,

% yield = (Mobs/Mcal) x 100

%yield = (0.305/0.875) x 100

% yield = 34.86 %

The % yield of the aldol condensation-dehydration reaction between 4-Methylbenzaldehyde and acetone gives 34.47 % yield of the product 4-p-Tolylbut-3-en-2-one.