Question

Crossed aldol condensation of 3.4-dimethoxybenzaldehyde and l-indanone your co m ments cholateOH -4.0 dehydration formation OH O CHOY carbonyl addition initial aldol product Ill mbwa 5,4-dimethylonyaldivide CHO a quantity demande 1.119 g/mol 0,25g. 1- indanone Cg H₂O . 132. 16 grool 0.20g. | 1.103 ImL. sodium hydrol NaOH 39.997 I pellet | 1.515 gine 10% aqueone Ha Ha 2mL | 280.3 glane 0.171g. 12-[(3,4-dimethoxy CH phenyl) mothylene] 12, 3-dihydro-1H- inden -t-one product
the top half of the image depicts the synthesis of 2-[(3,4-dimethoxyphenyl)methylene]-2,3-dihydro-1H-inden-1-one from 1-indanone and 3,4-dimethoxybenzaldehyde. The bottom half of the image attached shows the table of chemicals used in the reaction. How would I calculate theoretical yield and furthermore, the percent yield? This experiment had a lot of chemicals used and I don't know where to start.

Answer 1

To find out the percent yield, first we will calculate the number of moles used in the reaction to determine limiting reagent in the reaction.

Number of moles of 1-indanone = 0.20/132.16 = 0.0015 moles

Number of moles of 3,4-dimethoxy benzaldehyde = 0.25/166.17 = 0.0015

Both the reactant are in 1:1 ratio and we can take any of the molecule for the calculation of percent yield (the limiting reactant should be taken for percent yield calculation)

Therefore, 166.17 gm of 3,4-dimethoxy benzaldehyde gives 280.3 gm of product

Then, 0.25 gm of 3,4-dimethoxy benzaldehyde will give x gm of product

Amount of product expected (theoretical yield) = (0.25*280.3)/166.17 = 0.42 gm

Amount of product obtained in the reaction = 0.171 gm

Percent yield of reaction = (0.171/0.42)*100 = 40.71%