Question

calculate % yield for this equation.

if start with .318g of P - Tolualdehyde, and end up with .452g as my product.

P-Tolualdehyde: 120.12g/mol

Procedure Cekaldehyder In a 25 ml rbf add 5 mL of saturated sodium_bicarbonate, 0.5 mL Marimethylphosphonoacetate and an equimolar amount of your assigned aldehyde. Reflux the mixture with vigorous stirring for 30 minutes. After cooling perform a liquid-liquid extraction into ethyl acetate and isolate the product.

this us a HEW reaction for organic chem 2

Answer 1

NaHCO3 на CH3 + H3C- Trimethyl phosphonoacetate 182.11g/mol density = 1.125 g/mL 0.5mL*1.125g/mL = 0.563g CH3 methyl (2E)-3-(4-methylphenylprop-2-enoate MW= 176.2 g/mol P- Tolualdehyde 120.15g/mol 0.318g number of moles of WHE reagent = 0.563g / 182. 1g/mol = 3.0917*10^-3 mole number of moles of p-tolualdehyde = 0.318g/120.2g/mol = 2.6456*10^-3 mole Thus the limiting reagent is p-tolualdehyde. Expected number of moles of product=2.6456*10^-3 mole Theoretical yield = 2.6456*10^-3mol*176.2g/mol = 0.466 g % yield = (practical/theoretical yield)*100 = (0.452/0.466)*100 = 97.00%