Question

how would you calculate the percent yield of lidocaine in the exp.,

N. HN CI CI CI сH,CO2H CHяCOONa Lidocaine Precursor 'N H .N 'N toluene 110 C Lidocaine IZ

if- the set up outlined 300 mg precursor, 2 mL diethylamine, and 4 mL toluene in reflux?

and experimental values were- 298 mg precursor, 3.9 mL toluene, and 2 mL diethylamine obtaining a dried sample of 86.5 mg?

Also-- a general question. if purification proceeded this reflux reaction, using diethyl ether, h2o, 3m ccl and 12m naoh, would that be included in yield calculations? or, since it wasn't in the reaction itself, it doesn't matter?

if it does-- 10 mL diethyl ether, 45 mL h2O, 30 mL HCl (aq) and about 3 12 M NaOH were outlined, whereas 11 mL diethyl ether, 46 mL h2O, 19 mL HCl and 6 mL NaOH were used.

Is this a good or bad percent yield/amounts used ?

Answer 1

СНз CH3 NH NH "CHз Нас "CHз NH toluene, 110°C CH3 "CHз "CHз Molecular Formula: CH2NC 234.2g/mol Molecular Formula: CHCINO 197.1 g/mol Molar mass 197.1g/mol Diethylamine73.1g/mol formulas used here: M+: Molar mass density moles mass 0.298 g 1.414g Precursor 1.512*10A-3 0.707g/mL 19.34*10A-3 Moles mass in g molar mass in g/mol mass densityvolume In our reaction we have only these two reactant (the second step of total synthesis). The other steps you mentioned are for sepration and purification of the product. Hence we need not to take in account for all those solvents and acids/bases for our yield calculation Precursor is the limiting reagent because number of moles of precursor are less than that of diethylamine. As per the balance equation 1 mole of precursor should give 1 mole of product. then 1.512*10A-3 moles should give 1.512*10A-3 moles of the product Theoretical yield 1.512*10A-3 mole * 234.2g/mole 0.354 g Thus if we obtain 0.354 g of product then we have 100% yield. So, practical % yield (0.0865/0.354)*100 = 24.44%

As you can see from the calculations, the practical % yield is 24.44%, which is considerably low yield. There are many factors such as reaction time, isolation of the product, purification of the product may have some error leading to such low yield.

Answer for a general question: We need not take into account for all those reagents for calculation of our yield. We just need to work around the moles and mass of our actual reagents that to particularly limiting reagent.