Question

Not sure how to do #1

Teuutiny ayen. H3C NaBH, 95 % EtOH ? OCH vanillin acetate Figure 1. The reaction you will be investigating. possible products: Dr OH and/or on and/or ic Hzcy НО and/or HO OCH OCH OCH 4-hydroxy-3-methoxybenzaldehyde 4-hydroxymethyl) -2-methoxyphenol 4-hydroxymethyl) -2-methoxyphenyl acetate Figure 2. Possible reduction products. Pre-lab task: 1.) Assume that vanillin acetate is the limiting reagent, and calculate the theoretical yield of product if all of the vanillin acetate is converted to: a. 4-(hydroxymethyl)-2-methoxyphenol b. 4-hydroxy-3-methoxybenzaldehyde c. 4-(hydroxymethyl)-2-methoxyphenyl acetate

Answer 1

I H- BH, totis OCH3 OCHZ Assuming 10g of limitant reactant Balanced Global Reaction 194.06g/mol 196.06g/mol C10H1004 + NaBH4 + EtOH → C10H1204 +BH3 + EtoNa Eto— H moles of limitant reactant = mass/molecular weight moles of limitant reactant = 10g/194.06g/mol moles of limitant reactant = 0.0515 COH According the equation, 1 mol of limitant reactant (LR) produce 1 mol of product (ratio 1:1), this is equal to 0.0515 moles LR produce 0.0515 moles of product. mass of product expected = (moles) (molecular weight) mass of product expected = (0.0515 moles) (196.06g/mol) mass of product expected = 10.97g (theoretical yield) OCH3

HBH3 OCH3 Assuming 10g of limitant reactant OCH3 HO EtOH OCH3 Balanced Global Reaction 194.06g/mol 152.05g/mol :0 C10H1004 + NaBH4 + EtOH C H2O3 + CH3O+BHz + EtoNa o moles of limitant reactant = mass/molecular weight moles of limitant reactant = 10g/194.06g/mol moles of limitant reactant = 0.0515 OCH3 According the equation, 1 mol of limitant reactant (LR) produce 1 mol of product (ratio 1:1), this is equal to 0.0515 moles LR produce 0.0515 moles of product. mass of product expected = (moles) (molecular weight) mass of product expected = (0.0515 moles) (152.05g/mol) mass of product expected = 7.8305g (theoretical yield)

o= II OCH Assuming 10g of limitant reactant OCH OH EtOHY co OCH3 OH H BH OCH OH Eto "HOY HO OCH3 OCH Balanced Global Reaction 194.06g/mol 154.05g/mol C10H1004 + 2NaBH4 +2EtOH • CgH003 + 2BH + 2EtONa + CHO moles of limitant reactant = mass/molecular weight moles of limitant reactant = 10g/194.06g/mol moles of limitant reactant = 0.0515 According the equation, 1 mol of limitant reactant (LR) produce 1 mol of product (ratio 1:1), this is equal to 0.0515 moles LR produce 0.0515 moles of product. mass of product expected = (moles) (molecular weight) mass of product expected = (0.0515 moles) (154.05g/mol) mass of product expected - 7.9335g (theoretical yield)