Question

this is a three part question, please.

Assume you performed the reaction in the above question using the amounts of reactants shown below and you obtained 1.14 g of reaction product. • 1-phenyl-1,2-propanedione (2.10 mL) • sodium borohydride (0.200 g) FOR THIS QUESTION, USE THE EXACT INFO BELOW IN YOUR CALCULATIONS OR YOU WILL OBTAIN A WRONG ANSWER AND NOT RECEIVE CREDIT atomic weights = C, 12.0; H, 1.01; 0, 16.0; Na, 23.0; B, 10.8; density of 1,2-phenyl-1,2-propanedione = 1.10 g/mL 1,2-Propanedia Show you How many moles of 1-phenyl-1.2-propanedione are used in this reaction decimal point, +/- 0.00005. ve places past the

Answer 1

1) Volume of 1phenyl -1,2 propanedione given 2.10 ml

Density = 1.10gm/ml

Mass= volume × density = 2.10× 1.10 = 2.31gm

Molecular mass of 1 phenyl - 1,2 propanedione = 148 gm

Moles of 1 phenyl -1,2 propanedione =2.31/148

= 0.01560

2) In the following reaction acetylbenzoyl contains two sites for nucleophilic attack of hydride from NaBH4.

One NaBH4 molecule contains 4H- .

H2O gives one proton to oxygen to form Hydroxyl group and 4 -OH attached to Boron.

Therefore the values of

OH, OH + 1. NaBrochu ! Naphy P 4 H2O = 2pn Ph! 2 Phpht mit

a = 2

b= 1

c= 4

d= 2

e= 1

3) The molecular mass of NaBH4 = 37.8gm/mole

The mass of NaBH4 used in the reaction = 0.200g

No of moles = 0.200/37.8

= 0.00529

No of moles of NaBH4 is 0.00529.