Question

EAS//Nitration of bromobenzene

Brief summary of procedure.

- Mix sulfuric acid +nitric acid, then cool in water bath. Add stir bar, condenser and reflux for 10 minutes while slowly adding bromobenzene. Don't exceed 55 degrees celcius

- Heat mixture for 15 minutes below 60 degrees celcius. Cooldown, then pour into 40 mL of cold h2o

- vacuum filtration to isolate crude bromonitrobenzene

- recrystallize with 95% ethanol (cool to room temp, then to 0 degrees celcius on ice)

- isolate product by vacuum and wash with ice-cold 95% ethanol.

- concentrate mother liquor under reduced pressure solvent removal.

Suppose your lab mate wants to make 2,4-dinitrobromobenzene from bromobenzene using a variation of the proceduredescribed in the textbook.

a) What changes should your lab mate make to the procedure to maximizetheir yield of 2,4-dinitrobromobenzene? (1.5 point)

b) What is the minimum volume of 12.3M HNO3 that is required to make themost 2,4-dinitrobromobenzene from 3.8 mL of bromobenzene?

Answer 1

(a): To maximize the yield of 2,4-dinitrobromobenzene, the reaction whould be carried out at higher temperature.

(b): The balanced chemical equation for the production of 2,4-dinitrobromobenzene from bromobenzene is

bromobenzene + 2 HNO3 + 2 H2SO4 -------> 2,4-dinitrobromobenzene

Hence 1 mol of bromobenzene reacts with 2 mol of HNO3.

Volume of bromobenzene taken, V = 3.8 mL

Density of bromobenzene, d = 1.495 g/mL

Hence mass of bromobenzene taken = Vxd = 3.8 mL x 1.495 g/mL = 5.681 g

molar mass of bromobenzene = 157.01 g.mol-1

Hence moles of bromobenzene taken =  5.681 g / 157.01 g.mol-1 = 0.03618 mol

1 mol of bromobenzene reacts with 2 mol of HNO3.

=> 0.03618 mol of bromobenzene that will react with the moles of HNO3

= 0.03618 mol of bromobenzene x ( 2 mol HNO3 / 1 mol bromobenzene)

= 0.072365 mol HNO3

Given the concentration of HNO3 = 12.3 M

=> Volume of 12.3M HNO3 required = 0.072365 mol / 12.3 mol/L = 0.005883 L = 5.883 mL (answer)