EAS//Nitration of bromobenzene Brief summary of procedure. - Mix sulfuric acid +
ID: 1053690 • Letter: E
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?
Explanation / Answer
(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)
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