Question

11. Complete the table and determine the theoretical yield of your product Br D NaNO, HO 2) KI 2-bromoaniline 2-iodobromobenzene Reagents MW Density Amount Moles Equivalents Remarks (e/mol) (g/ml) 69 166 2-bromoaniline NaNO 1721.56 2.0 mL Bp 229 C 1.9용 0 mL HCI (ag) KI excesscorrosive 0.0 mL 2-iodobromobenzen 283 22 mL grams Mp 9.0 C Show your work here

Answer 1

Mass = volume * density

No of moles = Given mass / Molecular weight

Mass of 2- bromoaniline= 2*1.56=3.12 g

No of moles of 2- bromoaniline= 3.12/172=0.018 moles

No: of equivalence of 2- bromoaniline= 0.018/0.018=1 eq

No of moles of NaNO2=1.9/69=0.027 moles

No: of equivalence of NaNO2=0.027/0.018=1.5

No of moles of KI= 3.6/166=0.0216 moles

No: of equivalence of KI=0.0216/0.018=1.2

Limiting reagent= 2-bromoaniline( least amount of reagent)

Theoretically 0.018 mol of 2-bromoaniline should give 0.018 mol of 2- Iodobromobenzene

therefore theoretical yield of 2- Iodobromobenzene = 0.018*283=5.094 g