Question

2) a) Beyond the method you used in lab and that you’ve learned in lecture, another way to install an iodine onto a benzene ring is to treat the benzene with n-butyl lithium and then I₂. Provide the mechanism of this reaction. (2 points)

b) n-butyllithium is usually sold as a solution in hexanes. What volume of a 2.5 M solution of n-butyllithium in hexanes would correspond to 1.15 equivalents if it were reacted with 4.26 g of benzene? (2 points)

Answer 1

a)

n-BuLi reacts with benzene to form Lithiated benzene. Then addition of Iodine results in removal of LiI and addition of I- on the benzne ring in a concerted single step.

b)

Molar mass of benzene = 78 g/mol

amount of benzene = 4.26 g

hence, number of moles of benzene in 4.26 gm of benzene is

hence, one equivalent of nBuLi is going to be 0.0546 mol.

1.15 equivalent =

1 L of 2.5 M nBuLi contains 2.5 moles of nBuLi.

Hence, the volume of solution that contains 1.15 equivalent or 0.0628 mol is

Hence, the volume of 2.5 M solution that contains 1.15 equivalent is 25.12 mL.