Question #7f105

1 Answer
Jun 17, 2017

The reaction would not be useful for the large scale manufacture of 2-bromohexane.

Explanation:

The reaction produces a mixture of 1-, 2-, and 3-bromohexane:

#"CH"_3"CH"_2"CH"_2"CH"_2"CH"_2"CH"_3 stackrelcolor(blue)("Br"_2, hνcolor(white)(mm))(→) "CH"_3"CH"_2"CH"_2"CH"_2"CH"_2"CH"_2"Br"#

#+ "CH"_3"CH"_2"CH"_2"CH"_2"CHBrCH"_3 + "CH"_3"CH"_2"CH"_2"CHBrCH"_2"CH"_3#

The selectivity of #"Br"# in free radical halogenation is

#1°:2°:3° = 1:82:1640#

There are

#6color(white)(m) "1° H atoms"# to produce 1-bromohexane,
#4color(white)(m) 2° "H atoms"# to produce 2-bromohexane, and
#4color(white)(m) 2° "H atoms"# to produce 3-bromohexane.

The relative yields of each product are

1-Bromo: #color(white)(ll)6 × 1 =color(white)(ml) 6#
2-Bromo: #4 × 82 = 328#
3-Bromo: #4 × 82 = 328#
#color(white)(mmmmm)"Total" = 660#

∴ The theoretical yield of 2-bromobutane is

#328/660 × 100 % = 50 %#

Unless the product is quite valuable, it would be uneconomical for a company to manufacture a product with a maximum yield of 50%.