Question

Question #7f105

Answer 1

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%.