Question 7f105

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:

$\text{CH"_3"CH"_2"CH"_2"CH"_2"CH"_2"CH"_3 stackrelcolor(blue)("Br"_2, hνcolor(white)(mm))(→) "CH"_3"CH"_2"CH"_2"CH"_2"CH"_2"CH"_2"Br}$

$+ {\text{CH"_3"CH"_2"CH"_2"CH"_2"CHBrCH"_3 + "CH"_3"CH"_2"CH"_2"CHBrCH"_2"CH}}_{3}$

The selectivity of $\text{Br}$ in free radical halogenation is

1°:2°:3° = 1:82:1640

There are

$6 \textcolor{w h i t e}{m} \text{1° H atoms}$ 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
$\textcolor{w h i t e}{m m m m m} \text{Total} = 660$

∴ 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%.