Question

What is the chief product obtained when $n-butane$ is treated with bromine in the presence of light at ${130}^{o}C$ ?

• A. $C{H}_3-C{H}_2-C{H}_2-C{H}_2-Br$
• B. $C{H}_3-C{H}_2-\underset{C{H}_3}{\underset{|}{C}}H-Br$
• C. $C{H}_3-\underset{C{H}_3}{\underset{|}{C}}H-C{H}_2-Br$
• D. $C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-Br$

Answer 1

Answer: (B)

$C{H}_3-C{H}_2-\underset{C{H}_3}{\underset{|}{C}}H-Br$

Br-Br bond cleavage forms 2 Br radicals. One radical reacts with n-butane, and a n- butane radicals is formed. This n-butane radical abstracts a Br atom from another Br-Br molecule, forminf 2- bromobutane and a new Br radical.

The carbon with the single electron in the n-butane is ${sp}^2$ hybridized. The single electron is an unhybridized p orbital and the new bond to Br can form on either side of carbon, thus giving a 1:1 racemic mixture of the R and S isomer.