Question

What will be the major product, when 2-methyl butane undergoes bromination in presence of light?
A.1-bromo-2-methylbutane
B.2-bromo-2-methyl butane
C.2-bromo-3-methyl butane
D.1-bromo-3-methyl butane

Answer 1

Hint: Bromination refers to any reaction in which bromine (and no other elements) is treated into a molecule. Bromination of an alkene takes place by electrophilic addition of $B{r}_2$ . Bromination of a benzene ring is done by electrophilic aromatic substitution reaction. Bromination of a benzylic position takes place by a free radical substitution reaction.

Complete step by step answer:
Free radical substitution takes place, in the presence of sunlight.
The free-radical bromination of this alkane produces four products:
A. 1-bromo-3-methylbutane
B. 2-bromo-2-methylbutane
C. 2-bromo-3-methylbutane
D. 1-bromo-2-methylbutane
wherever bromine attacks it becomes highly regioselective.
The relative rates of attack at the different types of hydrogen atom are given as:
 $3^0:2^0:1^0=1640:82:1$
In order to calculate the relative amounts of each product, we have to multiply the numbers of each type of atom by the relative reactivity. So, we will get:
A: $1^0:3\times 1\Rightarrow 3$
B: $3^0:1\times 1640\Rightarrow 1640$
C: $2^0:2\times 82\Rightarrow 164$
D: $1^0:6\times 1\Rightarrow 6$
The percentages of each isomer can be calculated as:
$$\begin{gathered} A={\displaystyle \frac{3}{1813}}\times 100\mathrm{\%}\Rightarrow 0.1\mathrm{\%} \\ B={\displaystyle \frac{1640}{1813}}\times 100\mathrm{\%}\Rightarrow 90.5\mathrm{\%} \\ C={\displaystyle \frac{164}{1813}}\times 100\mathrm{\%}\Rightarrow 9.0\mathrm{\%} \\ D={\displaystyle \frac{6}{1813}}\times 100\mathrm{\%}\Rightarrow 0.3\mathrm{\%} \end{gathered}$$
And the most stable $3^0$ free radical forms in intermediate,
so, 2-bromo-2-methyl butane forms a major product.
Hence option B is correct.

Note:
There is a different mechanism of bromination also that reaction is called anti-addition. In anti-addition the two substituents are added to opposite sides of a double bond or triple bond that means from the opposite faces. It results in a decrease in bond order and increase in number of substituents.
When phenol is treated with bromine it is known as bromination of phenol. Phenol reacts with bromine in the presence of Carbon disulphide and produces a mixture of o-bromophenol and p-bromophenol.
Bromination reactions are crucial in today's chemical industry due to the versatility of the formed organic bromides making them suitable building blocks for numerous syntheses. So, due to the use of the toxic and highly reactive molecular bromine these brominations become very challenging and hazardous.