Question

2-Methyl butane reacting with ${Br}_2$ in sunlight mainly gives:
(A.) 1-Bromo-2-methyl butane
(B.) 2-Bromo-2-methyl butane
(C.) 2-Bromo-3-methyl butane
(D.) 1-Bromo-3-methyl butane

Answer 1

Hint: You should know that 2-Methyl butane reacting with ${Br}_2$ in sunlight mainly gives 3-degree haloalkane as a product. Now try to figure out the answer accordingly.

Complete step by step solution:
Now, we will look at the stepwise answer to this question.
First, let’s see the mechanism of Chemical reaction involved in the Halogenation of alkane -
When alkane reacts with halogen in the presence of UV light, it gives free radical substitution reaction. Here is the complete mechanism for this reaction -

This is the structure of 2-Methyl butane -

Here Bromine gets substituted to hydrogen, where carbon has the least number of hydrogen atoms.
The carbon where -${CH}_3$ group is attached is three-degree carbon. Radical formed on this carbon will be stable so H-radical will be replaced with Br-radical. The final product will be 2-Bromo-2-methyl butane.
This is the complete reaction involved in this process -

Option A) 1-bromo­-2- methyl butane is incorrect.
Option B) 2-bromo­-2- methyl butane is correct.
Option C) 2-bromo­-3- methylbutane is incorrect.
Option D) 1-bromo­-3- methylbutane is incorrect.

Therefore, we can conclude that the correct answer to this question is option B.

Note: We should also know that Alkanes are far less reactive than alkenes. They will only react with bromine water in the presence of UV light. As we have already discussed alkanes undergo substitution reactions with halogens under these conditions, and will slowly decolorize bromine water.