Question

The major product X in the following reaction is:
 ${(C{H_3})_2} - CH - C{H_2} - C{H_3}\, + \,B{r_2}\,\xrightarrow{{h\nu }}\,X$
A) ${(C{H_3})_2} - CH - C{H_2} - C{H_2} - Br$
B) ${(C{H_3})_2} - CH - CH(Br) - C{H_3}$
C) ${(C{H_3})_2} - C(Br) - C{H_2} - C{H_3}$
D) $C{H_2}(Br) - CH(C{H_3}) - C{H_2} - C{H_3}$

Answer 1

Hint:In order to answer this question, you must recall the concept of Free Radicals and Free Radical Reactions. And use this reaction to evaluate the product formed. Follow the mechanism properly and follow the rules in order to calculate the major product, and select the correct option.


Complete answer:
Free Radicals and Free Radical Reactions:Free radicals are chemical species that contain a singly occupied orbital. They are neutral and tend to be highly reactive. These are formed by homolytic cleavage of covalent bonds. These have a high tendency to pair the unpaired electron. These readily combine with each other or with other molecules to pair their electrons.
In a free radical reaction, free readily is formed as reaction intermediate. It is the reaction which involves free radicals. These reactions take place mainly in the presence of UV light.
To evaluate the product, follow the below stated procedure:
Step 1: By the action of UV(sunlight), the bromine will split into free radicals by homolytic cleavage of covalent bonds:
Step 2: Then the bromine free radical will attack on ${(C{H_3})_2} - CH - C{H_2} - C{H_3}\,$ and there are four possible products:
1-bromo-3-methylbutane (A)
2-bromo-3-methylbutane (B)
2-bromo-2-methylbutane (C)
1-bromo-2-methylbutane(D)
But bromine is highly regioselective in where it attacks.
The relative rates of attack at the different types of $H$ atom are \[3^\circ :2^\circ :1^\circ {\text{ }} = {\text{ }}1640:82:1\] .
 And therefore, The major product obtained in the bromination of ${(C{H_3})_2} - CH - C{H_2} - C{H_3}\,$ is ${(C{H_3})_2} - C(Br) - C{H_2} - C{H_3}$

Hence the correct answer is option C.


Note:Homolytic cleavage is the breaking of a covalent bond in such a way that each fragment gets one of the shared electrons. The word homolytic comes from the Greek homoios, "equal", and lysis, "loosening".