Question

Explain why chorination of n butane in the presence of light gives a mixture of 72 % of 2 chlorobutane and 28 % of 1 chlorobutane while bromination gives 98 % of 2 bromobutane and 2 % of bromobutane

Answer 1

The explanation can be provided as follows:

• Based on the question, we can write the reaction as :

CH3CH2CH2CH3 → CH3-CH-CH2-CH3 + CH3CH2CH2CH2-Cl

                                             I                                                                                    

                                            Cl

n-butane                           2-Chlorobutane (72%)                1-Chlorobutane (28%)

• If we know the the number and type of hydrogens and their relative rates of substitution, then we can calculate the relative ratios of 2-Chlorobutane and 1-Chlorobutane.
• 1-chlorobutane/ 2-Chlorobutane=

No.of 1 H/No.of 2 H x reactivity of 1 H/Reactivity of 2 H

• 6/4 x 1/3.8
• 28 percent/72 percent

Answer 2

Explanation:

Hope this helps you !!!!!!!!!