Why n-butane undergoes chlorination at `25^(@)C` in the presence of light to form 72% 2-chlorobutane and 28% 1-chlorobutane?

Explain why chlorination of n-butane in presence of light at 298 K gives a mixture of 72 % 2-chlorobutane and 28 % of 1- chlorobutane.

If relative rates of substitution of 1^(@) and 2^(@) H are in the ratio 1 : 3.8, show that in the presence of light at 298 K, the chlorination of n-butane gives a mixture of 72% 2-chlorobutane and 28% 1-chlorobutane.

Statement-I: Free radical chlorination of n-butane gives 72% of 2-chlorobutane and 28% of 1-chlorobutane though it has six primary and four secondary hydrogens. Because Statement-II: A secondary hydrogen is abstracted more easily than the primary hydrogen.

Statement-I: Free radical chlorination of n-butane gives 72% of 2-chlorobutane and 28% of 1-chlorobutane though it has six primary and four secondary hydrogens. Because Statement-II: A secondary hydrogen is abstracted more easily than the primary hydrogen.

Statement-I: Free radical chlorination of n-butane gives 72% of 2-chlorobutane and 28% of 1-chlorobutane though it has six primary and four secondary hydrogens. Because Statement-II: A secondary hydrogen is abstracted more easily than the primary hydrogen.

Write the equations for the preparation of 1-iodo-butane from- (2) 1-chlorobutane

The product formed on reaction of n-butanol with SOCl_(2) in presence of pyridine is : chlorobutanol, 1-chlorobutane, chlorobutanone, 2-chlorobutane

2-chlorobutane obtained by chlorination of butane will be .