1.

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

Answer»

Solution :According to the questions,
`underset("n-Butane")(CH_(3)CH_(2)CH_(3)) underset("light")overset(Cl_(2),298K)to underset("2-Chlorobutane (72%)")(CH_(3)-underset(Cl)underset(|)(C)H-CH_(2)CH_(3))+underset("1-Chlorobutane (28%)")(CH_(3)CH_(2)CH_(2)CH_(2)-Cl)`
The RELATIVE ratios of these two isomerric chlorobutanes can be easily calculated by knowing: (i) the number and type of hydrogens (i.e., `1^(@),2^(@)` or `3^(@)`) to be substituted and (ii) their relative RATES of substitution (i.e., 1:3.8:5.0 for `Cl_(2)` at 298K). THUS,
`("1-Chlorobutane")/("2-Chlorobutane")=("No. of "1^(@)H)/("No. of "2^(@)H)toxx("Reactivity of "1^(@)H)/("Reactivity of "2^(@)H)=(6)/(4)XX(1)/(3.8)=(6)/(15.2)=(28%)/(72%)`


Discussion

No Comment Found

Related InterviewSolutions