Question

Halogenation is a substitution reaction, where halogen replaces one or more hydrogens of hydrocarbon

Chlorine free radical make `1^(@), 2^(@), 3^(@)` radicals with almost equal ease, whereas bromine free radicals have a clear preference for the formation of tertiary free radicals. So, bromine is less reactive, and more selectrive whereas chlorine is less selective and more reactive.
The relative rate of abstraction of hydrogen by `Br^(**)` is
`underset(1600)(3^(@)) gt underset(82)(2^(@)) gt underset(1)(1^(@))`
`underset(5)(3^(@)) gt underset(3.8)(2^(@)) gt underset(1)(1^(@))`
`CH_(3)-overset(CH_(3))overset(|)(CH)-CH_(3)overset(Cl_(2)//hv)(to)` product
Major product in the above reactio is:
A. `CH_(3)-overset(CH_(3))overset(|)(CH)-CH_(2)-Cl`
B. `CH_(3)-overset(CH_(3))overset(|)underset(Cl)underset(|)(CH)-CH_(3)`
C. `CH_(3)-CH_(2)-CH_(2)-Cl`
D. `CH_(3)-underset(Cl)underset(|)(CH)-CH_(2)-CH_(3)`

Answer 1

Correct Answer - a
`o x-C-overset(C)overset(|)(C)-overset(C)overset(|)(C)-C`
`CH_(3)-overset(CH_(3))overset(|)(CH)-CH_(2)-Cl underset(1^(@))("Relative formation is") = 9xx1`, relative formation of `3^(@) = 5xx1`
`:.` %yield `= 9//14xx100`