Question

Halogenation takes place by free-radical mechanism. The values of $E$ for chain initiation and chain propagation steps and enthalpy of overall reaction of halogens with alkane are given. Match the values with the type of halogens.

Answer 1

(A) The halogenation reaction is as given below.
Chain-initiation step - $F_2\to 2\stackrel{.}{F}$
Chain propagation step -

i. $\stackrel{.}{F}+C{H}_4\to C{H}_3+HF$
ii. $C{H}_3+F_2\to C{H}_{3}F+\stackrel{.}{F}$
The values of E for chain initiation and chain propagation steps and enthalpy of overall reaction of halogens with alkane are 1591, 5.1 and 426.8 respectively.
(B) The halogenation reaction is as given below.
Chain-initiation step - $C{l}_2\to 2\stackrel{.}{C}l$
Chain propagation step -

i. $\stackrel{.}{Cl}+C{H}_4\to HCl=C{H}_3$

ii. $C{H}_3+C{l}_2\to C{H}_{3}Cl+\stackrel{.}{Cl}$
The values of E for chain initiation and chain propagation steps and enthalpy of overall reaction of halogens with alkane are 242.7, 15.9 and 102.5 respectively.
(C) The halogenation reaction is as given below.
Chain-initiation step - $B{r}_2\to 2Br$
Chain propagation step - i. $Br+C{H}_4\to C{H}_3=HBr$ ii. $C{H}_3+B{r}_2\to C{H}_{3}Br+Br$
The values of E for chain initiation and chain propagation steps and enthalpy of overall reaction of halogens with alkane are 192.5, 77.8 and 31.4 respectively.
(D) The halogenation reaction is as given below.
Chain-initiation step - $I_2\to 2\stackrel{.}{I}$
Chain propagation step - i. $I+C{H}_4\to C{H}_3+HI$
ii. $C{H}_3+I_2\to C{H}_{3}I+\stackrel{.}{I}$
The values of $E$ for chain initiation and chain propagation steps and enthalpy of overall reaction of halogens with alkane are 150.7, 140.2 and +46.0 respectively.