Question

The chlorination of propane proceeds as a radical chain reaction. Sort the 7 reaction steps (you may need to scroll down to see them all) into categories of:

Answer 1

Given below are the reactons corresponding to the initiation, propagation and termination in free radical chlorination of propane.

The reaction begins with an initiation step, which is the separation of the halogen (X₂) into two radicals (atoms with a single unpaired electron) by the addition of uv light. This is called the initiation step because it initiates the reaction. The next step is propagation wherein, the first step is hydrogen abstraction by a chloride radical. A primary and secondary carbon radicals are formed. The carbon radical then combines with Cl2 to form alkyl halide and regenerating Cl radical. In the termination step two radicals of carbon or two radicals of Cl can combine to terminate the process. Additionally a carbon radical may combine with a Cl free radical to terminate the process.

The detailed reactions are shown below in the figure.

Answer 2

Concepts and reason

This problem is based on the concept of free radical substitution reactions.

Free radical substitution reaction is a type of reaction which substitutes a halogen atom in place of hydrogen from an alkane with the formation of a radical. This is an example of a photochemical reaction.

Fundamentals

Free radical reactions take place in various steps which includes the chain initiation step followed by chain propagation step and then the final chain termination step.

Chain initiation step of chlorination of propane is written as follows:

${\rm{C}}{{\rm{l}}_2}\longrightarrow{{hv}}2{\rm{C}}{{\rm{l}}^ \bullet }$

Chain propagation step of chlorination of propane is written as follows:

$\begin{array}{l}\\{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_3} + {\rm{C}}{{\rm{l}}^ \bullet } \to {\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{CH}}_2^ \bullet + {\rm{HCl}}\\\\{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{CH}}_2^ \bullet + 2{\rm{C}}{{\rm{l}}^ \bullet } \to {\rm{C}}{{\rm{H}}_3}{\rm{CHClC}}{{\rm{H}}_3} + {\rm{C}}{{\rm{l}}^ \bullet }\\\\{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_3} + {\rm{C}}{{\rm{l}}^ \bullet } \to {\rm{C}}{{\rm{H}}_3}\mathop {{\rm{CH}}}\limits^ \bullet {\rm{C}}{{\rm{H}}_3} + {\rm{HCl}}\\\end{array}$

Chain termination step of chlorination of propane is written as follows:

$\begin{array}{l}\\{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{CH}}_2^ \bullet + {\rm{C}}{{\rm{l}}^ \bullet } \to {\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_2}{\rm{Cl}}\\\\{\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{CH}}_2^ \bullet + {\rm{C}}{{\rm{H}}_3}\mathop {{\rm{CH}}}\limits^ \bullet {\rm{C}}{{\rm{H}}_3} \to {\rm{C}}{{\rm{H}}_3}{\rm{CH}}\left( {{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CHC}}{{\rm{H}}_3}} \right){\rm{C}}{{\rm{H}}_3}\\\\{\rm{C}}{{\rm{l}}^ \bullet } + {\rm{C}}{{\rm{l}}^ \bullet } \to {\rm{C}}{{\rm{l}}_2}\\\end{array}$

Ans:

The seven reactions step for the chlorination of propane is arranged as follows:

Answer 3

Given below are the reactons corresponding to the initiation, propagation and termination in free radical chlorination of propane.

The reaction begins with an initiation step, which is the separation of the halogen (X₂) into two radicals (atoms with a single unpaired electron) by the addition of uv light. This is called the initiation step because it initiates the reaction. The next step is propagation wherein, the first step is hydrogen abstraction by a chloride radical. A primary and secondary carbon radicals are formed. The carbon radical then combines with Cl2 to form alkyl halide and regenerating Cl radical. In the termination step two radicals of carbon or two radicals of Cl can combine to terminate the process. Additionally a carbon radical may combine with a Cl free radical to terminate the process.

The detailed reactions are shown below in the figure.