Question

During the early stages of chlorination of propane, the reaction mixture is composed mainly of equal molar amounts of propane and $C{l}_2$(the products are still minor components )The irradiation of the mixture constantly produces a small concentration of $C{l}^{\ast }$ radicals Based on thermodynamic considerations ( and neglecting statistical factors) which of the following reactions is most likely to occur under these conditions?

• A. $C{l}^{\ast }+C{H}_3-C{H}_2-C{H}_3\to HCl+(C{H}_3{)}_{2}H{C}^{\ast }$
• B. $C{l}^{\ast }+C{H}_3-C{H}_2-C{H}_3\to HCl+C{H}_3-C{H}_2-H_2{C}^{\ast }$
• C. $C{l}^{\ast }+C{H}_3-C{H}_2-C{H}_3\to C{H}_{3}C{H}_{2}C{H}_{2}Cl+H^{\ast }$
• D. $C{l}^{\ast }+C{H}_3-C{H}_2-C{H}_3\to (C{H}_3{)}_{2}CHCl+H^{\ast }$

Answer 1

The correct option is A $C{l}^{\ast }+C{H}_3-C{H}_2-C{H}_3\to HCl+(C{H}_3{)}_{2}H{C}^{\ast }$

Most stable alkyl radical is formed in the first step of propagation. Free radicals increase in stability as the number of carbon substituents increases, from $methyl<primary<secondary<tertiary$ free radical. Therefore, the propagation step will be

$C{l}^{\ast }+C{H}_3-C{H}_2-C{H}_3\to HCl+(C{H}_3{)}_{2}H{C}^{\ast }$