Chlorination of other alkanes

When alkanes larger than ethane are halogenated, isomeric products are formed. Thus chlorination of propane gives both 1-chloropropane and 2-chloropropane as mono-chlorinated products. Four constitutionally isomeric dichlorinated products are possible, and five constitutional isomers exist for the trichlorinated propanes. Can you write structural formulas for the four dichlorinated isomers?

[CH_3CH_2CH_3 + 2Cl_2 ightarrow  ext{Four} ; C_3H_6Cl_2 ; ext{isomers} + 2 HCl]

The halogenation of propane discloses an interesting feature of these reactions. All the hydrogens in a complex alkane do not exhibit equal reactivity. For example, propane has eight hydrogens, six of them being structurally equivalent primary, and the other two being secondary. If all these hydrogen atoms were equally reactive, halogenation should give a 3:1 ratio of 1-halopropane to 2-halopropane mono-halogenated products, reflecting the primary/secondary numbers. This is not what we observe. Light-induced gas phase chlorination at 25 ºC gives 45% 1-chloropropane and 55% 2-chloropropane.

CH₃-CH₂-CH₃   +   Cl₂   →  45% CH₃-CH₂-CH₂Cl   +   55% CH₃-CHCl-CH₃

The results of bromination (light-induced at 25 ºC) are even more surprising, with 2-bromopropane accounting for 97% of the mono-bromo product.

CH₃-CH₂-CH₃   +   Br₂  →  3% CH₃-CH₂-CH₂Br   +   97% CH₃-CHBr-CH₃

These results suggest strongly that 2º-hydrogens are inherently more reactive than 1º-hydrogens, by a factor of about 3:1. Further experiments showed that 3º-hydrogens are even more reactive toward halogen atoms. Thus, light-induced chlorination of 2-methylpropane gave predominantly (65%) 2-chloro-2-methylpropane, the substitution product of the sole 3º-hydrogen, despite the presence of nine 1º-hydrogens in the molecule.

(CH₃)₃CH   +   Cl₂   →  65% (CH₃)₃CCl   +   35% (CH₃)₂CHCH₂Cl

It should be clear from a review of the two steps that make up the free radical chain reaction for halogenation that the first step (hydrogen abstraction) is the product determining step. Once a carbon radical is formed, subsequent bonding to a halogen atom (in the second step) can only occur at the radical site. Consequently, an understanding of the preference for substitution at 2º and 3º-carbon atoms must come from an analysis of this first step.