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15.6 Chlorination versus Bromination

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    28313
  • A Free Radical Substitution Reaction

    This page gives you the facts and a simple, uncluttered mechanism for the free radical substitution reaction between methane and bromine. This reaction between methane and bromine happens in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light.

    \[CH_4 + Br_2 \rightarrow CH_3Br + HBr\]

    The organic product is bromomethane. One of the hydrogen atoms in the methane has been replaced by a bromine atom, so this is a substitution reaction. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by bromine atoms.

    The mechanism

    The mechanism involves a chain reaction. During a chain reaction, for every reactive species you start off with, a new one is generated at the end - and this keeps the process going.The over-all process is known as free radical substitution, or as a free radical chain reaction.

    • Chain initiation: The chain is initiated (started) by UV light breaking a bromine molecule into free radicals.

    Br22Br

    • Chain propagation reactions: These are the reactions which keep the chain going.

    CH4 + BrCH3 + HBr

    CH3 + Br2CH3Br + Br

    • Chain termination reactions: These are reactions which remove free radicals from the system without replacing them by new ones.

    2BrBr2

    CH3 + BrCH3Br

    CH3 + CH3CH3CH3

    Selectivity

    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?

    \[CH_3CH_2CH_3 + 2Cl_2 \rightarrow \text{Four} \; C_3H_6Cl_2 \; \text{isomers} + 2 HCl\]

    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 suprising, 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

    Contributors

    Jim Clark (Chemguide.co.uk)

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