Question

Why are there multiple organic or hydrocarbon products produced from the halogenation of alkanes via ultraviolet light? I am grading pretty big on the why part, and it is not only about the number of H's present.....(10 pts.)

Answer 1

The halogenation of alkanes in presence of ultraviolet (UV) light gives the multiple alkyl halide and hydrocarbons as products. The whole reaction proceeds through the free radical mechanism chain reaction via three steps like 1. radical initiation, 2. chain propogation and 3. chain termination.

In the free‐radical chain reaction of halogenation of alkane, initially by homolytic cleavages of weak halogen bond occure to form free radicals and alkane. As the reaction proceeds, the number of free radicals increases, while the number of alkane molecules decreases. At the end of the reaction, more free radicals exist than alkane molecules. The termination steps become the predominant reactions.

All of the free radical halogenation (bromination and chlorination) mechanism reactions occur very rapidly to form the products within microseconds.

The reactivity order: free-radical halogenation generally proceeds in the following order

Alkane R-H relative reactivity order : benzylic > tertiary > secondary > primary > methyl.

Halogen reactivity F₂ > Cl₂ > Br₂ > I₂ , free-radical iodination is usually not possible because iodine is too unreactive to form a radical.

The below is an example of the free radical bromination of methane in presence of UV light and its reaction mechanism explain the whole process and will address the why part of the question after clear understanding of the last step 3 of the process as chain termination. In the last step of chain termination, all the radicals formed are utilized in order to form the products like alkyl halides (methyl bromide), Br₂ and the higher hydrocarbon as ethane.

1. Radical initiation step:

Under UV light, the weak halogen bond to undergo homolytic cleavage to generate two bromine radicals and starting the chain process.

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2. Chain propogation step:

A bromine radical abstracts a hydrogen to form HBr and a methyl radical, then the methyl radical abstracts a bromine atom from another molecule of Br₂ to form the methyl bromide product and another bromine radical, which can then itself undergo reaction 2(a) creating a cycle that can repeat.

H. -H HOW

3. Chain termination:

Various reactions between the possible pairs of radicals allow for the formation of ethane, Br₂ or the product, methyl bromide. These reactions remove radicals and do not perpetuate the cycle.

CH CH, - CH-CH