Question

help please

OPT, 10PT EA): Provide the mechanism for the to nanism for the following transformations gewollt Br2, hv

Answer 1

The reactant alkane react with Br₂ by radical substitution under photochemical or UV light, the alkyl halide and HBr are formed. In general the alkane free radical reactivity order is tertiary > secondary > primary > methyl. The reaction proceeds via radical chain mechanism which involves radical intermediates following two key steps mechanism.

Step 1: Radical initiation process

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

Step 2: Product formation process

A bromine radical abstracts a hydrogen from the starting reactant alkane to form HBr and the tertiary alkyl radical which abstract a bromine atom from another molecule of Br₂ to form the desired product as alkyl bromide and another bromine radical.

The reaction mechanism of the given reaction is explained in the below image.

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