Br^{bullet} will abstract which of the hydrog | vedclass

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(A) The bromine radical $(Br^{\bullet})$ is highly selective in hydrogen abstraction reactions.It preferentially abstracts the hydrogen atom that lead

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$a$

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(A) The bromine radical $(Br^{\bullet})$ is highly selective in hydrogen abstraction reactions.It preferentially abstracts the hydrogen atom that leads to the formation of the most stable free radical.Comparing the positions:- Hydrogen '$a$' is on a tertiary carbon atom. Removal of this hydrogen results in a $3^{\circ}$ free radical.- Hydrogen '$b$' is on a secondary carbon atom.- Hydrogen '$c$' is on a primary carbon atom.- Hydrogen '$d$' is on a secondary carbon atom.Since the $3^{\circ}$ free radical is the most stable among the options,the bromine radical will most readily abstract the hydrogen at position '$a$'.

$Br^{\bullet}$ will abstract which of the hydrogen most readily?

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