The correct order of the rate of dehydrohalog | vedclass

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(C) Dehydrohalogenation involves the elimination of a hydrogen atom and a halogen atom from adjacent carbon atoms to form an alkene in the presence of

Vedclass · Accepted Answer

$III > I > II$

Vedclass · Answer

(C) Dehydrohalogenation involves the elimination of a hydrogen atom and a halogen atom from adjacent carbon atoms to form an alkene in the presence of a strong base like alcoholic $KOH$.For a series of alkyl halides with the same alkyl group,the rate of dehydrohalogenation depends primarily on the strength of the $C-X$ bond and the leaving group ability of the halide ion.The leaving group ability follows the order: $I^- > Br^- > Cl^- > F^-$.Since the iodide ion $(I^-)$ is the best leaving group among the given halides,the alkyl iodide $(III)$ will undergo dehydrohalogenation the fastest.The chloride ion $(Cl^-)$ is the poorest leaving group among the three,so the alkyl chloride $(II)$ will react the slowest.Therefore,the correct order of the rate of dehydrohalogenation is $(III) > (I) > (II)$.

The correct order of the rate of dehydrohalogenation of the following halides in the presence of alcoholic $KOH$ is:

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