For the reaction, $C_{2} H_{5} O H + H X Z n X_{2} ⟶ C_{2} H_{5} X$ , the order of reactivity is:

H I > H C l > H B r

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H I > H B r > H C l

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H C l > H B r > H I

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H B r > H I > H C l

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Solution

The correct option is B $H I > H B r > H C l$
Because the reaction follows $S_{N} 1$ mechanism and hence the halide formed is stable when it is of a larger size and size increases down the group. $I > B r > C l$

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