Arrange the following compounds in decreasing order of their reactivity for hydrolysis reaction.

I > I I > I I I > I V

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I V > I I > I > I I I

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I I I > I V > I I > I

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I V > I I I > I I > I

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Solution

The correct option is B $I V > I I > I > I I I$
$I V$ has the highest reactivity due to formation of most stable carbocation.
Among all, $I I I$ has least reactivity due to formation of aromatic intermediate.
$I I$ has higher rate than $I$ due to formation of more stable carbocation (benzylic and $2^{0}$ carbocation).
So order is $I V > I I > I > I I I$ .

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