Increasing order of equilibrium constant for the formation of a hydrate is:

i < ii < iii < iv

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iv < ii < i < iii

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ii < iv < iii < i

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iv < ii < iii < i

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Solution

The correct option is D iv < ii < iii < i
In the formation of hydrate, a compound reacts with water to form a product. $N O_{2}$ has $- I$ effect and will decrease the electron density on oxygen. Hence, $(i v)$ will be less reactive towards hydration. $O C H_{3}$ also has $- I$ effect at meta position. So, $(i i)$ will also be less reactive for the formation of hydrate. $(i i i)$ does not have an electron withdrawing group attached to it, therefore, it will be more reactive than $(i i)$ and $(i v)$ . In $(i i)$ , $O C H_{3}$ is present at para position and will increase the electron density on the oxygen atom, hence will increase its reactivity towards hydration.
Compounds which are more reactive will have a high value of the equilibrium constant.

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