Maximum hydration takes place of :

C F_{3} O | | C C F_{3}

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C H_{3} O | | C C H_{3}

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C H_{3} C | C l H - O | | C C H_{3}

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C_{6} H_{5} O | | C C_{6} H_{5}

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Solution

The correct option is B $C F_{3} O | | C C F_{3}$
The reaction is reversible and equilibrium shifts left, gives back stable aldehydes but the floral hydrate is a stable crystalline substance. The removal of $O H$ is difficult because of presence of electron withdrawing $C F_{3}$ group.
$O H$
$\Rightarrow C F_{3} - | C | - C F_{3}$
$O H$

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Similar questions

C H_{3} O ∥ C C H_{3}

C H_{3} O ∥ C O H

C H_{3} O ∥ C C H_{3} I_{2} / N a O H - ----- \to △, H^{+}

C H_{3} O ∥ C C H_{3} C r_{2} O_{7}^{2 -} / H^{+} - ------ \to

C H_{3} O ∥ C C H_{3} A g (N H_{3})_{2}^{+} - ----- \to

C H_{3} C H_{2} C H O

C H_{3} C | | O C H_{3}

C H_{3} - C H = C (C H_{3}) - C H (B r) - C H_{3}

C H_{3} - C (C H_{3}) = C H - C O O H

C H_{3} - C (C l) = C (C H_{3}) - C H (C_{2} H_{5}) - C H_{2} - C \equiv C H

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