Arrange the following in the order of increasing value of the equilibrium constant for hydration, $K_{h y d}$ (smallest value first).

2 < 1 < 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 < 1 < 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 < 2 < 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 < 3 < 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $3 < 1 < 2$
As the polarity of $C O$ bond decreases hydration decreases. In compound 3 polarity is least due to $+ I$ effect of tert-butyl. Compound 2 is most polar among the given compounds as $+ I$ effect will be less in 2 compared to 1. So the order would be $3 < 1 < 2$

Suggest Corrections

Similar questions

K_{h y d .}

\frac{1}{2}

\frac{2}{2}

\frac{1}{3}

\frac{2}{3}

+ I

1^{\circ} Alkyl Group, 2^{\circ} Alkyl Group, 3^{\circ} Alkyl Group and Methyl Group

10 D q .

[C r (e n)_{3}]^{3 +}

[C r (o x)_{3}]^{3 -}

[C r F_{6}]^{3 -}

[C R (H_{2} O)_{6}]^{3 +}

N a^{+}, M g^{2 +}

A l^{3 +}

Join BYJU'S Learning Program