For the compounds,
I) HCOOH II) $C H_{3} C O O H$ III) $C H_{3} C H_{2} C O O H$ IV) $C_{6} H_{5} C O O H$
The correct order of decreasing acidity is :

I > II > III > IV

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IV > I > II > III

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I > IV > II > III

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IV > II > III > I

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Solution

The correct option is D I > IV > II > III
To see the strength of acid here need to compare stability of conjugate bases of these acids. Higher $+ I$ effect of alkyl group $- C H_{3} C H_{2}$ makes $C H_{3} C H_{2} C O O^{-}$ least stable.

Presence of more electronegative atom stabilizes the negative charge.

$s p^{2}$ hybridized carbon shows a weaker $+ I$ effect as compared $s p^{3}$ carbon. Hence phenyl group is more electronegative as compared to $- C H_{3}$ (methyl group) but still less electronegative than hydrogen.

So, summarizing, the electronegativity order of the three substituents is $H > P h > C H_{3} > C H_{2} C H_{3}$ .
Therefore the acidity order is $H - C O O H > C_{6} H_{5} - C O O H > C H_{3} - C O O H > C H_{3} C H_{2} - C O O H$ .

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Similar questions

The correct order for decreasing stability of the following free radicals is

Order of decreasing acidity of

$(i) H C O O H (i i) C H_{3} C O O H (i i i) C l_{2} C H C O O H (i v) C F_{3} C O O H i s$

Order of decreasing acidity of the following is

(I) HCOOH

(II) ${C H}_{3} C O O H$

(III) ${C l}_{2} C H C O O H$

(IV) ${C F}_{3} C O O H$

H C O O H,

C H_{3} C O O H,

C l_{2} C H C O O H,

C F_{3} C O O H

(i) H C O O H (i i) P h - C O O H (i i i) P h - O H

(i v) C H_{3} C O O H

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