Arrange the following in the decreasing order of their acidic strengths. Explain the arrangement. $C_{6} H_{5} C O O H, F C H_{2} C O O H, N O_{2} C H_{2} C O O H$

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Solution

Factors affecting acidic strength

Acidic strength is governed by the stability of the conjugate base. So, the higher is the stability of the conjugate base, the higher will be its acidic strength.

Generally, electron-withdrawing groups increase the stability of the conjugate base, while electron-donating groups decrease the stability of the conjugate base.

Conjugate base of the compounds

$C_{6} H_{5} C O O H \to C_{6} H_{5} C O O^{-} + H^{+}$

$F C H_{2} C O O H \to F C H_{2} C O O^{-} + H^{+}$

$N O_{2} C H_{2} C O O H \to N O_{2} C H_{2} C O O^{-} + H^{+}$

Order of stability of conjugate bases of the given compound

Electron-withdrawing ability $(- I$ effect) of $N O_{2}, F a n d C_{6} H_{5}$ are as follows: $- N O_{2} > - F > - C_{6} H_{5}$

The order of stability of the conjugate bases is: $C_{6} H_{5} C O O^{-} < F C H_{2} C O O^{-} < N O_{2} C H_{2} C O O^{-}$

Decreasing order of acidic strengths of the given compounds

Since acidic strength depends on the stability of the conjugate base, the decreasing order of acidic strengths is: $N O_{2} C H_{2} C O O H > F C H_{2} C O O H > C_{6} H_{5} C O O H$

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