Which of the following represents the correct order of the acidity in the given compounds?

Which of the following represents the correct order of the acidity in the given compounds?
  1. $\mathrm{FCH}_2 \mathrm{COOH} > \mathrm{CH}_2 \mathrm{COOH} > \mathrm{BrCH}_2 \mathrm{COOH} > \mathrm{ClCH}_2 \mathrm{COOH}$
  2. $\mathrm{BrCH}_2 \mathrm{COOH} > \mathrm{ClCH}_2 \mathrm{COOH} > \mathrm{FCH}_2 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{COOH}$
  3. $\mathrm{FCH}_2 \mathrm{COOH} > \mathrm{ClCH}_2 \mathrm{COOH} > \mathrm{BrCH}_2^2 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{COOH}$
  4. $\mathrm{CH}_3 \mathrm{COOH} > \mathrm{BrCH}_2 \mathrm{COOH} > \mathrm{ClCH}_2 \mathrm{COOH} > \mathrm{FCH}_2 \mathrm{COOH}$

Solution

Electron withdrawing substituent increases the acidity by increasing the ionic character of $-\mathrm{OH}$ by inductive effect. Electronegativity decreases in the order.
$\mathrm{F} > \mathrm{Cl} > \mathrm{Br}$
-I effect also decreases in the same order. Thus, the correct sequence is
$\begin{aligned}
& \mathrm{FCH}_2 \mathrm{COOH} > \mathrm{ClCH}_2 \mathrm{COOH} > \\
& \mathrm{BrCH}_2 \mathrm{COOH} > \mathrm{CH}_3 \mathrm{COOH}
\end{aligned}$
Related Theory
I effect is a permanent effect & generally represented by an arrow on the bond.

Asked in: NEET 2007

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