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Which of the following represents the correct order of the acidity in the given compounds?
A
$FCH_2COOH \gt CH_3COOH \gt BrCH_2COOH \gt ClCH_2COOH$
B
$BrCH_2COOH \gt ClCH_2COOH \gt FCH_2COOH \gt CH_3COOH$
C
$FCH_2COOH \gt ClCH_2COOH \gt BrCH_2COOH \gt CH_3COOH$
D
$CH_3COOH \gt BrCH_2COOH \gt ClCH_2COOH \gt FCH_2COOH$
Detailed Solution
An electron withdrawing substituent increases the acidity of a carboxylic acid: by its $-I$ (inductive) effect it increases the polarity of the O-H bond and stabilises the carboxylate anion formed.
The halogen atoms are electron withdrawing, so all three haloacetic acids are stronger than acetic acid, in which $CH_3$ is an electron releasing group.
Electronegativity decreases in the order: F > Cl > Br
Hence the $-I$ effect also decreases in the same order.
Therefore the correct order of acidity is: $FCH_2COOH \gt ClCH_2COOH \gt BrCH_2COOH \gt CH_3COOH$
(The $pK_a$ values are about 2.6, 2.9, 2.9 and 4.8 respectively.)
The halogen atoms are electron withdrawing, so all three haloacetic acids are stronger than acetic acid, in which $CH_3$ is an electron releasing group.
Electronegativity decreases in the order: F > Cl > Br
Hence the $-I$ effect also decreases in the same order.
Therefore the correct order of acidity is: $FCH_2COOH \gt ClCH_2COOH \gt BrCH_2COOH \gt CH_3COOH$
(The $pK_a$ values are about 2.6, 2.9, 2.9 and 4.8 respectively.)