Question

Consider the methyl substituted benzoic acids.

$1.$ PhCOOH

$2.$ $o-CH_{3}C_{6}H_{4}COOH$

$3.$ $p-CH_{3}C_{6}H_{4}COOH$

$4.$ $m-CH_{3}C_{6}H_{4}COOH$

The correct sequence of acidity is

• A. 1 < 2 < 3 < 4
• B. 2 < 3 < 4 < 1
• C. 3 < 4 < 1 < 2
• D. 3 < 4 < 2 < 1

Answer 1

Answer: (C)

Both p- and m-methyl groups are electron-donating and acid weakening. Hyperconjugation of $p-CH_{3}$ produces a little extra $\delta^{-}$ on the ring carbon, bearing the $COO^{-}$ . This results in more effective electron-donation and acid-weakening. This makes $p-CH_{3}C_{6}H_{4}COOH$ less acidic than $m-CH_{3}C_{6}H_{4}COOH$ .

$o-CH_{3}C_{6}H_{4}COOH$ is a stronger acid than benzoic acid, due to ortho-effect.

$p-CH_{3}C_{6}H_{4}COOH$ < $m-CH_{3}C_{6}H_{4}COOH$ < PhCOOH < $o-CH_{3}C_{6}H_{4}COOH$