Question

The correct order of decreasing acidity of nitrophenols will be

• A. $m$-nitrophenol $>\, p$-nitrophenol $>\, o$-nitrophenol
• B. $o$-nitrophenol $>\, m$- nitrophenol $>\, p$-nitrophenol
• C. $p$-nitrophenol $>\, $m- nitrophenol $>\, o$-nitrophenol
• D. $p$-nitrophenol $>\, $o-nitrophenol $>\, m$-nitrophenol

Answer 1

The Correct Option is D

$- NO _{2}$ group at o- and $p-$ position withdraws electrons of the $O - H$ bond towards itself by the stronger $-R$ effect while the $- NO _{2}$ group at $m$-position with draws electrons of the $O - H$ bcnd by the weaker -I effect. Thus, $o$ - and $p$ - nitrophenols are more acidic than $m$ - nitrophenol. Among $o$- and $p$ nitrophenols, o- nitrophenol is little less acidic than $p$-nitrophenol due to intramolecular H-bonding which makes loss of a proton little more difficult.

$\therefore $ The order of acidity of nitrophenols is
$p$- nitrophenol > $o$- nitrophenol > $m$- nitrophenol