Question

The decreasing order of acidity of the following phenol derivatives is

• A. $\mathrm{P}>\mathrm{Q}>\mathrm{R}>\mathrm{S}$
• B. $\mathrm{S}>\mathrm{R}>\mathrm{P}>\mathrm{Q}$
• C. $\mathrm{S}>\mathrm{R}>\mathrm{Q}>\mathrm{P}$
• D. $\mathrm{Q}>\mathrm{S}>\mathrm{R}>\mathrm{P}$

Answer 1

The correct option is D

$\mathrm{Q}>\mathrm{S}>\mathrm{R}>\mathrm{P}$

The explanation for the correct option:

Option (d): $\mathrm{Q}>\mathrm{S}>\mathrm{R}>\mathrm{P}$

• $\mathrm{Q}>\mathrm{S}>\mathrm{R}>\mathrm{P}$ is the order of acidity.
• Electron donating group decreases acidity.
• $-\mathrm{I}$ effect increases acidity.
• This is due to the fact that the conjugate base of the acid, $\mathrm{RCOO}-$, is stabilised by delocalisation of the generated negative charge if $\mathrm{R}$group is electron-withdrawing.
• If $\mathrm{R}$ group had donated electrons, the conjugate base would be destabilised due to inter-electronic repulsions.

• $+\mathrm{M}$ decreases acidity.
• When electrons or pi electrons are transported from one group to another, increasing the electron density of the conjugated system, this is referred to as the ($+\mathrm{M}$) effect or positive mesomeric effect.
• In P), $-{\mathrm{OCH}}_3$ shows $-\mathrm{I}$ and $+\mathrm{M}$
• In Q) $-{\mathrm{OCH}}_3$ shows $-\mathrm{I}$ only
• In R) $-{\mathrm{OCH}}_3$ shows $-\mathrm{I}$ (less than that in (P) ) and $+\mathrm{M}$
• In S) no effect happens.

Thus the correct order is $\mathrm{Q}>\mathrm{S}>\mathrm{R}>\mathrm{P}$.