Arrange the following in decreasing acidic strength.

Solution

Chemistry ma’am · 3 steps

Ask what each group does to the phenoxide ion's negative charge: pull it away, or push more in.

1. The rule. A phenol is more acidic when its phenoxide ion is more stable. Electron-withdrawing groups spread the negative charge and raise acidity. Electron-donating groups crowd more charge onto the ring and lower it.
2. Position matters. The resonance effect (R) reaches the oxygen only from ortho and para. From meta, only the inductive effect (I) acts, and it fades with distance.

   Phenol	Effect acting on O⁻	Net result	pKₐ
   A: para-NO₂	-R and -I	charge spread onto NO₂, most stable	7.1
   B: meta-NO₂	-I only	strong pull, but no resonance	8.4
   C: meta-OCH₃	-I only (no +R from meta)	weak pull, slightly above phenol	9.6
   D: para-OCH₃	+R beats -I	charge pushed in, least stable	10.2

3. So. Both nitrophenols beat both methoxyphenols, and in each pair the meta isomer sits between. The order is A > B > C > D, option (a).

The idea to keep. For substituted phenols, check the position first: ortho and para feel both resonance and induction, meta feels induction alone. OCH₃ donates from para but withdraws weakly from meta.

Answer from the NTA final answer key. Solution by PrepWiser.