Q.Ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.

The key idea is that the nitro group stabilises the phenoxide ion through resonance, but only when it is ortho or para to the oxygen — meta nitrophenol is not more acidic than phenol for this reason.

Reasoning:

1. Deprotonation of phenol gives the phenoxide ion, where the negative charge is delocalised into the ring (ortho and para positions carry partial negative charge).
2. A nitro group at the ortho or para position can directly accept this negative charge via resonance, forming additional stable structures where the negative charge is on the highly electronegative oxygen atoms of the nitro group. …

Why this happens -- the concept

Acidity is all about the stability of the conjugate base. For phenols, the conjugate base is the phenoxide ion. Phenol itself is weakly acidic because the negative charge on oxygen can be delocalised into the benzene ring. But when a nitro group ($-\mathrm{NO}_2$) is present, it is a strongly electron-withdrawing group -- it pulls electron density away from the ring. This further stabilises the phenoxide ion, making the corresponding phenol more acidic.

The key is where the nitro group is attached. The nitro group withdraws electrons both by induction and by resonance. The resonance effect is especially powerful when the nitro group is at the ortho or para position because the negative charge on the phenoxide oxygen can be delocalised onto the nitro group itself. At the meta position, this direct conjugation is not possible -- the negative charge cannot reach the nitro group through resonance.

Step-by-step reasoning

1. Start with the phenoxide ion from phenol itself. The negative charge on oxygen can be delocalised into the ring, giving resonance structures where the charge appears at the ortho and para positions (relative to the $-\mathrm{O}^-$ group). This is why phenol is more acidic than a simple alcohol -- the charge is spread out.

2. Now consider ortho-nitrophenol. The nitro group is at the ortho position relative to the $-\mathrm{OH}$ group. When the phenol loses a proton, the phenoxide ion forms, and its negative charge can be delocalised onto the nitro group through resonance:

   • One resonance structure has the negative charge on the phenoxide oxygen, with the ring drawn in its normal alternating-bond form.
   • A second resonance structure moves the negative charge onto the ring carbon that bears the nitro group (the ortho carbon), with the ring's double bonds shifted accordingly.
   • A third, key resonance structure pushes that charge further onto one of the two oxygen atoms of the nitro group itself, with the nitrogen now bearing a formal positive charge (as in the nitro group's own normal resonance form, $-\overset{+}{N}(=O)(-O^-)$).

   The result: the negative charge is spread over three oxygen atoms (the phenoxide oxygen and the two oxygens of the nitro group). This is much more stable than the phenoxide ion from phenol, where the charge is only on the ring carbons and one oxygen.

3. Now consider para-nitrophenol. Exactly the same logic applies, but now the nitro group is at the para position. The resonance delocalisation works just as well -- the negative charge travels through the ring (via the para carbon this time) and ends up on the nitro group's oxygens, again spread over three oxygen atoms in the most stabilised resonance form.

4. Now consider meta-nitrophenol (for contrast). Here, the nitro group is at the meta position. When you try to draw resonance structures that put the negative charge on the carbon bearing the nitro group, you find it is impossible -- the meta position is not connected to the oxygen by a conjugated path that allows the charge to reach the nitro group. The nitro group can still withdraw electrons inductively (through sigma bonds), but the powerful resonance stabilisation is absent. So meta-nitrophenol is less acidic than ortho- and para-nitrophenols, though still more acidic than phenol itself.

The resonance structures of the phenoxide ions, described …

Method: Resonance Stabilisation Analysis of Conjugate Bases

This method explains why ortho- and para-nitrophenols are more acidic than phenol by comparing the stability of their conjugate bases (phenoxide ions) through resonance.

Step 1: Draw the resonance structures of phenoxide ion (the conjugate base of phenol)

The phenoxide ion has the negative charge delocalised into the ring:

    O⁻
    |
    ══╗
   ║   ║
    ══╝

Resonance structures (only the major ones):

1. Original structure with negative charge on oxygen.
2. Charge delocalised to ortho position (C-2).
3. Charge delocalised to para position (C-4).
4. Charge delocalised to the other ortho position (C-6).

Key observation: In phenol, the negative charge is delocalised only onto ortho and para carbons — not onto the meta carbon.

Step 2: Draw the resonance structures of ortho-nitrophenoxide ion

The nitro group ($-NO_2$) is a strong electron-withdrawing group by both inductive and resonance effects. When present at the ortho position, it can directly accept the negative charge via resonance.

Resonance structures (focus on the key extra stabilisation):

• Negative charge on oxygen delocalises into the ring as in phenol.
• Additional structure: The negative charge moves onto the nitro group oxygen:

    O⁻
    |
    ══╗
   ║   ║
    ══╝
    |
    N⁺
   / \
  O   O⁻

This structure places the negative charge on an electronegative oxygen of the nitro group — this is highly stabilising.

Step 3: Draw the resonance structures of para-nitrophenoxide ion

Exactly analogous to ortho — the nitro group at the para position can also accept the negative charge via resonance:

• Same delocalisation as phenol.
• Additional structure: Negative charge moves onto the nitro group oxygen at the para position.

Critical point: In meta-nitrophenol, the nitro group cannot accept the negative charge through resonance because the negative charge never reaches the meta position. Hence, meta-nitrophenol is less acidic than ortho- and para-nitrophenols.

Step 4: Compare stability of conjugate bases …

Here is a breakdown of the common mistakes students make on this specific Electrophilic Aromatic Substitution (EAS) concept, along with how to avoid them.

Mistake 1: Drawing the wrong resonance for the phenoxide ion

The Error: Students often draw resonance structures for the neutral phenol molecule instead of the phenoxide ion (the conjugate base). The question explicitly asks for the phenoxide ion.

Why it’s wrong: The acidity of phenol is determined by the stability of its conjugate base (the phenoxide ion). The negative charge on the oxygen can be delocalized into the ring. If you draw the neutral molecule, you miss the entire point of charge stabilization.

How to Avoid:

• Read the question carefully. Underline "phenoxide ions."
• Start with the correct structure: Draw the benzene ring with an $\ce{O-}$ (negative charge) attached.
• Remember the rule: The negative charge moves into the ring via resonance. The oxygen atom becomes a double bond to the ring, pushing the negative charge onto a carbon atom (usually ortho or para to the oxygen).

Mistake 2: Forgetting the nitro group's role in the phenoxide ion

The Error: Students draw the same resonance structures for ortho-nitrophenoxide and para-nitrophenoxide as they do for plain phenoxide. They fail to show how the nitro group ($\ce{-NO2}$) specifically stabilizes the negative charge.

Why it’s wrong: The nitro group is a strong electron-withdrawing group (EWG) via both induction and resonance. Its key role is to directly accept the negative charge from the ring. If you don't show the negative charge on the nitro group's oxygen atoms, you haven't explained why ortho/para nitrophenols are more acidic.

How to Avoid:

• Identify the special resonance: When the negative charge from the phenoxide ion lands on the carbon ortho or para to the nitro group, it can be delocalized onto the nitro group itself.
• Draw the extra step: Show the double bond from the ring carbon to the nitrogen of the $\ce{-NO2}$ group, breaking one of the N=O bonds and placing the negative charge on an oxygen of the nitro group.
• Key insight: This creates a quinoid-like structure with the negative charge on a highly electronegative oxygen atom. This is a major contributor to stability.

Mistake 3: Not showing the complete set of resonance structures

The Error: Students draw only 2-3 resonance structures and stop. They miss the crucial structure where the negative charge is on the nitro group.

Why it’s wrong: The exam expects you to show all significant contributors. The structure with the charge on the nitro group is the most important one for explaining the increased acidity. Missing it means you lose marks and fail to demonstrate the core concept.

How to Avoid:

• Use a systematic method:
  1. Start with the phenoxide ion (charge on O).
  2. Move the charge to the ortho and para positions of the ring (3 structures for plain phenoxide).
  3. For ortho-nitrophenoxide: When the charge is on the carbon ortho to the nitro group, draw the extra resonance onto the nitro group.
  4. For para-nitrophenoxide: When the charge is on the carbon para to the nitro group, draw the extra resonance onto the nitro group.
• Count them: For para-nitrophenoxide, you should have 5 significant resonance structures. For ortho-nitrophenoxide, you should have 5 as well (including the one on the nitro group).

Mistake 4: Confusing ortho and para directing effects with acidity

The Error: Students think that because the nitro group is a meta-director in EAS, it cannot stabilize a negative charge in the ortho or para position. …