Q.A solution contains 1 g mol. each of p-toluene diazonium chloride and p-nitrophenyl diazonium chloride. To this 1 g mol. of alkaline solution of phenol is added. Predict the major product. Explain your answer.

The key idea is Electrophilic Aromatic Substitution — specifically, the coupling reaction of diazonium salts with phenol, where the more electrophilic diazonium cation reacts preferentially.

Reasoning:

1. Both diazonium salts are electrophiles, but the nitro group ($-\text{NO}_2$) is strongly electron-withdrawing. This makes the $p$-nitrophenyl diazonium cation a stronger, more reactive electrophile — the electron-withdrawing nitro group pulls electron density toward itself and stabilises the developing negative character in the transition state, making the adjacent $-\text{N}_2^+$ carbon more electrophilic. The methyl group on the $p$-toluene diazonium cation is electron-donating and does the opposite, making that diazonium salt a weaker electrophile.
2. Phenol is highly activated toward electrophilic attack at the para position (relative to $-\text{OH}$), but only 1 mol of phenol is available for the 2 mol of diazonium salt present — so the two diazonium salts compete for it. …

Concept and Intuition

This is a classic azo coupling reaction — a special case of electrophilic aromatic substitution (EAS). In EAS, an electron-deficient species (the electrophile) attacks an electron-rich aromatic ring. Here, the electrophiles are diazonium cations ($\ce{Ar-N2^+}$), and the aromatic substrate is phenol.

Why phenol? Because the –OH group is strongly activating and ortho/para-directing. It donates electron density into the ring via resonance, making the ring highly nucleophilic — especially at the para position (and to a lesser extent, ortho). This is why phenol couples readily with diazonium salts under mild alkaline conditions.

Now, the twist: we have two different diazonium salts — one with an electron-donating methyl group ($p$-toluene diazonium chloride) and one with an electron-withdrawing nitro group ($p$-nitrophenyl diazonium chloride). The nitro group makes the diazonium cation more electron-deficient (more electrophilic) because it pulls electron density away from the $\ce{-N2^+}$ group through resonance and induction. The methyl group does the opposite — it slightly stabilizes the positive charge, making the diazonium cation less reactive.

So, in a competition for the single mole of phenol, the more reactive electrophile wins. Let's work through the numbers.

Step-by-Step Reasoning

1. Identify the reactants and their amounts.

   We have 1 g mol. each of two diazonium salts:

   • $p$-toluene diazonium chloride: $\ce{CH3-C6H4-N2^+ Cl-}$
   • $p$-nitrophenyl diazonium chloride: $\ce{O2N-C6H4-N2^+ Cl-}$ And 1 g mol. of phenol ($\ce{C6H5OH}$) in alkaline solution. So the molar ratio of total diazonium salt to phenol is 2:1.

2. Understand the coupling mechanism.

   In alkaline medium, phenol exists partly as the phenoxide ion ($\ce{C6H5O-}$), which is even more electron-rich than phenol itself. The diazonium cation attacks the para position of the phenoxide ring (or ortho if para is blocked, but here it's free). The product is an azo dye — a compound with the $\ce{-N=N-}$ bridge.

   The reaction is:

Only one mole of phenol is available, so only one mole of diazonium salt can react. The other mole remains unreacted (or may undergo side reactions like decomposition, but that's not the focus).

3. Compare the electrophilicity of the two diazonium salts.

   The reactivity of an aryldiazonium cation depends on substituents on the aromatic ring:

   • Electron-withdrawing groups (like $\ce{-NO2}$) increase the positive charge on the $\ce{-N2^+}$ group, making it a stronger electrophile.
   • Electron-donating groups (like $\ce{-CH3}$) decrease the positive charge, making it a weaker electrophile.

   So, $p$-nitrophenyl diazonium chloride is more reactive than $p$-toluene diazonium chloride. …

Method: Diazonium Coupling Reaction (Electrophilic Aromatic Substitution)

This is a classic electrophilic aromatic substitution where the diazonium ion acts as the electrophile and the phenol acts as the nucleophilic aromatic ring.

Steps to predict the major product:

1. Identify the electrophiles

   • Two diazonium ions are present:
     • p-toluene diazonium chloride (has an electron-donating –CH₃ group)
     • p-nitrophenyl diazonium chloride (has an electron-withdrawing –NO₂ group)

2. Identify the nucleophile

   • Phenol ($C_6H_5OH$) is added in alkaline medium.
   • In alkaline solution, phenol is converted to phenoxide ion ($C_6H_5O^-$), which is a much stronger activating group than –OH.
   • The phenoxide ion activates the ortho and para positions strongly.

3. Compare electrophile reactivity

   • The diazonium ion with the –NO₂ group is more electron-deficient (more electrophilic) because –NO₂ withdraws electrons by both inductive and resonance effects.
   • The diazonium ion with –CH₃ group is less electrophilic because –CH₃ donates electrons, reducing the positive charge on the diazonium nitrogen.

4. Determine the major product

   • The more reactive electrophile (p-nitrophenyl diazonium) will attack the activated phenoxide ring first and preferentially. …

Here’s a clear breakdown of the common mistakes students make on this specific Electrophilic Aromatic Substitution (EAS) problem, and how to avoid each.

🧪 The Core Concept: Competing Diazonium Coupling

This is a diazonium coupling reaction (a type of EAS). The key idea: two different diazonium salts compete to attack the same phenol. The major product is determined by which diazonium ion is the stronger electrophile.

✗ Mistake #1: Ignoring the Effect of Substituents on the Diazonium Salt

What students do wrong:

They treat both diazonium salts as equally reactive. They assume the product is a mixture of both possible azo dyes in equal amounts.

Why it’s wrong:

The substituent on the benzene ring of the diazonium salt dramatically changes its electrophilicity:

• Electron-withdrawing groups (EWG) like $-\text{NO}_2$ increase the positive charge on the diazonium nitrogen, making it a stronger electrophile.
• Electron-donating groups (EDG) like $-\text{CH}_3$ decrease the positive charge, making it a weaker electrophile.

How to avoid:

Always check the substituent on the diazonium salt.

• $p$-Nitrophenyl diazonium chloride ($\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{N}_2^+$) → stronger electrophile (EWG).
• $p$-Toluene diazonium chloride ($\text{CH}_3-\text{C}_6\text{H}_4-\text{N}_2^+$) → weaker electrophile (EDG).

Correct reasoning: The $p$-nitro derivative will couple faster and dominate the product.

✗ Mistake #2: Forgetting the Reaction Conditions (Alkaline Phenol)

What students do wrong:

They treat phenol as if it reacts in neutral or acidic conditions.

Why it’s wrong:

In alkaline solution, phenol is converted to phenoxide ion ($\text{C}_6\text{H}_5\text{O}^-$). The phenoxide ion is much more reactive than phenol itself because the negative charge is delocalized into the ring, making it a stronger nucleophile for EAS.

How to avoid:

Remember:

• In alkaline medium, phenol → phenoxide ion → enhanced reactivity.
• The coupling occurs at the para position (or ortho if para is blocked) of the phenoxide ring.

Correct reasoning: The reaction is fast and selective because of the activated phenoxide.

✗ Mistake #3: Predicting a Mixture of Both Products

What students do wrong:

They think since both diazonium salts are present in equal moles, both will react equally, giving a 50:50 mixture.

Why it’s wrong: …