Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid give

Q: Why doesn't the diazonium ion attack the nitrogen of aniline?

While attack on nitrogen can happen to form a diazoamino compound, in an acidic medium, this intermediate rapidly rearranges to the more thermodynamically stable C-coupled product (p-aminoazobenzene).

Q: Why is the para product preferred over the ortho product?

The diazonium group is very bulky. Approaching the ortho position causes severe steric repulsion with the $-\text{NH}_2$ group, making the para position the kinetically and thermodynamically favored site.

Q: What happens if the medium is strongly acidic?

In a strongly acidic medium, the $-\text{NH}_2$ group of aniline gets protonated to form an anilinium ion ($-\text{NH}_3^+$). This group is strongly electron-withdrawing and deactivates the ring, stopping the coupling reaction.

Q: Why do we need an acidic medium at all?

A mildly acidic medium (pH 4-5) keeps the diazonium salt stable and prevents it from reacting with hydroxide ions to form a diazotate ion, which cannot act as an electrophile.

Q: What is the color of the final product and why?

The product, p-aminoazobenzene, is a yellow dye. The extended conjugation between the two benzene rings through the $-\text{N}=\text{N}-$ double bond lowers the energy gap for electronic transitions, allowing it to absorb blue light and appear yellow.

Question

Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives

Sigma Percentile · Accepted Answer

### The Setup: Meet the Reactants

Imagine you are in a laboratory, mixing two fascinating aromatic compounds: **benzene diazonium chloride** and **aniline**. The environment is kept mildly acidic using dilute hydrochloric acid. This setup is the classic recipe for an **azo coupling reaction**, a process that has historically revolutionized the dye industry.

### The Electrophile and The Nucleophile

In this chemical dance, the diazonium ion ($	ext{Ph}-	ext{N}_2^+$) acts as our electrophile. However, it is a relatively weak electrophile. To make the reaction happen, it needs a highly motivated partner—a strong nucleophile.

Enter aniline. The $-	ext{NH}_2$ group on aniline is a powerful electron-donating group due to its $+R$ (resonance) effect. It pumps electron density into the benzene ring, making it highly activated and eager to attack an electrophile. Because of this resonance, the electron density is highest at the **ortho** and **para** positions.

### The Battle of Positions: Ortho vs. Para

Now, the diazonium ion has a choice: should it attack the ortho position or the para position of aniline?

Here is where geometry plays a crucial role. The diazonium group is quite bulky. If it tries to approach the ortho position, it crashes into the electron clouds of the $-	ext{NH}_2$ group. This **steric hindrance** makes the ortho attack energetically unfavorable. Therefore, the electrophile takes the path of least resistance and targets the spacious **para** position.

### The Coupling Mechanism

The actual coupling is a beautiful electrophilic aromatic substitution. The pi electrons from the para position of aniline reach out and form a bond with the terminal nitrogen of the diazonium ion.

During this process, the aromaticity of the aniline ring is temporarily broken. To restore its stable aromatic state, the intermediate quickly loses a proton ($	ext{H}^+$), which pairs up with the chloride ion ($	ext{Cl}^-$) to eliminate a molecule of $	ext{HCl}$. The result is our final product: **p-aminoazobenzene**.

### The pH Catch: Why Dilute HCl?

You might wonder why we specifically use *dilute* $	ext{HCl}$ to maintain a mildly acidic medium (pH 4-5). There is a brilliant chemical reason for this.

If the solution were too alkaline, the diazonium ion would react with hydroxide ions to form a diazotate, destroying our electrophile. On the flip side, if the solution were too strongly acidic, the $-	ext{NH}_2$ group of aniline would grab a proton to become an anilinium ion ($-	ext{NH}_3^+$). The anilinium ion is strongly electron-withdrawing and completely deactivates the ring, bringing the coupling reaction to a dead stop.

By keeping the pH just right, we ensure both the electrophile and the nucleophile are in their optimal states to create the vibrant yellow dye, p-aminoazobenzene.

Animated Solution for Chemistry - Organic Chemistry: Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives

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Visualized Solution

$R e a c t an t s$

$R e a c t i o n T y p e$

$N u c l eo p hi l e$

$S t er i c H in d r an ce$

$M ec hani s m$

$F ina l P r o d u c t$

$K ey C o n d i t i o n$

The Sigma Insight: Amines

The Setup

The Electrophile and The Nucleophile

The Battle of Positions

The Coupling Mechanism

The pH Catch

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