Q.Predict the product of reaction of aniline with bromine in non-polar solvent such as $CS_2$.

The key idea here is that aniline’s amino group is strongly activating and ortho/para-directing, but in a non-polar solvent like $CS_2$, bromination occurs without the usual formation of the tribromo derivative.

Reasoning steps:

1. In polar solvents, aniline reacts with bromine to give 2,4,6-tribromoaniline because the $-\text{NH}_2$ group activates the ring so strongly that all three positions get substituted. …

The key here is understanding how the solvent changes the reaction pathway. Aniline ($C_6H_5NH_2$) is a highly activated aromatic ring because the lone pair on nitrogen donates electron density into the ring through resonance. In polar solvents (like water or acetic acid), this lone pair is fully available, making the ring so reactive that bromination happens rapidly at both ortho and para positions, often giving a tribromo product (2,4,6-tribromoaniline). But in a non-polar solvent like carbon disulfide ($CS_2$), the story is different.

Why? Because in non-polar solvents, the amino group's lone pair is not protonated and remains free, but the solvent does not stabilize any charged intermediates. More importantly, the reaction is carried out under controlled, mild conditions — typically using bromine in $CS_2$ at low temperature. This slows down the reaction and allows the most stable monosubstitution product to form.

Let's walk through the reasoning step by step.

1. Activation by the amino group

   The $—NH_2$ group is a strong ortho/para director. Its lone pair conjugates with the benzene ring, increasing electron density at the ortho and para positions. This makes the ring much more reactive than benzene toward electrophilic attack.

2. The role of the solvent

   In polar protic solvents, the amino group can get protonated to $—NH_3^+$, which is a strong deactivating group. But in $CS_2$, no protonation occurs. However, the non-polar solvent does not help stabilize the highly polar transition state of electrophilic substitution. This means the reaction is slower and more selective.

3. Steric hindrance at ortho positions

   The ortho positions are adjacent to the bulky $—NH_2$ group. In a slow, controlled reaction, the electrophile ($Br^+$) preferentially attacks the less hindered para position. The ortho attack is disfavored due to steric crowding.

4. Monosubstitution is the goal …

Concept: Activating and Directing Effects of Substituents in Electrophilic Aromatic Substitution

Aniline has an –NH₂ group, which is a strong activating and ortho/para-directing group. However, in polar solvents, bromine reacts so vigorously that it leads to tribromination (2,4,6-tribromoaniline). In a non-polar solvent like CS₂, the reaction can be controlled.

Method: Controlled Electrophilic Aromatic Substitution in Non-Polar Solvent

Steps:

1. Identify the substrate and reagent

   • Substrate: Aniline (C₆H₅NH₂)
   • Reagent: Bromine (Br₂)
   • Solvent: Carbon disulfide (CS₂) — non-polar, aprotic

2. Recognize the directing effect

   • The –NH₂ group donates electrons via resonance, making the ortho and para positions highly reactive toward electrophiles.

3. Role of the solvent

   • In CS₂, the reaction is mild — no excess Br₂ or polar medium to force complete substitution.
   • Only monobromination occurs at the para position (less steric hindrance than ortho).

4. Write the product …

The Core Concept

Aniline ($C_6H_5NH_2$) is highly activating for electrophilic aromatic substitution because the $-\text{NH}_2$ group donates electrons via resonance. Normally, in polar solvents (like water), bromination happens at the ortho and para positions — and so vigorously that tribromination occurs, giving 2,4,6-tribromoaniline.

But here, the solvent is non-polar ($CS_2$). That changes everything.

Common Mistake #1: Predicting tribromination (2,4,6-tribromoaniline)

Why students do this:

They memorise that aniline + bromine water gives 2,4,6-tribromoaniline (white precipitate). They apply this blindly without checking the solvent.

How to avoid:

Always check the solvent before predicting the product.

• In polar solvents (water, alcohol): tribromination occurs.
• In non-polar solvents ($CS_2$, $CCl_4$): only monobromination occurs — and at the para position (steric hindrance at ortho).

✓ Correct product: p-bromoaniline (4-bromoaniline)

Common Mistake #2: Predicting ortho-bromoaniline as the major product

Why students do this:

They know ortho/para directing, but forget that the bulky $-\text{NH}_2$ group and the incoming bromine both cause steric hindrance at the ortho position.

How to avoid:

Remember: ortho positions are sterically hindered by the amino group. In a controlled, mild reaction (non-polar solvent, low temperature), para product dominates due to less steric clash.

✓ Major product: para-bromoaniline (with a tiny amount of ortho, but exam expects para)

Common Mistake #3: Forgetting to protect the amino group

Why students do this:

They think the reaction proceeds exactly like in water, but in $CS_2$, the amino group is not protonated (no acid present). So it remains a strong activator.

How to avoid:

• In water with $Br_2$, the $-\text{NH}_2$ gets protonated to $-\text{NH}_3^+$ (meta-directing, deactivating) — but that's not the case here.
• In $CS_2$, the amino group is free, so it's strongly activating and ortho/para directing.

✓ So the reaction is faster and milder than in water — but still only monobromination.

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