Question

Identify the major product from the following reaction sequence:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

In aromatic substitution, \(-CH_3\) is ortho-para directing and gives mainly para product during nitration due to steric reasons. After reduction, \(-NH_2\) strongly activates the ring and bromination occurs at available ortho positions to \(-NH_2\).

Answer 1

The Correct Option is D

Step 1: Nitration of toluene.
Toluene contains a methyl group \((-CH_3)\), which is an activating and ortho-para directing group.
When toluene is treated with \[ HNO_3/H_2SO_4 \] at \[ 288K \] nitration takes place mainly at the para position due to less steric hindrance.
Hence, the major product is \[ p\text{-nitrotoluene} \]

Step 2: Reduction of nitro group.
The nitro group \((-NO_2)\) is reduced by \[ Sn+HCl \] to an amino group \((-NH_2)\).
Thus, \[ p\text{-nitrotoluene} \rightarrow p\text{-toluidine} \] So, the compound now has \[ -CH_3 \] and \[ -NH_2 \] groups at para positions.

Step 3: Bromination with bromine water.
The \(-NH_2\) group is a strongly activating ortho-para directing group.
In \(p\)-toluidine, the para position to \(-NH_2\) is already occupied by the \(-CH_3\) group.
Therefore, bromination occurs at the two ortho positions with respect to the \(-NH_2\) group.
These positions are \[ 2 \] and \[ 6 \] relative to \(-NH_2\).
Thus, the product formed is \[ 2,6\text{-dibromo-}4\text{-methylaniline} \]

Step 4: Final conclusion.
Therefore, the major product is \[ \boxed{2,6\text{-dibromo-}4\text{-methylaniline}} \] Hence, the correct option is \[ \boxed{(4)} \]