Question

Bromination of Aniline gives –

• A. Monobromoaniline
• B. Benzene nitrile
• C. Tribromoaniline
• D. Benzene isonitrile

Hint:

The –NH$_2$ group is strongly activating and directs substitution to ortho and para positions, leading to tribromoaniline in bromination.

Answer 1

The Correct Option is C

Step 1: Recall the effect of –NH$_2$ group.
Aniline (C$_6$H$_5$NH$_2$) contains an amino group (–NH$_2$), which is a strong activating group. It strongly activates the benzene ring towards electrophilic substitution, especially at ortho and para positions.

Step 2: Reaction with bromine.
When aniline is treated with bromine water, it undergoes bromination very easily, even without a catalyst. Instead of monobromination, it directly gives 2,4,6-tribromoaniline.
\[ C_6H_5NH_2 + 3Br_2 \; \longrightarrow \; C_6H_2Br_3NH_2 + 3HBr \]

Step 3: Identify product.
The product formed is tribromoaniline (2,4,6-tribromoaniline), which is a white precipitate.
Step 4: Final Answer.
Thus, bromination of aniline gives tribromoaniline.
\[ \boxed{\text{Tribromoaniline}} \]