# Monobromination of napthalene derivative

How does this reaction take place? Why does the bromine attach to that one particular location?

I do know that

• since is $\ce{FeBr3}$ is present only 1 $\ce{Br}$ will attach

• non-substituted areas of benzene will get $\ce{Br}$

• Observe the carbonyl group. It is important – Waylander May 22 '18 at 13:23
• Does it do resonance? i mean the carbonyl group – suyashsingh234 May 22 '18 at 13:33
• It does with the NH, but it also deactivates the right hand right – Waylander May 22 '18 at 14:04
• oh now i understand. On right hand side there is more negative charge so bromine prefers left side. – suyashsingh234 May 22 '18 at 15:33
• But why it not attach to place adjacent to ch3? Is it because of steric hindrance? – suyashsingh234 May 22 '18 at 15:34

You have electronically asymmetric rings. The right ring is deactivated by the carbonyl group that acts like an acyl group (remember that acyl groups are strong meta-deactivators). The left ring is activated by both $\ce{-CH3}$ and the nitrogen substituent acting as an amine (activating ortho/para directing), so bromination will occur on the left ring (since it is more nucleophilic).
Monobromination is not guaranteed experimentally. Although the right ring is deactivated under the right condition (at higher temperatures) you can end up with more than one $\ce{Br}$ being added. Maybe this question is just showing that a bromination will occur probably in that particular position.