# Aromatic Apparent Substituion Reaction Mechanism

The first molecule reacts with $\ce{NaOH}$ to form the second molecule. I am told that it does't undergo via a substitution mechanism. I am asked to draw the mechanism for it.

I am thinking that the double bond will attack the OH group (addition reaction) so that it will add onto the meta carbon. Then the bromine group will leave (elimination reaction) which then the OH group attacks the para carbon.

• Why not a usual aromatic nucleophilic substitution? And what are the reaction conditions. The same reaction occurs with 1-bromo,4-nitro benzene at around 443K. So taking into account that ketonic groups are less electron withdrawing than nitro groups, may be at around 470-480 K the reaction may occur by aromatic nucleophilic substitution. See : en.wikipedia.org/wiki/Nucleophilic_aromatic_substitution – Varun Feb 16 '16 at 13:00
• @Varun Well the question states that it is not a substitution reaction but an addition reaction followed by an elimination reaction. All the question says is that it reacts with NaOH. – Nanoputian Feb 16 '16 at 20:00

The question is slightly misleading. It is usually referred to as a substitution; nucleophilic aromatic substitution to be precise; but it is actually an addition-elimination. What the question is trying to say is that it is not a regular $S_N1$/$S_N2$ substitution (which I think you correctly realised anyway).