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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.

I am not really sure if that is correct. Could someone please help me out. Thanks.

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    $\begingroup$ 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 $\endgroup$ – Varun Feb 16 '16 at 13:00
  • $\begingroup$ @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. $\endgroup$ – Nanoputian Feb 16 '16 at 20:00
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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).

The mechanism involves attack at the carbon attached to the bromine to form an intermediate anion which is stabilised by the electron withdrawing acetyl group, para to the site of attack.

I'll let you draw the arrows and come up with a complete mechanism.

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  • $\begingroup$ thanks for the help. I think I got the mechanism for the reaction. Am I correct in saying that there is a negative charge on the meta carbon when the OH adds on the para carbon? If so, I not really sure how it stabilises the molecule because doesn't the electron withdrawing group form positive charges on the ortho and para carbons, so would it be better to have a negative charge on those carbons? $\endgroup$ – Nanoputian Feb 16 '16 at 20:22
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    $\begingroup$ @Nanoputian Take a look at the mechanism here for a very similar reaction. Note the ability of the electron withdrawing group to stabilise the negative charge on the para position. $\endgroup$ – bon Feb 16 '16 at 20:30
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    $\begingroup$ @Nanoputian i.imgur.com/70T9RDT.png $\endgroup$ – orthocresol Feb 16 '16 at 20:32

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