# Acid promoted Enol ring formation Mechanism

I know this has to be an enol mechanism but which carbon gets attacked by the $\pi$ electrons?

• Why does it have to go via an enol? In fact, it doesn’t. The electron-rich arene is nucleophilic (you can surely figure out which positions) and the ketone is electrophilic. – orthocresol Apr 30 '18 at 0:50

This reaction proceeds in a few steps. In acidic media, the upper carbonyl oxygen is proton. This results, in greater electron deficiency of the upper carbonyl carbon, which strongly attracts nucleophilic attack. Due to strong activation by the methoxy substituent, the arene's $\pi$ electrons can attack the electron-deficient carbon in the carbonyl group. This nucleophilic addition to the carbonyl group gives a tetrahedral intermediate with a positive charge delocalised in the neighbouring ring.
A chloride ion now comes in to remove the proton from the para position of the arene. We then get back the aromatic ring with the methoxy group. In the next phase of the reaction, we now want to remove the $\ce {-OH}$ that we have created on the previous carbonyl carbon. This would entail us first protonating it, to make it a good leaving group. How do we create the double bond then?
Well... It is now important to note that $\alpha$ hydrogens in carbonyl compounds are relatively easily lost due to the electron-withdrawing effect of the carbonyl oxygen. Furthermore, the negative charge generated from this loss of the proton can also be delocalised into the the neighbouring aromatic ring on the right of the figure, enhancing the stability of this conjugate base, increasing the acidity of this hydrogen. What happens now is simply elimination to form the double bond. The protonated $\ce {-OH}$ first leaves as water to generate a tertiary carbocation. Then, a chloride ion acts as a base to remove the acidic $\alpha$ hydrogen, with the electrons going into forming a $\pi$ bond between the two carbons. This is essentially an $\ce {E 1}$ process. The desired product is now obtained.