[Heli0x]

Hi everyone.

I'm having trouble completing a reaction where Hydroquinone becomes 1,4 Dimethoxybenzene through a reaction with Methanol, Methyl Iodide and Sodiom Hydroxide.

(I'm very bad at this so please bear with me).

From what I understand, the NaOH should be the nucleophile, Methanol the solvent, and the Methyl Iodide should provide the halogen.

What really confuses me is where, and how the Iodide enters the reaction.

So far I've only come to the conclusion that NaOH first deprotonates both oxygens leaving them negative. Wouldn't the sodium want to form NaI after that or? Furthermore I can't see what the MeOH does..

As you can see I'm fairly off here, so if someone could shed some light on this I would be very happy.

[mschelthoff]

You're looking at the reaction from the wrong angle: the Methyl Iodide is the source of the electrophile; the Iodide anion itself isn't participating.

Your conclusion on the deprotonation by hydroxide is correct however.

Try to follow the reaction based on charges: you have a deprotonated oxygen, and an electrophillic methyl group. Just follow it through to the product.

[Heli0x]

Well, I know that the reaction should yield two new oxygen-methyl formations. But why do they they write the halogens as a part of the reaction in my handout from school, if the halogens don't actually participate directly? Is the electrophile a methyl group from the MeOH or MeI?

In the handout however, the halogen is Bromine and not Iodine. When the reaction is complete, the nucleophile has switched places with the Bromine. I guess they left out the part of the reaction when the Bromine bonds with the molecule since that isn't the defining part of the substition reaction.

[mschelthoff]

Ions are almost always included in reactions just to keep track of charges. And they may not be participating directly but they are still involved in the reaction, hence their inclusion.

As for determining the source of the electrophile: you already know this is an SN2 reaction, therefore some leaving group is being kicked-out. Now ask yourself, of the two possible leaving groups, Iodide anion and Hydroxide anion, which would make a better leaving group?

[Heli0x]

Well, I guess the Iodide is the better LG. Does that mean that the Me from MeI would be the electrophile since it's positively charged?

[mschelthoff]

Well, I guess the Iodide is the better LG. Does that mean that the Me from MeI would be the electrophile since it's positively charged?

It's not a full positive charge, but yes, it is the more electropositive species.