[SVXX]

To get straight to the point....I have this bet with my pal who says I can't solve this question. He says I've made a mistake somewhere in my mechanism, and I have no other way of finding out.
I'm reacting succinic anhydride with BrMg(CH₂)₄MgBr, and then acidifying and heating after the Grignard addition. I just want to confirm whether I got it right...so I'm posting my mechanism. Please go through it and tell me where I'm wrong, if I am.



PS : Pardon my handwriting . The end product is an 8-membered diketone ring, as per me.

[Doc Oc]

Your friend is right, the mechanism isn't correct.  Think of the succinic anhydride as being an ester.  What happens when you react a Grignard with an ester?

[SVXX]

But it's a double sided Grignard reagent...well when we react it with an ester we get a tertiary alcohol usually, and that too when we use excess of GR. What's wrong with this way?
Ok, I think I got it now. Will post in a while.


How about this? This is how GR's usually attack esters...it is imperative to use excess GR because esters are not as reactive. But the problem is, how do I know whether in the second step instead of nucleophilic attack, the GR decides to attack the other carbonyl? This is where I'm confused.
PS : R is -(CH₂)₄

[nj_bartel]

I don't know the rate of anhydride cleavage from the sp3 intermediate, but it seems plausible to me that it's slower (or possibly comparable in rate to) the rate at which the opposite end of the grignard would attack the other carbonyl.

Idk.  Your first mechanism seems plausible to me, but I'm no chemist.

[SVXX]

Well the first mechanism is more stylish, and organic chemistry loves style . I guess we need some divine intervention on this thread if we're to get the answer..no problem, I can wait the whole day.

[Doc Oc]

You're almost there.  Grignard attack on an ester does not yield a tertiary alcohol, the alkoxide leaves and you are left with a ketone.  But you're right, excess Grignard will yield a tertiary alcohol once it adds again to the ketone.  Now look at your molecule and see if you can figure out what a double sided Grignard will do (hint: intramolecular reactions are faster than intermolecular reactions).

[SVXX]

Do you mean one of the RMgBr's which have been attached already will attack the other C=O?

[Doc Oc]

Bingo.

[SVXX]

Aight thanks mate...the end product is still an 8 membered ring but with some minor differences then..

[Doc Oc]

No.  Not an 8-membered ring.  Sorry, I misunderstood what you typed.  What's more reactive, a ketone or an ester?  Remember the product from reaction of esters and excess Grignard.

[SVXX]

The second case I've drawn is exactly for excess Grignard!

[Doc Oc]

That's true, but you're forgetting that intramolecular reactions are faster than intermolecular reactions.  In this case, having a di-Grignard reagent is equivalent to having excess.

[SVXX]

I'm failing to get the point here...could you please give me a diagram of the final product?

[Doc Oc]

I don't have time to draw out a diagram and scan it, but I'll explain it.

The first Grignard addition opens the succinic anhydride.  That leaves you with a carboxylate anion (shift the MgBr to stabilize this if you want) and the newly formed ketone where the Grignard attacked.  Now the other end of your Grignard will attack the ketone, giving you a 5 membered ring with a tertiary alcohol.  So the final product after workup is a cyclopentanol with a carboxylic acid on it.

[SVXX]

But that case is not one of excess GR...it's the normal case, where the alkoxide ion leaves(which means ring opening here as you've pointed out). You just said a double sided GR is as good as excess GR!
Okay...I'm beginning to understand what you mean. Let me draw, scan and post it. But..after that there is addition of heat too. In that case a lactone will form as well. Gosh...what a reaction.