Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: SVXX on June 15, 2010, 04:02:14 PM
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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.
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PS : Pardon my handwriting ;D. The end product is an 8-membered diketone ring, as per me.
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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?
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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.
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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
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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.
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Well the first mechanism is more stylish, and organic chemistry loves style :P. 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.
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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).
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Do you mean one of the RMgBr's which have been attached already will attack the other C=O?
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Bingo.
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Aight thanks mate...the end product is still an 8 membered ring but with some minor differences then..
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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.
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The second case I've drawn is exactly for excess Grignard!
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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.
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I'm failing to get the point here...could you please give me a diagram of the final product?
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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.
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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.
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If you add excess Grignard to an ester you end up with double addition product to get the tertiary alcohol. In this case, you have a cyclic ester reacting with two Grignards built into the same molecule, which is equivalent to adding excess.
At any rate, yes, your new mechanism looks correct.
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Yeah..thanks a bunch. I won the bet :P
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On a side note, are these di-Grignard reagents easy to make? Making it from a dihaloalkane seems impossible. Deprotonation of a diol with 2 equivalents of a Grignard? Do you not just get the bis-alkoxy anion coordinated to a single Mg2+? Now I'm curious!
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@SVXX, I agree with the intermediates in your second mechanism. Obviously, the final product is wrong as you removed a carbon. (Your use of curved arrows is also errant. In my class, I would require you to maintain the C-Mg bond as no curved arrows show a change in those electrons nor can you start a curved arrow with an atom, especially Mg2+.)
@johnnyd, if the halogens are geminal or vicinal, then you cannot form the Grignard. With longer chains, then you can.
If you deprotonate a diol with a Grignard reagent, with small diols, I expect you would get the chelate as suggested. If the chains are long, and concentrations high, then intermolecular salts are likely to be prevalent. I would use rates of ring formation to be a guide for making this type of prediction.
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@SVXX, I agree with the intermediates in your second mechanism. Obviously, the final product is wrong as you removed a carbon. (Your use of curved arrows is also errant. In my class, I would require you to maintain the C-Mg bond as no curved arrows show a change in those electrons nor can you start a curved arrow with an atom, especially Mg2+.)
For someone who has authored an organic chemistry book published by the "Curved Arrow Press", I'd say that was expected aye :P.
Yes, I noticed that part later. It's actually two five membered rings fused at only one carbon, I made a silly mistake in my excitement there. But I rectified it when I showed it to my mate, forgot to rectify it here though...thanks for pointing it out!
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Not fully understanding you orgopete.
"if the halogens are geminal or vicinal, then you cannot form the Grignard. With longer chains, then you can." (Sorry not sure how to use the proper quote thing)
I would presume that once the Grignard is formed on one end of the chain and the halogen is present on the other end, either ring closure or intermolecular SN2 reactions (depending on chain length) would proceed? I'm missing something here I reckon. Can you clarify?
Is it a case that Grignard formation (by that I mean insertion of magnesium into a carbon-halogen bond) occurs with a more readily than the unwanted SN2 reactions in which case the di-Grignard could be formed at cold temperature?
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Anyone?
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Not fully understanding you orgopete.
"if the halogens are geminal or vicinal, then you cannot form the Grignard. With longer chains, then you can."
Because the original statement was general (di-Grignard … seems impossible), I was simply limiting it. If the halogens are on the same carbon (geminal), the Grignard will decompose to form a carbene (I presume). If the halogens are on adjacent carbons (vicinal), then the Grignard will eliminate the second halogen to form an alkene. This reaction is often done with other metals than magnesium, but magnesium will give the same result. If more carbons are present, then dimagnesium salts can form (I didn't look up an references, but I seem to recall seeing this).
I understand how something like this may be difficult to research as no one would try to make a Grignard from dibromomethane, for example, it is then difficult to find what might happen if you did. Incidentally, I believe the Simmons-Smith (Zn + CH2I2) may be an example of a (semi) stable zinc equivalent of the intermediate one would get in the magnesium reaction. Because magnesium salts may be considered more ionic, that ionic character leads to the decomposition to magnesium bromide and a carbene. The zinc salts are more stable and thus one can find reactions that may be considered carbenes or carbanions. That is a whole different question though.
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Hi SVXX,
It seems that your second scheme is right, though the final product structure is wrong, it should be a spirolactone.