Predicting Grignard Rx product (Identifying leaving groups)

[SaintJude]

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The question asked me to determine the product of this rx:

Well, if you work it out and have the grignard attack the carbonyl, you'll form this intermediate.

Now the deprotonated oxygen will displace the "phenoxide" a good leaving group. 😕 What should have led me to understand that phenoxide is a good leaving group? I thought you could just do the acid work up next to turn the oxygen into an alcohol--and end it there.

 

[chiddler]

if i ask you what happens when I add sodium methoxide to a benzoyl chloride. what do you get? of course not an alcohol because Cl is a very good leaving group, right?

similarly for this question, lets compare quality of leaving groups. A grignard is functionally a C anion. The other possible LG, the phenol, is a pka 10 alcohol

 

[balambfishie]

The ability of a leaving group to leave is correlated with its pKa. The group with the lower pKa (that is, the group with the stronger conjugate acid) is the better leaving group.

Take for example, an alkoxide group, which normally isn't a good leaving group. However, the bond energy of a carbonyl group is so high (179 kcal/mol) that there's a powerful driving force for the expulsion of the alkoxide and the formation of a carbonyl group in this type of reaction. If you compare the pKa of methanol (pKa ~15) and that of phenol, the conjugate acid of phenoxide, (pKa ~10), you'll see that the phenoxide group can indeed act as a leaving group in this situation.

 

[SaintJude]

Is there a way to understand this a person who has not memorized the pKa for something like phenoxide?

If not, that's fine, but I mean I'm confused. Are you suggesting that the only way I would have been able to figure these problems out is if I've memorized the pka?

 

[chiddler]

Is there a way to understand this a person who has not memorized the pKa for something like phenoxide?

If not, that's fine, but I mean I'm confused. Are you suggesting that the only way I would have been able to figure these problems out is if I've memorized the pka?

no not necessary. you have to ask yourself: is it possible for an alcohol to be a leaving group?

sure. you need a pretty strong nucleophile for that to happen though.

is the grignard a strong nucleophile?

damn right. like i wrote before, it is functionally a carbanion. therefore, the alcohol is a good leaving group in this context.

 

[syoung]

Is there a way to understand this a person who has not memorized the pKa for something like phenoxide?

If not, that's fine, but I mean I'm confused. Are you suggesting that the only way I would have been able to figure these problems out is if I've memorized the pka?

Plus here's another way to look at it.
The O- will come in to make the double bond and displace the phenoxide. Why? Because the Phenoxide can resonance stabilize the - charge on O within the ring.
This is in addition to all the other pKa reasons. Or maybe because of it. Or the cause of it...
The Carbanion (from the Grignard reagent) couldn't get kicked back off by the O, since it is a stronger nucleophile than the O.
And really, the Alcohol can't leave. It could get protonated by the acid, so you would get PheC(CH2Phe)(OH)O-Phe (sp3 carbon). But I don't think this is the case.