[joyb]

What happens when Dibenzoylmethane is boiled with NaOH?



I'm guessing it must be similar to reverse Claisen... any hints?

[Dan]

I'm guessing it must be similar to reverse Claisen... any hints?

That would also be my guess

[orgopete]

I wasn't in class, so I don't know what was discussed.

"How does Dibenzoylmethane react with NaOH?" It will form the enolate anion.

Boiled? This implies something different must be happening. It would seem that if a retro-Claisen were to occur, then the acetophenone and benzoic acid should react further. I don't think that is the answer.

I think another possibility is that even though enolate formation should protect the dione from further reaction, I don't think it does in this case and an aldol condensation may result. I think that product may do a retro-Claisen. I think this will be a different product than an aldol of the dione and acetophenone.

Just guessing, but this seems like there should have been more information to this problem to actually come up with the product.

[joyb]

The actual question is:
What would happen if PhCOCH₂COPh were boiled with aqueous NaOH? Explain why
PhCOCH₂COPh gives an iodoform reaction.

[orgopete]

That is helpful. I want to change my answer. I think NaOH simply makes the enolate. With iodine present, it will make the diiodo-dione. It cannot enolize. Now, a retro-Claisen will give benzoic acid and diiodoacetophenone. The diiodoacetophenone can react further with NaOO/I2 to give benzoic acid and iodoform.