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Topic: Gabriel synthesis part 2: removal of phthalimide  (Read 3687 times)

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Offline Arnoldus

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Gabriel synthesis part 2: removal of phthalimide
« on: May 27, 2015, 02:27:09 PM »
Hi,

I am removing a phthalimide with ethanolic hydrazine and it works well.
In the same molecule, there is also a benzamide. I was wondering why the hydrazine reacts with the phthalimide (with formation of phthalhydrazide) and not with the benzamide. Formation of phthalhydrazide is probably energetically favored, but why is the hihgly nucleophilic hydrazine not attacking the carbonyl group of the benzamide?

Thanks

Offline pgk

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Re: Gabriel synthesis part 2: removal of phthalimide
« Reply #1 on: May 27, 2015, 02:52:08 PM »
Which compound is more easy to prepare, phthalimide potassium salt or benzamide potassium salt, and why?

Offline Arnoldus

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Re: Gabriel synthesis part 2: removal of phthalimide
« Reply #2 on: May 27, 2015, 02:58:41 PM »
Phthalimide is acidic. When it loses a proton, the negative charge can delocalize to the carbonyl groups. So, phthalimide potassium salt is easier to prepare than benzamide potassium salt.
Thanks for posing another question, but I don't see the link with the reactivity of hydrazine with the carbonyls (in either phthalimide or benzamide, being present in the same molecule)?

Offline pgk

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Re: Gabriel synthesis part 2: removal of phthalimide
« Reply #3 on: May 27, 2015, 08:01:35 PM »
Phthalimide anion is more stable compared to benzamide anion because it has two carbonyls instead of one and thus, a more extended anionic amide conjugation. But phthalimide extended and stable conjugation can also be cationic (and it is) by the nitrogen electron pair, that helps the hydrazine attack to one carbonyl (amide structure remains) that finally forms phthalhydrazide (the conjugated form of the aromatic phthalazine-1,4-diol).

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