Tacho
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Aromatic dialdehydes?
If I formylate hydroquinone long enough with chloroform and NaOH, will it become a dialdehyde? Or a single aldehyde group de-activates the ring so bad
it stops there?
Wouldn't such aromatic be highly polymerizable?
Aromatic dialdehydes seem to be rare, just a few hits on google. None very informative.
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unionised
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orthophthalaldehyde (1,2 benzene dicarboxaldehyde) is coming into use as a bactericide. You might be able to track some down under the trade name
cidex.
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ziqquratu
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Doing that formylation on hydroquinone SHOULD yeild the dialdehyde given time, although yeilds most likely suck due mostly to polymerization.
If I may suggest off the top of my head (in other words, I have nothing to back what I'm about to say up!!), a better route may be to
di-bromomethylate the hydroquinone, then oxidize that to the dialdehyde using (I think) manganese dioxide.
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Trolf
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It is quite possible that u can make the deprotonated dihydroquinone attack the formaldehyde, but u wont get the aldehyde, u will get (firstly) an
hemiacetal (see the attachment)
If u oxidize that u will get (theoretically) the ester of hydroquinone and formic acid, but most likely u will just then get the quinone and formic
acid.
If u bromomethylate the hydroquinone, u will get the ether Ar-O-CH3 this will be hard to oxidize (and wont oxidize to the aldehyde)
Think u should start from something else if u want to make the aromatic dialdehyde.
Attachment: hydroquinone.doc (74kB)
This file has been downloaded 746 times
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Turel
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Reimer Tiemann
Formylation using the Reimer Tiemann reaction on 1,4-dihydroxybenzenes is well known to with 2,5-dialdehydes. Use a potassium salt to increase yield
of the isomer.
Hydroquinone is quite activated and it is in fact difficult to halt the formylation at only a single substitution.
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Trolf
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Sry Turel is completely right. I missed the fact that chloroform was actually the reactant.
beats me how the dichlorocarbene is formed though. If anyone knows that plz post.
The reaction i wrote is by the way reversible and wont have any influence if a competing reaction is present.
My bad
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unionised
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Deprotonation of, followed by loss of chloride from, chloroform gives dichlorocarbene.
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Turel
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Carbene Formation
Carbenes are formed in two chief ways: either by elimination, or by a compound substitution that never reaches completion.
Deprotonation by a strong base (in fact the hydroxide is strong enough in this case, which makesit very cheap) to deprotonate the chloroform, leaving
a carbanion.
The carbanion is rather unstable due to Cl's strong electronegtivity, and the increased bond length caused by formation of the carbanion.
Chlorine is able to withdraw an electron in addition to its own from the covalent bond, forming a chloride ion. The carbanion simply shifts an
electron, facilitating the removal of Cl-, and forming dichlorocarbene.
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