Does the aldehyde group react at all in the Fehling test?
Hello, I've been wondering for a long time how it's possible that the Fehling test detects aldehydes (positive = red precipitate due to reduced copper). However, according to the latest findings, the aldehyde group doesn't react to form a carboxyl group, but rather to glucosone. How do you then know that you have an aldehyde? And above all, they say that sugars must be in an open-chain form for this test. But since the aldehyde group doesn't react, is that irrelevant?
I simply can't classify this anymore because all sources are still based on the old version (reaction to carboxylic acid) and would be very happy to receive an answer 🙂
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These findings are not so new. You know that for almost 100 years.
The maling sample knows reducing sugars (i.e. aldoses and ketoses), aldehydes and α-hydroxyketones.
If you want to read about it quite accurately, also with regard to schooling, let’s take this Link in this “Search;
Thank you for the first time, but shouldn’t that work in the cyclic form of, for example, glucose, because the aldehyde group still remains unchanged?
I do not know the exact mechanism, but it may be that the open-chain form is at least relevant to the extent that the OH groups must stand correctly for a reaction and that does not go in ring form. Saccharose does not, for example, suggest the faulty sample because it is not reducing or it cannot open. And even though it contains glucose.
Thank you, that really helped me.
That’s right. You can also reduce Cu(II) with hydroquinone.
Typically, one makes such a test as the maling sample if one has a concrete suspicion of a special group or just wants to prove something (i.e. that sucrose is not reducing).
Of course, nothing else can be there, which distorts. And for this reason, a positive failure test is not yet a proof of an aldehyde group, but only for the reducing property of the tested species.
Okay, thank you very much, one last question: can’t everything that has a kind of reducing effect lead to CuO being created? Or why exactly “reducing sugars (i.e. aldoses and ketoses), aldehydes and α-hydroxyketones”?
In addition to the answer already given, the mistake of thinking may be to equalize reducing sugars and aldehydes. Not every aldehyde reacts positively in the mising sample, for example benzaldehyde does not react.
The (first) reaction product with glucose is Glucoson, you have already mentioned. The mechanism was published in 1960 and seems to run over an Enolat. That is, the aldehyde must be enolisable. Also stands out of the chemocon in the paper linked below.