Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Rutherford on September 11, 2012, 08:08:37 AM
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In my book it is said that aldehydes react with Phehling's and Tollens' reagents while ketones don't react. In another book it is said that fructose reacts with Phehling's and Tollens' reagents. How is this possible when fructose is a ketohexose?
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Ketoses can isomerise to aldoses (and vice versa) under basic conditions. Can you suggest a mechanism for this isomerisation?
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Fructose can isomerize. What about sucrose?
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Sucrose, a disaccharide, does not give positive
result to Tollen's or Benedict's or Fehling's test.
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Ketoses can isomerise to aldoses (and vice versa) under basic conditions. Can you suggest a mechanism for this isomerisation?
Then, in the presence of a base, the isomerisation takes place. Thanks for the answer.
I can't think of the mechanism.
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I can't think of the mechanism.
If you treat a ketone with a base, what are the possible reactions?
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It can produce an enolate or aldol maybe.
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It can produce an enolate.
Yes. Let's look at a simplified case:
Consider 1,3-dihydroxypropanone, and the corresponding enolate it forms on treatment with base. Remember this is an equilibrium and enolates are constantly protonating and reforming carbonyls in protic solvents - can you see how you could get glyceraldehyde (2,3-dihydroxypropanal) from this enolate?
Now extend this idea for a ketohexose ::equil:: aldohexose equilibrium.
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"can you see how you could get glyceraldehyde (2,3-dihydroxypropanal) from this enolate?"
I see that I can remove the pi bond by adding H2/Pt, I have 2 hydroxy groups now and one -O- group, but why would be only 1 hydroxy group oxidized by oxidation to form aldehyde?
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No. No other reagents, it is a base-catalysed isomerisation. Can you draw a mechanism for:
1,3-dihydroxypropanone ::equil:: enolate ::equil:: 2,3-dihydroxypropanal
under basic conditions? Just an aqueous base, no other reagents.
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I suppose that you meant the mechanism I attached, but I don't know how the second step happens. Maybe aldol addition?
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Sucrose, a disaccharide, does not give positive
result to Tollen's or Benedict's or Fehling's test.
True, but maltose, which is also a disaccharide does give a positive test. @Raderford, why is that?
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I suppose that you meant the mechanism I attached, but I don't know how the second step happens. Maybe aldol addition?
No, there is no aldol in the isomerisation, just carbonyl-enol equilibria.
Let's try working from both ends. You have drawn 1,3-dihydroxypropanone and its enolate, fine. We will call that enolate 1.
Now draw 2,3-dihydroxypropanal, and draw the enolate you would get from that by deprotonation. We will call that enolate 2.
How are enolates 1 ad 2 related and how might they interconvert? There is an equilibrium between them:
Ketone ::equil:: enolate 1 ::equil:: enolate 2 ::equil:: aldehyde
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Sucrose, a disaccharide, does not give positive
result to Tollen's or Benedict's or Fehling's test.
True, but maltose, which is also a disaccharide does give a positive test. @Raderford, why is that?
In succrose both groups (aldehyde and ketone) are blocked, naltose shows one group free (hemiacetal).
http://en.wikipedia.org/wiki/Maltose
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Got this. Seems that the enolates are same only turned on other sides, or maybe resonance forms. Is it okay?
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Structures 2 and 3 look identical to me.
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Got this. Seems that the enolates are same only turned on other sides, or maybe resonance forms. Is it okay?
Essentially yes. Though you have now switched to enols instead of enolates, but conceptually you are there. As has been said, the two enols you drew are identical.
Now you need to put in curly arrows and you will have a mechanism for the isomerisation reaction.
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I am not skilled at drawing mechanism, but I ended up with the one attached here. I didn't use enolates because I am not sure what H atom from what OH group would be replaced. I wrote previously that it is the H atom on the central OH group accidentally, but how to predict what H atom will be removed?
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No, that mechanism is fundamentally incorrect. The arrow starts from electrons in a lone pair or bond.
See: http://en.wikipedia.org/wiki/Keto-enol_tautomerism
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Huh, I got a hard time doing the one I attached now (I should've maybe lower the number of steps). Hope it's okay.