unscored FL #16 aamc

[akimhaneul]

For this question why can't D give positive result for tollen's test?




Also, on a related note, can ketones give positive result for tollen's test and benedict's test?

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[theonlytycrane]

A tollen's test checks for whether something can be oxidized. If it can, the silver is reduced and deposits as the shiny mirror. The question is basically saying: Which is a non-reducing sugar? (That term may be more familiar).

All sugars with free -OH on anomeric carbons can get be oxidized by opening up from ring form to straight chain form. For choices A-C, check the right-most anomeric carbons to see that they have free -OH groups. These are all reducing sugars because they can be oxidized and reduce something else.

In sucrose, the -OH groups on the anomeric carbons of glucose and fructose are the ones that form the glycosidic linkage. This means that there isn't a free -OH group on an anomeric carbon so sucrose is a non-reducing sugar (it can't be oxidized).

When looking for the anomeric carbon on each unit, look for the carbon with 2 attachments to oxygen atoms.

 

[LIC2015]

....please don't tell me this question is to be expected on the new exam.

 

[akimhaneul]

A tollen's test checks for whether something can be oxidized. If it can, the silver is reduced and deposits as the shiny mirror. The question is basically saying: Which is a non-reducing sugar? (That term may be more familiar).

All sugars with free -OH on anomeric carbons can get be oxidized by opening up from ring form to straight chain form. For choices A-C, check the right-most anomeric carbons to see that they have free -OH groups. These are all reducing sugars because they can be oxidized and reduce something else.

In sucrose, the -OH groups on the anomeric carbons of glucose and fructose are the ones that form the glycosidic linkage. This means that there isn't a free -OH group on an anomeric carbon so sucrose is a non-reducing sugar (it can't be oxidized).

When looking for the anomeric carbon on each unit, look for the carbon with 2 attachments to oxygen atoms.

Thanks! But can't you open the glucose sugar ring on the left and get aldose?

Is the oh group necessary to open up the ring?


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[theonlytycrane]

Thanks! But can't you open the glucose sugar ring on the left and get aldose?

Is the oh group necessary to open up the ring?


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In sucrose, the oxygen atom that would bring electrons down to form an aldehyde at the anomeric carbon on glucose is busy in a glycosidic bond with fructose so it can't open up.

 

[akimhaneul]

In sucrose, the oxygen atom that would bring electrons down to form an aldehyde at the anomeric carbon on glucose is busy in a glycosidic bond with fructose so it can't open up.

Really? Aren't there extra electrons on that oxygen that are not involved in the glycosidic bond? Why can't those electrons go down? Just wanted to clarify things


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[aldol16]

Really? Aren't there extra electrons on that oxygen that are not involved in the glycosidic bond? Why can't those electrons go down? Just wanted to clarify things

Then you would have three bonds to an oxygen, giving it a positive 1 formal charge and oxygen doesn't like charges. That's not even to mention the strained geometry that would have to occur for that to happen. The oxygen is happy in a glycosidic bond and doesn't want to open up. Opening up is not a favored resonance form.

 

[theonlytycrane]

@kaygeebee @theonlytycrane yo, remember the passage about photographs.... no joke there was one question where I was looking at it and it said that "freedom equates to photos" and I was thinking THIS MAKES NO SENSE NOR IS IT IN THE PASSAGE.... haha

Really? Aren't there extra electrons on that oxygen that are not involved in the glycosidic bond? Why can't those electrons go down? Just wanted to clarify things


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If the electrons go down then O now has 3 bonds and 2 electrons with a formal change of +1. It's unhappy 🙁

Usually the hydrogen from the -O-H is picked off, and the electrons from that bond are the ones that fall down to form the aldehyde.