How are carbons numbered in fructose?

[rocketbooster]

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Why when cyclic fructose is drawn by itself the carbons are labeled like normal from right to left going clockwise, but when fructose is drawn in sucrose the structure is flipped making an alpha-1 glucose, beta-2 fructose disaccharide? for consistency sake, I don't see why it wouldn't be drawn normal, thus making it an alpha-1, beta-5 disaccharide?

Anyone know why the fructose is flipped when drawn in sucrose?

 

[canjosh]

I think I follow your question...look for the longest possible carbon chain and then number. I think you're missing the carbon sticking off the ring.

 

[rocketbooster]

I think I follow your question...look for the longest possible carbon chain and then number. I think you're missing the carbon sticking off the ring.

what? no....haha obviously.

I know how to label carbon chains. I want to know why fructose is INVERTED when draw in the sucrose structure. You won't even notice this when you see the structure of sucrose unless you also see the alpha-1, beta-2 linkage listed with it. In sucrose, fructose is labeled from left to right rather than standard right to left format for carbohydates.

Anyone know why it's flipped?

 

[Ari1584]

what? no....haha obviously.

I know how to label carbon chains. I want to know why furanose is INVERTED when draw in the sucrose structure. You won't even notice this when you see the structure of sucrose unless you also see the alpha-1, beta-2 linkage listed with it. In sucrose, fructose is labeled from left to right rather than standard right to left format for carbohydates.

Anyone know why it's flipped?

I'm actually wondering the same thing?? it always confuses me! 😕

 

[computerdorkdan]

Sucrose is beta 1-2 linkage...some books flip the fructose below the glucose it to make it easier to draw (I guess). I think they do this because we get used to numbering carbohydrates from the molecule to the right of the oxygen (in the ring) as 1...etc. So, in sucrose, the anomeric carbons are connected by the ether linkage - right next to each other - unlike all other disaccharides.

Hope that helped...

 

[rocketbooster]

Sucrose is beta 1-2 linkage...some books flip the fructose below the glucose it to make it easier to draw (I guess). I think they do this because we get used to numbering carbohydrates from the molecule to the right of the oxygen (in the ring) as 1...etc. So, in sucrose, the anomeric carbons are connected by the ether linkage - right next to each other - unlike all other disaccharides.

Hope that helped...

umm...not really, well maybe? so basically, sucrose isn't actually a straight-chain then. it's more like a glucose molecule under a fructose molecule? ehh or something like that?

 

[BerkReviewTeach]

umm...not really, well maybe? so basically, sucrose isn't actually a straight-chain then. it's more like a glucose molecule under a fructose molecule? ehh or something like that?

Basically, it's the same numbering system you would use on fructose if it were a straight chain structure. If you consider the six carbons on the Fisher projection of the sugar, you'll note that the carbonyl carbon assigned #2 rather than #5, because you want to give the highest priority carbon the lowest possible number. It ends up that the carbonyl carbon is the most oxidized off all six carbons, so the carbonyl carbon is top IUPAC priority. The rules stay the same in the cycic structure. The carbonyl carbon becomes the anomeric carbon during cyclization, so that carbon retains two bonds to oxygen (keeping it as the top IUPAC priority). So if we consider the longest carbon chain from both ends, the anomeric carbon is going be either the second carbon or the fifth carbon. By IUPAC conventions, the hemiketal carbon (formerly the carbonyl carbon in the straight chain) gets assigned a 2 rather than a 5.

BTW, I referenced the BR organic chemistry book for this, so if you happen to have a copy around, there is a pretty good drawing of fructose.

 

[rocketbooster]

Basically, it's the same numbering system you would use on fructose if it were a straight chain structure. If you consider the six carbons on the Fisher projection of the sugar, you'll note that the carbonyl carbon assigned #2 rather than #5, because you want to give the highest priority carbon the lowest possible number. It ends up that the carbonyl carbon is the most oxidized off all six carbons, so the carbonyl carbon is top IUPAC priority. The rules stay the same in the cycic structure. The carbonyl carbon becomes the anomeric carbon during cyclization, so that carbon retains two bonds to oxygen (keeping it as the top IUPAC priority). So if we consider the longest carbon chain from both ends, the anomeric carbon is going be either the second carbon or the fifth carbon. By IUPAC conventions, the hemiketal carbon (formerly the carbonyl carbon in the straight chain) gets assigned a 2 rather than a 5.

BTW, I referenced the BR organic chemistry book for this, so if you happen to have a copy around, there is a pretty good drawing of fructose.

right. I use BR, too. 😀 yea, it makes sense now. either left or right is anomeric, and the convention is carbon 1 being the anomeric one, not the most rightward one.

btw, BR is amazing. it has taken me from last year's 8-9s on the PS to easily 11-12s now. i think i can get even higher than that with more practice and avoiding fewer careless mistakes. the amount of practice in BR books is what makes you rape the real test because you have seen each topic in sooo many different scenarios. 👍

 

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