I am studying carbohydrates in organic chemistry and I am confused a bit on what they are and how you recognize whether a molecule is a carbohydrate or not. For example, will a carbohydrate always have an $\ce{OH}$ group with either an aldehyde or a ketone group? Can it also contain other atoms like nitrogen or bromine? Can it have ether, ester?

  • 1
    $\begingroup$ Some sugars do have nitrogen in them. See en.wikipedia.org/wiki/Amino_sugar $\endgroup$ – Pritt Balagopal May 11 '17 at 16:30
  • $\begingroup$ Ok then. How can you recognize a carbohydrate molecule, without running into so many exception? $\endgroup$ – TLo May 11 '17 at 16:41

You've asked a great and challenging question, especially for people just learning this material. The other answers are useful and provide good background. They also illustrate how the great structural diversity of carbohydrates quickly muddies the waters. I'll offer an answer that is less technical and gets at the "recognition" part.

The trouble with simple carbohydrates (5- or 6-carbon monosaccharides specifically, but we should avoid that term since we haven't defined it here) is that they can be drawn in an an open chain form (as a Fischer projection) or in a cyclic form (as a Haworth projection, or in some cases as a "chair"). So one commonly encounters two or even three "pictures" of the same thing, which hardly helps (examples below).

In my courses I nearly always use the cyclic forms since for 5- and 6-carbon sugars they are the dominant and biologically relevant forms. If one sticks to this, a pretty simple rule of thumb can be stated: carbohydrates are compounds with a 5- or 6-membered ring containing oxygen, and decorated extensively with hydroxyl (or alcohol) groups. This "scales" to polysaccharides, and accommodates deoxy and amino sugars, so it works a great deal of the time (and is enough for non-experts in many cases). When and if someone needs to understand more, it's not too hard to transition to the Fischer projections by focusing on opening (breaking open) the ring.

drawing styles


The name "carbohydrate", which literally means“carbon hydrate,” arises from their chemical composition, which is roughly $\ce{(C.H2O)_n}$, where $n\ge 3$. The basic units of carbohydrates are monosaccharides.

Monosaccharides or simple sugars are aldehyde or ketone derivatives of straight-chain polyhydroxy alcohols containing at least three carbon atoms. Such substances, for example, D-glucose and D-ribulose, cannot be hydrolyzed to form simpler saccharides.

Examples of monosaccharides:



Monosaccharides are classified according to the chemical nature of their carbonyl group and the number of their C atoms.

If the carbonyl group is an aldehyde, as in glucose, the sugar is an aldose. If the carbonyl group is a ketone, as in ribulose, the sugar is a ketose. The smallest monosaccharides, those with three carbon atoms, are trioses.

Those with four, five, six, seven, etc., C atoms are, respectively, tetroses, pentoses, hexoses, heptoses, etc. These terms may be combined so that, for example, glucose is an aldohexose, whereas ribulose is a ketopentose.

However, there are similar molecules with different atoms such as nitrogen which are also considered carbohydrates. A class of polysaccharides known as glycosaminoglycans (unbranched polysaccharides of alternating uronic acid and hexosamine residues).Practical examples are Hyaluronic acid(linked disaccharide units that consist of D-glucuronic acid and N-acetyl-Dglucosamine) and Heparin (sulfated disaccharide units):

a b

Nevertheless, when carbohydrates bond with other molecules to form other newer bonds, they are given different names and are not necessarily carbohydrates but are given new names - a good example are glycoproteins (proteins covalently associated with carbohydrates.)


Biochemistry Voet and Voet, Grisham

  • 1
    $\begingroup$ Thanks for the reply. The reason why I ask this question was because the books definition of carbohydrate was too basic. For instance, do carbohydrate have to follow the strict ratio of 1:2:1 ratio for C:H:O atom. IF not then can 3hydroxybutanal be considered a carbohydrate? It has a aldehyde and oh group? Also, in carbohydrate, does every carbon atom in the carbohydrate molecule have to be attached to 1 OH and 1 H atom? Can you have exception where one carbon atom is attached to both 2 OH group and other carbon atom attached to 2 H atom as in CH2? These are just my curiosities. $\endgroup$ – TLo May 11 '17 at 16:38
  • 3
    $\begingroup$ Very good question! Only the simple sugars, or monosaccharides, fit this formula exactly. The other types of carbohydrates, oligosaccharides and polysaccharides, are based on the monosaccharide units and have slightly different general formulas, as for the other part, its not necessarily each carbon is bonded to OH, H , a good example to explain this is deoxyribose , it has a carbon bonded to H,H $\endgroup$ – xavier_fakerat May 11 '17 at 16:48
  • $\begingroup$ ahh... Got it thanks... If i could ask just one more question, since the previous comment wasn't allowing me, I was reading other resources and was reminded that carbohydrates are polyhydroxy aldehydes and ketone, which thus means that 3 hydroxybutanal cannot be carbohydrate since it does not have multiple OH groups. So my other question is that can simple sugars have poly-adehydes or ketone groups, or can each molecule just have one of the other? Or can a simple carbohydrate contains both aldehyde and ketone groups together in one molecule? Sorry if this seems a lot. But thanks anyways. $\endgroup$ – TLo May 11 '17 at 16:55
  • $\begingroup$ I'm sorry, the last part i'm not really sure so I can't answer now, would it be fine if I just confirm and get back to you later ? $\endgroup$ – xavier_fakerat May 11 '17 at 17:10
  • $\begingroup$ Sure no problem. $\endgroup$ – TLo May 11 '17 at 17:11

A carbohydrate is a biological molecule consisting of carbon (C), hydrogen (H) and oxygen (O) atoms, usually with a hydrogen–oxygen atom ratio of 2:1 (as in water); in other words, with the empirical formula Cm(H2O)n (where m could be different from n).This formula holds true for monosaccharides. Some exceptions exist; for example, deoxyribose, a sugar component of DNA,has the empirical formula C5H10O4.Carbohydrates are technically hydrates of carbon; structurally it is more accurate to view them as polyhydroxy aldehydes and ketones.

Carbohydrates often display chemical groups such as: ''N''-acetyl (e.g. chitin), sulfate (e.g. glycosaminoglycans), carboxylic acid (e.g. sialic acid) and deoxy modifications (e.g. fucose and sialic acid).This is from Wikipedia

They can also be defined as optically active polyhydroxy aldehydes and ketones.

Now about the general formula $\ce{C_m(H2O)_n}$ also have exceptions like $\ce{CH3-COOH}$($\ce{C2(H2O)2})$ but it is not a carbohydrate. Similarly, rhamnose($\ce{C6H12O5})$ and deoxyribose($\ce{C5H10O4})$ as mentioned in the above article does not fit into this general formula, but then too they are classified as carbohydrates.

Apart from that carbohydrates are also ether.

enter image description here


This is the open chain structure of glucose. Some of the properties of glucose was not able to be explained by this structure.

enter image description here

These are the two hemiacetal cyclic forms $\alpha$ and $\beta$ of glucose also called as Haworth structures which remain in equilibrium with the open chain structure of glucose. It is an ether as you can see in the structure. Other examples of ether include cellulose, starch, sucrose, etc. Sucrose is a disaccharide which gives a glucose and fructose by hydrolysis. Starch and cellulose are polysaccharides. Starch is a branched polymer of $\alpha$ glucose units and cellulose is a straight chain polymer of $\beta$ glucose units.

  • $\begingroup$ Ok got it. So together with multiple OH groups, can simple sugars also have multiple aldehyde and ketone groups? Or can both aldehyde and ketone groups be found in simple sugars? $\endgroup$ – TLo May 11 '17 at 17:10
  • 1
    $\begingroup$ I don't think so, I have never heard of such molecule. You can ask this as a different question. @TLo $\endgroup$ – user237650 May 11 '17 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.