Hemi-acetal formation in aldose
The chemical reaction that occurs during the ring formation of monosaccharides leads to several important changes. Most notably, the key point is that this process results in the formation of a hemiacetal. In chemistry, a hemiacetal is defined as a compound formed by the reaction between an aldehyde group (-CHO) and an alcohol group (-OH).
So what are aldehydes and alcohols? An aldehyde is a compound in which a carbonyl group (C=O) is bonded to a hydrogen atom, while an alcohol refers to a compound that contains a hydroxyl group (-OH). Monosaccharides have an aldehyde group located at the end of the carbon chain (for aldoses, in the middle for ketoses) and have multiple hydroxyl groups along the chain. If we recall from the previous article about the ring formation process that it was the spontaneous internal chemical reaction between an aldehyde group and a hydroxyl group that resulted in the formation of the ring structure, we can immediately realize that the cyclization of a monosaccharide is the formation of a hemiacetal.
In general, hemiacetals are considered unstable intermediates before being converted into acetals, and they often undergo an additional reaction with another alcohol group (-OH) to ultimately form an acetal. The prefix “hemi-” means “half,” implying that the compound is not yet a complete acetal. However, in the case of monosaccharides such as glucose, even in the hemiacetal state, they form a very stable structure such as a five-membered (furanose) or six-membered (pyranose) ring. These configurations are so stable that there is no pressing need for further conversion into acetals — the hemiacetal structure can stably exist on its own.
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What does it mean for a monosaccharides to form a hemiacetal?
Glucose in its open linear state has four chiral centers. Carbons 2, 3, 4, and 5 are the chiral centers. As can be seen in its Fischer projection, each of these carbons is all bonded to four different atoms or groups, satisfying thus conditions necessary for chirality. However, the first carbon, which bears the carbonyl group, is involved in a double bond and thus does not meet these criteria, making it non-chiral in the open-chain form.
However, the double bond was broken when nucleophilic alcohol (OH) attacked the electrophilic carbon of carbonyl (C=O). And carbon 1 is now connected to four different substituents and at this point a hemiacetal is formed. And as a result, carbon number 1 became a new chiral center and now is called an 'anomeric carbon'. The anomeric carbon is a special chiral center that is newly created through cyclization of monosaccharides.
This chiral center allows for the creation of two possible stereoisomers. The anomeric carbon is significant because it is the key site where one monosaccharide can bond with another to form various carbohydrates. When monosaccharides join to form disaccharides or polysaccharides, it is the anomeric carbon that reacts with another monosaccharide’s hydroxyl group (-OH), resulting in the loss of a water molecule and the formation of an acetal — a glycosidic bond. In other words, the hemiacetal becomes an acetal during polysaccharide synthesis, making the hemiacetal the central point of polysaccharide formation.
This chiral center allows for the creation of two possible stereoisomers. The anomeric carbon is significant because it is the key site where one monosaccharide can bond with another to form various carbohydrates. When monosaccharides join to form disaccharides or polysaccharides, it is the anomeric carbon that reacts with another monosaccharide’s hydroxyl group (-OH), resulting in the loss of a water molecule and the formation of an acetal — a glycosidic bond. In other words, the hemiacetal becomes an acetal during polysaccharide synthesis, making the hemiacetal the central point of polysaccharide formation.
α-anomer and β-anomer
Let’s focus on D-glucose for the sake of easier understanding. In separate articles on chirality, we talked about homochirality, which refers to the fact that biological systems almost exclusively favor one form of enantiomer. In carbohydrates, the D-form is dominant, while in amino acids, the L-form is predominant. Thus, in this article — which aims for practical understanding — we will only consider D-glucose.
Now, let’s look at the two stereoisomers that form at the newly created chiral center. During ring formation, we imagine the Fischer projection of glucose lying on its side (rotated to the right), with the substituent of the 5th carbon slightly rotated to place the OH group at the end, positioning the CH₂OH group above the ring upon cyclization.
When a carbon is bonded to three atoms or groups, it adopts sp² hybridization, forming three single bonds and one pi bond perpendicular to the plane. This gives it a trigonal planar structure with 120° bond angles. However, when the hemiacetal is formed, the carbon becomes sp³ hybridized, forming four covalent bonds at 109.5° angles — a tetrahedral structure. Since the OH group on carbon 5 (C5) is attacking the carbonyl carbon (C1), which lies on a plane, the direction of attack can occur from either above or below the plane. As a result, the newly formed OH group on the anomeric carbon (C1) can end up either above or below the plane.
So how do we name these two possible structures? We use the position of the CH₂OH group (carbon 6) as the reference point. If the OH group on the anomeric carbon is on the same side as the CH₂OH group, it is called the β-anomer; if on the opposite side, it is the α-anomer. Technically speaking, if the groups are on the same side, it’s cis; if on opposite sides, it’s trans. Thus, for D-glucose — where CH₂OH always points upwards — if the anomeric OH is also up, it’s β; if down, it’s α. In the case of L-glucose, the opposite would be true — if the OH is in the same direction as CH₂OH, it’s always β; if in the opposite direction, it’s α.
This may sound confusing, but since we are focusing primarily on D-glucose, here’s a fun trick to remember it easily. This is the best way I picked up some time ago: “β looks like a bird, flying up; α looks like a fish, swimming down.” It’s a brilliant and memorable way to remember the difference!
Epimers: α and β anomers
The α- and β-anomers differ only in the orientation of the OH group at the anomeric carbon. In every other respect, their structures are identical. Thus, these two are a type of stereoisomers called epimers — isomers that differ at only one chiral center.
In the names of cyclic monosaccharides, you can often find all the necessary information: whether the structure is an anomer, whether it’s D or L, and how many atoms are in the ring. While the names may seem long, they are in fact helpfully descriptive (Let’s not be intimidated — 😆!).
Mutarotation: α ↔ β anomeric interconversion
Both the interior and exterior of human cells are aqueous environments. In such a physiological aqueous environment, there exists a dynamic equilibrium between the open-chain linear form and the cyclic form of monosaccharides. “Equilibrium” here means that interconversion between these two forms occurs freely and dynamically, maintaining a stable ratio.
That is, the α-form can revert to the linear form and then become the β-form, or vice versa. Since the internal cyclization reaction is reversible, the ring can repeatedly open and close, and each time the hemiacetal is reformed, the position of the OH group at the anomeric carbon can change. This phenomenon is called mutarotation, from the Latin “mutare,” meaning “to change.”
Importantly, this reaction does not require enzymatic catalysis — it occurs naturally and spontaneously in the body. In water, the equilibrium typically settles at about 37% α-anomer and 63% β-anomer. This process only occurs in aqueous environments — in non-aqueous conditions, one form may dominate and become fixed.
Even if the proportion of the open-chain form is very low, mutarotation still occurs continuously. Also, although less common, five-membered ring structures can form again during this process. It’s a vivid, dynamic picture — like a document being written, erased, and rewritten continuously.
Most important of all is the fact that the aldehyde group is repeatedly exposed as the ring opens — this aldehyde plays a key role in the various reactions that monosaccharides undergo.
Reducing sugar
In chemical reactions, reduction refers to the process in which a substance donates electrons and is itself oxidized. A reducing agent is a substance that performs this function, and a reducing sugar is any sugar capable of acting as a reducing agent. The key to this reducing ability in monosaccharides lies in their aldehyde group. In aqueous solutions, the aldehyde group donates electrons and becomes oxidized from a –CHO group to a –COOH group. In other words, the aldehyde group is oxidized to a carboxylic acid, thereby reducing another substance in the process.
This reducing property is crucial in biological systems. Within cells, sugars play an essential role in redox reactions, such as the reduction of NAD⁺ to NADH during metabolic pathways leading to ATP production. However, this same property can also have detrimental effects. Reducing sugars can react non-enzymatically with proteins, initiating glycation reactions that form harmful compounds known as Advanced Glycation End Products (AGEs). Additionally, they can bind to hemoglobin in the blood, forming glycated hemoglobin (HbA1c), a key marker in diabetes monitoring. The oxidation of sugars may also generate reactive oxygen species (ROS), contributing to oxidative stress.
Even ketoses—sugars with a ketone functional group—can act as reducing agents after they isomerize into an aldehyde form. Since the aldehyde group is the site of reactivity, this highlights the importance of mutarotation. Although monosaccharides primarily exist in cyclic forms, mutarotation enables the dynamic interconversion between cyclic and open-chain (linear) forms, exposing the reactive aldehyde group. Therefore, as long as a sugar can freely revert to its linear form through mutarotation, it retains its reducing properties.
This principle applies even when monosaccharides are joined via glycosidic bonds. If the anomeric carbon of the terminal sugar in a polysaccharide remains unbound—i.e., not involved in a glycosidic bond—it is free to open up into the linear form and thus retains reducing ability. In other words, a sugar is considered a reducing sugar if its anomeric carbon is free. If the anomeric carbon is involved in a glycosidic bond, it loses this freedom and the sugar is classified as non-reducing.
Many sugars are reducing sugars. Examples include the monosaccharides glucose, galactose, mannose, and fructose, as well as the disaccharides maltose and lactose. Polysaccharides such as starch, glycogen, and cellulose are composed of numerous monosaccharides linked through glycosidic bonds and are generally considered non-reducing. However, the terminal monosaccharide at one end of the chain usually retains a free anomeric carbon, which allows for a minimal level of reducing activity—though this is often negligible.
This concept will become clearer when we examine the formation of disaccharides in the following section. For instance, sucrose, which we commonly refer to as table sugar, is a disaccharide composed of glucose and fructose. Uniquely, in sucrose, both anomeric carbons are involved in the glycosidic bond, meaning there is no free anomeric carbon left. As a result, sucrose is non-reducing. The detailed process of disaccharide formation and glycosidic bonding will be addressed in the next article.
Hemiketal formation in Ketose
Until now, we have mainly focused on the formation of hemiacetals in aldoses using glucose as an example. However, it is equally important to examine the formation of hemiketals in ketoses. You may recall from the first article on carbohydrates that monosaccharides are categorized into aldoses and ketoses based on the position of their carbonyl group: if the carbonyl group is an aldehyde located on the first carbon, the sugar is an aldose; if it is a ketone located internally (typically on carbon 2), the sugar is a ketose.
Since glucose contains an aldehyde group, our discussion centered around aldoses and the formation of hemiacetals. Just as a hemiacetal is formed when an aldehyde reacts with an alcohol group, a hemiketal is formed when a ketone reacts with an alcohol group. While the cyclization process for ketoses is fundamentally the same as that for aldoses, in ketoses, the carbon of the ketone group breaks its double bond with oxygen and becomes the anomeric carbon, forming a hemiketal in the process.
To understand this more clearly, let’s examine the cyclization of fructose—a six-carbon ketose (ketohexose).
Cyclization process of fructose
When looking at the chair conformation of glucose, all the bulky substituents—such as the hydroxyl (–OH) and the hydroxymethyl (–CH₂OH) groups—are arranged equatorially in β-D-glucopyranose. This minimizes steric hindrance and intramolecular repulsion, making β-D-glucopyranose the most stable of all D-aldohexoses. This structural stability is likely one reason why β-D-glucopyranose is the most prevalent sugar form found in nature. Fructose follows a similar pattern, and for the same reasons, β-D-fructofuranose is the dominant form of fructose in nature.
The primary difference in the cyclization of fructose compared to glucose is the size of the ring that forms. While glucose typically forms a six-membered ring (pyranose), fructose predominantly forms a five-membered ring (furanose). Theoretically, just as glucose could cyclize using different hydroxyl groups (e.g., at carbon 4 or 6), producing rings of various sizes (like five- or seven-membered rings), fructose could as well. However, while five-membered furanose rings do exist, they are generally less stable than pyranose rings and occur in smaller proportions. Seven-membered rings are highly unstable and rarely exist in nature. Hence, our primary focus has been on the most stable and common form—six-membered pyranose rings.
Now, what about fructose, where the carbonyl group is not at the end, but located at carbon 2, as is typical for ketoses? In this case, fructose achieves its most stable ring structure when the hydroxyl group on carbon 5 performs a nucleophilic attack on the carbonyl carbon. Because the ketone is at position 2, the resulting ring is a five-membered furanose.
To summarize, the cyclization of glucose and fructose follows essentially the same mechanism. Just as an aldehyde reacts with an alcohol group to form a hemiacetal, a ketone reacts with an alcohol to form a hemiketal, producing a new anomeric carbon. This carbon can take two stereoisomeric forms: α and β anomers. Of these, the β-anomer is typically more stable.
The process of fructose cyclization can be illustrated as follows
Having now thoroughly examined the cyclization processes of glucose and fructose, we have completed our overview of monosaccharides. Next, we will explore how these monosaccharides link together to form more complex molecules. The upcoming article will address the formation of disaccharides.
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