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Cyclic hemiacetals and hemiketals

Before we get into the discussion of cyclic hemiacetals and hemiacetals, let’s just quickly recollect how they are formed. They are formed when an alcohol oxygen atom adds to the carbonyl carbon of an aldehyde or a ketone. This happens through the nucleophilic attack of the hydroxyl group at the electrophilic carbonyl group. Since alcohols are weak nucleophiles, the attack on the carbonyl carbon is usually promoted by protonation of the carbonyl oxygen. When this reaction takes place with an aldehyde, the product is called a ‘hemiacetal’; and when this reaction takes place with a ketone, the product is referred to as a ‘hemiketal’.
reaction creating hemiacetal
The above reaction exemplifies the formation of an intermolecular hemiacetal. These are intrinsically unstable and tend to favor the parent aldehyde.
Molecules (aldehyde or ketone), which contain both an alcohol and a carbonyl group, can instead undergo an intramolecular reaction to form a cyclic hemiacetal/ hemiketal. These, on the contrary, are more stable as compared to the intermolecular hemiacetals/hemiketals. Stability of cyclic hemiacetals/hemiketals is highly dependent on the size of the ring, where 5 & 6 membered rings are generally favored.
reaction creating cyclic intramolecular hemiacetal
Intramolecular hemiacetal and hemiketal formation is commonly encountered in sugar chemistry. Just to give you an example: in solution, ~ 99% of glucose exists in the cyclic hemiacetal form and only 1% of glucose exists in the open form.
honey bee indicating honey is sweet because it contains glucose and fructose

Cyclization of glucose to its hemiacetal form

Let’s first draw a molecule of glucose (Cstart subscript, 6, end subscriptHstart subscript, 12, end subscriptOstart subscript, 6, end subscript). The simplest way to do so is by using the Fischer Projection as shown below
fischer projection of glucose highlighting its aldehyde group and hydroxyl group
Glucose has an aldehyde group and five hydroxyl groups. Does that ring a bell? Yes, glucose can form an intramolecular cyclic hemiacetal. Let’s now show the formation of hemiacetal of glucose starting from its open structure (Fischer projection).
diagram showing how to form a hemiacetyl of glucose from its original open form, seen in the fisher projection.
So why doesn’t the hydroxyl attached to C-4 react with the carbonyl group? Why does the carbonyl group react with the hydroxyl attached to C-5? C-4 hydroxyl attacking the carbonyl group will lead to the formation of a 5-membered ring, while the attack of C-5 hydroxyl at the carbonyl group will generate a 6-membered ring (as shown in the above figure). In the case of glucose, a 6-membered ring is thermodynamically more stable than a 5-membered ring, thus favoring the formation of a 6-membered ring over a 5-membered ring.
Now let’s shift our focus to the hemiacetal of glucose (Haworth projection). If you notice this cyclization process creates a new stereogenic center, C-1, which is referred to as the anomeric carbon. Glucose can exist as an α or a β isomer, depending on whether the OH group attached to the anomeric carbon (C-1) is on the same side as the CH2OH group or is on the opposite side. These two forms are referred to as anomers of glucose.
PS: when you move from a Haworth projection to a chair conformation, the groups pointing upwards in the former become equatorial and the groups pointing downwards become axial respectively in the latter.
In aqueous solution, glucose exists in both the open and closed forms. These two forms always exist in equilibrium. In the process of converting from closed to open form and then back to closed form, the C-1→ C-2 bond rotates. This rotation produces either of the two anomers. We term this phenomenon of opening of the ring, rotation of the C-1→ C-2 bond and the subsequent closing of the ring as mutarotation. So as a result of mutarotation, both the α and β anomers are present in equilibrium in solution. In the case of glucose, β anomer is more predominant than α anomer. This may not be the case with all the monosaccharides.
diagrams of monosaccharides

Cyclization of fructose to its hemiketal form

Now let’s change gears and apply the same principles (as applied to glucose) to a molecule of fructose. Fructose has a ketone group and five hydroxyl groups. So, fructose should also be able to cyclize to form an intramolecular hemiketal.
diagram of D-fructose
There are in fact two ways in which a molecule of fructose can cyclize. The first is as illustrated below
diagram of fructose cyclizing
Here, as you can see, the hydroxyl attached to C-5 attacks the carbonyl group, yielding a 5-membered ring (furanose form).
In the second scenario (as shown below), the hydroxyl attached to C-6 attacks the carbonyl group, resulting in a 6-membered ring (pyranose form).
reaction creating alpha-D-fructopyranose

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