Organic chemistry (Essentials) - Class 12
- Formation of hydrates
- Factors determining the stability of hydrates.
- Formation of hemiacetals and hemiketals
- Acid and base catalyzed formation of hydrates and hemiacetals
- Formation of acetals
- Acetals as protecting groups and thioacetals
- Cyclic hemiacetals and hemiketals
- Hemiacetals and Acetals
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’.
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.
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.
Cyclization of glucose to its hemiacetal form
Let’s first draw a molecule of glucose (C
H O ). The simplest way to do so is by using the Fischer Projection as shown below
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).
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.
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.
There are in fact two ways in which a molecule of fructose can cyclize. The first is as illustrated below
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).
Want to join the conversation?
- The article says "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", but in the diagram, the H on the green carbon in the Haworth projection is pointing up but is axial in the chair conformation, and the OH on the green carbon in the Haworth projection is pointing down but is equatorial in the chair conformation.(15 votes)
- You're right. For these transitions, my professor always said "up is always up" and "down is always down." Meaning in the Haworth, the green H is "up." When moving to the chair conformation "up is still up" which for the green carbon, happens to be axial. For the red carbon the OH group points up. We keep this group up when moving to the chair conformation, and in this case "up" happens to be equatorial. The rule they give in the P.S. is just wrong as far as I can tell.(8 votes)
- The article explains the preference of C5 over C4, but why doesn't C6's hydroxyl group react with the carbonyll?(8 votes)
- Does somebody knows if both structures (alpha and Beta fructose) are present in the nature? or there is a specific structure, so to say in the fruits or for the disaccharide sucrose
PS. I have read that only Beta-fructoe is present in the sucrose, but I haven't fount anything about Alpha-fructose, it is not clear(2 votes)
- in the last diagram, a-fructopyranose is shwown. it is a six membered ring, I was wondering why fructose is normally shown as a five membered ring, and if a-fructopyranose bonded with glucose to form sucrose, would it be a hemiacetal and therefore a reducing sugar?(1 vote)
- Are hemiacetals and hemiketals considered to be a type of ester or ether? In my book it says that these are kind of ethers, but I was wondering whether the authors had made mistake?(1 vote)
- Is there anyway we can have 7 member ring in the cyclization of the hemiacetal. If 7-membered ring is not possible, why . if it is too how do we do it(1 vote)
- Although it is possible using C6, it is much less stable than a 6 member ring, so does not occur to a significant extent. This is the same reason that is given for comparing C5 and C4(1 vote)
- in identifying alpha or beta do we only look at the wether CH2OH and OH are on the same side or you ca consider the up or down positions of the OH.(1 vote)
- Although comparing the position of CH2OH and OH is the primary way to determine which isomer is formed, I have read in some places how the up or down positions of the OH can be used to do the same. However, I would use the comparison as the more reliable method since that is the more commonly used method I have come across in the scientific community (from almost all, if not all, sources I have read/studied).(1 vote)
- Where on this platform can I learn more about naming of polysaccharides and identifying whether a sugar is reducing or can mutarotate(1 vote)
- During cyclization, why doesn't the c=o connect to the CH2OH instead of the farthest CHOH?(1 vote)