Monosaccharides - Haworth Projections: Videos & Practice Problems

Haworth projections are simplified representations of sugars in cyclic forms, crucial for understanding sugar chemistry. The anomeric carbon is always positioned at the far right, with clockwise numbering for carbons. For D-glucose, the beta anomer is favored in equilibrium. Identifying anomers involves recognizing the orientation of hydroxyl groups; if both are up, it's beta (cis), while if one is down, it's alpha (trans). This method streamlines the visualization of monosaccharides, enhancing comprehension of their structural properties and reactivity.

Monosaccharides - Haworth Projections

Now we're going to talk about another type of representation that's very important for sugar chemistry, and that's the Haworth projection. So guys, Haworth projections are simplified drawings of sugars in a cyclic form, imagining that your ring is planar. Now we know that it's not planar. We know that it's in a puckered chair conformation according to the rules of cyclohexane that we learned in Orgo 1. But we use these simplified structures a lot because they're just easier to draw. Okay?

Now, what do you need to know about converting a monosaccharide chain into a Haworth? Guys, it's exactly like a cyclohexane, just easier. The anomeric carbon, note, is always at the far right in a Haworth projection, which is exactly the same as cyclohexane. Your anomeric position, your one position should always be there. So that means this carbon is the same as that carbon there. And in the same way, you're going to number going clockwise. So it's going to be 2, 3, 4, 5, and remember this is actually 6. Okay? So let's say this is true for both pyranose and furanose rings. So in this case, these are pyranose rings, but if there were furanose rings, your anomeric carbon should still go to the right.

Now, this practice problem is pretty straightforward. It says properly number the following Haworth projections, identify each anomer, and predict the anomeric equilibrium. Well, guys, I already started on this problem before we even began because I started showing you guys how your right carbon should be number 1, right? It should be your anomeric. So that means that this is your 1 and that you should just go clockwise like you would with this regular cyclohexane, meaning that this is also 1, 2, 3, 4, 5, 6. Okay?

Now, can you guys explain to me why carbon 6 is going up? Because once again, we have a D-glucose, which is like a D-stereo isomer, and D always goes up with the stereo descriptor always goes up, okay? Now, how can we identify each anomer? So notice that this O is facedown and this O is faced up. So how do we figure out which one's alpha and which one's beta? Well, notice that for both of these, the stereo descriptor is faced up meaning that this is a trans and this is a cis. So guys, the same rules are going to apply. This is the alpha anomer. And this is the beta anomer. Okay?

Now guys, I told you you're not going to be responsible to know when alpha is more prevalent over beta, at least not now. That you need more knowledge for that. But I did already show you guys what the anomeric equilibrium was specifically for D-glucose, and remember which one was favored. Remember it was the beta, right? It was actually the beta. So I'm going to go ahead and draw that the anomeric equilibrium is more towards the beta and less towards the alpha simply because I already showed you the same exact molecule. Okay? Just because it is this way for mannose it might be different. But for glucose, we know that one specifically, we know that it's going to go to the right. Okay? Awesome guys, that wasn't so bad right? Now you have another even easier way to represent monosaccharides. Let's move on to a practice problem.

Draw the Haworth Projection that would be expected from cyclization of D-Fructose into α-D-Fructofuranose

• A.
• B.
• C.
• D.

Answer: (D)

Here’s what students ask on this topic:

What is a Haworth projection in organic chemistry?

A Haworth projection is a simplified way to represent the cyclic structure of monosaccharides. It assumes the sugar ring is planar, even though in reality, it adopts a puckered chair conformation. This projection is particularly useful for visualizing the spatial arrangement of atoms in the ring, especially the anomeric carbon, which is always positioned at the far right. The carbons are numbered clockwise, similar to cyclohexane. This method helps in understanding the structural properties and reactivity of sugars.

How do you convert a monosaccharide chain into a Haworth projection?

To convert a monosaccharide chain into a Haworth projection, follow these steps: First, identify the anomeric carbon, which will be at the far right. Number the carbons clockwise starting from the anomeric carbon. For D-sugars, the CH₂OH group on carbon 6 will point up. Next, determine the orientation of the hydroxyl groups (OH) on each carbon. If the OH group on the anomeric carbon is down, it is the alpha anomer; if it is up, it is the beta anomer. This method simplifies the visualization of the sugar's cyclic form.

What is the difference between alpha and beta anomers in Haworth projections?

The difference between alpha and beta anomers in Haworth projections lies in the orientation of the hydroxyl group (OH) on the anomeric carbon. In the alpha anomer, the OH group is trans (down) relative to the CH₂OH group on carbon 6. In the beta anomer, the OH group is cis (up) relative to the CH₂OH group. This distinction is crucial for understanding the reactivity and properties of different sugars.

Why is the beta anomer of D-glucose more stable than the alpha anomer?

The beta anomer of D-glucose is more stable than the alpha anomer due to the spatial arrangement of the hydroxyl groups. In the beta anomer, the OH group on the anomeric carbon is equatorial, reducing steric hindrance and making the molecule more stable. In contrast, the OH group in the alpha anomer is axial, leading to increased steric strain. This stability difference results in the beta anomer being more prevalent in equilibrium.

How do you identify the anomeric carbon in a Haworth projection?

In a Haworth projection, the anomeric carbon is always positioned at the far right of the ring. It is the carbon that was originally the carbonyl carbon (C=O) in the open-chain form of the sugar. This carbon is crucial for determining the alpha or beta configuration of the sugar. Number the carbons clockwise starting from the anomeric carbon to correctly identify its position.