Can you quickly turn this D-Glucose into its Haworth Projection? What about this D-fructose? Want to learn a quick shortcut? Simple carbohydrates such as D-Glucose can be presented as a Fischer Projection, a Chair Conformation, or a flattened Haworth Projection even though many call it Hay-worth. Let me show you how converting between them does not have to be as tricky as your professor made it out to be. But first, Fischer Projections allow us to easily represent a 3D Chiral molecule in two dimensions as I teach in the series below. Glucose has 4 chiral carbons. We start with this skeleton and 4 lines, a carbonyl at the top and an achiral CH2OH at the bottom. Now for the chiral carbons, I remember it as right, left, right. The lowest chiral carbon will have the OH on the right for D-Glucose and on the left for L-Glucose. Here we’ll put it on the right for D-Glucose. The other side just gets a Hydrogen. MCAT students, you do have to know this. Glucose is an Aldohexose, that’s Al- for the Aldehyde, hex for 6 carbons and -ose, letting me know that it’s a sugar. And since Aldehyde is the highest priority, we number from the top. D-Fructose is a Ketohexose, that’s a 6-carbon sugar with a ketone instead of an Aldehyde. We start with this skeleton and add the ketone on position 2 an achiral CH2OH at the top and bottom. And then we’ll follow the same pattern. D-Glucose was Right, Left, Right, and this one to the right for D. Fructose also follows the pattern of Right. Left, Right where the lowest OH goes right because it’s D-Fructose. And then you fill in the Hydrogens, and since ketone is the highest priority, once again we number from the top. Sugars will naturally inter-convert between their linear and ring forms. Can you see why Glucose prefers to be in the cyclic or chair conformation? Let me know in the comments below. But chairs are so annoying to draw, luckily you get to draw the flat ring or Haworth projection. So, let’s look at the mechanism, the conventional way to draw it and yeah, I also find it confusing, then my time-saving shortcut. Ready? The mechanism is a simple nucleophilic attack where the lone pairs on the nucleophilic Oxygen on carbon number 5 attack the partially positive Carbonyl Carbon putting way too many bonds on carbon kicking up the pi electrons which ultimately get protonated giving me an OH on 1. Why Carbon 5? Because this will give me a stable 6-membered ring, And then, you likely learned to re-draw the molecules sideways trying to rotate it and visualize what goes where...So what does go where? Instead of trying to visualize it, we’ll look at it head on starting with carbons 2 and 3 and then complete the 6-membered ring making this position 1, 4, 5, and 6. Since the Oxygen on carbon 5 attacked carbon 1, 6 is not in the ring. Instead, Oxygen takes that position. Let’s make the Oxygen red and add it in. Again, that’s the Oxygen on carbon 5 attacking carbon 1. And we’ll show the Oxygen between carbon 5 and carbon 1, where this represents the new bond from the attack. What about everything else? This is where it can get tricky, or does it? The easiest way to go from linear to horizontal is to take a look at your Fischer and one, drop it right down, two, the rest is left up. If we drop it right down, everything on the right is drawn down, the rest being left up means everything on the left goes up. So let’s drop these right down on 2, 3, and 4 and these will be left up on 2, 3, and 4. Wait, did I say that was tricky? No, no, carbon 5 is tricky! Since the OH swung around to attack, we can’t just drop it right down, the entire chiral carbon twisted. It would have been down on carbon 5 with the green hydrogen going up, but that would put the CH2OH going to the right, but that is not the Oxygen that attacked. But if you picture the carbon rotating, you’ll see the CH2OH going up and the green hydrogen going down. Wait, what? Or you can just remember this, when starting with the D-Glucose, C6 goes up. If you’re starting with an L-Glucose, C6 goes down. We’re almost there. All that’s left is the anomeric carbon, that’s carbon number 1. That’s this one right here which turns into an OH after the pi bond breaks. Since the carbonyl carbon was Sp2 hybridized, an sp2 is trigonal planar or flat. The attack could’ve come from the top or the bottom. If we draw the OH down we get the alpha product, if we draw it up we get beta. The way I remember it is alpha reminds me of a fish swimming down in the sea, and beta like a bird, as my toddler says, flying high in the sky! This is a pyranose ring with 6 atoms in the ring. What about Fructose? Since fructose has a carbonyl at position 2, attacking with the Oxygen on carbon 5 gives me a smaller furanose or 5-membered ring. If carbon 6 attacks, we get a pyranose and yes both can occur. But since we already saw the pyranose, let’s see what happens when carbon 5 attacks the carbonyl and forms a 5-membered ring. We’ll apply the same trick as before with the 5-membered ring with two major differences. 1, we have an oxygen at the top , and 2, we put carbons 3 and 4 at the bottom. Ready to fill it in? Since carbon 5 did the attack, and carbon 2 got attacked, only 3 and 4 remain unchanged and so we drop them right down, and the other side gets left up. Just like with the pyranose, the achiral position will be opposite the attacking OH. Since the attacking OH is right down, carbon 6 will be left up. Hydrogen just goes to the other side. As a reminder, D-Fructose, carbon 6 is up, L-Fructose, carbon 6 is down. As for Carbon 2, since the carbonyl is sp2 and flat, once again we get alpha down or beta up specifically looking at the position of OH. And carbon 1 just goes opposite. What if your sugar was written like this? Or if you’re asked to convert the Fischer to 3D? That’s exactly what I cover in the next Fischer Projection video which you can find on my website along with my entire Fischer Projection series. Practice quiz and cheat sheet at leah4sci.com/fischer