Glucose is the most abundant monosaccharide. From memory, draw...

Question

Glucose is the most abundant monosaccharide. From memory, draw glucose in

(a) the Fischer projection of the open chain.

(b) the most stable chair conformation of the most stable pyranose anomer.

(c) the Haworth projection of the most stable pyranose anomer.

Accepted Answer

Alright. Hi, everyone. So this question says that galactose is the C four epimer glucose. Draw the structure of D galactose in a fissure projection. B chair confirmation of the beta Pinos Anno NC Hayworth projection. So for this example, I'm actually going to go ahead and start with part B with the cheer confirmation because galactose is the C four epimer of glucose. And what that means right is that the structures of galactose and glucose only differ at carbon four or C four. So if I go ahead and I proceed with drawing the chair confirmation of the glucose, I would simply have to modify it as needed to reflect the structure of galactose. So the first thing I'm going to do, right is I'm going to draw the chair confirmation of the glucose and then change it as necessary because recall that when drawing a cheer confirmation, right? First of all, you must always draw a pucker. And in addition to that the anomic carbon or C one is always going to be depicted on or at the bottom right corner with the oxygen in the ring on the top right corner. So here when it comes to drawing glucose, recall it, glucose has all of its substituents, particularly hydroxy groups on alternating sides of the ring. So when drawing the chair confirmation for glucose, you always go ahead and you put your hydroxy groups in the equatorial position. And from there you go ahead and you add your corresponding hydrogens as well. So now what I also want to mention here is the fact that the cheer confirmation that's being asked of us must be of the beta Pinos ener. Now, when a chair confirmation is in the beta configuration, then the hydroxy group of carbon one, which is the anomic carbon must be in the equatorial position. If it was in the alpha or if it was the alpha ener, then this hydroxy group on carbon one would be in the axial position instead. But now, what we've done is we've gone ahead and we've actually drawn D glucose but because galactose is the C four epimer glucose, what I do in this case is I locate carbon four and I essentially flipped the configurations of both the hydroxy group and the hydrogen. So D galactose, the hydroxy group is axial and the hydrogen is equatorial instead. So that concludes our chair confirmation, right? Instead of having D glucose, this is now D galactose. So now we can go ahead and convert our, our cheer confirmation to a Hayworth projection. So here the Hayworth projection is going to be the same six memberm ring except the ring itself is a little bit more, it's a little bit more flat, right? So instead of drawing this sort of curve structure characteristic of a chair confirmation, I'm going to draw a flatter six member ring. However, I am going to shade the bottom half of the ring itself just to show some three dimensionality, right. So I'm going to shade this bottom half right here. So as a convention, right, the anomic carbon carbon one is going to be on the very far right side. So now that I've gone ahead and I've drawn the the ring of the Hayworth projection. What you can do is you can actually match each hydroxy group to each respective carbon in the Hayworth projection. So here, if I look at the chair confirmation of the galactose, right, the hydroxy group of carbon one is equatorial pointing upwards. So in my Hayworth projection, the hydroxy group is also going to be pointing upwards. Where is the hydrogen points downwards? So now I'm going to do the same thing with carbon two. And in carbon two, the hydroxy group is pointing downward. So that's what I'm going to reflect in my Hayworth projection. In carbon three, it's pointing upwards and in carbon four, it's also pointing upward and I almost forgot my hydrogen on carbon five here. But essentially, right, you would draw the, the Hayworth projection by drawing the ring itself and adding hydroxy groups depending on what direction it's pointing in in the chair confirmation, right, if it's pointing downwards, whether it's axial or equatorial, then it will point in the same direction in the Hayworth projection. So at this point, we can go ahead and actually draw the Fisher projection because the Fisher projection is the representation of D galactose as a straight chain as opposed to a ring. So D galactose recall is an all dose, which means that in the fisher projection, right, as a straight chain, the anomic carbon one is going to be an aldehyde. Hence the abbreviation cho. So one way that you can imagine converting between a Fisher projection and a Hayworth projection. For example is if you imagine taking the fissure projection and laying it down on its right side, and the reason I'm bringing this up is because it helps you decide where to put the hydroxy groups on each position. Because if I look over on carbon too, right? In my Hayworth projection, the hydroxy group of carbon two is pointing downwards. And if I was to imagine laying down my Fisher projection on its right side, if the hydroxy group was on the right, then it would be pointing downwards in my Hayworth projection. And likewise, right, because the hydroxy group is pointing upwards on carbon three and carbon four for that matter, then their hydroxy group should be on the left. So at this point in the fish projection, right, we have the first four carbons, but carbon five is a little bit tricky because recall that the hydroxy group of carbon five is what creates the ring in both the chair confirmation and the Hayworth projection. So the configuration of the hydroxy group is a little bit iffy. It's not too clear, but we have to recall that we're drawing right here. D galactose. So because it's in the dec configuration, the hydroxyl group of carbon five must be to the right. And that leaves me with carbon six and there you have it. Here are the three representations of the galactose. So A was the chair confirmation, excuse me, the Fisher projection because B was the chair confirmation and then C was the Hayworth projection. So by imagining the differences between D galactose and D glucose, we were able to draw the chair confirmation by simply changing the configuration of carbon four. And then from there, we converted the cheer confirmation into the Hayworth projection. And then we could use either of the two that we did first to generate the Fisher projection. So if you stuck around to the end, thank you so very much for watching. And I hope you found this helpful.

Glucose is the most abundant monosaccharide. From memory, draw glucose in
(a) the Fischer projection of the open chain.
(b) the most stable chair conformation of the most stable pyranose anomer.
(c) the Haworth projection of the most stable pyranose anomer.

Key Concepts

Fischer Projection

Chair Conformation

Haworth Projection

Watch next