In Problem 15.24, you were given the structure of the free aldehyde form of glucose. Try to draw the two cyclic hemiacetal forms of glucose you would get if (a) the OH on C4 formed the ring and (b) the OH on C3 formed the ring.

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All right. Hi, everyone. So this question says that the free aldehyde structure and cyclic hemi aal formed at C five of galactose are shown below. Draw the cyclic hemi aal form of galactose. If one, the ring is formed via the O on C four and two, if the ring is formed via the O on C three note, refer to the psych the chem acetyl show. So the first thing worth mentioning here is the fact that a he acetyl is the product of an aldehyde and an alcohol combining together. So an aldehyde is a compound that contains a car bottle or a carbon double bonded to an oxygen. But in order for this car bottle to be considered an aldehyde, there has to be at least one hydrogen attached to the car bottle carbon. So when the aldehyde combines with an alcohol, the carbon oxygen is converted to a hydroxy or a H group instead, and the oxy group or, or group of the alcohol attaches itself to the car bono carbon directly via its oxygen atom. So when a given molecule contains both an alcohol and an aldehyde functional group, a cyclic hem acetyl can form where there is an oxygen atom inside of the ring. So let's go ahead and start with part one. But before beginning to draw the specific structure, for the sake of answering this question, we're going to follow a few guidelines for the structures. The first guideline is that the O group or the hydroxy group of carbon one is always going to point downwards. Now, the second guideline has to do with the location of the remaining hydroxy groups in the free aldehyde structure. If a given hydroxy group points to the left side, then it is going to be pointing upwards in the ring structure. If it is to the right, then it is going to point downwards instead. So for part one here, let's go ahead and imagine a ring forming between the hydroxy group of carbon four and the carbon carbon at position one. So the first structure or the first ring I should say is going to contain four carbons and the hydroxy oxygen of position four, meaning that for part one, our ring structure is going to have five members in total. So that is the first thing I am going to draw here and I'm adding some shading here to showcase three dimensionality, right. So here carbon one is depicted on the right side. So its hydroxy group is going to be pointing downwards. Whereas the hydroxy group of carbon two is also going to be pointing downwards because it is to the right side in the free aldehyde structure. Now, because the hydroxy group of carbon three is to the left side in the free aldehyde structure in my ring, it's going to instead point upwards no, on carbon four, I have to take into consideration the remaining carbon atoms that are not part of the ring structure. So from carbon four in the ring, I'm going to go ahead and draw carbon five which has one hydrogen and one hydroxy group and then branching from carbon five goes carbon six with its own hydroxy group. So that concludes the ring structure for part one. Now lets talk about part two. So if I scroll up here to view the question, part two, imagines a ring forming between the hydroxy group on carbon three and the car bono carbon on position one. So in this case, the ring is going to contain three carbons and the hydroxy oxygen of position three. So that means that we're going to have a four member ring, meaning that I'm going to go ahead and draw a rectangle do here. I have a rectangle and once again, I am going to add some shading the show three dimensional. So in this case, the hydroxy group, a rather carbon one of the ring is depicted on the top right corner and its hydroxy group is going to point downwards per our guidelines. And so will the hydroxy group on carbon too. Now, carbon three is going to be attached to the remaining carbons in the structure. So in addition to drawing the hydrogen atom attached to carbon three, I'm going to make some space for the remaining carbons in the structure. So attach the carbon tree goes carbon fork. And because the hydroxy group of carbon four is pointing left, I'm going to go ahead and place it facing upwards in my structure. This gives way to carbon five whose hydroxy group should instead point downwards. And last but not least I have carbon six attach directly to carbon five with its own hydroxy group and there you have it. So here we have the two new cyclic hem acetyl structures for galactose, specifically part one created a five member ring and part two created a four membrane. So if you stuck around to the end of this video, I thank you so very much for watching and I hope you found this helpful.

In Problem 15.24, you were given the structure of the free aldehyde form of glucose. Try to draw the two cyclic hemiacetal forms of glucose you would get if (a) the OH on C4 formed the ring and (b) the OH on C3 formed the ring.

Verified Solution

Key Concepts

Aldehyde Structure of Glucose

Cyclic Hemiacetal Formation

Anomeric Carbon