Draw the structure that corresponds to the name provided.
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Question

Draw the structure that corresponds to the name provided.

(b) (2R,3R,4S)-heptane-2,3,4-triol

Accepted Answer

Hello everyone. Let's work through this problem. It says provide the structure that is compatible to the IU pack name, three S four S five R octane, 345 trial. So when we are drawing a structure from a given Iupac name, I'd like to read the name from right to left and piece together the structure in that order. So we do this because at the end of the name is going to be the root name. So we want to identify that parent structure from the root name, then we will move to the left and identify substituent and their locus and add those substituent to the parent structure. And lastly, we will apply any configuration given to us like EZ configuration or RS configuration. And in this case, we see we are given R or S configuration. So how do we determine whether we have R or S configuration? Well, we know that we assign this stereo chemistry to chiral carbons and chiral carbons have four different groups on them. So the first step is to assign priorities of the four groups. And we do this in terms of atomic mass, right. So the higher the atomic mass, the higher the priority, which means hydrogen is always going to be the lowest priority, right? If it is present and if there is a tie between two atoms, let's say there are two carbon groups attached, then we simply keep looking at adjacent atoms until they differ. OK. Then once we have our priorities assigned, we need to check the orientation of group number four. Is it dash or is it which because this changes the criteria for R and S configuration? So we take note of the orientation of group number four, then we connect groups 12 and three in a circle. If group number four is dash and that circle goes counterclockwise, that is s orientation clockwise is our orientation. And the opposite is true for if group number four is wedge, counterclockwise is R clockwise is S OK. So let's apply these steps. 23, S four, S five are octane 345 trial. So looking towards the end of the name, I see octane, which means we have an eight carbon chain, 12345678. Let me draw that better.  aN:aN:000NaN and 345. Try all. Now we know the ending all is for an alcohol. Try means we have three alcohols and they are on carbons 34 and five. So one alcohol on carbon 31 on four and one on five. OK. And now we just have to make sure we are representing our stereo chemistry correctly. So when we are doing this backwards and by backwards, I mean, we are given the stereo chemistry and then need to draw it since hydrogen is going to be our group number four on each of these chiral centers, right? And we know hydrogen is our lowest priority. I'm going to ignore the hydrogen for now assign priorities of the other three groups, see which direction they go in and then make sure that that direction matches the one that is given to us. And if not, we will change up the orientation, right, the direction that it goes in will determine how we assign our wedges and dashes. OK. So looking at carbon three, our alcohol group is highest priority because of that oxygen. And that is going to be the case for all three of those carbons. Now looking at the groups attached to carbon three, we have ch two ch three or we have cho H cho H so that carbon that has the alcohol group is going to be the second highest priority. Then the carbon with two hydrogens and another carbon will be third priority. And we know that hydrogen is fourth priority, right? So we connect groups 12 and three, which in this case, are going counterclockwise. So for counterclockwise to be s right, the problem says three is S stereo chemistry. We need group number four to be dashed So let's make that hydrogen dash in the alcohol wedge. I'm going to redraw our chain and start applying these stereo chemistries that we assign. OK. Moving on to carbon four, we have S configuration as well and we already have alcohol first priority, hydrogen, fourth priority. Let me change the color. So now we have two carbons on either side and both of those carbons have alcohol groups, right? So if we look at the rest of the chain, there are just CH two S and at the end of the chain, ch three, if we go to the right, we have more carbons in that alka chain, right? It's a longer alk chain. So that will have a higher atomic mass. So carbon five and onward, that group is going to be second highest priority. And carbon three, that group is going to be third priority connecting groups 12 and three, this goes clockwise. So in order to be s orientation, number four has to be wedge which means our alcohol will be dash on carbon four. OK? And lastly, carbon five, alcohol is first priority. Hydrogen is fourth priority. And the carbon with that alcohol will be second priority. Carbon six, that plain alcohol group is going to be third priority connecting groups 12 and three that is going clockwise. So for it to be our orientation, we need group number four to be dash which means that alcohol is going to be which OK. So we have an eight carbon chain with three alcohol groups. The alcohol groups on carbons three and five are wedge and the alcohol group on carbon four is dashed. That's it for this problem. I hope this video is helpful and thanks for watching.

Draw the structure that corresponds to the name provided.
(b) (2R,3R,4S)-heptane-2,3,4-triol

Key Concepts

Stereochemistry

Functional Groups

IUPAC Nomenclature

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