Draw the correct structure from the following IUPAC names:
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Question

Draw the correct structure from the following IUPAC names:

(c) (1S,4R)-4-bromocyclohex-2-en-1-ol

Accepted Answer

Welcome back everyone. What is the structure of the molecule? With an Iupac name? One R four S four chloral cyclops two in 10. Let's begin solving this problem. As always, our first step is to identify the parent. At the end of the name, we see that the structure is cyclops. Cyclops essentially means that we have a five member grain, right? Because we have pent and all means that it is an alcohol. So we're going to add an alcohol group, which also defines carbon number 01, just as the name suggests, we're going to add carbon number one and we can continue from here. Our next step is to identify the substituents. The name tells us that we have two in it's technically not a substituent. It tells us that there is a double bond present in coming from Elkins. And then we have for chloral. So carbon number four is bonded to chlorine. Let's suppose that we label our carbon atoms clockwise, we have 12345. OK. And two E means that the double bond starts at carbon number two and extends to carbon number three. So we're going to add that double bond for carbon. Number four, let's add the chloral substituent. Well done. Our next step is to define chirality as the name suggests, this is one R four S carbons, one and four. They are chiral, they have four unique substituents. And we want to take that into account. So let's start with carbon number one, it is bonded to oxygen. There's also an implicit hydrogen atom as well as two carbon atoms. So we want to utilize the Khan Ingle Prilo rules to assign wedges and dashes. Right. First of all, oxygen will have the highest priority because it has the greatest atomic number. Hygen has the lowest atomic number. So it gets the lowest priority in between the two carbon atoms. We have to think about the atoms that are bonded to each carbon. Carbon. Number five has two, implicit hydrogen atoms as well as one carbon from the rink. Now, carbon number two has a double bond to another carbon. So we can identify that as two single bonds for the K OL pre law rules and there's also one implicit hydrogen. So to determine which carbon gets a higher priority, we're going to exclude exactly the same hydrogens and carbons. And now we are differentiating between hydrogen and carbon. So that means the carbon atom with a double bond gets a higher priority priority two M priority three. So now essentially we can just guess let's suppose that the alcohol group is winding up on a wedge. So that priority number four is pointing down on a dash according to the K Ingle pre rules, we can assign a rotation directly. So now we have one for oxygen four for hydrogen, two, for the double bonded carbon three for the single bonded carbon. And if we trace our path from 1 to 2 to 3, right, we get a clockwise rotation which is R and this is indeed what we have in the name. It's one R, meaning we have assigned the configuration for carbon number one correctly. It must be a wedge for that carbon, a wedge between carbon and oxygen. And now let's think about carbon number four, right? Because this is our next chiral carbon atom, that specific carbon is bonded to chlorine, it's also bonded to an implicit hydrogen. Once again, we have two carbon atoms. Similarly, we have a double bonded carbon atom and a single bonded carbon atom. So as we know the double bonded carbon atom, well, get a higher priority. So let's go ahead and label priorities. Chlorine gets priority one. It has the greatest atomic number, hydrogen gets the lowest priority. The double bonded carbon gets priority number two and single bonded carbon gas, priority three. So now we can just redraw our sketch to make it easier to visualize. OK. So what we're going to do is just add that double bond. We now have a wedge for carbo number one. Let's simply assume that the chloral group is pointing up on a wedge as well because we want to have priority number four pointing down to assign the rotation directly. And now let's see what we have. Well, chlorine has priority one, hedging gets priority four. The double bonded carbon gets priority two and the single bonded one gets priority three. So if we go along 123, if we trace our path, we get a counterclockwise rotation, which is s and this is indeed what the name suggests. Therefore, we have our structure. Let's simply redraw it for complete clarity. So first of all, we are drawing a five number ring. OK. Now what we're going to do from here is just add the alcohol group to carbon number one pointing up on a wedge. Now we are adding a double bond between carbons two and three and carbon number four gets a chloral substitution pointing up on a wedge. And that would be our file structure. Let's go ahead and label it. That would be it for today. And thank you for watching.

Draw the correct structure from the following IUPAC names:
(c) (1S,4R)-4-bromocyclohex-2-en-1-ol

Key Concepts

IUPAC Nomenclature

Stereochemistry

Cyclohexene Structure

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