Given the name, draw the structure of the following compounds....

Question

Given the name, draw the structure of the following compounds.

(c) (3S,6Z)-8-ethyl-3-iododeca-1,5-diene

Accepted Answer

Welcome back, everyone. Draw the structure of the compound with the name three S five E three, Chloro seven methyl Octa 15 di So let's break down our problem or so let's identify our parent. And if we carefully analyze the name and if we look at the ending of our name, we have Octa dyin if we exclude the coefficients. Now, what does that mean? It means that the continuous carbon chain has eight carbon atoms and we have two double bonds die in, in means a double bond die means two and those double bonds are located at positions one and five. So what we're going to do is just draw an eight number chain 12345678. And we are going to enumerate that. Let's say we enumerate that from right to left, then we have 123, 4567 and eight. Well done. What we want to do is just add two double bonds, starting at carbon number one and then starting at carbon number five. Perfect. We have our parent number two. Let's consider our substitutions. Looking at the name, we have a Chloro substituent bonded to carbon. Number three. And then we have a methyl substituent monitor to carbon number seven. So we're gonna see add another structure where we draw the substituent carbon. Number three has a chloral group. Carbon. Number seven has a metal group. So we're just going to draw a solid line because the carbon atom is implied. Number three, we are nearly done. We just have to consider three S five E and we're going to start with five E. It basically tells us that carbon number five can exhibit cyran isomerism or E and Z isomerism. And E means that the groups, in this case, all kill groups. But in general, it means that the highest priority groups R on opposite sides of the double bond. The reason why we're considering E and trace in this case as synonyms is because carbons five and six have to implicit hydrogens. And indeed, if we consider our double bond, the two alki groups starting with carbon seven and carbon four, they're on opposite sides of the double bond. So we have thrown it properly, well done. And finally, we want to consider three as it means that carbon number three is Kyle. So if carbon number three is chiral, we have to make sure that we assign the S configuration to that carbon by adding a dash or a wedge or the chloral group. In order to do that, we simply want to essentially guess and check if we are correct because we only have two possibilities, right? It can be R or S. So what we're going to do is just replace the solid line with a wedge so that the chloral group is planning up. And then that means the implicit hydrogen atoms pointing down on a dash. We want to assign the priorities to our substitutions. We have two carbon atoms, one, it's carbon number three hydrogen and chlorine. So the Chloro substi gets priority. Number one. Actually, let's use another color so that we don't get confused. Let's suppose black, the Chloro substituent gets the highest priority number one because it has the greatest atomic number, hygiene gets priority number four because it has the lowest atomic number and uh between the two alka groups because we have two carbon atoms. It is important for us to think about the other atoms that are bonnet to them. So, Carbo number four has two implicit hydrogen atoms and then it is bonded to carbon. Number five. On the other hand, Carba number two has an implicit the hydrogen atom because it's double bonded and then because it's double bonded to carbon number one, we can just state that there are two carbon carbon single bonds. This is definitely not correct, but it will help us to assign the correct priorities. So whenever we have a double bond, we can say that that double bond corresponds to two single bonds. And if we exclude our exactly same hydrogen atoms exactly the same carbon atoms. We are comparing hydrogen against carbon and carbon wins because it's greater in atomic number than hydrogen. So we can say that the group that has a double bond will be higher in priority than the one than the alki group starting with carbon number four. So it gets priority number two. And then the siege two group followed by our group gets a lower priority. So that'd be priority number three, well done. And now according to the K Engel pre rules, we can assign the rotation directly because the lowest priority group is pointing down. And if we do that, and if we try to assign the rotation and trace our path from 1 to 2 to three, we get an R configuration, which is not as meaning if we simply want to get an S configuration, we are just going to reverse our wedge into a dash and that will give us us. So we are done. Let's go ahead and draw the file structure carbon atoms. We're going to add our double bonds, carbon number one and carbon number five. Then of course, we are adding our methyl group. And finally, we said that our guess was incorrect. Instead of having a wedge, we must have a dash for the chloral group so that the configuration becomes S instead of R. And that's how we get our final structure. Thank you for watching. And I hope to see you in the next video.

Given the name, draw the structure of the following compounds.
(c) (3S,6Z)-8-ethyl-3-iododeca-1,5-diene

Key Concepts

Stereochemistry

Geometric Isomerism

IUPAC Nomenclature

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