Using IUPAC rules, name the following molecules.
(e)

Question

Using IUPAC rules, name the following molecules.

(e) Molecular structure with a bromine (Br) and thiol (SH) group attached to a zigzag carbon chain.

Accepted Answer

Welcome back, everyone. Give the Aipac name of the following AYC style. We're given our structure and we want to name it. First of all, let's identify our parent to identify our parent. We have to use a form of Alcaine. Thy specifically what that means is that our parent will correspond to the can itself. So we want to identify the longest continuous carbon chain. And what we noticed is that we have a total of six carbon atoms in the parent chain and sex corresponds to X. So our parent will be hack sane because that's an Alcaine tho. However, what we want to modify here is the position of the thol group, our sh group or the thyle group belongs to carbon number two, we are going to assign it a low count of two because if we go in an opposite direction, that would be a low count of five, which is greater than two. And recall that we want to minimize the low count which carries the group with the highest priority. So two is the lowest possible Logan for us, meaning we modify our parent into Hexane two A L which tells us that the thy group is bonded to carbon. Number two, well done. Now, let's identify our substitutions and we only see one substituent bonded to carbon. Number four, C four has AC L substituent which we call a Chloro group. So within our name, we will have four Chloro. This is our one and only substituent number three, we want to assign the configurations to our chiral carbon atoms. So let's identify each of them starting with carbon. Number two, we are simply going to draw our substitutions. We have our age group. Let's actually identify that the S age group is winding down. So we're going to draw a dashed bond, then we will absolutely have that wedge for the hygiene atom, right? It is implicit, then we have a metal substituent. So that's our CH three group. And then on the right, we have CH two and we can move along our chain. The other carbon will be CHCL CH bonded to CL. And of course, we can state that the chain continues. So what we're going to do here is just assign priorities. Sulfur has the greatest atomic number. It gets priority. Number one, Igen gets the lowest priority based on its lowest atomic number. And then comparing the two alki groups, the one that is shorter gets a lower priority because we don't have any branching. When we consider our cie H three versus Sieh two, we notice that within the CH three group carbon is binded to three hys. But then the CH two C group has two hygienes and one carbon. So that one carbon is greater in priority than one hyen. Meaning the right side gets priority number two. And the left side gets priority number three. Then we assign our path going from 1 to 2 to 3. We notice that we're going clockwise, that would be an R configuration. But according to the Ingle pre rules, since priority number four is not pointing down, we're actually going to assign S right. We have to flip our assignment to make sure that the lowest priority group is pointing down. So the actual configuration is on carbon number two, let's say that carbon C has an S configuration. Now, what about carbon number four? We're going to draw another carbon atom which is bonded to the Chloro group than an implicit hydrogen atom that must be pointing up, right. So we are already keeping in mind that we will be reversing our assignment on the right side. What do we have? We simply have an ethel group that will be CH two CH three. On the left side, there is a necessity to think about all of our atoms. But first of all, we have CH two and now this is really important, we have two equivalent CH two groups on each side. So we want to see what goes next. After that CH two group, we have CHS, right? And then the chain continues, there's another carbon. So now let's think about the priorities. The chloral group will get priority number one, based on its highest atomic number, hygiene gets priority number four. Then let's simply exclude the two s two substitutions. And what we are comparing is the carbon carbon bond, right? Carbon sulfur bond and carbon hydrogen bond versus three carbon hydrogen bonds if we exclude one hydrogen, so that we will have two carbon hydrogen bonds versus carbon carbon and carbon sulfur bond, right? So any of them is hiring priority. So we can say that the substituent on the left gets a higher priority and the substituent on the right gets a lower priority. Now, we assign our path 123, that's counterclockwise, but now we flip it to R because the lowest priority group is pointing up, meaning carbon number four gets are well done. Now, first of all, we're going to think about the name starting with our one and only substituent, we have four chloral, right? And then our parent is Hexane Sheth Isle. That would be our name, but we definitely want to include our configurations. We want to make sure that our configurations are written in an ascending order. So we have two S and then four R. So the final name of the given structure would be two S four R, four chloroxine two. Tho Thank you for watching.

Using IUPAC rules, name the following molecules.
(e)

Key Concepts

IUPAC Nomenclature

Functional Groups

Locants and Substituents

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