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

Question

Using IUPAC rules, name the following molecules.

(a) Structural representation of organic molecules illustrating cis and trans isomerism with IUPAC naming instructions.

Accepted Answer

Welcome back, everyone use the Iupac rules and write the name for the molecule given below the structure that we are given is a hydrocarbon with one double bond. So first of all, we want to define the class of that compound and because there's one double bond present, we define the compound as an Keen number two, we want to identify the parent chain which contains the longest continuous carbon chain. If we go from left to right, we have carbon atoms. So we can say that our parent is heine hex means six. And in means that we have one double bond in our structure. Well done. Now, let's enumerate our parent shame properly. We have two possibilities. We can go from left to right or right to left. So let's go from right to left and green. And we're going to have 1 to 3 or 56. First, what we notice is that in each case, the double bond starts at carbon number three, we always want to make sure that first of all, we minimize the coefficient for the double bond in each case, that's the same. So that means we have to carefully analyze our substitutions. If we go from left to right, we have substituent on carbon two, substituent on carbon three and substituent on carbon number five. But if we go from right to left, then we have substituent bonded to carbon's number two, four and five. And we want to make sure that we minimize the coefficients for our substitutions, right. So we're going to use 235, meaning the red pathway is preferred because we end up with lower low cons or our substitutions. Specifically, we are going to use the right pathway and enumerate our chain from left to right because that gives us lower coefficients or the substitutions. And what is the most important part here is to understand that the double bond starts on carbon number three. So we're going to have X three in which is the correct name of the parent because it clearly states where the double bond is, it starts at carbon. Number three, step number four is to identify the substitutions. We noticed that all of our substations are exactly the same. There are three solid lines which indicate the presence of three methyl substitutions because we have three of them. We're going to use a prefix, try and that we want to make sure that we understand where they are. One of them is bonded to carbon two, the other one is bonded to carbon three and the final one is bonded to carbon number five. So we can say that we have two comma three comma five. Those are the positions dash try, which tells us that there are three me groups. So we're going to say tri methyl, well done. And now we want to make sure that we also include stereo chemistry. If applicable. Whenever we have a double bond, we want to make sure that we understand if the structure is E or Z or none of them to do that, we are, first of all going to start with carbon number four, it has an implicit higher channel and carbon number four, which is double bonded has hydrogen substituent. And then if we go down, it has carbon bonded to it, right? Because hydrogen is lower in priority, it gets priority two and carbon is greater in priority, it gets priority number one due to the fact that carbon has a greater atomic number than hydrogen. And we once performed the same analysis for the carbon atom labeled as carbon number three. However, in this case, it's a bit more complex because it's bonded to the metal group. So we have carbon bonded to three hydrogen atoms and the other substituent which is an isopropyl group is carbon that is bonded to one implicit hydrogen and two carbon atoms from the adjacent methyl groups. So if we exclude our exactly same hydrogen atoms, we notice that the isopropyl group is greater in priority because we have two cc bonds versus two C H bonds and carbon has a greater atomic number than hydrogen. So that means for carbon number three, the isopropyl group is greater in priority. So it gets priority number one, damn, the metal group which gets priority number two. Now, because the highest priority groups are on opposite sides of the double bond. That means we are going to give it an E configuration. Well done. We have all of the information we need and we are ready to give it a name. We start with E and prime CS. Our next step is to add our substitutions 235 tri methyl. And because we are done with our substitutions, we want to finish our name by adding the parent hax three in. So let's label our final answer and state that the name of the given Lke is E 235 trime hex three in. Thank you for watching.

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

Key Concepts

IUPAC Nomenclature

Functional Groups

Stereochemistry

Watch next

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

Key Concepts

IUPAC Nomenclature

Functional Groups

Stereochemistry

Watch next

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