Give the IUPAC name for each of the following:

Question

d.

Accepted Answer

Hi, everyone. Lets look at our next question. It says, what's the correct Iupac name for the following compound. And we have a Pentagon shape or a ring with a double bond in it. And a methyl group on the carbon adjacent to that double bond. Our answer choices are a five methyl, one cyclops, B five methyl, two cyclops, C three methyl, one cyclen or D three methyl, two cyclops. So the first step, whenever we're looking for a name for a compound is to find the longest carbon chain. In this case, we've got a ring and it only has a methyl substituent. So the ring is clearly the longest chain. So we see it as 12345 carbons. So that will be our longest carbon chain. It's a ring. So it will have the name cyclo and with five carbons, it will be pent and then with the double bond in it pine. So cyclo Pantene, now we need to number the carbon so we can give the location of this double bond. So when we're numbering carbons, we will start with the carbon closest to the double bond and we number the carbon so as to give the lowest possible substitute numbers. So in this case, we'd first start by thinking, do we even need to give the location of the double bond? Does it matter sometimes in rings? Um, it doesn't really matter because all the carbons are identical. But in this case, since one has a methyl group, they are not all identical because each carbon is a different distance away from the methyl group and the double bond. So we need to start on the closest carbon to the double bond. Well, since we're in a ring, we can start at the double bond. However, there's two directions we can go since we're in a ring. And we need to then keep in mind that we need to give the lowest possible substituent numbers. So we need to number our double bond, carbon one will be one of the carbons in the double bond and number it so that we get the lowest possible number for the methyl group. So that would mean starting on the carbon farthest away from the methyl group, calling that number one, then carbon two is the other side of the devil bond. And that means that the methyl group will be on carbon three, then we keep going around four or five. If we started on the other carbon, we have to include the double bond. So number one and number two would have to be the two carbons in the double bond that would send us the other direction around the ring and put the methyl group at carbon five. So that would be a higher substituent number. So we know that our cyclops team will have a one in front of it, one hyphen cyclops since our double bond starts at carbon one. Now, we just need to name the substituent group. We can double check, looking over at our answers to see if there's any answers we can eliminate, even just with what we've discovered so far. Choice A has one cyclopia, but choice B has two Cyclin. So again, we don't want to start the uh numbering at the second carbon. We'll call carbon one. The first one in that double bond. Choice C has one cyclops and choice D has two cyclen. So we can eliminate B and D for having the wrong number to identify our double bond. So now we just need to look at our meth methyl substituent, which again, we've learned is on the third carbon. So it will be a three, there's only one of them. So just three hyphen methyl and then a hyphen and then one hyphen Cyclin. So let's match that with our answer. Choices. Choice A says five metal. So choice A is not correct. That would be again if we numbered it in the other direction, but we wouldn't choose that since that gives us a higher substituent number. And finally, we just have choice C left, which does have three methyl one Cyclin, which is indeed the correct Iupac name for our compound. So the answer here is choice. C see you in the next video.

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Naming Alkenes practice set

Give the IUPAC name for each of the following:

d. <IMAGE>