How many of the following names are correct? Correct the incor...

Question

How many of the following names are correct? Correct the incorrect names.

d. 1-ethyl-1-pentene

e. 5-ethylcyclohexene

f. 5-chloro-3-hexene

Accepted Answer

Welcome back, everyone identify whether or not the provided names are correct. If not give the correct names to those with incorrect ones. And we're given three names to consider. Let's start with the first one or four di ethyl Cyclo one in as the parent suggests, we have Cyclo one in. So we want to draw a six member ring, right? And we want to add a double bond at any position. Let's recall that the double bond always starts at carbon number one. So we can immediately label one of our double bonded carbons as one and we can enumerate the structure. Let's suppose we go clockwise so that we have carbon 23 and four, carbon four has two ethyl groups, right? Because it says die ethyl. So we simply are going to add two ethyl groups to carbon number four because it says 44. So let's go ahead and add two ethel groups. And now let's think if this name is correct. Well, first of all, our parent is Cyclo hex one E and now our ethyl groups are at carbon number four. We want to make sure that we minimize that coefficient. So what we're going to do is see if we can actually go counterclockwise. Well, we cannot do that right. Because whenever we label our double bond, it always starts at carbon number one and always extends to carbon. Number two, we cannot now go counterclockwise and say that the two ethyl groups are bonded to carbon three instead. And that would be the correct name because it minimizes the locus simply because this is not allowed. And we must go from carbon one to carbon two. And specifically in this case, clockwise, because the two carbons define the double bond all the time. So that means the name is correct because we have the correct parent. The double bond always starts at carbon number one and extends towards carbon number two. And then our substituents are two ethyl groups bonded to carbon number four as a result of the OPEC numbering. What about number two? If we look at the parent of the name three chloral one isopropyl but one in, we understand that the parent chain is but one in. So we simply want to draw a four member chain. The double bond starts at carbon number one and extends to carbon number two. We also see that there's a chloral substituent bonded to carbon number three. And also it says a one iso pro pill. So that means we're going to take carbon number one and draw an isopropyl substituent and bonded set. Now we want to check if this name is correct. First of all, what we can immediately notice is that the given structure has a longer parent than a four member chain. So what we want to do is thus, we want to find the longest continuous carbon chain. And we noticed that we have 123456 carbon atoms. So first of all, it's hexen. But now how do we enumerate that? Well, if we go from left to right, the double bond starts at carbon three. If we go from right to left, we get one to once again, that will be carbon number three. And it also minimizes the position of the chloral substituent, right? We want to minimize our lo so we're actually going to choose the direction from right to left instead of left to right because this gives us our chloral substitute with a lower low count. So in this case, we're going to say that first of all, we have one of our substituents as two chloral and also there's a methyl group B as a carbon five. So we're going to say five methyl. OK. And the parent is hex, the level bond starts at carbon three. So that'll be hex three in. Now, are we asked to include E or C A summarize? Well, we aren't, but we can see that we noticed that in this case, actually, we cannot sell what exact iser we're considering, right? Because we can make it E or C and we aren't given any hint regarding which item are we're considering. So we're not going to worry about the trans, let's simply add the name without E or C assignment. So we went to, we went to arrange the substituents alphabetically C goes before M. So the name would be to Kloo five Muscle Hex three in and the original name given to us was incorrect. So we had a correct name for number one and an incorrect name for number two. The correct name is now given to us two chloral five methyl hex three in what about the third structure? But the third structure were given a 11 day ethyl hex one in. So first of all, we're going to draw hacks one E A six member chain 123456. The double bond starts at carbon one and then carbon number one also has two ethyl groups, dy ethyl, one comma one, right. So essentially, it implies that carbon number one has two equivalent ethyl substituents, you know, we can simply draw them out. OK. And now let's consider this name once again, we immediately noticed that the parent is longer than hex one in. Instead, if we look at the number of carbon atoms in the longest continuous chain, we have 12345678. So that would be octin, right? And we also want to make sure that the double bond, the position of the double bond is minimized as a lon. So we did the right choice by enumerating our parent from left to right because going from right to left would give us 12345, right. So the double bond cannot start at the higher low count. So the lowest possible low count for the double bond is carbon three and we can sell that. First of all, do we have any subscriptions? Yes, we do. Carbon three has an ethyl group. So that'd be three ethyl. OK. And that's it. That's the only substitution. We can say that the correct name would it be three SL? And now our parent is because we have eight carbon atoms in the longest continuous chain. So three ethel act three ian because the double bond starts at carbon three and the original name was incorrect. It didn't considered the longest continuous carbon chain. Now, let's recap starting with the first structure. What do we see? Well, essentially the name was correct. We had the correct parent, the correct locus and also the cracks substituents, right? Two ethyl groups bonded to carbon four. The second structure had an incorrect name to begin with and we have corrected that name to two chloral five methyl hex. And now for the third structure, the name was incorrect and the correct name should have been three ethyl oct three in and that would be it for today. Thank you for watching.

How many of the following names are correct? Correct the incorrect names.
d. 1-ethyl-1-pentene
e. 5-ethylcyclohexene
f. 5-chloro-3-hexene

Key Concepts

IUPAC Nomenclature

Alkenes and Their Naming

Cycloalkenes

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