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

Question

(b) ((Z)-1-cyclohexyl-2-methylhept-2-ene

Accepted Answer

Welcome back, everybody. Here's our next problem. It says provide the structure of the alkene with the name Z three bromo two, Cyclo Butty oct two E and we have four answer choices labeled A through D. So we don't need to freehand draw the structure. We just have to figure out which of these structures is correct. So our first step will be to look for the size of the main chain. And if we were drawing freehand, we would draw our chain first. We have the root OCT and we also have OCT in. So we know it has a double bond with that two in the middle, a double bond on carbon two. So let's look at the lengths of our chains here in choice. A starting from left to right. 12345678. So that's got the right number of carbons in B. Now, it's a little bit confusing because we've got this Cyclo butane ring. So we might be trying to think, do we start in the ring? But we see in our name that Cyclo Butyl is a chain, a side chain. So we would not want to count carbons in the ring, we know it's attached to our main chain. So left to right, we have that Cyclo bel and the bromine attached to the first carbon. So we already know that's incorrect because we've seen the numbers up there. We've got three bromo cyclo beetle. So in one sense, just a quick glance will tell us it's incorrect, even without needing to count the carbons in the chain. Because both the bromine and the Cyclo butty. First of all, they're on the same carbon despite having numbers three and two and they're on that first. So they're on carbon one. So they're on the wrong carbon. We can eliminate choice B choice C. Let's double check our chain length. Now note we have the way it's drawn. It looks as if we have a methyl group on our double bond. So we might be tempted to ignore that and start counting at the double bond. But remember we look for the longest continuous chain so that methyl group is the beginning of the chain. So counting carbons starting from the methyl group, 12345678. So C also remains a possibility. And then in choice D again, a quick look will tell us it's got the same problem as choice B, the bromine and the Cyclo butane are on carbon one. So they're in the wrong place just right off. So we can just eliminate that. So we've got two possibilities. A and C. So let's keep going through our name and check our structures here. So after doing the main chain and putting on the double bond at carbon two from carbons two to carbons three, let's look at our substituents. So we have two cyclo Butal. So our cyclo butane ring needs to be on carbon two. Well, we know that carbon two is the first carbon of the double bond. And both of these do indeed have that cyclo butane ring on carbon two. So A and C still remain a possibility and they need to have bromine on carbon three. So the other side of the double bond, which they do indeed both have bromine and carbon three. So the only thing left is to figure out our isomer here. We haven't labeled Z recall that we have the possibility of naming uh alkenes, either cis or trans or Z or E. And the ze naming system is better when we have multiple chains here. None of which are identical when you've got two identical substituents, cis and trans is a quick shorthand. So let's recall that Z means the top priority groups are on the same side of the double bond Z and E come from the German words for together an opposite. The way I like to remember it is Z and S, you can think sort of the and S similar letters. So you pair Z and Z and then E which has a vowel and opposite, which has a vowel. So that can help you remember which is which. So we know our, according to our name, we have the Z isomer. So our top priority groups need to be on the same side. So let's look at molecule a well, which is priority number one. We've got the bromine and then we've got this hydrocarbon chain. Well, bromine will be priority one on the other side, we have our methyl group and our Cybele group. Well, that cycle beetle group is going to be the top priority because it has, when you look at the atoms that it's bonded to, we have a carbon and then another carbon, whereas our methyl group is carbon and then hydrogen. So we see in molecule a, the top party groups are on the opposite side. So this is the e isomer. Therefore, a is not correct. So by process of elimination C is our answer. And indeed, we can see that the bromine which is priority one is on the same side as the Cybele group. So we have same side. So this is RZ actually, I should write that in blue. So it doesn't look confusing. So again, our proper structure for Z three bromo two cyclo butyl tune is choice C. We've got that eight carbon chain. It has a double bond from carbons 2 to 3. A cyclo beetle group on carbon two B roaming on carbon three and as a Z isomer the top pity groups are on the same side of the double bond. See you in the next video.

Given the name, draw the structure of the following alkenes.
(b) ((Z)-1-cyclohexyl-2-methylhept-2-ene

Key Concepts

Alkene Structure

Cis-Trans Isomerism

Cycloalkane Substituents

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