Disregarding stereoisomers, draw the structures of all alkenes...

Question

Disregarding stereoisomers, draw the structures of all alkenes with molecular formula C₅H₁₀. Which ones can exist as cis–trans isomers?

Accepted Answer

Welcome back, everyone draw the five constitutional isomers of C 4h 8, which can exist as cist trans isomers. Whenever we want to identify the possible structures of a given molecular formula, we can first of all calculate the hydrogen deficiency index as one half multiplied by the two multiplied by the number of carbon atoms plus two plus nitrogens minus hydrogens minus halogens. This will give us an idea regarding the presence of any pythons or rings. So we have one half multiplied by two multiplied by four carbon atoms plus two plus zero nitrogens. We subtract eight hydrogens and zero halogens. And this gives us. Now we have eight minus eight plus 22, divided by two, gives us one. Now, what does it tell us? It tells us that we either have one rank or one pipe band, right? Meaning if we have one pi bond, it's just one double bond. So let's start with our structures and let's identify those that have a ring in them. If we have four carbon atoms, the first structure we can draw is just cyclo butane, four carbon atoms, eight hydrogens. Now we can make our ring smaller. So let's move on to a three membered ring instead. And all that we have to do is just add a metal group to one of our carbon atoms, right. So we have one, we have methyl cyclopropane, that's pretty much it, right? We can add that metal groups to any of our carbon atoms and that carbon atom will be identified as carbon number one of the three member ring. So that's our second structure. We cannot have any two member rank, right? So we can stop here and we can move on to a linear or open chains. So specifically what we want to do to begin with is draw a three membered parent chain. So let's go ahead and do that 123 carbon atoms, right? We can of course add four carbon atoms in the parent. So let's actually start with the longest possible chain just as we did with the rings, right? It will make it more consistent. So in this case, we have a four membered parent chain and we want to think about the possibilities of adding a double bond, we can add that double bond just at the end or between carbons two and three, right? We have two possibilities. In this case, let's go ahead and draw both of them. OK. So this gives us structures three and four. Now, the reason why we cannot shift the pipe bond between carbon three and four is simply because it will give us structure three, right? We will just enumerate it differently or we can simply say we can flip it and get structure number three. So we only have two possibilities to move our double bond within a four member chain. Now, what about the fifth structure? It says five of them, right? So that means we can essentially shorten our parent to let's say three carbon atoms. What we're going to do is just add a methyl substituent, add carbon two, right? And now we want to add a double bond, we can add a double bond anywhere because we have three equivalent metal groups bonded to a central carbon, meaning we can simply add it at the top. And this gives us our fifth and last possible structure. Now, what about cyst trans isomerism? Whenever we want to consider cis trans isomerism, we want to understand that we have a double bond and we want to make sure that each double bonded carbon atom has two different substituents. So let's say if we have two, exactly the same subscriptions bonded to an sp two carbon, there will be no cyran isomerism because to begin with, we want to make sure that the two substitutions bonded to our double bonded carbon are different. For example, structure three, it doesn't exhibit cyran is amorous because there are two implicit hydrogen, right? So you can say number three, no Crans what about five? Once again there, their noses trans isomers because we have two exactly the same hydros bonded to a double bonded carbon atom. On the other hand, we can also notice that there are two equivalent metal groups but structure four has a trans se. How do we know that? Well, essentially, first of all, we notice that there's an implicit hydrogen and a metal group. So those are two different substituents on each double bonded carbon atom. So what we want to do is just draw its trans isomer. We're not required to draw it in this problem. But we can, what we're going to do is draw the given double bond. And since we have drawn a trans isomer, right, because the two hydrogen atoms or the lowest priority groups are on opposite sides of the double bond, we can simply draw its cis form where the two implicit hydrogen atoms or the lowest priority groups are on the same side of the double bond. So that would be cis and we have drawn a trance Heiser. So we can say that technically, there will be six isomers, right? We have drawn five of them ignoring stereo chemistry. But if we take into account stereo chemistry structure, number four can actually have translate some isomers, right? And this gives us an additional cyst structure. So we can get six of them. So we can label the cyst form as the sixth one if we want, right, if we are asked to include stereo chemistry. So let's go ahead and label our answers. That would be it for today. And thank you for watching.

Disregarding stereoisomers, draw the structures of all alkenes with molecular formula C₅H₁₀. Which ones can exist as cis–trans isomers?

Key Concepts

Alkenes

Cis-Trans Isomerism

Molecular Formula and Structural Representation

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Disregarding stereoisomers, draw the structures of all alkenes with molecular formula C5H10. Which ones can exist as cis–trans isomers?

Key Concepts

Alkenes

Cis-Trans Isomerism

Molecular Formula and Structural Representation

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Disregarding stereoisomers, draw the structures of all alkenes with molecular formula C₅H₁₀. Which ones can exist as cis–trans isomers?