Using the alcohol shown, draw all the possible alkenes that might be formed on its dehydration. Which alkenes can exist as cis–trans isomers? Draw them, in both condensed and line structure, and identify each as cis or trans. Explain your choices.

Question

Using the alcohol shown, draw all the possible alkenes that might be formed on its dehydration. Which alkenes can exist as cis–trans isomers? Draw them, in both condensed and line structure, and identify each as cis or trans. Explain your choices.

Accepted Answer

Welcome back, everyone. Our next problem says, consider the structure of the alcohol shown below. Sketch, all the possible alkenes that can be formed from the dehydration of the compound in both line and condensed structure and identify each as cis or trans if applicable. So we have an alcohol here where it from the point of view of the carbon that is connected to the alcohol, which I'll highlight in blue, it has an ethyl group on one side and then a propal group with a methyl branch on the other side. So we have a secondary alcohol since there are two R groups on that Carmen. And we need to look at all the possible alkenes I can form. So let's recall that in a dehydration reaction, we have the hydroxy group removed from one carbon and a hydrogen removed from an adjacent carbon. So let's identify the adjacent carbons that can be involved. So they would be the ones on either side of our alcohol carbon. So I'm highlight, I've highlighted the alcohol carbon in blue and the adjacent carbons in red to keep track here. So let's look at the possible alkins that can be formed now, we know we can also end up with both cysts and trans isomers. And that's a possibility. So let's start with our first alkene on the left side on that ethyl group. So we have our, I'm gonna label the carbons from left to right, not the way we would number them if we were naming our structure. But just because it's easier to follow along on the structure from left to right. So I start with my first carbon, I've now hit my carbon that's going to participate in the double bond. So then I have a double bond between the 2nd and 3rd. So the second carbon here was the red carbon and then in blue, the al formerly the alcohol carbon. And then we go on to our other our constituent, which is propal group with a methyl branch. So if we were to number it properly, we have a six carbon chain here with the double bond on carbon number two. So we have a hexen with a methyl group on carbon five. So that's our first one and then our other possibility for a double bond. So we'll look at citizen translator first, we'll just look at the two different types of alkenes with the two different adjacent carbons. And our other possibility, we'll put that ethel group there this time not participating in the double bond. Then we've reached our blue alcohol carbon, which is now in a double bond on the other side. So from that third to that fourth carbon, so there's our red carbon that gets its hydrogen eliminated and then our purple group with its metal branch. So again, we have a hexen here. But in this case, our double bond is on carbon three, still with our methyl group on carbon five. So these are two basic options for the location of the double bond. Now, we need to see do we have a assist and trans isomers? Well, yes, we do because in each case, we have on either side of our double bond, we have two unique R groups. So in each case, we have a hydrogen on hydrogen and then a an alkane or an alkyl branch. So let's look at our first structure that we drew, we have the hydrogens on opposite sides and the alkyl branches on opposite sides. So this is a trans isomer. So we need to draw this cis isomer. So we put the alky branches on the same side of the double bond and the hydrogens on the same side. So let's draw that structure. So carbon number one, and then we have our double bond. So I'm gonna turn this around otherwise I'm going out of space. So apartment number one, our double bond and then coming down in the same direction to make a cis isomer. So now I have 1234 carmens and then the rest of our branch structure there. So that's our cis isomer. And then our second structure that we drew is also trans, since our hydrogens are on opposite sides and our keel branches on opposite sides. So we'll just redraw that as a cis isomer. So 123, we're on carbon three, it's a double bond. And then again, we'll make this next branch go down. So it's on the same side of the double bond and then it's branch structure there. So again, this is the cis isomer as both alkyl branches are on the same side of the double bond and hydrogens on the same side. So we needed to sketch the possible alkenes that could be formed and identify a cis or trans. So we've drawn both the cis and trans versions. We've drawn line structures. We also need a condensed structure. So one important thing to note is that the condensed structure for the trans and cis isomers will be the same because the order of the bonds and the number of hydrogens and things like that will all be the same. It's just their arrangement and space around the double block. So we only need to write two condensed structures. So let's write the condensed structure for our first molecule where our double bond is from carbon two to carbon three. So we'll start with carbon one. Let's again, number the carbon. So we can keep track of how many CS we're writing here. So we'll number everything. 123456. So we start with carbon one which is a terminal carbon. So ch three, now carbon two is involved in a double bond. So it has only one hydrogen. So c carbon three in a double bond also has just a single hydrogen. So another ch carbon four has two hydrogens. So c two, carbon five has a methyl branch on it and that gives it three bonds. So it will have a single hydrogen. So we'll have CH and then in parentheses, ch three. However, when we look at this, essentially carbon six is another methyl branch. So if we put subscript two, after that, CH three, in parentheses, that expresses the structure of this in a condensed form. So we have CH three, Chchch two, CH and then in parenthesis, ch three subscript two. So let's just double check and make sure we have the correct number of carbons 123456 in the main chain and seven with the branch. So we have our kind of structure and we have our trans and cis isomers again, that forms the double bond between carbons two and three. So now we have one more condensed structure to write for our second um set of alkenes, the trans insist isomers when we have the double bond between carbons three and four. So we start with carbon one, which is CH three. In this case, carbon two has two hydrogens. So ch two, now we're on to carbon three, which is participates in the double bond. So it has a single hydrogen, ch carbon four, also a single hydrogen, another ch and then carbon five has three bonds. So it's CH as well. And then again, we have the methyl branch and the last CH three. So we'll write that in parenthesis, CH three and then subscript two. So let's double check our number of carbons 123456 for the main chain and seven for the branch. So we have CH three, CH two, Chchch and then in parenthesis, CH three subscript two. So that's our second condensed structure that shows the structure of our cis and trans isomers of our second alkene possibility with a double bond between carbons three and four. See you in the next video.

Using the alcohol shown, draw all the possible alkenes that might be formed on its dehydration. Which alkenes can exist as cis–trans isomers? Draw them, in both condensed and line structure, and identify each as cis or trans. Explain your choices.

Verified video answer for a similar problem:

Key Concepts

Dehydration of Alcohols

Cis-Trans Isomerism

Condensed and Line Structures