Write structural formulas for compounds that meet the following descriptions:
a. An alkene, C₆H₁₂, that cannot have cis–trans isomers and whose longest chain is 5 carbons long

Question

Write structural formulas for compounds that meet the following descriptions:

a. An alkene, C₆H₁₂, that cannot have cis–trans isomers and whose longest chain is 5 carbons long

Accepted Answer

Welcome, everyone. Here's our next problem, provide the structural formula for an alkene with a molecular formula of C 10 H 20 does not exhibit a cist trans isomerism thick carbons as its longest chain and is symmetrical. We've got a lot of conditions to fulfill here. Let's just take them one at a time and fill in our structural formula. Well, the simplest place to start, we have an keen, so we know we've got one cc double bond. So lets fill that in our molecular formula of C 10 H 20 tells us that we have only one double bond. As we know, we can look at the formula CNH two N plus two for an alkane. And in this case, that would be C 10 H 22. If it were an Alcaine, it's only H 20. So we have a single, either double bond or ring. We know where it's an alkene. We know we have a double bond, but there will only be one will then go to six carbons as its longest chain. So let's start with and we need it to be symmetrical. So we need it to be the same on either side. So let's look at putting a bond on either side and adding some more carbons. So we'll add a carbon on either side, vit tour double bond carbons. Now we have 1234 carbons. So each one just needs one more. Again, we have to add one on each side to keep this being a symmetrical molecule. So now we have six carbons, we cannot get any longer with this chain. It's symmetrical on both sides. However, we've still got two more conditions to fill. We need four more carbons and we cannot have cyst trans isomerism. That means that on at least one of our sides, the two R groups must be identical. But since our molecules to be symmetrical are remaining to uh the groups coming off of our double bond must be identical on both sides as well as on the same side. Since we have four carbons left, the simple solution is two carbons on either side. So we essentially have a double bond between two carbons and our four groups are all ethyl groups. So this gives us no cistron isomerism completely symmetrical and six carbons is our longest chain. We essentially have 26 carbon chains here. So now we just need to fill in our hydrogens to get a structural formula. So each of our terminal hydrogen, terminal carbons, excuse me gets three hydrogens note that now we have 12 hydrogens and then our four single bonded internal carbons will get two hydrogens each, which will give us the remaining eight hydrogens, our carbons in the double bond. Of course, don't have any more bonds and they have no hydrogens. Let me just fill in the rest of the hydrogens. So we've now drawn the structural formula or a molecule that fills all of these parameters. We have a hexen, six carbons, that would be a three hexen. We're just going to write the names to give a good description of this. A three hexen and then two ethyl groups coming off of carbons, three and four. So it would be three comma four die ethel, the hexen being the molecule we're looking for. See you in the next video.

Write structural formulas for compounds that meet the following descriptions:
a. An alkene, C₆H₁₂, that cannot have cis–trans isomers and whose longest chain is 5 carbons long

Verified video answer for a similar problem:

Key Concepts

Alkene

Cis-Trans Isomerism

Longest Carbon Chain

Write structural formulas for compounds that meet the following descriptions:a. An alkene, C6H12, that cannot have cis–trans isomers and whose longest chain is 5 carbons long

Verified video answer for a similar problem:

Key Concepts

Alkene

Cis-Trans Isomerism

Longest Carbon Chain

Write structural formulas for compounds that meet the following descriptions:
a. An alkene, C₆H₁₂, that cannot have cis–trans isomers and whose longest chain is 5 carbons long