In contact with a platinum catalyst, an unknown alkene reacts ...

Question

In contact with a platinum catalyst, an unknown alkene reacts with three equivalents of hydrogen gas to give 1-isopropyl-4-methylcyclohexane. When the unknown alkene is ozonized and reduced, the products are the following:

Deduce the structure of the unknown alkene.

Ozonolysis reaction products of an alkene with carbonyl groups shown.

Deduce the structure of the unknown alkene.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to learn how to solve the following problem. An unknown Elkin reacts with three equivalents of hydrogen gas to give isopropyl cycle hexen in the presence of a platinum catalyst. Deduce the structure of the unknown elkin. If the following are the products obtained when it is organized and reduced. Now, one of the most important facts in this problem is that the unknown acts, let's label it as acts for reference reacts with three equivalents of hygiene gas in the presence of platinum to produce isopropyl cycle hexane, Isopropyl cycle hexen essentially implies that we have a six member cyclo hexane with an isopropyl group, bonnet said the isopropyl group would be ch with two method groups bonded to it. Right, that's isopropyl. And looking at the conditions of that reaction, we understand that this is a hydrogenation reaction. There are three equivalents of hydrogen added X is an al kin, meaning because there are three equivalents of hydrogen absorbed, there must be three double bonds. So our goal is to distribute these three double bonds across that molecule to get that unknown L king And with the help of ops analysis reaction will be able to do that. Let's see how that works, recall that The general model for any of this analysis reaction. Is this one If you draw a double bond with different LKOH substitutions, let's suppose you have our one here hydrogen for diversity and two other ones, R two and R three. If you want to perform an analysis reaction because that's what it says, it's also nice and reduced. Which implies that you are adding number one ozone at a low temperature negative 78 degrees Celsius and number two. The muscle sulfide in the presence of water. You're going to break your molecule apart. You have the carbon carbon double bond cleavage. So you may think about your products as two structures. You're going to draw these two fragments that you get when you break that bond and then on the left side you have two substations, R three and R two On the right side are one and hydrogen hye jin and are one. And to each end of these remaining double bonds, you're just going to add auction atoms and you are done. This is how this works. So, using this model, we are just going to think about the compounds we have here and where we can place those double bonds. Starting with the elder Hyde. This is probably the most important part that helps us a lot. The only way to produce an album hide is to make sure that the two alcohol substitutions here are two hydrogen atoms because the alga hide here is for formaldehyde. Right? So we have two hydrogen atoms, meaning there must be two hydrogen atoms bonded to the double bond. And what are the ways to do that? Really? We can only use any of these two methyl groups to turn it into a series to group. There's no other position of the double bond that could produce formaldehyde. Right? So this allows us to assign our first double bond because the two methyl groups are equivalent to each other. Their bond, it's of the same carbon atom. It doesn't really matter whether you choose to add it here or here, that's perfectly fine. So let's just choose the upper carbon to be double bonded and then leave this as a metal group. Okay, so we have a sign our first double bond. We need two more. And now this metal group is really important for us notice that there is no more alcohol substitutions bond into the ring and there's only one metal group over here, which is sticking out of the carbon that is forming a double bond. So this metal group must be our carbon number one. We will just label this as carbon number one because that key tone has a metal group bonded to it and all of the other carbonell groups. In particular, if you look at the external parts, they have hydrogen atoms. This is the only one that has a muscle group and it is forming a double bond. So definitely that metal group must be representing carbon number one over here. If we understand that this is carbon, number one. then this must be carbon number two. Right. Part of that double bond makes sense. Right, Because it forms a keystone derivative. Now we already said that the top part of the double bond represents carbon number three. So carbon number two is directly bonded to carbon number three, They're forming that carbon carbon double bonds. When we say that this is carbon number three And now we may think about carbon # four. Carbon number four Is a carbon steel group ride and we may notice that this is our carbon number four because it forms a carbon eel group, it must imply that this carbon is double bonded as well in acts. It doesn't really matter if you place the double bond here or here because again, the molecule is symmetrical so far. Right? So let's suppose we go down. So we added double bonds is the presence of the carbonell group and we simply want to think about which carbon is bonded to And because we're breaking that bond, that carbon will be seen in the remaining structure. Structure number three. And because it's symmetrical, we can just say that carbon number five will be this one. It doesn't matter. You can say this one as well. It's basically again it has the line of symmetry. Right? So this will be carbon number five. And it makes total sense. Right? Because if you look at carbon number five, it has a hydrogen atom and an L kill group. So that's where this hydrant comes from. It's bonded to carbon number five meaning going down, we have our siege to group just as received. Right? That's carbon number six, isn't it? That's carbon number six hour. Siege to group carbon number seven has an AL. The high group meaning it must be a double bonded carbon or carbon number seven will have a double bond. And we are done. We understand that after the cleavage of these T bonds we will form to all the hides. Right? There will be the cleavage of that bond auction over here, hydrogen sees two carbon, oxygen, hydrogen. So that's our full structure. Now from the top view, we will essentially see all of this remaining fragment. We already said that this is carbon number four. Right? So if we go up, that would be carbon number eight, we can just say that this is number eight which is a siege to group carbon number nine, which makes sense. Once again, if you go from the top you or 89489. Siege to nine has an AL to hide group because there's a hydrogen atom here. So that makes perfect sense. We have a sign our three double bonds and we understand that this must be our final structure because our goal was to identify the positions of those three double bonds based on the o cyanosis reaction here, we may simply conclude that the correct answer choices. A small pool choice question is a However, if you have any questions don't have to leave your comment down below in the comment section and thank you for watching

In contact with a platinum catalyst, an unknown alkene reacts with three equivalents of hydrogen gas to give 1-isopropyl-4-methylcyclohexane. When the unknown alkene is ozonized and reduced, the products are the following:

Deduce the structure of the unknown alkene.

Key Concepts

Hydrogenation of Alkenes

Ozonolysis

Cycloalkanes and Substituents

Watch next

In contact with a platinum catalyst, an unknown alkene reacts with three equivalents of hydrogen gas to give 1-isopropyl-4-methylcyclohexane. When the unknown alkene is ozonized and reduced, the products are the following:Deduce the structure of the unknown alkene.

Key Concepts

Hydrogenation of Alkenes

Ozonolysis

Cycloalkanes and Substituents

Watch next

In contact with a platinum catalyst, an unknown alkene reacts with three equivalents of hydrogen gas to give 1-isopropyl-4-methylcyclohexane. When the unknown alkene is ozonized and reduced, the products are the following:

Deduce the structure of the unknown alkene.