When treated with hydrogen and a platinum catalyst, an unknown...

Question

When treated with hydrogen and a platinum catalyst, an unknown compound X absorbs 5 equivalents of hydrogen to give n-butylcyclohexane. Treatment of X with an excess of ozone, followed by dimethyl sulfide and water, gives the following products:

Propose a structure for the unknown compound X. Is there any uncertainty in your structure?

Propose a structure for the unknown compound X. Is there any uncertainty in your structure?

Accepted Answer

Hey everyone and welcome back in this video, we are going to tackle the following problem which reads an unknown compound. Acts absorbs five equivalents of hydrogen to give and beautiful cyclops tensing when it is treated with hydrogen gas in the presence of a platinum catalyst. The following products are obtained when Acts is treated with an excess of ozone followed by the metal sulfide and water proposed a structure for the unknown compound. Acts. Is there any uncertainty in the structure that you proposed? We may start this problem by looking at the first bit of information, it says that whenever Acts is treated with five equivalents of hydrogen in the presence of platinum, it produces and beetle cycle fencing. So we are drawing a cycle pension ring and beetle means that the beetle group is normal or straight chain. So we are adding four carbon atoms. We understand that this is a hydrogenation reaction because as you recall whenever we are adding hydrogen over platinum, palladium or nickel, we are reducing double and triple bonds into single bonds. So we are removing any kind of saturation and we are producing al canes out of Elkins and all kinds. So number one, if you remember every equivalent of hydrogen reduces one pi bond. So it implies that there must be five buy bonds. Right? So if I have equivalents of hydrogen are used, there must be five bonds and X that's the first part of information that we have and we also have some sort of a skeleton of Acts. We will simply wish to distribute these pi bonds around this structure. Now to do that, we are going to think about oz analysis And going back to those analysis, we are going to introduce a few concepts that will be important for us. Wasn't always this is based on the reaction mentioned in that question. It's when we are using ozone followed by the metal sulfide and then water essentially we will have two cases whenever it comes to pi bonds. A if we perform analysis on AL kinds, we produce assets. On the other hand, the other type of double bonds is seen in al kins here, we have double bonds and the products off and the ozone noses reaction would be Aldo hides or ketones. Now, knowing this, we can simply go back to our products and we can think about this analysis as applied to this problem in particular, if we try to find our assets, we understand that, okay, this group is an AL to hide, right, C H O AL to hide Aldo hide again, Seiko, Seiko al to hide all to hide acid. Al to hide on the left and on the on the right, we have acid. But in general, this is just a single compound. It only has one carbon group. So it just makes sense to say overall this is an asset. It's formic acid, right? It's just that instead of an hour group, it has hydrogen. So yeah, now we understand that there are two assets, right? We have to carve oxalic acid groups and al kings. My apologies, AL kinds produce assets. So that implies these two carbon atoms in blue because they are a part of the car book silic acid group. They form a triple bond. So we may say that this carbon number one and carbon number two, they form a triple bond and that's really important for us. However, what is more important is the presence of that hydrogen and former cassette, which implies that due to the absence of an al kill chain, it must be a terminal AL kind. So this implies number two, that this acts must be a terminal al Qaeda, meaning there must be a triple bond with a C. H group. Right? There will be another carbon with an L. Kill chain. But they're strictly must be ch and no alcohol chain here due to the presence of this hydrogen. And now, if we think about the skeleton of ACTS, the only way to make it a terminal AL kind is actually to create a triple bond. A disposition. Otherwise all the carbons have neighbors. What we can do Now we can start by drawing our structure of ACTS and already filling this information in. So we redraw the five member ring And we will add four carbon atoms. We already understand that it must be terminal alkaline. So we are going to add hydrogen and disposition to make it a terminal alkaline and then we will make these two carbons triple bond. It. This is our carbon number one, just as we said before, and this is the carbon number two. Going back to carbon No two, there is some more information that we wish to conclude from here. Carbon number two is bonded to carbon number three directly. And because carbon number three has an alga hide group, it implies that it comes from an L. King, Meaning that carbon # three must have a double bond. So we are going to add a double bond over here due to the fact that it's not a keto group, it's an AL to hide group. There also must be hydrogen bonded. Said, Great job! So now if we count the number of bonds we have therefore for this carbon for for this one, for for this one, we're good with our bonds and we're ready to move on. Now, we have analyzed these two structures. Now, we simply wish to think about the remaining ones. First of all, Can this carbon number three be bonded to this carbon or that carbon? And the answer is actually no. Because if you think about these two groups, they're both al to hide. So they would both belong to al keen carbons. Is it really possible to draw a double bond here and then an adjacent double bond? It's not because that would make this carbon five bonded. So we understand that we are excluding this molecule is not linked to carbon number three. We are going to look at this structure which is indeed linked to carbon number three In particular. We also have in mind that this is our carbon # three. And we simply wish to find carbon number four, which as you can see, is also double bonded. So any of these two doesn't really matter which one you take. They are al to hide groups. So let's suppose that this one is carbon number four because it's an AL to hide group. You already know that this is carbon number four. We already showed a double bond from carbon number three to carbon number four. So it makes perfect sense for us. We understand that. Okay, this is an AL to hide group. So this is our carbon number four. And just the presence of an AL to hide this hydrogen tells us that there must be a hydrogen atom bonded. Said, right, so we are good. We now notice that okay, we have 1234 bonds here. So we are done. And now we're thinking about the position of carbon number five. Now, carbon # five will essentially be this one. And we may go any route we really want because the ring is symmetrical, right? We may draw a vertical line here and that would be a line of symmetry. And we may think about it as follows if we started with five pi bonds, we have 123 assigned we need two more within the ring itself. So what are the ways to assign them? Now? We understand that there will be no adjacent double bonds. And first of all, we wish to think about it using our compound number two, it tells us that we have to al hi groups bonded. In other words, we will be separating two carbon atoms and forming alga hide. So each of them must have one hydrogen atom. Now this is super important for us because it tells us that there must the system must be conjugated, so there must be two double bonds linked. So what if we try to draw it this way? What if we draw a double bond in this position and then draw a conjugated form. In other words, droid, a double bond at this position. Would that work for us? We may simply check if we break these two double bonds, we will form these two carbon atoms and we may label them as for consistency, six and seven. So if that's six and seven, breaking these bonds will indeed form to all the high groups because there there are two hydrogen atoms here dodo. And now when we break it apart, that would give us this al to hide from the other side, we would get an elder Hyde group. Right? There's one more hydrogen than a siege to group that works for us Ceo group. Carbon eel which also works for us. Right? There are no hydrogen is here. This carbon has no hydrogen. So that works for us perfectly. And then see each group which corresponds to that double bonded carbon. Now this makes social sense. So we have shown that this is our final structure. We may label this is our final answer to this problem. And now let's think about the second part of the problem. Do we have any uncertainty regarding its identity? The answer is yes. There is a bit of uncertainty. If you recall CIS and trans exam arr is um notice that there is in particular one double bond here which may exhibit this trans is amateurism. Whether these two high johns are on the same side of the plane or on opposite sides of the plane. We don't know. So we may say that we are not certain wetter, this is CIS or trans. We are only certain regarding the positions of our 55 bombs, but we are not certain regarding stereo is MERS and whether this should be assist or a trans I. Zimmer, however, that would be it. We have successfully solved our problem. And the correct answer to this multiple choice question is C However, if you have any questions, don't hesitate to leave your comment down below and thank you for watching

Key Concepts

Hydrogenation

Ozonolysis

Structural Isomerism

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