Show how you would accomplish the following synthetic transfor...

Question

Show how you would accomplish the following synthetic transformations. Show all intermediates.

j. trans-hex-2-ene → cis-hex-2-ene

Accepted Answer

Hey everyone and welcome back in today's video, we are going to learn how to solve the following problem. Show how the following synthetics transformation can be accomplished and give the structures of all intermediates Trans B two informs says B two in and if we draw the two structures, we may just understand that the trans version of B two in forms, this is I Zimmer. So the main difference between them is not really the atomic arrangement, it's actually the fact that they are the hysteria emerging. So the two method groups are on the same side of the double bond in the cis sizemore and on opposite sides within the trans is more now, this is not really easy to accomplish because there is restricted bond rotations in the presence of the pi bond. So let's think about this reaction in general. What if we go in reverse And what comes to your mind when you think about forming a cis alki? Unless there's a reaction that allows us to do that. And yes, there is essentially what we want to do here. We want to use an AL kind. The triple bond must must start at carbon number two. So we need to metal groups on the left and on the right or essentially beat to iron. Right, and we are going to use one equivalent of hydrogen in the presence of landlords catalyst. Recall that landlords catalyst allows us to reduce al kinds to al kings and exclusively form sis Elkins. That's why we're going with this reaction? We're going to think about formation of the cis version using the landlord's catalyst. So this works perfect for us. But now we have to think about how are we forming our AL kind? Keeping in mind that the starting material is an al keen. Right? We can use a double elimination for the double elimination. We are going to show that to form this al kind. We first of all, one to eliminate one hydrogen and one grooming and then the other one. So apparently the second intermediate and that reaction would be and al kin where the bromo substitute print is in anti arrangement to the hydrogen atom that we are eliminating and then that we would have two methyl groups remaining. The basis we're using might be hydroxide we may use but as a hydroxide or sodium hydroxide, the metal here is not so important. You can use group one a element and we are also promoting elimination using heat. And because we are pretty insisted that this is a double elimination, we are going to use that reaction once again. And if we add hbr across this bomb in particular, we will notice that we are forming our that and I'll kill Hayley. So there will be four carbon atoms With two bromine atoms at positions two and 3. And you may notice that this is our double D hydro elimination. First of all, we remove this hydrogen atom forming a double bond full by the loss of the leaving group. And then once again removing this hydrogen forming a triple bond full by the loss of the leaving group. So this scheme works for us. Right? So now we are going to think about it in terms of the falling, we want to form our L kill hey light, how can we do that? And to do that? We're essentially thinking in a reverse. If we start with an alc in we may simply add blooming in the presence of carbon tetrachloride, this is our solvent. If we add bro. Mean you may wish to recall that we are forming the bra ammonium bridge and then we will have a backside attack to form our alcohol. Hey light. So that makes perfect sense for us essentially the intermediate that we get here would be our bridge. So let's draw that bar ammonium bridge. We call that the double bond attacks one of the roaming adam. So we may just draw it to get our intermediate correctly. When the molecule of roaming approaches, it gets slightly polarized. So this is partially positive and and this is partially negative and the double bond attacks the lone parent room in a text back we're forming our bridge with a formal positive charge and we have a loss of bromide. Now the two method groups are on opposite sides meaning we are going to draw one of them pointing away another one pointing up. Then there is a hydrogen atom pointing away and then another hydrogen is pointing up. So that's how ammonium bridge. And then as you remember the broadway Emma that has left will essentially attack the electra, fill a carbon atom. They are essentially equivalent. Right? So it doesn't matter if you attack this one or that one, you will get a pair of financial marks and then we will have the loss of the leaving group opening the rain and forming our product. So now let's see if that works for us. Let's see if this logic was good. Now we are starting with our trend zelkin, we're adding blooming across the double bond. We're forming our bro ammonium as our intermediate. We have a backside attack opening the rain and forming our alcohol. Hey light performing a double elimination to form an al Qaeda and then using little nurse catalyst to form the cis al king. So yes, this definitely works for us. We see that all the reactions that we generally know were really useful for us in this problem. This is our general synthesis scheme. Just to summarize that, we can just say that the reaction conditions were the following ones if we start with our trans version and then draw the system Number one, was the addition of blooming in the presence of carbon tetrachloride. Number two. We performed that double elimination. So we said, but that same hydroxide two equivalents, right, because we needed two of them to make sure that we are producing our final al kym In the presence of heat reaction. # three was addition of hydrogen gas in the presence of limb blurs catalyst. Now, this is our general reaction scheme. Of course, the problem is asking us to give the structures of all intermediates. All of the intermediates are seen here. This is intermediate number one hour, Ramon E. Umbridge our alcohol. Hey light now, once again another alcohol. Hey, light number four. That's our alka informed. And this is our final product. And I've seen all of this scheme and all of the intermediate. We may just conclude that the correct answer to this Multiple choice question is the However, if you have any questions don't have states levy are comin down below in the comment section and thank you for watching.

Show how you would accomplish the following synthetic transformations. Show all intermediates.
j. trans-hex-2-ene → cis-hex-2-ene

Key Concepts

Geometric Isomerism

Hydrogenation Reaction

Stereochemistry and Reaction Mechanisms

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