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

Question

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

d. trans-hex-2-ene → hex-2-yne

Accepted Answer

Hi everyone and welcome back in today's video, We're looking this problem which asked us to show how the following synthetics transformation can be accomplished. Give the structures of all intermediates. The starting material as we can see is trans bt in. We are ready to draw it. It's an elk in which has a double bond position number two. We made through positions number two and three double bond starts over here, position number two because its beauty and its entrance arrangement, we will need two more carbon atoms. One of them will be here. That's our carbon number one and we need an additional carbon atom and trans arrangement so that be a methyl group. So there are these two alcohol groups are on opposite side of the double bond, meaning it's trans. So we're good. And now we are ready to draw our final product which is built to be two. Iron is an alkaline because as you notice in its name, it has wine. We may draw a triple bond starting at position number two, meaning there will be an additional carbon atom over here. That's position number one and we need one more carbon atom because the parent is beat sign, that's for carbon atoms total. If we compare the two structures, we notice that they both have the same number of carbon atoms four and four. The only difference between them is that the starting material has a double bond and the final product has a triple bond. Now we may think about this reaction a bit because it's a bit complex. It's not really easy to form a triple bond out of a double bond immediately. We essentially want to form an al kill hey light and use the basics of elimination later to form an AL kind. So there's no direct route in which we can perform this transformation in a single step. So the efficient synthetics transformation would be as follows. In the first stop, we are going to form an al que hay light out of it and stood that we're going to add a molecule of some halogen Elizabeth. We're adding bro mean to that double bond, we are going to use electra filic edition of bromine. So let's indicate that we're using bromine and the common solvent. And this reaction is carbon tetrachloride in this reaction. As you recall, the bromine molecule will approach that double bond and it will get slightly polarized, suits the presence of these electrons making the left romaine partially positive and the right one partially negative. The polarized molecule will then participate in that electra fill a condition in which the pi bond will attack bromine and bromine will attack back to form brah monium bridge. As you know. So we are ready to draw that intermediate because we are asked to draw all the intermediate ramon E. Umbridge would have brimming with a formal positive charge because now it forms two bonds and we noticed that the two methyl groups are pointing in opposite directions. So we are ready to draw one of them on a wedge and another one on a dash. So that's how it looks like and there will be a hydrogen atom, a disposition and another hygiene in that position, that's our intermediate. And now the brunette and I and that we have formed here. Let's not forget that additional error. My apologies. There was a leaving group here. Right? So the negative charge from roaming would make it a weak nuclear file and that means it would attack a more substituted position. But because they are equally substituted and the compound is symmetrical, it doesn't really matter which one we attack. So let's suppose we attack this one. And when the bromide and ion attacks that carbon atom, we would have a backside attack in which we would get inversion of configuration. So let's not worry about cereal chemistry and let's just make sure that we draw an anti relationship between beta hydrants and bromine atoms. So we may draw our next structure which would essentially have were carbon atoms, Bromine atoms at position two and 3. And now we are ready to think about this alcohol. Hey light recall that Our goal is to use the basics of an elimination reaction to form the final al kind and to that we are going to identify our beta hydrogen. So let's suppose that we are choosing one of our room means that will be our alpha carbon atom and the beta high genes are here and here. So let's label them for complete clarity. This is one of the beta hye jin's and this is the other one. If we think about this being our alpha position, If we expand one of our beta hydrogen and in particular we want to eliminate this beta hydrogen because we're using an E two reaction. And let's choose our strong base, a common strong base that we may use as hydroxide. So why don't we use? But as soon hydroxide or sodium hydroxide, you may choose the majority of group one a elements here, bonnets hydroxide and we also want to promote elimination using heat. Now this is an E two reaction. So we want to make sure that our major product is the stage of product which which would be a more substantive talc in so that proton will be removed. Indeed, the hydroxide an ion would attack that proton. We would form a double bond due to the transfer of these bonding elections into a pi bond. And we would have the loss of the leaving group. So now we are ready to draw our next intermediate which is an elk in right. So that might be a bit strange because we started with with an AL keen. We got another AL keen but the main difference between these structures, the original one and the one that we have just got is that we have a perfect living group now. So this is the rationale while we why we are performing all of these steps, we wanted to get that leaving group to form an al kind because notice that if we use that structure, we still have one beta Haijun remaining in antico plane arrangement. So let's expand that. There's this beta hydrogen which is anti to booming and it essentially can be removed in another step of any two reaction. So why don't we perform that step? The next step? It's pretty much our final stop. Right? So we will get our final structure if we use a strong base once again. So let's suppose we're using but estimate hydroxide. Why not? And heat the hydroxide and Ironwood attack that hydrogen. This is the beta hydrogen. We would have the formation of a triple bond. Now there's another pipe bond form and simultaneously there's a loss of the leaving group to form the final product to form the final alkaline. So essentially these are our steps for the overall transformation. We noticed that if we recall all of the steps that we performed, the first step was the addition of blooming and carbon tetrachloride. This is our intermediate A. This is our intermediate product. B. Now the second step was addition of potassium hydroxide and applying heat getting our intermediate C. And then finally we had our third reaction where we repeated that process which is equivalent to number two. And we got our final product. These are our steps. So we may label this as our complete transformation with all the intermediates and the region's needed, and we may now simply conclude that the correct answer choice to this multiple choice question is C. However, if you have any questions, don't have States, leave your comment down below. Thank you for watching this video.

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

Key Concepts

Elimination Reactions

Hydrohalogenation

Synthesis of Alkynes

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