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

Question

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

a. 2,2-dibromobutane → but-1-yne

Accepted Answer

Hey everyone and welcome back in this video. We are going to tackle the following problem. Show how the following synthetic transformation can be accomplished. Give the structures of all intermediates. If we look at our problem, the starting material is to to dye bromo plantain. So we may draw a five member chain to indicate plantain right? And position number two contains two bromo substitutes. We are adding to roman Adams at position number two. According to the problem. The product that we are producing is .19 to the presence of iron. We have to draw a triple bond And that triple bond starts at position number one we have a total of five carbon atoms. So let's add three more. That'd be 123 that spent one on our final product. We want to propose a set of reagents that performs this conversion. Well recall that because we are starting with an L. Kill Hey light. This is actually a dye bromo al kill he lied. And the final product that's alkaline production of double or triple bonds can be performed using an elimination mechanism E two in particular. So we are mainly interested in an E two reaction where the beta hydrants are removed to give us double and triple bonds because we have good leaving groups and beta hydrants present. We can see that for sure. So let's draw our structure once again and analyze it. So that's our starting material in an E. Two elimination. We can essentially find our beta hydrants that we can remove because this is the alpha position, the beta hydrants are the ones that are adjacent to it. The neighboring positions are beta. So these two are our beta hydrogen because we're not producing an AL keen but an al Qaeda instead. We want to perform this reaction twice. So we may say that this would be a double D hydro halogen ation. Since we are removing hydrogen and roaming in this case twice. Since we have two living groups, two bromine atoms. That makes perfect sense. We can perform an E two reaction twice. The only thing that we want to determine now is that which of the two hydrants should we remove we call that whenever we are producing Elkins, we follow this eight of schools. So it would make sense to deport innate this hydrogen to produce a more substituted elkin. But that would only be true. If we used hydroxide as our base in alcohol, then we would fold this pathway and we would not be able to produce a terminal alkaline. Right? This is a terminal alcott and that's really important to understand because the triple bond starts at position number one, meaning we cannot really use hydroxide, it would produce an internal al Qaeda and a triple bond at that position. However, we want to remove this hydrogen study and if you recall the way to remove this hydrant instead is to use a strong base such as a might and a mite can produce terminal al kinds, right? So we have to be very careful in choosing which base we are going to use in this reaction. And because the final product is a terminal AL kind, our choice is to use a mid. So now that now that we understand what's happening, let's just think about the steps of this reaction and any possible intermediate. If we withdraw our starting material once again, we already understand that. Our goal is to produce a terminal AL kind. So we are going to expand the beta hydrogen that we are removing. And because we're using sodium Amytal, we will attack the proton, transfer these bonding electrons into the first pi bond. And we will have the loss of the leaving group to form an elk in. So the first product that we're forming is an elk in there will be double bond at position number one. There would still be one bro mean remaining. And that's perfect for us because we want to repeat that. Stop here, we added sodium a mine and we are just going to repeat that step. We call that the hydrant that we are eliminating in the step must be anti to booming. That's one of the requirements of the E two elimination and there are two hydrants available. We're going to remove this one and we are adding another equivalent of a mite and H two minus will attack that hydrogen form a triple bond. Now followed by the loss of the final leaving group. Now this would be our product now. So we may say that we added another equivalent of sodium Amytal. But let's not get super excited. Even though we already forming that triple bond with three More carbons. Right? Because we are forming a nine. The problem is that we always use an excess of base and because there is an acidic proton over here, as you recall, there is this acidic hydrogen bonded to that carbon. It can be easily D protein ated further. So a meid essentially D protein. It's the result on the terminal al kind to produce a satellite iron. And that would be our intermediate product. We may say that a satellite is our intermediates. We may label this is our intermediate and because we want to get our al Qaeda and stud, we simply want to protein ate it in our next reaction. The question is which base should we use? My apologies. Which assets should we use because assets are proton donors. Right? So we want to use an asset which would protein ate it? And we are going to use water simply because the protein eating water would restore the basic conditions and we would form in a wage upon the protein nations that O H minus. Right? So the proton from water would be removed and it will be attached to satellites to form our final product Pant one iron. Now, this is really important. So we have our reaction scheme, we understand that we added three equivalence of sodium Amytal, right? We added one more over here. In general, we can just say we had an access every step generally requires heat because we want to promote elimination. And finally, in the final step we had to protein ate a satellite to form our final AL kind. We may now summarize all these steps in our reaction scheme. So number one, we are adding sodium Amytal access heat. Right? Number two, we need to proton it our structure with water. These two are separate steps indicating that first of all, we're performing elimination and dipoto nation. And then when we get a satellite, we are essentially adding water to protein, ate it and get our terminal AL kind. So we may label this is our final answers. These are our final conditions and final regions for this reaction and conclude that the correct answer choice to this multiple choice question is a. However, if you have any questions, don't hesitate to leave your comment down below and thank you for watching

Show how you would accomplish the following synthetic transformations. Show all intermediates.
a. 2,2-dibromobutane → but-1-yne

Key Concepts

Elimination Reactions

Alkyne Formation

Reaction Intermediates

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