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

Question

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

b. 2,2-dibromobutane → but-2-yne

Accepted Answer

Hey everyone and welcome back in today's video, we are going to tackle the following problem which reads show how the following synthetic transformation can be accomplished. Give the structures of all intermediates. The starting material is to to dye bromo plantain and we are forming pencil line notice that this is an alcohol highlights. We are ready to joy pain plantain chain. So that's a five member chain. We are going to draw five carbon atoms and there are two bromine atoms at positions two in particular, there's one at position C. And there's another one. We are forming pent to iron. So let's show our final product. We understand that this is an al Qaeda because there's a presence of two Ein and the triple bond starts a position number two. Just as the name suggests, meaning there must be an additional carbon over here. That indicates position number one. And because the parent has five carbon atoms, we need to add two more. That'd be 12. We are going from the original structure, which is an alcohol. Hey light to an al kind. The main changes that even though we still have five carbon atoms, we started with five, we ended up with five. The two bromo substitute prints are essentially eliminated and we are forming a double bond and my apologies a triple bond at that position. So we need to eliminate them. Recall that we may use an elimination mechanism. The first step in an elimination reaction in particular, we are going to use an E two elimination to make sure that we are forming a zetas product in this case is very important to form a more substituted al keen at the word first step. And because if we think about this as being our alpha carbon, we identify our beta hydrogen. So we have a source of hydrogen at this position. This is our first beta hydrogen and this is the second position. This is our other beta hydrogen because in an E two elimination with a small base, we can remove any of them to form a mixture of products. We essentially want to make sure that the major product is a more substantive al keen removing this hydrogen atom in an E two elimination. So that's why we are going to use an E two reaction with a small strong base which is hydroxide. We may use hydroxide as our base for the negative charge, but we may also indicate that as an ionic compound with and with an eye on spectator such as potassium or sodium. So let's suppose that we're using potassium hydroxide and definitely we may apply heat to make sure that we're favoring elimination versus substitution because this is a strong and small base. The first step in this reaction is that the negative charge on oxygen attacks the proton and then these bonding electrons are transferred into a double bond. Afterwards we have the loss of the leaving group which is one of the two bromine atoms and we that would be our major product because that would be more substituted and this possible elkin. And this is why we're not worried about this beta hydrogen at all. It would form a minor product. So we are ready to draw our intermediate product that we get in the first step. And we already understand that this would be an L keen since we have formed a double bond. So let's throw it horizontally and across the double bond, the remaining hydrogen and the bromine atom that is still there. There will be an anti co planer arrangements. So we may essentially state that we have bromine. There's a metal group pointing down and the hydrogen atom that is remaining here is pointing down. Whereas the ethyl group is pointing up now because bromine and hydrogen they are in antico planer arrangement. It's perfect for us to repeat the process and eliminate the remaining hydrogen to form the final al kind. Let's see if that works for us. If we repeat that process and use another reaction in which but assume hydroxide is a strong base. We are applying heat to favor elimination hydroxide, which is a small and strong base would attack the proton. We would form a triple bond now and we would have the loss of the leaving group which indeed forms the triple bond at that position. We have that metal groups, we may label it as 12345, correspondingly we have 12345. Everything is consistent. We get an al Qaeda and the second step. So we are done. This is essentially our synthetic transformation and we have the structure of our intermediate, The intermediate formed in the first step of an E two reaction. Is this al king? Now, afterwards, The final step is just the repetition of the first one, which is another E two reaction to produce an al Qaeda. And this is a classical way to form an AL kind out of a to to die bro, no pension or you know, even if you have a different halogen these positions, we essentially might form an internal alchemy in this way using a strong basin and an E. Two reaction. We may now label this is our final synthetic transformation with all the steps included and the intermediate given. And the correct answer to this multiple choice question is the However, if you have any questions, don't have states to leave your comment down below in the comment section, thank you for watching. And I'll see you in the next video

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

Key Concepts

Elimination Reactions

Alkyne Formation

Reaction Mechanisms

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