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

Question

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

e. 2,2-dibromohexane → hex-1-yne

Accepted Answer

Hey everyone and welcome back in today's video, we are going to tackle the following problem. Show how the following synthetics transformation can be accomplished and give the structures of all intermediate. If we look at our problems starting materialists to to die bro mo butane. So we are ready to draw a four member chain with two bromo substitution at position number two. There are two bromine atoms over here. That's how we're starting material. Well done. And now if we look at our product, we are forming beauty 19 due to the presence of iron. We understand that this is an AL kind which has a triple bond starting at position number one and the parent has four carbon atoms. So we need to add two more to have four total. We first of all understand that the starting material is a dye bromo al killed. Hey, light white. And the final product is an alkaline because it has a triple bond. Now, how do we produce double or triple bonds out of alcohol highlights? We essentially need to use an E two mechanism. We need to use an elimination reaction. That elimination reaction tells us that we may remove beta hydrogen, followed by the loss of the leaving group to form double bonds and then we can repeat the process to form triple bonds because we have two leaving groups right? We can perform a double D hydro elimination. Now, what does that mean? It just means that we are going to apply that E two reactions wise if we withdraw our starting material, we will start the analysis of our structure and we will think about the beta hodgins that we can remove. This is our alpha position. Because we have our leaving groups monitored and the adjacent positions are called the beta positions. So we have two different beta hydrants that we can eliminate. Now the question is, which one do we want to remove? Do we want to remove this one or that one? And if we look at our structure because we are producing an al Qaeda that is terminal meaning the triple bond starts at the first carbon. We are interested in removing that hydrogen atom, This beta hydrogen atom instead of that one. And for this purpose, if you notice the removal of this beta hydrant would produce us with a less substituted elkin. Right? If we remove this one that would be more substituted Elkin. So how does it work? It basically works this way. If we choose an O. H. In alcohol for anne to it would definitely follow the standard elimination mechanism in which we want to produce the seeds of product right? We would form a more substituted al king. But to remove this beta hygiene we have to choose a different base and that base that we are using in production of terminal al kinds of sodium Amytal. So keep in mind that for the purposes of this reaction because we want to produce a terminal alcott and we are going to use sodium Amytal as our base. Now let's erase. Or maybe let's just redraw the structure once again. Right. Because we have our analysis and I think that's pretty much important information so we can just continue and re troll that structure to see what's happening step by step now that we clearly see which basically want to use. So if we retry it once again, we already understand that the beta hygiene that we are about to remove is this beta hygiene because we are going to use sodium amid NH 2 - is very strong base which prefers to stick hydrogen. So we may simply eliminate this hydrogen transfer these bonding electrons into a pi bond followed by the loss of the leaving group. And this is how we are producing our first intermediate. That would be a double bond in this position. There is still one grooming left, right, and that's pretty much it. This is our first product. When we are adding the first equivalent of sodium a mites, we may say that this was the addition of sodium a Mine is our first step however, we want to produce a triple bond and we still have these acidic hydrogen is that could be removed by NH two and one of them which is an anti arrangement. Recall there is a requirement of anti arrangement. So we want to expand that implicit hygiene which is anti to booming. It will be attacked by another equivalent of an H two minus. So we are going to attack it. Form a triple bond now and we will have the loss of the leaving group. This will give us our AL kind. So we may now draw and actually indicate that there's another equivalent of sodium a mite involved. We are producing that triple bond, two more carbon atoms. Now let's be extra careful and not let's not get overly excited. Yes, this is indeed our AL kind. Right. But because we're not measuring the exact story geometric ratio of our starting material and base generally we have an excess of base. There will be the third equivalent of sodium A. Might use two D. Protein. It that acidic hydrogen we call that all kinds have this terminal acidic hydrogen that can be further removed. So what happens is that the fact that we're generally using an access of base. We are going to D. Protein ate it. And that will be one of our intermediates. The negative charge would attack it. And we would produce a negative charge here, which is a satellite. Right? So we may simply draw our structure after the addition of the third equivalent of sodium Amytal And a an H. two. That would be a satellite. And now we essentially want to protein it. That we want to make sure that we get our alcohol and our terminal AL kind and we may simply use water because it would give a proton and it would restore those basic conditions. So that works perfect for us. If we use water, water is a proton donors. So the negative charge attacks one of the hydrogen atoms. We have the loss of hydroxide and we are forming our final product, which is Beauty. One iron. Right? So we may now look at this analysis and we may simply conclude that The reaction stops that we have to perform here. They would be number one sodium A mine. We may also add heat to promote elimination. Right? So often times you will see heat added in that reaction, clearly indicating that elimination takes place. Right? Heat promotes elimination and number two, addition of water to pretty much protein eight that a satellite form. Right? Because we're generally using excess of sodium A might. So it's inevitable to get a proton being still attached to that stop. Right. Unless we measure precisely just the equivalence of base which generally is not done in chemistry. So, because we're using access, we are definitely forming this and I on which we want to protein it later. That's why it's very important to make sure that you indicate that step number two to get the alcohol and you wanted to get. So we may now label these are finally answers. These are our regions that we want to use in this reaction to form butte 19 out of two to die bro mo Beauty. And the correct answer to this multiple choice question is D If you have any questions, don't hesitate to leave a comment down below in the comments section and thank you for watching.

Show how you would accomplish the following synthetic transformations. Show all intermediates.
e. 2,2-dibromohexane → hex-1-yne

Key Concepts

Elimination Reactions

Alkyne Synthesis

Reaction Mechanisms

Watch next