Show how hex-1-yne might be converted to
a. 1,2-dichlor...

Question

Show how hex-1-yne might be converted to

a. 1,2-dichlorohex-1-ene.

b. 1-bromohex-1-ene.

c. 2-bromohex-1-ene.

Accepted Answer

Hey, everyone and welcome back in today's video, we are going to solve the following problem show how you can make these compounds from Bete one, I number 112 dibromm number 32 BROMO. But one in now, if you look at this problem, our starting material in each case starting with problem number one is be one I, so we are going to draw a four member chain, right? Because we have, but and due to the presence of one, I, we understand that this must be a triple bond at high carbon or basically an alk right at position number one. So we have our triple bond starting as carbon number one, a total of four carbon atoms. So we need to add two more carbon atoms. This is how we starting material. And for the first problem, if we think about this reaction is the problem suggests we want to form 12 di bromo but one, right? It is basically an LK, we have a double bond at position number one and positions one and two have Romo substituent. So you may say we went to farm bro and Bromine over here, there'll be a hydrogen atom which is implicit. And because we have a total of two carbon atoms, we need four total, right? So we need to add two more. There will be siege two C three group. So that's 12 di bromo, but one in that makes sense, right? This is a, in particular, if we want to form that product and starting with an alkin, we understand that instead of having a triple bond, now we have a double bond and we are performing a basic edition, we are using one of the two pi bonds to perform an electrolytic edition. It's also important to understand that the two bromine atoms will be added in trans arrangement. So it's really important to make sure that the product that you draw here is a trans arrangement product because in an electrolytic addition mechanism, the two bromine groups with one equivalent will be added on opposite sides. So that's exactly what we're doing here. We're simply adding bromine, right. This is an trophic addition of one equivalent of bromine. One of the pi bonds is used in an addition of bromine. So the two bromine atoms are added in trans arrangement. So that's what we get here. We simply need to add bromine in, in an electro addition mechanism. Regarding number two, our starting material is exactly the same. So we can draw beat one I and once again, that's our starting material. And we want to propose a set of regions to produce one bromo B one in. So again, let's throw the double bond at position. Number one position. Number one also has a BROMO substation. That's what it says, right? So you may simply draw bromine at position number one for that implicit hydrogen. And we need two more carbon atoms because it's butane. So ch two, CH three and that implicit the hydrogen at this position. Now, if you think about this reaction, and if we compare our starting material to the final product, we have used one of our pie bonds. And instead of having that pi bond, we now see that there are two new atoms bonded, there is bromine at position number one. So we are adding bromine over here and hydrogen at the second position. Notice that bromine is added to a less substituted position. So this left position is less substituted and the right one is more substitute at R because that's the case here, we're adding hydrogen to a more substituted position. We have to keep in mind that this is the and see more cognitive edition because according to the Markov Kov rule, it states that hygiene is added to the less substituted position while bromine would be added to a more substituted position. So in this trophy addition, we want to reverse that and we want to make sure that the region chemistry is anti Marconi. So instead of simply adding HBR because that's what we need to add right, we want to add HNBR across that original triple bond to make it a double bond with two new substi to make sure that this is anti cognitive radio geochemistry. Remember that, first of all, we want to make sure that it's one equivalent since we are only using one of the two pi bonds. And second of all, let's use peroxides, Roor, which is a general form for, for peroxides. It allows us to, to achieve the anti cognitive free geochemistry, which works for us, right? So we are done with the second one and let's look at the final reaction reaction. Number three for reaction. Number three, starting with our alk butte one, I, we want to propose a set of regions that could produce the following product. And according to the problem, the product that we have is two bromo but one in. So we may draw a two bromo but one in again, starting with the double bond. The main difference between structure and number two, right? This is our product for number two. And the structure in number three is that bromine is now at position number two. So we will have two hydrogen atoms over here. And instead of having bromine at position one, we're simply adding it to position number two. So there it goes and then there is that ethyl group. So now if you compare reactions two and three, you will essentially notice that they're very similar. But here we're adding HBR across the triple bond in Markov Ko's radio chemistry noticed that there has been one hydrogen atom added on the left side and bromine added on the right side. So, hygiene has been added to the position that is less substituted, right? And bromine has been added to the position that is more substituted. Meaning now this is the more cognitive electro addition, right. So we are adding H pr we may simply say we need to add hbr hydrobromide. And again, one equivalent is really important for us because we're only using one of the two pi bonds present in the triple bond. We want to preserve that remaining pi bond in the structure and that would be it, right? So we may simply label our three reactions. They are corresponding regions and products. We may simply summarize our thought process and say that for the first reaction, the region that we used was one equivalent of bromine because we have the production of a trans alkin that was one comma two di bromo substituted. For reaction number two, we used hydro brom acid one equivalent in peroxides to make sure that hydrogen and bromine are added in and Antar cognitive radio chemistry. And for, for reaction number three, we wanted to make sure that the product follows the more cognitive rego chemistry. So that we simply used hydro Brom acid one equivalent, no peroxides, right? We wanted to make sure that the more cognitive product is formed to conclude the correct answer to this multiple choice question is c however, if you have any questions, feel free to leave them down below in the comment section. And thank you for watching.

Show how hex-1-yne might be converted to
a. 1,2-dichlorohex-1-ene.
b. 1-bromohex-1-ene.
c. 2-bromohex-1-ene.

Key Concepts

Alkyne Reactivity

Halogenation of Alkenes

Regioselectivity in Organic Reactions

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