Predict the major product(s) of the following reactions:

Question

a. phenylacetylene + 2 HBr

b. hex-1-yne + 2 HCl

Accepted Answer

everyone and welcome back to this video, we're going to tackle the following problem. Right? The products for the following al kind hydro elimination reactions. A ethanol cycle hexen plus two modes of hcl and B. Cycle lock time plus two modes of HBR. We may start drawing our reactions in particular if we start with A it states that we have ethanol cycle hexen. So we may draw a cycle hexen ring which is a six member ring and there's an ethanol group bonded at pretty much any position. It doesn't matter which one you choose to be the first one, which means that we have two carbon atoms triple bonded. So that's our starting material and we're adding two modes of hcl in particular. We may say that we're adding one mole at a time. So why don't we just add Hcl as our first small and we have to recall that because we have a hydro hydrogenation reaction. We want to make sure that we're following the mark optical stroll. This is more comic of addition, meaning hydrogen will be added to a less substituted position. So we'll be adding it over here where we already have one hydrogen and chlorine will be added to a more substituted position or basically over here. So we get our first product and we're using only one of the two pi bonds because we're only adding one mole so far. We're getting our first product which is an L. Keen. So this would be our ring and now the two carbon atoms that we have in particular. They will have a close substitution at position number one. And now there will be two hydrogen atoms bonded to that position. There's also a double bond now because one of the pi bonds had been used in the addition. Now if we perform that step once again because we want to add two moles of hcl, we're adding another hcl and again, we're following the more comical rule. Right? So hydrogen will be added to a less substituted position that would produce us with three hydrogen atoms total and chlorine will be added to a more substituted position. So we will have a dick laurel compound. We may start by drawing our structure. So the final product that we get for reaction A would have that ring two carbon atoms and now there would be two chloral substitution ints at that carbon atom. Well done. So we are done with our first product. We may essentially state that because the question was asking us to get our product, this is product number one for A and let's move on to part B for part B. We have cyclone stockton. Right? Essentially because we have a suffix of iron. That implies a triple bond. So we made troy triple bond and it's a cyclic alkaline. So we have a total of eight carbon atoms in it due to the presence of octane. We may say that because we have to we will draw two more and then we need four more. So we may say, okay, that's one, two, three and then that would be four. We may now bond these two together. So that's cyclone oxen. We essentially just want to make sure that this is not misleading. There are no carbon atoms in between. So let's make it a straight line. So that's cycle oxen and we are adding to most of hbr. We are going to add An arrow, indicating that first of all we will add one mole of HBR And just as we did it before we have to recall that this would be the more cosmic of addition, hydrogen will be added to a more substituted position and roaming will be added to a less substituted position. But in this case both ants are equally substituted. So it doesn't really matter where you add hydrogen and where you at bro Ming, let's suppose that we're adding hydrogen to the white carbon and running to the left. So if we do that, we are going to form an eight member ring But now it will not look so linear. Right, these are sp carbons. We'll be forming SB two which means that we may draw a ring as follows. A total of eight carbon atoms over here. And then we said that we are going to add blooming at this position. Now the second equivalent of H p R would be added if we repeat the process. And if we recall that we have another more cognitive edition because so far we have used only one of the two pI bonds. Now we're going to use another one remaining. So hydrogen goes to a less substituted position in particular. This is the one that has more hydrogen atoms. Now this position has one hydrogen atom, so definitely we'll add another hygienic disposition since the other position is more substituted, it also Hasbro mean and it has no hydrogen atoms. So we are going to add hydrogen to the position where we already have more hydrogen, adding hydrogen over here and another blooming at this position would essentially break that I bon and we would produce our L kill hey light. So we are ready to draw our structure. And if we do so, we may start by drawing that eight member grain and now what we have. We have another bro mean being attached to that same position, right? Because we added one more hydrogen here. Now we have to total over here and we are done. This is our product for this reaction. And we may simply label it as our final product for reaction be. And now look at these two structures. We may conclude that the correct answer to this multiple choice question is C However, if you have any questions, don't have states, leave your comment down below and thank you for watching

Predict the major product(s) of the following reactions:
a. phenylacetylene + 2 HBr
b. hex-1-yne + 2 HCl

Key Concepts

Alkyne Reactivity

Markovnikov's Rule

Hydrohalogenation

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