Propose a mechanism for the reaction of pent-1-yne with HBr in...

Question

Propose a mechanism for the reaction of pent-1-yne with HBr in the presence of peroxides. Show why anti-Markovnikov orientation results.

Accepted Answer

Hey everyone and welcome back in today's video, we are going to learn how to solve the following problem proposing mechanism that explains the ante Markovic of re geochemistry of the product of the reaction of one pro pine with hydrochloric acid in the presence of hydrogen peroxide. Now, because we're using hydrogen peroxide in our reaction as well as hydrochloric acid, we may recall that such a reaction implies a free radical formation. And whenever we are dealing with free radicals, we have three different steps, initiation and propagation and termination. So let's start by drawing our mechanism with the first step which is initiation during the initiation step. We're always starting with peroxide because it's easier to break that oxygen oxygen bond to form two radicals instead of breaking the chlorine chlorine bond. So let's see how that looks like. First of all we're using hydrogen peroxide and we're going to expand that oxygen oxygen bond because that's where the harm political leverage takes place. Now. It's important to use light in this reaction because light from most information of the two radicals and what we're doing, we're essentially breaking that oxygen oxygen bond and we are uncaring the two bonding electrons so that two radicals are formed. Right? So this is called a hom elliptic clear veg, essentially that implies the t original bonding electrons are now shared equally as unpaid electrons on each of the remaining two fragments were forming two equivalent radicals that will be two hydroxyl radicals are formed. Now this is really important because the presence of hydroxyl radicals essentially allows us to form chlorine radicals more easily. And the next step in the initiation step is that the hydroxyl radical, what then react with hydrochloric acid present in that mixture. So we may essentially expand the hygiene chlorine bond to clearly show how the mechanism looks like. And what's happening here is that the radical the hydroxyl radical will essentially attack and one of the electrons within the hydrogen chlorine bond will be used in the formation of that hydrogen oxygen bond. Right, we are going to form water while the other electron will be placed on chlorine to form that chlorine radical which later on participates now reaction. So we may say that in the step we are producing too fragments. One of them Is a water molecule and the 2nd 1 is the chlorine radical. This is where initiation step is over. We have successfully initiated the reaction because we have the chlorine radical that can react with an actual material given them the problem, meaning we are ready to look at a propagation stop and this is where our starting material becomes important. The problem says one pro pine. So we are going to draw a triple bond. These are two carbon atoms. So we need one more. Right. The problem says that the triple bond starts at position number one and pro pine implies that there are three carbon atoms. So this is our starting material. What is important here is that the chlorine radical formed in the previous stop will attack it radicals are reactive. So what's happening here is that the chlorine radical attacks. And one of the electrons of one of the pi bonds is used in the formation of that carbon chlorine bond. While the other electron will be UNP aired and placed onto the adjacent carbon atoms to form a new radical. We are breaking that carbon carbon pi bond, right? We're not worried about the bottom pipe bond, that's perfectly fine. So because we're breaking one of the bonds, we are going to form an elk in radical and let's see how that looks like we are going to draw three carbon atoms first and then we understand that position. Number one. Now has that carbon chlorine bond. We still have the presence of the double bond between carbons one and two. So there's a double bond here but there is that additional electron here. Because look at this arrow, right. Were formally radical at this position meaning now this radical will react with a molecule of hcl. Right? We want to make sure that radicals react with molecules and the molecules that can react with it is the molecule of hydrochloric acid. So we're going to take that radical. We're going to redraw it. There's an unpaid electron which will attack H C. L. And we have another home a little cleavage essentially the attack takes place and one of the electrons within that bonding pair is used in the formation of the new bond. We are going to form that carbon hydrogen bond while the other electron will form a new chlorine radical. So these are our three arrows required. And we may now show that we are forming our final product. That's would be our final product. Because now we have paired that electron with one of the electrons within the hydrogen chlorine bond and now we have successfully added hydrogen at this position, also forming a chlorine radical. And because this is our final product in particular, we may say that this is our final product. We are nearly done. We got our final product and we may simply terminate our reaction When we see our final product we can move on and say there is that final termination step in which we want to make sure that reaction stops and there are no radicals that promote the reaction to proceed further so we can call it termination. And there are multiple ways to combine radicals in this reaction because there are multiple radicals. Right. However, for spectrometry reasons, we can say that one of the two hydroxyl radicals formed in the initiation stuff will be used in determination stuff. So the hydroxyl radical will react with that chlorine radical and the two electrons will be paired to form H O Cl hipAA Cloris asset. Right, We're forming that oxygen chlorine bond because the un paired electrons are on oxygen and chlorine respectively. So we are forming our final substance in the termination step. This is our mechanism. We were asked to propose the mechanism and we may just conclude that if we think about the overall reaction for complete clarity, the overall reaction can be written as follows. We said that we are starting with pro pine, so that's a triple bond, our metal group. And then we are saying, okay, we are forming our final product. We have determined the identity of our product using hydrochloric acid in the presence of hydrogen peroxide. So in addition to seeing our mechanism, we can also say that this is our reaction scheme for the scenario that was described in the problem. Looking at this mechanism and understanding the products of the reaction, we may just conclude that the correct answer choice. This multiple choice question is B. However, if you have any questions, don't hesitate to leave your comment down below in the comment section and thank you for watching

Propose a mechanism for the reaction of pent-1-yne with HBr in the presence of peroxides. Show why anti-Markovnikov orientation results.

Key Concepts

Alkyne Reactivity

Radical Mechanism

Anti-Markovnikov Orientation

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