Predict the product of formula C₇H₁₃BrO ...

Question

Predict the product of formula C₇H₁₃BrO from the reaction of this same unsaturated alcohol with bromine. Propose a mechanism to support your prediction.

Accepted Answer

Welcome back everyone in today's video, we are looking at the problem which asks us to determine the product of the reaction of the fulling compound with booming. If the formula of the product is C eight H 15 B R O. Propose a mechanism for the formation of the product. Now, first of all, we noticed that this reaction takes place between an elk in which also has an alcohol group and the region that we are using is so we are going to add roaming and let's take a look at what we have here bro. Ming is a di atomic molecules which involves two halogen atoms. These are two bromine atoms and we have a double bond over here. This is an al king and we can say that we are dealing with an electra filic edition of grooming across the double bond recalled that the world first step in an electra filic edition is that when the blooming molecule approaches that double bond it gets slightly polarized. Generally romaine is non polar, but due to these pi electrons, they are essentially shifting these electrons towards the right grooming. So this roaming becomes slightly partially positive and this booming becomes slightly partially negative. So the double bond essentially attacks booming blooming attacks back and we are forming a ramon EUM bridge. So we may indicate our first intermediate in this mechanism. We are going to read through everything that we have so far. Now, first of all, we are going to throw our chain and at this position, there will be no more double bond. We are forming a bar, ammonium bridge. So there are no pi electrons here anymore. We have a sigma bond and we are forming remain in a bridge with a formal positive charge because now we see there are two bonds that edge to it. So this is our first step. We may essentially think about our next step. We have our bro ammonium bridge, we can call that a ammonium. And the next step whenever we're dealing with Elektra filic edition is the nuclear attack stop in particular here because there's oh age, which is a better nuclear file than the negatively charged bromide that we have formed. We will think about the lone parent oxygen acting as a nuclear file here instead of another bromide. Now the lone pair might attack two different positions. Either the right one or the left one. And we have to recall that because the left carbon is more substituted. It has a greater electra filler character. This is the position that will be attacked by the lone parent oxygen. So we are going to attack that position because once again it's more substituted. We have three substitutions, three carbon substations to be precise. And here we have only one carbon substitution. Therefore we are checking this position and simultaneously we're opening the ring, we have the loss of the leaving group and we made drool our intermediate in this step. You have to be careful because notice that during this attack were forming a ring. So it makes sense to draw the result in the ring. There will be oxygen which still has that hydrogen and it forms an additional bond. And what that means is that oxygen has a form of positive charge. Now we are going to complete our ring. And now the remaining two substitutions are an ethyl group. So we're going to draw a siege to C. Three group as well as this carbon bonnets blooming so that we cease to be our So we have an additional substitute which essentially looks like this. Now the world lost step is the deep rotation step. We have a protein eight intermediate. So why don't we expand at O. H. Bond? To clearly indicate what's happening in the mechanism. And we need to use a base two D. Protein. It. That the base that we are going to use is just the bra might. And I informed in the first step and we are going to take the bra might and iron which attacks that proton. And simultaneously these bonding electrons are placed in the form of a lone parent oxygen. And that's how we will get our final product. We may now show that our product is form. Let's redraw the structure. Everything will look exactly the same. Except from the fact that it will not have that proton. So our rain with oxygen. Then we have an ethyl group CH two B. R. That's our mechanism. That's our product. We are also given a hint that the product has a molecular formula of CH C. Eight. My apologies. 8 15 B. R. O. So let's count the number of atoms in that structure. How many carbon atoms do we have? We have 123-456-78. We're good. 8 15 3 plus 25 plus two. That's 72 more. 92 more. 11, two more. That's 13 and two more from here. That's 15 br yes, we do see there is one grooming atom and one oxygen atom. So our proposed product is indeed consistent with the molecular formula given originally this is our final product and all these steps below. You may see our mechanism that was used to arrive at that product. The correct answer to this Multiple choice question is C. And in case you have any questions, feel free to leave that leave them down below. Yeah.

Predict the product of formula C₇H₁₃BrO from the reaction of this same unsaturated alcohol with bromine. Propose a mechanism to support your prediction.

Key Concepts

Unsaturated Alcohols

Electrophilic Addition Mechanism

Halohydrin Formation

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Predict the product of formula C7H13BrO from the reaction of this same unsaturated alcohol with bromine. Propose a mechanism to support your prediction.

Key Concepts

Unsaturated Alcohols

Electrophilic Addition Mechanism

Halohydrin Formation

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Predict the product of formula C₇H₁₃BrO from the reaction of this same unsaturated alcohol with bromine. Propose a mechanism to support your prediction.