Propose a mechanism for the addition of bromine water to cyclo...

Question

Propose a mechanism for the addition of bromine water to cyclopentene, being careful to show why the trans product results and how both enantiomers are formed.

Accepted Answer

Hey everyone and welcome back in today's video, we're looking at the following problem which reads suggest a mechanism for the addition of foaming water to psycho Heche seen. Your proposed mechanism should show why the bromo and hide Troxel substitue ints have anti relationship in the addition product and how each in and humor is formed in the reaction. So let's begin our mechanism by drawing cyclo hexen because that's what the problem suggests. We have a six member ring with a double bond and now we are adding bro mean in water. That's how reaction we may essentially state that our final product will be a halo high trend since recall that we are using bro mean in water, this indicates the formation of a halo hydrant and we will have an anti um repair. Now if you look at the mechanism the Well, first step in this mechanism is the electra filic edition of roaming the bromine molecule approaches the double bond which makes it slightly polarized. So the pi electrons essentially polarized. This molecule, which naturally is non polar and the left bro mean becomes partially positive and the right romaine becomes partially negative. The double bond attacks roaming the lone pair on romaine attacks back and we are forming a row monium bridge. So it may indicate that we have our ring. What bro ammonium formed? We may indicate these two bonds in sin arrangement. Let's suppose that they are both wedges. So blooming is pointing up, it doesn't really matter. There's a line of symmetry over here. So you may indicate two wedges or two dashes that's perfectly fine And we have formed our ceremony um recall that the ammonium here has a formal positive charge because now remain forms T bonds. And then the next step we are thinking about the nuclear attack. We have a water molecule which will act as a nuclear file in this reaction. Now we have to be very careful and we have to understand that we will have two possibilities for this reaction to occur. Also notice that we may indicate these implicit hydrogen is pointing down and now the water molecule generally attacks a more substituted position but they are both equally substituted. That means we'll have two possibilities. Let's draw them out. So we will have two arrows, one of them in green, another one in blue. Let's suppose that the water molecule undergoes the green path and essentially we will attack this carbon, This is a backside attack. Recall that this is a backside attack. So we will have an inversion of configuration at this point that this is very important for the purposes of this problem. And simultaneously we will have the loss of grooming we made through our intermediate now and what we get is the following. We start by drawing the ring. Now notice because this is a backside at second roman was pointing up the water molecule would be pointing down making hydrogen pointing up. So we may say that on a dash we have an always too which is protein ated. It will have a formal positive charge on oxygen. We may not worry that much about the implicit hydrogen even though it's on a wedge. But now we have the loss of roaming roaming will retain its configuration at this position because there was no reaction at this point, meaning we may simply indicate blooming on a wedge. And the world last step is the deep throaty nation step in which we are deep protein eating that water molecule. We may use either bromide itself or another water molecule. I suppose that we're using another molecule of water. So a molecule of water will act as a protein except er the lone pair on oxygen will attack one of the two hydrogen and the bonding electrons will be transferred in the form of lone pair and oxygen. And we will form our first product. We may indicate that product over here because that's our main reaction. And we noticed that our wage group is pointing down while the bromo group is pointing up, which suggests that there is indeed and see relationship between O. H. And B. R. This is G to the backside attack of this stuff, recalled that this stuff was the backside attack. Now what if we look at the blue pathway, what if another water molecule attacks this position instead? We will troll an appropriate mechanism. So H. T. O. The lone pair on oxygen. A text. This position we have the loss of bromine or in particular in this case will be still blooming bonded to that remaining bond. So this is another backside attack. And at this position we will have inversion of configuration so that if roman was pointing up now water will be pointing down. We are ready to draw our protein ated intermediate. So now starting at this position, we noticed that water will be pointing down. It's still protein ated. So we have a formal positive charge on oxygen. And similarly bro, mean will still be pointing up because there is no change in configuration at the center. So it may simply state at bro mean this position is pointing up now the world last step is the deeper automation step. We wish to use a base, we may use another molecule of water. The lone pair on oxygen will attack one of the acidic protons and we will transfer these bonding electrons in the form of a lone pair onto oxygen. That's how we will form our next product. And our next product again could be shown above in our next product. We have romaine pointing up and now O. H. The alcohol group pointing down. Now we have shown the mechanism we have explained why we have the anti relationship between O H N. B. R. C. To the backside attack stuff and inversion of configuration. And now we have got our two products notice that essentially if you flip this structure with respect to that horizontal line, you will clearly see that these two structures are in an tumors because you may simply redraw it as follows to make sure that we understand that they're in and immerse, we may redraw the second structure in blue with blooming pointing down and then the alcohol group would be pointing up. We have opposite configurations, dashes and wedges for O. H. As well as for blooming. Notice that we also have a wedge and a dash. So we have two opposite configurations, meaning these two are indeed in anti MERS. So these are our final products for this reaction. The mechanism is shown below. We have discussed all of the steps and now we may simply conclude that the correct answer to this multiple choice question is a However, if you have any questions, don't hesitate to leave them down below in the comments section and I'll see you in the next video

Propose a mechanism for the addition of bromine water to cyclopentene, being careful to show why the trans product results and how both enantiomers are formed.

Key Concepts

Electrophilic Addition Mechanism

Stereochemistry and Trans Product Formation

Enantiomers and Chiral Centers

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