Provide arrow-pushing mechanisms for Assessments 9.10(b) and 9...

Question

Provide arrow-pushing mechanisms for Assessments 9.10(b) and 9.10(c) that rationalize the regioselective and stereospecific formation of each halohydrin.

(b) Chemical reaction diagram showing the addition of bromine and water to an alkene, illustrating halohydrin formation.

(c) Chemical structure showing bromine and water reacting with an alkene, illustrating halohydrin formation.

Accepted Answer

Welcome back, everyone provide an arrow pushing mechanism for the given reaction that explains how the halo hydron is formed in a regional selective and stereo specific manner. The starting material is E two methyl butte, two en one L benzene which reacts with bromine and water to produce a pair of halo hydrants, which are in this case, in an humors. So we want to understand the mechanism of this reaction. And what we're going to do is just redraw our starting material. And we are going to remember the first step of the halo Hyde transformation mechanism. We're going to use a molecule of bromine. When that molecule of bromine approaches the pi bond, it gets slightly polarized so that the pi bond can attack bromine, which has a long pair and it tags back to form the bro bridge followed by the loss of the living group. So now what we have here is that we are forming our BRO bridge and we have to understand that there are actually two ways to do this. We can add bromine above the plane or below the plane. So one way for us is to, of course, first of all break each pi bond. And now we are going to draw two wedges. This is our first possibility. Then we're going to add bromine, it will have a formal positive charge. Since now, we have three actually two bonds for bromine. And we can also add it on two dash bonds, meaning we are adding it below the plane. As a result, we have to change the configurations for the two metal groups. We're going to add two wedges for the second structure and two dash bonds for the first one. So let's do that as well. Well done. So we have our ammonium cat ions and then the next step, we will perform the nucleic attack. Now, what we noticed is that the products have an arrangement of the alcohol group and bromine. So when we consider stereo specificity, we understand that we will have a backside attack for that inversion of configuration at the electrolytic position. And regarding the chemistry, let's recall that water will attack the more substituted carbon because it can form a more stable carbo ion. So now let's perform that backset attack, we will have an aversion of configuration, right? And of course, we are breaking our carbon bro bond to open that ring. So now let's think about the products formed. So let's see what we have. We're going to take the first one. And now of course, here we are going to open the wing. So let's see that if we are opening our ring. First of all, we have to keep in mind that the bromo group will still be pointing up on a watch because it's the leaving group. And now we will have an inversion of configuration. So if we had a wedge for bromine that we'll have a dash or water, so let's add water, oxygen will get a formal positive charge because it's we bonded. Now, the structure looks a bit odd, right? Because there's an implicit hydrogen on a solid line. So we can perform bond rotation, meaning we will simply replace our methyl group with a solid line and then that hydrogen well, b on a dashed bond. So let's see that as well. We are going to first we add our metal group and now regarding bromine, it must be an an arrangement to that a wedge, right? So we're going to add that wedge for bromine. So that looks better. And similarly, we're going to observe exactly same thank for our second structure, except from the fact that now water will be pointing up on a wedge to that backside attack, right? We have that inversion of configuration and the promo group will be antsy on the dash that we are seeing. So let's add a dashed bond for B roaming so far. So good, our final step will be the de protonation step. So we definitely want to see an additional step for this mechanism. What we're going to do is just illustrate that stuff. And for the de protonation step, we can just use the bromite anion formed in the first step acting as a base. Now let's add the bromite and ion to both of our intermediate. And now let's illustrate that proton capture. So now we are eventually ready to draw the two results on products. Now let's draw our products and we can simply do that by removing that one hydrogen from our structure so that oxygen loses its formal positive charge in each case. And for simplicity, we can even condense our oh into a single symbolic form. So we don't need to show that bond, right? And that's it. We clearly see how we got the two desired products. The mechanism explains the region selective manner, right? Water attacks the more substituted carbon because that carbon atom can form a more stable carbo ion. And regarding ster specificity, we had to explain why the products have that an arrangement. That's because our rank has two bonds pointing on the same side. And we're performing the backside attacks of the inversion of configuration allows us to achieve the NC arrangement. Thank you for watching.

Provide arrow-pushing mechanisms for Assessments 9.10(b) and 9.10(c) that rationalize the regioselective and stereospecific formation of each halohydrin.
(b)
(c)

Key Concepts

Arrow-Pushing Mechanism

Regioselectivity

Stereospecificity

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