Predict the product(s) that would result when the alkenes are ...

Question

Predict the product(s) that would result when the alkenes are allowed to react under the following conditions: (iii) Br₂, H₂O ; (iv) Cl₂, CH₃OH

(f) Structural representation of an alkene with potential halohydrin products from reactions with Br2 and Cl2.

Accepted Answer

Welcome back everybody. Our next question says, predict the products that will be formed in the following reaction conditions. We have two reactions, one and two, the reactant molecule and alkene with several substituents. We'll look at that in more detail in a minute. In number one, it's reacting with the L two in water in reaction. Number two, it's reacting with I two in methanol. So we have here a reaction of an alkene with a diatomic halogen. And there's a reason why our problem helpfully tells us to pay attention to the reaction conditions. We might be tempted to say that the product will be a die Halide where we add the halogen atom to each carbon in the double bond. However, that will only occur as the major product in an inert solvent. And we have a polar product solvent in both cases, since the water and the methanol, in this case, will be a more active base than our Halide iron that will be act as the nuclei preferentially. So instead of having a di Halide, we're going to end up with a halo hydran uh as our product in this case. So now we just need to think about the specifics of this reaction. We know that we're gonna end up with a halogen atom added to one carbon and either oh or och three on the other side. But how, what's the specifics? Where will it add? What's s chemistry? Well, let's recall, the first step will have the formation of this bridged intermediate with one of the one halogen atom and that will result in a RM mixture. So step one, we'll have a RM mixture of our bridged intermediate. So we'll have our Carmen Carmen single bond and then whatever the group will label are in our prime as the groups there. And then we'll have each carbon that was in the double bond having a bond to R halogen atom. So X there. So I can either put it as two solid wedges and then we'll end up with this positive charge on the ring, which is why this ring is so unstable or two dashed wedges just how to fix my solid wedges there as I drew them a little bit incorrectly. So we have this isa mixture of our bridge intermediate. And then of course, our second step involves the ring opening after a nucleic attack. And what do we need to remember about that? Well, one key is it's got to be done in anti fashion because the presence of the ring will provide steer kindreds preventing the nuclear file from attacking the carbon. So it's going to come in from the back. So a backside attack leading to the two new substituents being anti to each other. Let's put anti edition. Yeah. And the other thing is if there's different substituents on the carbons, there, there will be, it will be a Markov Noko addition, the water or the methanol will attack on the most substituted carbon. And then our final step that doesn't really affect our stereo chemistry in any way would be dep protonation. So we end up with end up with water acting as a nucleo and having this positively charged 02 plus and when one of those protons will be taken away. So let's think about our possible products. So I'm just going to paste up the kind of basic structure here. So I don't have to redraw it every time. Let's look more carefully at our alkene. In more detail, we have this long carbon chain, it has nine carbons. I'm just going to number them. Our double bond goes from carbon four to carbon five. And then we have methyl substituents on carbon 25 and then two methyl groups on carbon eight. And it's the same molecule in both cases. So we think about, we have this reca mixture, let's think about it's easier to start with which carbon are nuclear foul attacks. So it has to be the most substituted carbon. Well, carbon five is a tertiary carbon because it, because it has that methyl group on it carbon four is a secondary carbon. So carbon five is going to be where a nuclear file will attack. So I've just pasted in the basic structure. I put a blue highlight um on that bond between carbons four and five just to keep myself in the right spot as I'm adding my substituents on, that's where my double bond was. So I know that my water Nile is going to attack carbon five. So that means I'll have an oh substituent on that carbon. And it doesn't matter which way I draw. First, I'll start with it on a solid wedge. We know that in one case, it will be in the front and in one case, it will be in the back due to the reca mixture of my bridged intermediate. So I've done a solid wedge to my 08 substituent and then I need to add on the methyl group. So that's added just as a dashed wi and then I know that my water needs to be anti to the chlorine. In this case, this right with cl two that formed in the first step. So then it, since the water was on a solid wedge or the oh is on a solid wedge, my chlorine is on a dashed wedge. So I have my halo hydrant, that's my first product. And then my second, I have the same basic structure, but in this case, the oh group will be on a dashed wedge. So the opposite stereo chemistry, the methyl group will be on a solid wedge and then the chlorine will be on a solid wedge on carbon four. So there's my two products that I would predict from the first reaction with CL two, I have the oh group added to carbon five. And when the oh group is on the solid wedge coming forward out of the plane of the paper, and the chlorine is on a dashed wedge when the oh group is on a dash wedge, chlorine is on the solid wedge so that they're anti to each other. And of course, my o age group was added to the more substituted carbon five. Now, reaction two is quite easy because I started with the same alkene reacted with I two. The only thing I need to be careful about is that my solvent is different, it's methanol instead of water. So instead of oh being added, I will add och three. So let's put up this sort of base structure. And I will add the och three group on a solid wedge. And then the methyl group on a dashed wedge on carbon five. And then carbon four has an iodine on a dashed wedge. And then finally, just the opposite theory chemistry for my second product. So in this case, the OC three group on a dashed wedge and the methyl group on a solid wedge on carbon five excuse me. And then my iodine on a solid wedge on carbon four. So there we go, there's all my products. So my two reactions very similar except chlorine and iodine. And in the chlorine reaction I had oh as my new substituent and iodine och three. But again, the important principles here that with that polar product solvent, we have a halo hydrant that we have an anti edition and we need to be looking for the Markov Niko edition of our nuclear file. See you in the next video.

Predict the product(s) that would result when the alkenes are allowed to react under the following conditions: (iii) Br₂, H₂O ; (iv) Cl₂, CH₃OH
(f)

Key Concepts

Electrophilic Addition Reactions

Regioselectivity

Solvent Effects in Reactions

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