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.

(e) Chemical structure of an alkene with highlighted hydrogenation sites for reaction prediction.

Accepted Answer

Hi, everyone and welcome. Our next problem says, predict the products formed in the following reactions. We have two reactions shown here, the same molecule. In both case cases, a long branch chain alkene in reaction one, it's reacting with iodine gas I two in a water solvent. In reaction two, it's reacting with bromine gas. Pr BR two in a methanol solvent. So what do we have? We have a reaction with a diatomic halogen like our I two RBR two in a nucleic solvent like water or methanol. Well, we're going to have the formation of a halo hydran where will add an alcohol group and the halogen to adjacent carmen from AC C double bond. And the reason why even though uh halogen in the form of an X minus is a better nucleo than water or methanol because they're the solvent, there's far more water or methanol around. So our product will have the alcohol far more often or in the case of methanol, the ether far more often because there's so much of it around. So let's think about our mechanism here. For our first reaction. We can draw our iodine, we have our two atoms of iodine and we have the electro pilic edition of one of them. So if we draw our electron movement electrons from the double bond common attached to one of the iodines, the electrons from the bond between them goes to the other generating I minus. And then the iodine also bonds with the other carbon to form a ring intermediate. So just out of the bottom, I'll draw the general form of this ring intermediate, not the whole long molecule, but just how that ring forms. So what was our double bond is now a single bond? I'm just gonna draw some wiggles for the other substituents there. And we just want to look at the general mechanism and then we have this partial bond to our iodine and then a partial bond to the other side there. I decided to redraw that with the dotted lines and iodine and red just to distinguish it from the wiggles the rest of my molecule. And we have a positive charge there and then our oxygen in the case of water 02. But the mechanism is the same with the alcohol comes along and performs a nucleic attack on the most substituted carbon. So it's a Markov Niko addition and then the electrons that were attached to the iodine in the ring go to the iodine and the iodine is now attached on the other side, this is also added in anti fashion. So the 02 plus group and the iodine will be on opposite sides. So we need to remember that in asteric chemistry. So let's look at our products here. We have two possible products in aic mixture since our oxygen can add in the back or in the front. So let's look at what they are. So I've just pasted up an incomplete structure here. So I don't have to redraw the whole thing. I have to add the rest of the groups to show the stereo chemistry. But I just wanted to highlight in blue, the bond that was the carbon carbon double bond just to keep oriented in this molecule. So our water needs to add in the most substituted side, which should be on the left side of our drawing because that has that side of the double bond has two branches on it. So we can draw two different ways the water can add in the front. So we can draw a solid. We that will initially be an 02 plus group that will get derotated by the I minus. But we weren't asked to draw the mechanism, just the product. So we won't worry about that. And then since that add in the front, then the ethyl branch that was there will now move to the back. So we draw a dashed wedge to our ethyl branch there. And then we know this must be an anti arrangement with the iodine on the other side. So if the waters on or the oh group, excuse me, it was a water that it added. It's now an alcohol is in the front, then the iodine must be in the back. So we put that as a dashed wedge and draw the iodine in the back. So that's one of our products. And again, this will be a bic mixture. So our other product, so I've just put up that same partial structure. So now if the oxygen adds in the back, so if we add a dashed wedge to the oh group, then we have a solid wedge to our ethyl group. So let's draw that a little longer there and draw that little ethyl branch there. All right, that doesn't look very good. And then if the alcohol group is in the back, our iodine must be in the front. So we draw the iodine on the adjacent carbon projecting forward. So there's our two products from our first reaction. Again, we have the first product, they both have an alcohol and an iodine added to the adjacent carbons. The alcohol group came from the nucleic attack of water to the more substituted carbon. So it's on the left side where it has two substituents on that carbon. In the one product, the alcohol is added in the front iodine in the back. In the other product, the alcohol is added in the back and the iodine in the front in reaction two, we have the same reactant molecule but we reacted with BR two and methanol. But methanol can also perform a nucleic attack in the same way. So our two final products are similar except for the addition of the methanol, which then gets deproteinate and the bromine. So let's look at what that looks like. So put up again a structure here and we can have the methanol group which now will make an ether out in the front. So I have a solid wedge and they will be added as O CH three. And then we'll have a dash switch to our hol branch. And then if the methanol group has added in front, the bromine will add on the adjacent carbon as a dash wedge in the back. And then our second product here, the methanol adds on a dash tw and we have the ethyl group in front. While on the adjacent carbon, we have the bromine in the front on a solid wedge. So, again, very similar, just the addition of that Methanol group, which makes the final product an ether there with that OC three. Oh I just realize I'm missing a three on that hydrogen there. And again, adding the och three to the more substituted carbon and the och three and the bromine add in an anti fashion. So there's our products. 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.
(e)

Key Concepts

Electrophilic Addition Reactions

Regioselectivity

Solvent Effects in Reactions

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