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: (v) mCPBA; (vi) 1. OsO₄ 2. NaHSO₃

(l)

Accepted Answer

Hello, everyone and welcome. Here's the next question. What are the products that will be formed in the following reaction conditions? We have two reactions starting with the same reagent, this branch chain alkene in reaction one, it's reacting with MC PB A lowercase M capital C PB A. And number two, it's reacting with in step one, OS 04 and in step two N A sodium hso three. So let's think about these two reagents and what's going on in each reaction and then we can predict the products. So our MC PP A is meta chloro paroxy benzo acid. And as you can tell by that paroxy, that's something that's going to introduce an oxygen. This reaction will be the oxidation of our kine. So we'll break the double bond and we're going to form an oxide ring where the two carbons of the double bond will be bound to an oxygen. This is a concerted reaction. So the stereo chemistry will be unchanged. So our double bond is an E arrangement. It will still be an E arrangement and any other chiral groups will have their stereo chemistry unchanged. And the other thing of note is that, that um oxygen can react and be, the ring can be formed on either side of the double bond. So we end up with a pic mixture of products. So let's look at the molecule, our alchem it's reacting here. I'm just going to think about the name of this to be able to talk about and describe the different groups. We have our longest chain, which we'll highlight here in blue, we have 1234, there's our double bond from C four to C five 678 carbons in our longest chain. And then we have three methyl groups attached to different points. So there's a, let's go ahead and number our carbon. So we can describe where things are attached. So 12345, our double bonds from carbon four to carbon 5678, we have a methyl group on carbon two, a methyl group on carbon three that's depicted as a dashed wedge. So projecting down behind the screen, a methyl group on carbon six and then there's a dashed wedge on the bond from carbon six to carbon seven or not, a dashed wedge, excuse me, a solid wedge making up the bond between carbon six and carbon seven. So that bond projects up towards us out of the plane of the screen. So those the dashed wedge, the solid wedge there, that story chemistry will be unchanged and around the double bond from carbons 4 to 5 we have an E arrangement. So the remainder of the chain is attached to opposite sides of that cc double bond. So let's draw our aic mixture here. First, I'm just going to paste in the basic structure of this molecule without the double bond. And I like to highlight in blue where my double bond originally was. So carbon is 4 to 5, it will now be a single bond that just helps me keep oriented here. And I remember having a pox side ring and that stereo chemistry is we have the ring being added on either side of this double bond. So the first product here, we draw the bonds to the oxygen there as solid wedges. So a bond from carbon four and carbon five to this oxygen. So that's our first product and then we'll have those bonds going the other direction. So our first direct first product, they're coming up towards us out of the screen. And in the second, let's look at the C four to C five bond and there's our ring with the two bonds to the oxygen shown as dashed wedges again, since that can be added on either side going to the oxygen. So we've converted our alkene into an oxide with C four and C five being bonded to oxygen. And we have a rhythmic mixture where the oxygen is being, the oxygen ring is coming towards you or away from you. And again, note that the stereo chemistry on the other chiral centers of the molecule is unchanged. So there is our answer for part one and then part two, we're also going to be oxidizing something but with a different reagent. So same molecules reacting with os 04. So in this case, the oso four for number two is going to form an intermediate where it's bound to our carbons from the double bond. And then the second step will cleave that either there leaving behind two oxygens that will then become alcohol groups with the addition of a hydrogen. So let's just draw the intermediate. So we get an idea here. So I had run our group with two R groups bonded to carbon, there's AC C bond there. And we have the formation of this intermediate and will draw solid wedges here to indicate that stereo chemistry, two oxygens bound those two carbons. And then as oxygens are bound to and osmium os and then coming from that, we have two co double bonds. So you can see how the product will be obtained as a vic mixture because this could be attacking from either side. So these could be dash switchs instead. So when we look about at our products, put up the basic structure of the molecule here, we know we'll have this ramic mixture here. Let's highlight our C four C five bond there. So in the first molecule here, I'll have the two alcohol groups and I will draw them as solid wedges. So they can add as oh with a solid wedge on C four oh with a solid wedge on C five. And then of course, we can just add those two alcohol groups as dash wedges oh with a dash wedge on C 40 with a dash wedge on C five. So there we have it. Those are our predicted products for reaction. Number two, we get this vicinal diol and our two alcohol groups are added in a sin fashion both on the same side of the double bond. So two solid wedges, two dash wedges there since it starts with this concerted reaction and the stereo chemistry of everything else in the molecule is preserved. See you in the next video.

Predict the product(s) that would result when the alkenes are allowed to react under the following conditions: (v) mCPBA; (vi) 1. OsO₄ 2. NaHSO₃
(l)

Key Concepts

Epoxidation

Dihydroxylation

Stereochemistry

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