What stereoisomers are obtained from the reaction of each of...

Key Concepts

Stereoselectivity in Oxidation Reactions

The inherent stereoselectivity of the osmium tetroxide oxidation reaction is significant because it determines whether a reaction will produce a single stereoisomer or a mixture. The rigid, cyclic transition state involved in the syn addition ensures that the resulting diol's stereochemistry is directly correlated to the geometry of the original alkene, emphasizing the predictability of the reaction outcome.

Chirality and Enantiomers

When new stereocenters are formed during the dihydroxylation reaction, the outcome can be a chiral molecule. In cases where the reaction creates two non-superimposable mirror images, these are known as enantiomers. The formation of enantiomeric mixtures and their separation is a key consideration in stereoselective synthesis.

Syn Addition Mechanism

The dihydroxylation reaction with OsO4 proceeds through a concerted syn addition mechanism, where both hydroxyl groups are delivered to the same face of the double bond. This mechanistic detail is crucial because it directly influences the stereochemical outcome of the reaction, leading to products with predictable two-dimensional and three-dimensional arrangements.

Stereoisomerism

Stereoisomerism refers to the existence of compounds with the same molecular formula and connectivity but different spatial arrangements of atoms. In the context of dihydroxylation, the syn addition to alkenes can generate chiral centers, resulting in enantiomeric pairs or meso compounds depending on the substitution pattern of the starting alkene.

Alkene Dihydroxylation

This reaction involves the addition of two hydroxyl groups to the carbon–carbon double bond of an alkene. When an alkene is treated with osmium tetroxide (OsO4) and then oxidized with hydrogen peroxide, a vicinal diol is formed. This process is a classic method in organic synthesis for introducing hydroxyl functionalities in a controlled manner.

Transcript

00:01 Okay, this problem is asking us to react each of these compounds with oso4 and aqueous h2 in order to form different compounds.

00:07 And we're supposed to show the different stereo isomeres of those compounds.

00:10 Okay, so let's go to this first one.

00:12 On this first one we have trans 2btine.

00:13 So it looks like this, it's a 4 carbon compound with my alken on the second carbon.

00:18 Okay, so right there.

00:19 And if we react this with oso4 and aqueous h2o2, it's going to add sin dials.

00:24 So we're going to have, on the same side, we're going to have two alcohols.

00:27 So right here, i'm going to have the drumming.

00:30 Of my alcohol right there on the front and then the alcohol on the front over here.

00:35 Okay, so that is a syn dial that is, or cis dial, that's where we have my alcohols on the same side, so they're both facing towards us.

00:43 So the question is, does this have a stereo isomer? so if we were to rotate this around, so let's just demonstrate a rotation like that, it would look like this.

00:53 So basically this side is going to be over here like that, and then we're going to get the connection to this alcohol.

01:00 That's going to be facing the back now.

01:02 Remember, we're doing a rotation.

01:04 Okay, and then we have this bond right there.

01:06 That's going to be over here.

01:08 And then we have the connection to this alcohol up here, which is now in the back.

01:11 And then finally, the connection to that methyl group.

01:14 Okay, so that is the rotation.

01:15 Okay, so is this compound? well, obviously, this one is the same exact as this one because we just underwent a rotation, but is the hypothetical an antimer the same as the rotation? so let's go draw that out.

01:27 So what i'm doing now is just adding the alcohols in the back now.

01:32 Okay, so this would be a syndial reaction in which we added dials with oso4.

01:38 So is this compound the same as this one? okay, and we got this because we reacted my oso4 with the trans tube butene.

01:45 Okay, and then we just added the alcohols on the same side.

01:47 Instead of the front, we add them in the back.

01:49 Okay, so is this the same as this one? let's see.

01:52 So in order to figure this out, what i would do is just go ahead and draw out my r's and s's.

01:57 So for example, i have on this one, or let's just demonstrate it on this one actually.

02:05 Okay, so on this one we have my hydrogen in the front and my hydrogen in the front right there.

02:10 Okay, so if my hydrogen is in the front, then i'm going to label this one one one, this side two, this one three, and we're going in this direction.

02:19 Okay, so that corresponds to s, but because this is, this hydrogen is in the front, that's going to be r.

02:24 So this alcohol is on r.

02:27 Okay, and now let's figure out this alcohol.

02:29 So this one, we're going to do the same thing where we have one, two, three that goes in this direction that corresponds to s as well, but that's going to correspond to r.

02:40 Okay, because the hydrogen is in the front.

02:43 Okay, so my r's correspond to that one because that's the exact same molecule.

02:47 It's just flipped.

02:49 So if this were the same molecule, this one right here, as the one that i have here, then it would have r's and r's.

02:56 So let's figure that out.

02:57 Okay, so if i have a hydrogen there and a hydrogen there, let's forget this out.

03:00 One, two, three, that's corresponding to that direction.

03:04 That corresponds to r, but because the hydrogen is in the front, it's going to correspond to s.

03:08 Okay, and then same with this one.

03:09 We have one, two, three, that's going in this direction.

03:14 That corresponds to r, but we're going to flip it because the hydrogen is in the front.

03:17 So s.

03:18 So we have two s's and those two r's.

03:21 These are not the same molecule, so i know i'm going to have an antemiris.

03:23 So one of them is going to be a case in which i have my alcohols in the front, and the other one is going to be a case in which i have my alcohols in the back.

03:31 Okay, this right here is just a flipped version of my alcohols in the front.

03:36 Okay, so not to this one.

03:38 Okay, moving on to this one.

03:39 Sys 2b team.

03:40 That is a four carbon compound like this, but we have the alken there demonstrating that we have my constituents on the same side.

03:48 Okay, so if i were to add my alcohols, it looked like this.

03:51 I erase that alken, add my alcohols to the same.

03:53 Side, just like that.

03:58 Okay, so does this have an anantamor? because if i were to add my alcohols from the back instead of the front, it would look something like this, right? but we realized that if we were to just flip this molecule around, it looked the exact same as if we were to draw that hypothetical an antimer...

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