What stereoisomers are obtained from the following reactions?

Question

Chemical reaction showing the hydrogenation of an alkyne with reagents and conditions listed for stereoisomer formation.

Accepted Answer

Hello everyone. So this video, we have to find these sterile isomers produced by the reactions indicated below. So we see here on the star material side, we have an alkin because of this triple bond here. So the first step of reaction we are using of course the sodium with liquid ammonia. So the N H three here with our alkin and this is a reduction reaction specifically a reduction to the trans alkene, the R groups and the hydrates will be on the different phase of the molecule. So you can go ahead and first draw up this double bond. So again, we're first starting off with step number one. So we of course saw this double bond here and we said we are doing a trans alkene. So on one side of the face, we will have a hydrogen because that is what we're adding here. And the same thing on the other and opposite end of this molecule, then the two groups that we're adding on is going to be the C H two and C H three three something. For the other side, we have the C H two, then the C H 33 All right. So we have reduced our alkin to an alkene, specifically a trans alkin. Now, working for our second step of this reaction, we have deter and palladium that's reacting with our trans alk, which again is another reduction reaction to give us a full reduction all the way to an ALCA. So this means we have broken all of our pie bons. Again, we're doing the second part of this reaction to give us a fully reduced molecule starting off with this single bond that we have here. Now, we know that we have the hydrate on opposite sides. Still, then we have of course, the C H two C H three groups again, on the opposite sides. Now, lastly, we're going to add our deter. All right. So using the wedge and dash configuration, we can get two different isomers. So let's go ahead and draw those isomers out again, starting off with our central line here, we can have car on a dash and then the deter on a wedge, of course, in the plane of a structure will be the bulky group, the C H two C H three three. So on the other side of the molecule, then this means that again, the hydrogen will be dashed, our deter will be wedged and then again, the bulky group will be in the plane of the structure or the paper. All right. So that's one Albert isomers. The other one that we get again, started off with the central single bond here, we now have on one side, deuterium pointing downwards or at us. Then we have our poke group is still on the plane was Santa towards the right. And then we have our proton or the hydrogen and a dash configuration send over to the left. We have hydrogen and a wedge or a dash bater over as a wedge. And in the plane we have the bulge group. So H two cc H 33. All right. So now we have to go ahead and cast these configurations into a fissure projection to get a better look at this. So we can go ahead and scroll down here. So again, here we're converting our, we in dash configuration to then give us our fisher projections. All right. So over on the left, of course, we start off with this central line here on opposite ends of the central line, we have these C H two C H three groups. All right, then we have the two centers. So we'll draw two horizontal lines. So over to the right, we have the deter then right below that we have a hydrogen over to the left. We are starting off on the top left corner being our hydrogen. And then right below that, we have deter. Now drawing the second fissure projection. Again, we have this vertical line and two horizontal lines on the opposite ends of the vertical line, we have our bulky group. We have the C H two C H 33, same thing on the bottom C H two C H 33 over to the right or top right corner. In this time for the horizontal lines, we have the hydrogen first and then the deter over to the left or top left, we have Ethereum right below that we have hydrogen. So we basically see that the two horizontal atoms have basically switched places. So take a look at these fissure projections, we can see that they are very similar groups and they have non identical, identical chiral carbons on opposite sides of the central line of projection, which is this big vertical line that we have here. And because of that, we have the exact definition of if we scroll back up, looking at answer choices A through E, we will have B being the correct answer for this problem. So again, we have reduced our alkin. So this trip a bond to an A trans alkene because it's on opposite sides of our molecule, then we did a further reduction over to an alkene. Then from this out came, we have used the wedge dash configuration and using the wedge dash configuration, we can go ahead and change this into the fisher projections to get a better look at our molecules. And again, we saw that we have similar groups to its non identical chiro carbons on opposite sides of our central line of objection. And this is how we got the answer of being B to B our isomers.

What stereoisomers are obtained from the following reactions?

Key Concepts

Stereoisomerism

Alkyne Hydrogenation

Reduction with Sodium Ammonia

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