Concept: Concept: Birch Reduction Mechanism5m
Hey guys! In this video, we're going to talk about a specific type reduction reaction that can happen with benzene. That's called the birch reduction.
Let's just take a look at the general reaction for a second. What a birch reduction does is it combines elemental sodium within an amine and alcohol to turn a benzene into what we call an isolated diene. Specifically, if this were to happen with an unsubstituted benzene like we have here, our product would be an isolated cyclohexadiene, two double bonds that are far apart from each other in a 1,4 position on a cyclohexane.
If you take a closer look at these reagents, they might look familiar because these are very similar to the reagents that we use on a dissolving metal reduction. This is a reaction from Orgo 1 that we learned a long time ago that worked with alkynes. It was a radical mediated mechanism. It turns out that this mechanism is really the same exact mechanism, except it’s going to work with benzene instead of with an alkyne. Let's get right into it.
The mechanism for this reduction is going to proceed through elemental sodium which means it’s going to possess just one electron. When that one electron donates to any of the carbons, we're going to have to break a bond. But this is going to be a mechanism where we have a combination up half-headed arrows and normal arrows just like the dissolving metal reduction, how there were some arrows that moved one radical and some arrows that moved a lone pair. When we make that bond, we have to break this bond in order to make room for the radical.
In order to keep these charges as far away from each other as possible or these intermediates as far away from each other as possible, this double bond is going to ionize into a lone pair on to the very bottom. Basically the furthest position possible from the radical, we're going to get an anion. Let’s go ahead and draw the product of this first step. What we're now going to get is a single radical at the top, double bonds on both sides and now a lone pair at the bottom which is going to be a carbanion.
This intermediate is called a radical anion which makes sense because that's what it is. It's a radical and it’s an anion. This is where our ethanol comes in. Our ethanol is going to serve as a propagating agent. Just so you know, ethanol isn’t the only alcohol you can use. Some text use tert-butanol. It doesn't matter. It’s the source of hydrogen. That's the biggest deal.
EtO-H my anion is going to grab the H and give a negative charge. What I’m going to get is a molecule that looks like this. I got my two double bonds. I still have my radical. But now I have two H’s at the bottom because I have one originally and now I just added a second which is the one that came from the ethanol.
At this point, I react with another equivalent of my elemental sodium. That elemental sodium is going to donate electrons to that same location. Now I'm going to get a lone pair anion. This is just a carbanion intermediate. This reaction just repeats itself. That's one thing about maybe dissolving metal reduction if your recall. It was the same thing twice.
Here, we would react again with another equivalent of ethanol. We would wind up getting our isolated diene because now I’ve got two H’s in the bottom. I've got two H’s on the top. I’ve got my isolated diene which is this molecule here.
For this reason and the fact that it reacts twice, sometimes you might see professors actually write ethanol times two or alcohol times two. It doesn't matter. It’s just going to have enough equivalents to make the reaction go to completion.
That's really it. That's the mechanism for Birch reduction. Now what we're going to do is we’re going to talk about specific regiochemistry that you have to consider with a Birch reduction.
Concept: Concept: Regiospecific products3m
Since this reaction always passes through an anion intermediate we can actually use activating groups and deactivating groups to direct the site of the isolated diene. How does that work? Well let's just take a look at the anion or the carbanion intermediates. This would be the point where we have the two double bonds, we have the two H's and we have a lone pair negative at the top. Let me ask you a question, if I add an electron withdrawing group to that anion, what do you think it does for stability? Do you think it makes that anion more stable or less stable? So hold that thought, now what happens if I add an electron donating group to that anion? So whatever I add to that is going to give more electrons to the negative, what does that do for the stability?
So the answer was that the first one is going to make it more stable because it pulls electrons away. An electron donating group is actually going to make it less stable because it's going to push more electrons into the anion so it turns out that these different groups are going to direct where the double bonds go. So as you guys can see withdrawing groups are going to what I say isolate themselves from the diene and I specifically chose that word for a reason because withdrawing and isolate kind of mean the same thing, if you're withdrawing from the crowd that means you're isolating yourself so a withdrawing group is going to be isolated from the double bond, it's going to be away from the double bond and why is that? It's not just because we memorised it, it's because you know that it's going to stabilise the negative charge. So it's going to want to be where the negative charge was whereas donating groups are going to attach themselves directly to the diene like in this situation.
Why? Because I have electrons going into the ring and I don't want it to be here because it was there it would make my anion less stable, so I'm trying to put it in a place where it's not going to affect the stability where it's going to be fine. So electron donating groups attached to the ring and withdrawing groups isolate from the ring. If you don't remember the mechanism you can at least remember the way that I'm telling you which is that withdrawing isolates, so you can think of just your isolating yourself from the crowd you're withdrawing or donating attaches which is basically the opposite. Awesome guys, so really that's it for this topic. Let's move on to the next.
Problem: Predict the major product from the Birch Reduction2m
Provide the major product for the following transformation.
Provide the major product for the following transformation.
Provide reagent to complete the following chemical transformation.
Predict the major organic product(s) for the following reaction. If more than one product is formed label the major product.
Predict the product of the following reaction.
Which species is reduced in the following Birch reduction?
a. None of them
Which of the following compounds is the major product of the reaction below:
Predict the major product obtained when the following compound is treated with Birch conditions:
Predict the major product for the following reaction paying attention to the regio- and stereochemistry.
Consider the detailed mechanism of the following reaction. Choose the structure below that is not a possible intermediate in the mechanism.
The following highly selective reaction featured in a synthesis of morphine by Rice and co-workers. Provide a complete mechanism.
Predict the major product for each of the following reactions paying attention to the region- and stereochemistry. If there is no reaction, write just “No Reaction.”
Birch reduction reduces aromatic compounds to isolated dienes. Substituents attached to the ring can affect the orientation of the double bonds.
How exactly does Birch reduction work? Good news! It uses reagents very similar to those in a reaction you’ve already learned: dissolving metal reduction (AKA metal-ammonia reduction of alkynes). Before we cover the effects substituents have, let’s cover the basics. Birch reduction uses two equivalents of lithium or sodium metal, two equivalents an alcohol, and liquid ammonia. The only major difference between this reagent set and dissolving metal reduction is the presence of alcohol.
The mechanism will look very similar to that of dissolving metal reduction, so strap in! The first step is sodium’s (or lithium’s) donation of an electron to the benzene, and that forms the radical anion. The resulting lone pair then pulls a hydrogen from the alcohol, resulting in a conjugated radical. Another equivalent of sodium donates an electron, and then the resulting lone pair pulls a hydrogen from another equivalent of alcohol. This mechanism produces an isolated diene, forgoing the more stable conjugated diene.
That’s all fine and dandy, but what happens when there are substituents on the benzene? Remember that benzene substituents can be divided into two categories: electron-donating groups (EDGs) and electron-withdrawing groups (EWGs). The methoxy group on anisole would be an EDG, and the chlorine on chlorobenzene would be an EWG. EDGs and EWGs will orient the double bonds differently. EDGs attach themselves to the diene, and EWGs
Above is the general reaction scheme with generic substituents. Below is the reaction scheme with toluene, aniline, nitrobenzene, and acetophenone.
So, that’s about it! Good luck studying. Remember that I’ve got tons of videos on this topic and everything else you need in Organic Chemistry.
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