Hello organic chemistry students. In this video we're going to cover the basics of writing organic reactions. Now we've already started to cover this when we've talked about acid and base chemistry, talking about a base attacking an acid and picking up the proton. So this won't seem that uncommon to us right now.
Now the general gist of writing a reaction is that we're going to start off here with some type of starting material and through one step or a series of steps we're going to obtain a product in a reaction. Now quite commonly we have some type of reagent that we put above the arrow and below the arrow we write something called the solvent. Now remember from general chemistry the solvent is the solute in the highest concentration or amount. So generally here our solvents in organic chemistry are going to be non-reactive. We use the solvent to solubilize the reaction to allow the reaction to proceed.
Now, this is if we have a simple one-step transformation. Now, we can also have reactions where starting material is going to be converted to product through multiple steps. We can have step one have certain conditions, step two more conditions, step three furthermore, and then step four doing other conditions.
Now, the one thing that we don't really say in a textbook, which we kind of just assume, is that after each one of these steps, after step one, we're going to do something called a workup. So after step one, there's a workup done. And what this workup entails is a very mild acidic solution. So basically anything that's negative becomes protonated.
Anything with lone pairs could possibly pick up a proton. And that's just to quench the reaction to stop it from proceeding any further. So if we look at the way this is written right now, this starting material undergoes the first step and worked up. This material then undergoes step two and then is worked up.
Step three, further transforming it, worked up. And then step four, and upon this workup is when we obtain the new final product at the end. And this is the basics of writing organic reactions.
Now we're going to go ahead and go over kind of a complex one right off the bat because this is just the best one to kind of show this multi-step process. So what we're going to do is going to go ahead and take, oops let's go ahead and put that in black for our standard nomenclature, cyclohexanone, which we don't know how to name just yet but we will, and we're going to elaborate it onto this molecule right here. This molecule that has an ester functional group and a two carbon long chain. Now the six carbons right here are going to come from our starting material.
Where does this oxygen go? Is it this oxygen here? Is it this oxygen? We have no idea.
Now the order of reactions here is we're going to have something called a Grignard react. What's a Grignard? We're going to see a lot of Grignards in this class. And then we're using THF, which is the solvent.
We're then going to use something called this functional group right here, which is what functional group again? That's right, an acid chloride. And that's what's going to form this reaction.
So now, in this topic of writing organic reaction mechanisms, what's happening? We're going to take that cyclohexanone, and we're going to draw the grignard. Now, in this grignard, we have a carbon and magnesium bond. A carbon is a nonmetal. magnesium is a metal.
So the way this truly exists in solution is a full negative charge on carbon, positive charge on magnesium. This is an ionic bond. So if it's an ionic bond, you can imagine this negative charge could attack this carbon of the carbonyl of the ketone. Why does it attack that carbon though?
The oxygen is partial negative, the carbon is partial positive. Now when we draw this, we have to have the carbonyl break open. because we'd have five bonds on carbon otherwise.
And that's our first arrow-pushing mechanism in this organic reaction. So now, after this step is done, we have an ethyl group added in, an oxygen with a negative charge, and magnesium bromide, positive charge here. And that can form an ionic bond.
That's the end of step one. We then work up the reaction, and anything that's negative gets a proton. Here's an OH.
There's the ethyl group. We have just made an alcohol from the ketone using a Grignard. Now remember a Grignard once again carbon is The non is a nonmetal, magnesium is a metal, the electrons are on the carbon, and the magnesium is deficient. So if we take a couple of minutes and just draw it in this way, it really shows our reactivity for the nucleophile.
What about step two? In step two, we take the product from step one. I'm just going to reverse the alcohol and the ethyl group right now. Here it's perfectly fine. We haven't talked about stereochemistry yet, but soon this will make a big deal.
And we're going to add in that acid chloride that we just talked about. Now, the acid chloride has a partial negative charge, partial positive charge. Could oxygen with this lone pair of electrons come in and attack this carbonyl and break open the double bond? Absolutely.
And we're going to form something called an intermediate in the reaction. Notice as this intermediate is present, we have a lot of negative and positive charge in this molecule. Now, this negative charge will donate down, pushing out the chlorine, giving us another intermediate, so close to what our final compound should be. But now, what's the next part about these workups?
It's a dilute acid, so it's mostly water. So if you have a molecule that's protonated, and it's a positive charge, the workup can actually take off that proton. in question. So the oxygen's proton gets removed by the workup, and we've now formed our final product.
And that's an arrow-pushing stepwise mechanism. Now the reaction types that we're going to be covering right now are going to be some basic ones. We're going to see substitution reactions. What are they? We'll give a little graphical distribution or representation of it.
We're going to see elimination reactions. And we're going to see addition reactions. Now before I get too far into this, the code for this video is the functional group ether. Now what's a substitution reaction?
A substitution reaction, you're going to have a carbon with some group on it. We're going to add something else in that's a nucleophile, and it's going to take its place. In essence, this Z substituted with the X on this molecule and these switched positions.
That's a substitution. An elimination reaction. Something's going to come in, we're going to see commonly a base here, ripping off the X allowing electrons to donate down and kicking out Y to form a double bond. So in elimination reaction we always form a double bond system.
We can take that double bond And then we can do chemistry where we add XY to it, some type of molecule that has both of them, and the X and the Y add across the former double bond. And that's an addition reaction. And we're going to cover a lot more of this when we start covering the alkene section in this class, chemical reactions with carbon-carbon double bonds. So in this video, we're just talking about the overview of how a chemical reaction occurs. It's a big picture here.
We can do a single step. We can do multiple steps. And that's what I was showing right here, a two-step reaction. They get much longer.
In this, you have step one, you have a solvent that's not being used or participating in the reaction. It's just used to solubilize it. The solvent is always non-reactive. I shouldn't say always, but 99% it is.
Step one proceeds, and then you work up the product of it. And that's how we get the alcohol from step one. Then that product from step one is subjected to the second conditions. And that is when we form our final product here.
And here are our three main types of reactions. That's not all of them, but the main type that we're going to be covering right now. Please don't forget that ether is our code word for this video. And I hope, I look forward to seeing everyone in lecture. And I hope each of you are doing well.