Overview of Organic Chemistry Reactions

In this video, we're going to focus on the four main types of reactions that you'll see in a typical organic chemistry course. So, let's say if we have an alkyne. If we convert it into an alkene, if you go from, let's say, a triple bond to a double bond, or from a double to a single bond, going in this direction, it's an addition reaction. And going from, let's say, an alkene to an alkene, it's an elimination reaction. Now, let's see why that's the case. So let's say if we have cyclohexane. In order to convert this into cyclohexane, we need to add two hydrogen atoms. So here we're going to add hydrogen gas using a metal catalyst. We're going to add two hydrogens across the double bond. And so this is an addition reaction because we're adding something to the molecule. We're adding two hydrogens to it. Now let's say we have this molecule, 2-butanol. And we're going to react with concentrated sulfuric acid with heat. So this is going to favor an elimination reaction. And so what's going to happen is we're going to remove or eliminate a hydrogen atom and a hydroxyl group, giving us an alkene. And so we're going from a single bond to a double bond. And whenever you see that, it's an elimination reaction. In order to go from a single to a double bond, we need to eliminate or remove two substituents. And when you remove hydrogen and an OH group, you're basically removing water. So this is also called a dehydration reaction. Now let's say if we have this molecule, 2-chloropentane, and let's react it with water. One of the products that we can get in this reaction will be 2-petanol. And what type of reaction do you think this is? Notice that we replace a chlorine atom with an OH group. So we've substituted the chlorine atom with an OH group. So this is called a substitution reaction. Now, there's one other type of reaction that you need to be aware of. So let's say we have a secondary carbocation. The reason why it's secondary is because the carbon that bears the positive charge, that carbon is attached to two other carbon atoms. Now notice that the secondary carbocation is adjacent to a more substituted tertiary carbon. Whenever you see that, a rearrangement will occur. And the reason why the rearrangement occurs is to form a more stable carbocation intermediate. Tertiary carbocations are more stable than secondary ones, and so this is a carbocation rearrangement. So those are the four main types of reactions that you need to be familiar with. Addition reactions, where you go from, let's say, a double bond to a single bond by adding something, by adding two groups across a double bond. There's the elimination reaction, which is the reverse of an addition reaction, going from a single to a double by removing two groups. And then you have substitution. where you replace a group with another group and then a rearrangement where the whole molecule just it just rearranges like nothing is added or subtracted it just things move around in the molecule or ion now let's go over some practice problems and I want you to determine the type of reaction that we're dealing with so on the left we have one butene and we're going to react with hydrobromic acid What do you think the product of this reaction will be? Feel free to pause the video, write the product of the reaction, and also write up a mechanism for it and determine what type of reaction we're dealing with. So in this reaction, the hydrogen atom will go on the less substituted carbon atom, that is the primary carbon, and the bromine atom will go on the more substituted carbon atom of the double bond. So we're going from... double bond to a single bond so we know that this is an addition reaction as you can see we are adding a bromine atom and a hydrogen atom now HBR is an electrophile Acids tend to be electrophiles, bases they tend to be nucleophiles. And so this is specifically called an electrophilic addition reaction because we're adding an electrophile to an alkene. Now let's talk about the mechanism for this reaction. So the first thing that's going to happen is the nucleophile, which is the alkene, is going to acquire the hydrogen atom, expelling the bromide ion. And so this will give us a secondary carbocation intermediate. because we've added the hydrogen to the primary carbon. And then in the last step, the bromide ion will combine with the carbocation. And so the product for this reaction will be 2-bromobutane. And we don't need to show the hydrogen. So that's the mechanism for this reaction. Now let's say if we have... cyclopentanone and we decide to react with sodium borohydride followed by H2O what's going to be the product of this reaction sodium borohydride is a reducing agent by the way feel free to pause the video as I go through these examples if you want to try it Now, we're going to get an alcohol in this reaction. So what type of reaction is this if we're going from a ketone to an alcohol? So notice that we had a double bond, and now it's a single bond. So just by seeing that, you know that this is an addition reaction. But what type of addition reaction are we dealing with? Well, we know it's an addition, first of all, because we've added a hydrogen to the oxygen. and we also added this invisible hydrogen to the carbon because carbon needs to have four bonds. So we've added H2 across the carbonyl group, so that's why it's an addition reaction. Now, sodium borohydride has a sodium ion and a boron atom surrounded by four hydride ions. Now the boron atom has a negative formal charge, but if we consider the boron-hydrogen bond, boron has an electronegativity value of 2.0 and hydrogen has a value of 2.1. So hydrogen is slightly more electronegative than boron, so it carries a partial negative charge, and boron carries a partial positive charge. Partial charges and formal charges are not the same thing. A partial charge is based on The electronegativity difference is between an element and what that element is bonded to. Whereas a formal charge is something that we assign to an element based on the number of bonds and lone pairs that it has. So they're not the same thing. Now, BH4-, the borohydride ion, is basically a boron attached to a hydride ion. And because this hydrogen has a negative partial charge, it's nucleophilic. And so we've added a nucleophile to this carbonyl group. So this is not just called an addition reaction, but it's also called, more specifically, a nucleophilic addition reaction. Because we're adding a nucleophile to the carbonyl group. So now let's write up a mechanism for this reaction. So in the first step, the borohydride ion will release a hydride ion to attack the carbonyl group. And so we're going to add the first hydride ion to the carbonyl carbon. So here it is. Now, we're going to have a single bond between the carbon and the oxygen. And so this oxygen now has a negative charge, at which point we can react it with water. So it's going to pick up a hydrogen from water. generating the hydroxide ion. And so now we've added a total of two hydrogen atoms across the carbonyl group, making it an addition reaction. And so that's the mechanism for the reduction of a ketone into an alcohol using sodium borohydride, which is a nucleophilic addition reaction. Now for those of you who might be studying for the organic chemistry final exam, I have a video that can help you. And it's on my Patreon page. If you go to patreon.com slash mathscienctutor, you can access that page. And if you scroll down, there's a lot of other videos I have here too, but... Let's say if you're taking the first semester of Organic Chemistry, I have a six-hour video that can help you with that, if you decide to become a patron. Now, on YouTube, I have a free two-hour trailer version of this video, but if you want the entire six-hour video, you can access it here, or on Vimeo as well. And for those of you who are taking the second semester of Organic Chemistry, I have an eight-hour video that you can access as well. And there's some other stuff here that you could find too. If you're taking Gen Chem or Physics, I have stuff on that as well. So that's it, just in case you're interested. Now what about this example? What's going to happen in this reaction? So notice that we have a secondary carbocation next to a tertiary carbon. And so we know what the end result is. a carbocadine rearrangement. But how does it happen? A hydride will move from the tertiary carbon to the secondary carbon, and so it's going to move here. And by doing so, the positive charge is going to move towards where the hydride left, and so we have a carbocadine rearrangement. So whenever you have a tertiary carbon adjacent to a secondary carbocadine, a hydride shift will occur. Now here's another example. of a rearrangement reaction. So this time we have a secondary carbocation next to a quaternary carbon. So instead of a hydride shift, we're going to get a methyl shift. So the carbon structure, the carbon backbone of this ion will change. So now we have a tertiary carbocation. So that's when a methyl shift will occur. You have to have a quaternary carbon next to a carbocation. Another type of rearrangement is the ring expansion. In this example, we have a 5-carbon ring, and so it's going to expand to a 6-carbon ring, which is more stable. And so, we can break either this bond or this bond, it really doesn't matter. Let's call this carbon 1, 2, 3, 4, 5, 6. So this is going to expand to a 6-carbon ring. Now where's the plus charge? Well first, let's put the methyl group somewhere. So this methyl group here is attached to carbon 1. So this has to be 1, let's call this 2, 3, and so forth. Now notice that we broke the bond between carbons 2 and 6, and then we're going to reform a bond between carbon 1 and 6. So carbon 2 lost the bond, but it didn't regenerate a new bond, so carbon 2 has the positive charge. So right now, we have a secondary carbocation next to a tertiary carbon. So what do you think is going to happen? Well, we know what's going to happen, a hydride shift. So now, not only do we have a more stable 6-membered ring, but we also have a tertiary carbocation. instead of a secondary carbocation and so the driving force for a rearrangement reaction is stability so if you have a five carbon ring it's going to expand to a six carbon ring if it can because it can reduce its potential energy becoming more stable and if you have a secondary carbocation it's going to rearrange to a tertiary carbocation if it can because tertiary carbocations are more stable they are lower in energy And so the driving force for any rearrangement reaction is usually stability. A molecule will try to rearrange itself in order to find its lowest energy state. Now let's say if we have a secondary alkyl halide, and we decide to react with water. And we're going to use heat as well. Now we can get a mixture of products here. But I'm going to focus on one of... the two main products that we can get. One of the products that we can get is an alkene. The other product we can get is an alcohol. So what type of reactions do we have here? If we get the alcohol, it's a substitution reaction because we've replaced the bromine atom with an OH group. But now let's focus on this product. What type of reaction do we have here? In order to form a double bond, we had to remove or eliminate the hydrogen anabromine atom. So this is an elimination reaction. Specifically, it's called the E1 reaction, a first order elimination reaction, because the rate of the reaction depends only on the concentration of the substrate and not the concentration of the base. Now let's go over the mechanism. So for the E1 reaction, the first step is that the leaving group leaves. And so we're going to get a carbocation intermediate. And then in the next step, the base, which is water, is going to abstract a proton, forming a double bond. And so that's how we can get the alkane. So remember, any time you go from a single bond to a double bond, it's going to be an elimination reaction. Now let's use the same alkyl halide, but this time, instead of using a weak base like water, Let's use a strong base like hydroxide, but dissolved in water. So what's going to be the major product in this example? The strong base will immediately go for the hydrogen, forming the double bond, kicking out the leaving group, all in one step. So this is a concerted reaction mechanism. No carbocation intermediates will form, and so we can't have any carbocation rearrangements. And this is going to go straight to the alkene. Now, in addition to getting trans-butene, we can also get cis-butene. We could also form the double bond here. However, this is going to be the major product, also known as the Zeta product. This product is the minor product, which is called the Hoffman product. But overall, this is an elimination reaction, because we're going from a single to a double bond, and we've eliminated the hydrogen and the bromine atom. So we've removed HBr from the equation. So this is called the E2 elimination reaction. It's a second order nucleophilic, I mean, it's a second order elimination reaction. The rate depends on the concentration of the substrate, that is the alkyl halide, and the concentration of the base, which in this case is hydroxide. So it's first order with respect to the substrate, first order with respect to the base. 1 plus 1 is 2, so it's second order overall. So it's a second order elimination reaction. Now let's say if we have a molecule that looks like this. So this is called a beta hydroxy ketone. With respect to the ketone, this is the alpha carbon and this is the beta carbon. So on the beta carbon, we have a hydroxy group. So that's a beta hydroxy ketone. And we're going to react it with a base, hydroxide, and we're going to heat it. Now the alpha hydrogen is relatively acidic. And the reason for that is because the conjugate base is stabilized by resonance. So in this reaction, the base is going to remove the acidic alpha hydrogen, giving us an enolate ion. So right now we have a carbon with a negative charge on it, but that negative charge is stabilized by resonance. We can take this lone pair, form a pi bond, and push a negative charge on the oxygen. And so this is called an enolate ion. Now in the next step, the oxygen can use one of its lone peers to reform a pi bond, causing this double bond to move here, expelling the OH group. And so in this reaction, we're kicking out a poor leaving group, and we're also getting rid of a hydrogen atom. So what type of reaction is this? So this is our final product. Notice that we formed a double bond between the alpha and the beta carbon by eliminating or removing H and OH. So this is an elimination reaction. Specifically, it's called the E1CB reaction, when you have to remove a poor leaving group. So our product is an alpha-beta unsaturated ketone. And so that is the E1CB reaction, another elimination reaction. Now let's move on to our next example. So let's say that we have one bromobutane. And let's react it with hydroxide. So what's going to happen in this reaction? So we have a primary alkyl halide. And so this is going to favor the SN2 reaction over the E2 reaction. If we had a secondary alkyl halide and we use a strong base, this would favor the E2 reaction. The SN2 reaction stands for second-order nucleophilic substitution reaction. So that tells us that we're going to replace or substitute the bromine atom with the OH group. And that's what's going to happen. The hydroxide ion will attack the carbon from the back, expelling the leaving group. And so we're going to get 1 butanol. So we've substituted the bromine atom with the OH group. And this is a nucleophile, so it's called a nucleophilic substitution reaction. And so that's it for this example. It's a second order nucleophilic substitution reaction. The rate depends on the concentration of the substrate and the concentration of the nucleophile. Now let's say that we have tert-butyl bromide and let's react it with methanol. Now there's two things that can happen here. We can get the SN1 reaction or we can get the E1 reaction. But we're going to focus on the SN1 reaction. What do you think will be the major product in this example? So let's go through the mechanism. The first step is that the leaving group is going to leave. We're going to get a tertiary carbocanine, at which point the methanol will attack the carbocanine, giving us an intermediate that looks like this. So we're going to have an oxygen attached to the tert-butyl group. It's going to have a hydrogen on it, a lone pair, the methyl group, and a positive charge. Whenever oxygen has three bonds, it's going to have a positive charge. And now we need to use another methanol molecule to take off that hydrogen. And so the end result is that we're going to get an ether. So notice that we replaced a bromine atom with an OTHG group. So this is a substitution reaction. And methanol behaved as a nucleophile because it attacked an electrophilic carbon. Bases, they abstract protons. They go for the hydrogen atoms. But a nucleophile... attacks an atom with a positive charge that is not a hydrogen atom. So this is going to be called a nucleophilic substitution reaction, because we're replacing the leaving group with a nucleophile. Now let's start with an alkane, in this case butane, and we're going to react it with Br2 using ultraviolet light. And what's going to happen is the hydrogen will be replaced with a bromine atom. And so this is another substitution reaction, but this reaction involves radicals, so it's called a free radical substitution reaction. Let's go over the mechanism for this process. So the first thing that happens is you have two bromine molecules, I mean two bromine atoms in a molecule, and they absorb energy from an ultraviolet photon, and they're going to break apart. So this bromine bond is going to cleave homolytically. So each bromine atom will gain one electron from those two electrons in that bond. So we have two bromine radicals. So this step, going from a neutral molecule to two radicals, is known as initiation. And then the alkane will react with the bromine radical in a propagation step. Now, one of the electrons in this carbon-hydrogen bond will pair up with the bromine radical, forming hydrobromic acid as one of the products. And the other electron in this carbon-hydrogen bond is going to go on the carbon, so we have a carbon radical. So, whenever you have a radical on the left side and on the right side, it's going to be a propagation step. Now, this radical will react with... the bromine molecule. And so one of the electrons in the bond between the bromine atoms will form a bond with the carbon atom. So we're going to get a carbon-bromine bond and then this other electron will go to the other bromine atom generating a new bromine radical. So this is another propagation step. But that's how we can get our alkyl halide. So the end result is we started with this and we replaced the hydrogen with a bromine by means of radicals. So it's called a free radical substitution reaction. Now let's say if we have benzene, what's going to happen if we react it with nitric acid and sulfuric acid? What's going to be the major product in this example? So this is nitration and the end result is that we're going to replace the hydrogen. with a nitro group known as an NO2 group. And so this is going to give us nitrobenzene. The nitro group is an electrophile. It looks like carbon dioxide, but we have a nitrogen atom at the center with a positive charge. It has the same linear geometry as carbon dioxide. So notice that we've replaced a hydrogen atom with an electrophile. And so this is... substitution reaction because we substituted this with NO2, but it's an electrophilic substitution reaction. And we're also dealing with an aromatic ring, so collectively this is called an electrophilic aromatic substitution reaction. Now let's go over the mechanism for a typical EAS reaction. So we're going to start with benzene. Now the first thing that's going to happen is that the benzene ring is going to react with a generic electrophile. And so this first step is an addition step because we're adding an electrophile to the benzene ring. And notice that the double bond will be converted to a single bond. So that is an addition step. In the second step, a base will come in, remove the proton, and regenerate the aromatic ring. So the second step is an elimination step. And so whenever you have addition and elimination combined, it doesn't matter the order, overall it's a substitution reaction. Because we started with a double bond and we ended with a double bond. So overall it's not addition or elimination, it's substitution. But if you take it step by step, the first step is addition and the second step is elimination. But overall it's substitution. So that's the basic mechanism for an electrophilic aromatic substitution reaction. Now what about the next example? So let's say if we have a benzene ring with a nitro group and a chlorine atom attached to it as well. What's going to happen if we react it with, let's say, hydroxide and if we heat the solution? In this reaction, the chloride leaving group will be replaced with an OH group. And so we're going to get a phenol derivative with a nitro group attached to it. So this is called paranitrophenol. Now, what type of reaction is this? So we're replacing a chloride group with a nucleophile. So this is going to be a nucleophilic substitution. And we have an aromatic ring, so collectively, it's called a nucleophilic aromatic substitution reaction. So now let's go over the mechanism for this reaction. So I need to draw out the nitro group, which looks like this. The nitrogen atom has a positive formal charge. And so what's going to happen is... The nucleophile is going to attack the carbon that has the leading group. And so this double bond is going to move here. This one is going to move to the nitrogen. And the double bond between the nitrogen and the oxygen will break, putting a negative charge on the oxygen. And so the nitro group, what it does is it basically stores the negative charge that you place on the ring when adding a nucleophile with a negative charge. And so right now we have two groups. Attached to this carbon we have a chlorine group and an OH group So we have a double bond here here and here now the negative charge that was stored on the nitro group will be released back Into the ring causing the chloride group to leave and so we're going to get our final product which will look like this. So here is our OH group. We now have a double bond here, here, and here. And we have a single bond between the carbon and the nitrogen group. We have a double bond O and a single bond O, and a positive charge on the nitrogen. And so this is going to be the product for this reaction. So this reaction overall is a substitution reaction. But in the first step, where we added the OH group and we still had the chloride group, In that step, that was an addition step. The second step was the removal of the chloride group, and so that was an elimination step. So this reaction is also known as an addition-elimination reaction because we added something first and then we removed something later. But overall, it's still a substitution reaction, but specifically a nucleophilic aromatic substitution reaction. Now I'm going to go over one more example. So let's say we have bromobenzene and let's react it with sodium amide, NaNH2. Now the sodium ion is a spected ion, so we could drop that off. So basically we're just reacting this with NH2-. The NH2-is a very strong base. What do you think is going to happen? So this is going to be another aromatic reaction. And what's going to happen is we're going to replace the leaving group with an NH2 group. So this is going to produce aniline. So we're replacing the leaving group with a nucleophile. So this is another nucleophilic aromatic substitution reaction, but it's different from the last one. In the last example, we had an addition-elimination reaction, but this one is going to be an elimination-addition reaction. Let's go over the mechanism. So the strong base is going to remove the hydrogen. We're going to get a triple bond kicking out the leaving group. And so this intermediate is called the benzyne intermediate. Some textbooks will say that this involves more like a diradical than a triple bond. But notice that this triple bond, it's not linear, and so it's very reactive. It's in a high energy state. In the next step, another NH2-ion will attack from either side of the triple bond, putting a negative charge on this carbon. So notice the first step was an elimination step. We eliminated... H and Br. Now the second step is an addition step because we're going to add NH2 and also we're going to add a hydrogen as well. You'll see it. So now we have this. So we regenerated the aromatic ring, but now we have an NH2 group in this position. And we have a lone pair on this carbon with a negative charge. So now this is going to react with ammonia. By the way, this reaction, the reactants are NaNH2, but it's to dissolve in liquid ammonia, which is around negative 33 degrees Celsius. Now, the carbon with the negative charge is going to grab the hydrogen from ammonia, regenerating the NH2-ion. And so the end result is that we get aniline. And so here is the hydrogen atom. So overall, this is a substitution reaction because we replaced a bromine atom with an NH2 group. But the first step was elimination because we removed or eliminated a hydrogen atom and a bromine atom. And the second step was an addition reaction because we added hydrogen and we added the NH2 group. So these two can be cancelled. And as you can see that the net result is that we replace a bromine atom with the NH2 group. So overall, it's a substitution reaction, but specifically a nucleophilic aromatic substitution reaction. And so that's it for this video. Hopefully it gave you a good understanding of addition reactions, elimination reactions, substitution reactions, and rearrangement reactions. So those are the main four reactions that you need to know. But there's different variations of those four reactions, like electrophilic addition reactions or nucleophilic substitution reactions. So there's a lot of different types of those four categories. But make sure you know those main four, addition, elimination, substitution, and rearrangements.

And going from, let's say, an alkene to an alkene, it's an elimination reaction. Now, let's see why that's the case. So let's say if we have cyclohexane.

In order to convert this into cyclohexane, we need to add two hydrogen atoms. So here we're going to add hydrogen gas using a metal catalyst. We're going to add two hydrogens across the double bond. And so this is an addition reaction because we're adding something to the molecule. We're adding two hydrogens to it.

Now let's say we have this molecule, 2-butanol. And we're going to react with concentrated sulfuric acid with heat. So this is going to favor an elimination reaction.

And so what's going to happen is we're going to remove or eliminate a hydrogen atom and a hydroxyl group, giving us an alkene. And so we're going from a single bond to a double bond. And whenever you see that, it's an elimination reaction. In order to go from a single to a double bond, we need to eliminate or remove two substituents. And when you remove hydrogen and an OH group, you're basically removing water.

So this is also called a dehydration reaction. Now let's say if we have this molecule, 2-chloropentane, and let's react it with water. One of the products that we can get in this reaction will be 2-petanol.

And what type of reaction do you think this is? Notice that we replace a chlorine atom with an OH group. So we've substituted the chlorine atom with an OH group.

So this is called a substitution reaction. Now, there's one other type of reaction that you need to be aware of. So let's say we have a secondary carbocation.

The reason why it's secondary is because the carbon that bears the positive charge, that carbon is attached to two other carbon atoms. Now notice that the secondary carbocation is adjacent to a more substituted tertiary carbon. Whenever you see that, a rearrangement will occur.

And the reason why the rearrangement occurs is to form a more stable carbocation intermediate. Tertiary carbocations are more stable than secondary ones, and so this is a carbocation rearrangement. So those are the four main types of reactions that you need to be familiar with.

Addition reactions, where you go from, let's say, a double bond to a single bond by adding something, by adding two groups across a double bond. There's the elimination reaction, which is the reverse of an addition reaction, going from a single to a double by removing two groups. And then you have substitution. where you replace a group with another group and then a rearrangement where the whole molecule just it just rearranges like nothing is added or subtracted it just things move around in the molecule or ion now let's go over some practice problems and I want you to determine the type of reaction that we're dealing with so on the left we have one butene and we're going to react with hydrobromic acid What do you think the product of this reaction will be?

Feel free to pause the video, write the product of the reaction, and also write up a mechanism for it and determine what type of reaction we're dealing with. So in this reaction, the hydrogen atom will go on the less substituted carbon atom, that is the primary carbon, and the bromine atom will go on the more substituted carbon atom of the double bond. So we're going from...

double bond to a single bond so we know that this is an addition reaction as you can see we are adding a bromine atom and a hydrogen atom now HBR is an electrophile Acids tend to be electrophiles, bases they tend to be nucleophiles. And so this is specifically called an electrophilic addition reaction because we're adding an electrophile to an alkene. Now let's talk about the mechanism for this reaction. So the first thing that's going to happen is the nucleophile, which is the alkene, is going to acquire the hydrogen atom, expelling the bromide ion. And so this will give us a secondary carbocation intermediate.

because we've added the hydrogen to the primary carbon. And then in the last step, the bromide ion will combine with the carbocation. And so the product for this reaction will be 2-bromobutane.

And we don't need to show the hydrogen. So that's the mechanism for this reaction. Now let's say if we have...

cyclopentanone and we decide to react with sodium borohydride followed by H2O what's going to be the product of this reaction sodium borohydride is a reducing agent by the way feel free to pause the video as I go through these examples if you want to try it Now, we're going to get an alcohol in this reaction. So what type of reaction is this if we're going from a ketone to an alcohol? So notice that we had a double bond, and now it's a single bond.

So just by seeing that, you know that this is an addition reaction. But what type of addition reaction are we dealing with? Well, we know it's an addition, first of all, because we've added a hydrogen to the oxygen. and we also added this invisible hydrogen to the carbon because carbon needs to have four bonds.

So we've added H2 across the carbonyl group, so that's why it's an addition reaction. Now, sodium borohydride has a sodium ion and a boron atom surrounded by four hydride ions. Now the boron atom has a negative formal charge, but if we consider the boron-hydrogen bond, boron has an electronegativity value of 2.0 and hydrogen has a value of 2.1.

So hydrogen is slightly more electronegative than boron, so it carries a partial negative charge, and boron carries a partial positive charge. Partial charges and formal charges are not the same thing. A partial charge is based on The electronegativity difference is between an element and what that element is bonded to. Whereas a formal charge is something that we assign to an element based on the number of bonds and lone pairs that it has. So they're not the same thing.

Now, BH4-, the borohydride ion, is basically a boron attached to a hydride ion. And because this hydrogen has a negative partial charge, it's nucleophilic. And so we've added a nucleophile to this carbonyl group. So this is not just called an addition reaction, but it's also called, more specifically, a nucleophilic addition reaction.

Because we're adding a nucleophile to the carbonyl group. So now let's write up a mechanism for this reaction. So in the first step, the borohydride ion will release a hydride ion to attack the carbonyl group. And so we're going to add the first hydride ion to the carbonyl carbon.

So here it is. Now, we're going to have a single bond between the carbon and the oxygen. And so this oxygen now has a negative charge, at which point we can react it with water.

So it's going to pick up a hydrogen from water. generating the hydroxide ion. And so now we've added a total of two hydrogen atoms across the carbonyl group, making it an addition reaction.

And so that's the mechanism for the reduction of a ketone into an alcohol using sodium borohydride, which is a nucleophilic addition reaction. Now for those of you who might be studying for the organic chemistry final exam, I have a video that can help you. And it's on my Patreon page. If you go to patreon.com slash mathscienctutor, you can access that page. And if you scroll down, there's a lot of other videos I have here too, but...

Let's say if you're taking the first semester of Organic Chemistry, I have a six-hour video that can help you with that, if you decide to become a patron. Now, on YouTube, I have a free two-hour trailer version of this video, but if you want the entire six-hour video, you can access it here, or on Vimeo as well. And for those of you who are taking the second semester of Organic Chemistry, I have an eight-hour video that you can access as well.

And there's some other stuff here that you could find too. If you're taking Gen Chem or Physics, I have stuff on that as well. So that's it, just in case you're interested. Now what about this example?

What's going to happen in this reaction? So notice that we have a secondary carbocation next to a tertiary carbon. And so we know what the end result is. a carbocadine rearrangement.

But how does it happen? A hydride will move from the tertiary carbon to the secondary carbon, and so it's going to move here. And by doing so, the positive charge is going to move towards where the hydride left, and so we have a carbocadine rearrangement. So whenever you have a tertiary carbon adjacent to a secondary carbocadine, a hydride shift will occur. Now here's another example.

of a rearrangement reaction. So this time we have a secondary carbocation next to a quaternary carbon. So instead of a hydride shift, we're going to get a methyl shift. So the carbon structure, the carbon backbone of this ion will change.

So now we have a tertiary carbocation. So that's when a methyl shift will occur. You have to have a quaternary carbon next to a carbocation.

Another type of rearrangement is the ring expansion. In this example, we have a 5-carbon ring, and so it's going to expand to a 6-carbon ring, which is more stable. And so, we can break either this bond or this bond, it really doesn't matter. Let's call this carbon 1, 2, 3, 4, 5, 6. So this is going to expand to a 6-carbon ring. Now where's the plus charge?

Well first, let's put the methyl group somewhere. So this methyl group here is attached to carbon 1. So this has to be 1, let's call this 2, 3, and so forth. Now notice that we broke the bond between carbons 2 and 6, and then we're going to reform a bond between carbon 1 and 6. So carbon 2 lost the bond, but it didn't regenerate a new bond, so carbon 2 has the positive charge.

So right now, we have a secondary carbocation next to a tertiary carbon. So what do you think is going to happen? Well, we know what's going to happen, a hydride shift. So now, not only do we have a more stable 6-membered ring, but we also have a tertiary carbocation. instead of a secondary carbocation and so the driving force for a rearrangement reaction is stability so if you have a five carbon ring it's going to expand to a six carbon ring if it can because it can reduce its potential energy becoming more stable and if you have a secondary carbocation it's going to rearrange to a tertiary carbocation if it can because tertiary carbocations are more stable they are lower in energy And so the driving force for any rearrangement reaction is usually stability.

A molecule will try to rearrange itself in order to find its lowest energy state. Now let's say if we have a secondary alkyl halide, and we decide to react with water. And we're going to use heat as well. Now we can get a mixture of products here. But I'm going to focus on one of...

the two main products that we can get. One of the products that we can get is an alkene. The other product we can get is an alcohol.

So what type of reactions do we have here? If we get the alcohol, it's a substitution reaction because we've replaced the bromine atom with an OH group. But now let's focus on this product.

What type of reaction do we have here? In order to form a double bond, we had to remove or eliminate the hydrogen anabromine atom. So this is an elimination reaction.

Specifically, it's called the E1 reaction, a first order elimination reaction, because the rate of the reaction depends only on the concentration of the substrate and not the concentration of the base. Now let's go over the mechanism. So for the E1 reaction, the first step is that the leaving group leaves. And so we're going to get a carbocation intermediate.

And then in the next step, the base, which is water, is going to abstract a proton, forming a double bond. And so that's how we can get the alkane. So remember, any time you go from a single bond to a double bond, it's going to be an elimination reaction. Now let's use the same alkyl halide, but this time, instead of using a weak base like water, Let's use a strong base like hydroxide, but dissolved in water. So what's going to be the major product in this example?

The strong base will immediately go for the hydrogen, forming the double bond, kicking out the leaving group, all in one step. So this is a concerted reaction mechanism. No carbocation intermediates will form, and so we can't have any carbocation rearrangements. And this is going to go straight to the alkene. Now, in addition to getting trans-butene, we can also get cis-butene.

We could also form the double bond here. However, this is going to be the major product, also known as the Zeta product. This product is the minor product, which is called the Hoffman product. But overall, this is an elimination reaction, because we're going from a single to a double bond, and we've eliminated the hydrogen and the bromine atom.

So we've removed HBr from the equation. So this is called the E2 elimination reaction. It's a second order nucleophilic, I mean, it's a second order elimination reaction. The rate depends on the concentration of the substrate, that is the alkyl halide, and the concentration of the base, which in this case is hydroxide. So it's first order with respect to the substrate, first order with respect to the base.

1 plus 1 is 2, so it's second order overall. So it's a second order elimination reaction. Now let's say if we have a molecule that looks like this. So this is called a beta hydroxy ketone. With respect to the ketone, this is the alpha carbon and this is the beta carbon.

So on the beta carbon, we have a hydroxy group. So that's a beta hydroxy ketone. And we're going to react it with a base, hydroxide, and we're going to heat it. Now the alpha hydrogen is relatively acidic.

And the reason for that is because the conjugate base is stabilized by resonance. So in this reaction, the base is going to remove the acidic alpha hydrogen, giving us an enolate ion. So right now we have a carbon with a negative charge on it, but that negative charge is stabilized by resonance.

We can take this lone pair, form a pi bond, and push a negative charge on the oxygen. And so this is called an enolate ion. Now in the next step, the oxygen can use one of its lone peers to reform a pi bond, causing this double bond to move here, expelling the OH group.

And so in this reaction, we're kicking out a poor leaving group, and we're also getting rid of a hydrogen atom. So what type of reaction is this? So this is our final product.

Notice that we formed a double bond between the alpha and the beta carbon by eliminating or removing H and OH. So this is an elimination reaction. Specifically, it's called the E1CB reaction, when you have to remove a poor leaving group.

So our product is an alpha-beta unsaturated ketone. And so that is the E1CB reaction, another elimination reaction. Now let's move on to our next example. So let's say that we have one bromobutane. And let's react it with hydroxide.

So what's going to happen in this reaction? So we have a primary alkyl halide. And so this is going to favor the SN2 reaction over the E2 reaction.

If we had a secondary alkyl halide and we use a strong base, this would favor the E2 reaction. The SN2 reaction stands for second-order nucleophilic substitution reaction. So that tells us that we're going to replace or substitute the bromine atom with the OH group. And that's what's going to happen.

The hydroxide ion will attack the carbon from the back, expelling the leaving group. And so we're going to get 1 butanol. So we've substituted the bromine atom with the OH group.

And this is a nucleophile, so it's called a nucleophilic substitution reaction. And so that's it for this example. It's a second order nucleophilic substitution reaction. The rate depends on the concentration of the substrate and the concentration of the nucleophile. Now let's say that we have tert-butyl bromide and let's react it with methanol.

Now there's two things that can happen here. We can get the SN1 reaction or we can get the E1 reaction. But we're going to focus on the SN1 reaction. What do you think will be the major product in this example?

So let's go through the mechanism. The first step is that the leaving group is going to leave. We're going to get a tertiary carbocanine, at which point the methanol will attack the carbocanine, giving us an intermediate that looks like this. So we're going to have an oxygen attached to the tert-butyl group. It's going to have a hydrogen on it, a lone pair, the methyl group, and a positive charge.

Whenever oxygen has three bonds, it's going to have a positive charge. And now we need to use another methanol molecule to take off that hydrogen. And so the end result is that we're going to get an ether. So notice that we replaced a bromine atom with an OTHG group.

So this is a substitution reaction. And methanol behaved as a nucleophile because it attacked an electrophilic carbon. Bases, they abstract protons. They go for the hydrogen atoms. But a nucleophile...

attacks an atom with a positive charge that is not a hydrogen atom. So this is going to be called a nucleophilic substitution reaction, because we're replacing the leaving group with a nucleophile. Now let's start with an alkane, in this case butane, and we're going to react it with Br2 using ultraviolet light.

And what's going to happen is the hydrogen will be replaced with a bromine atom. And so this is another substitution reaction, but this reaction involves radicals, so it's called a free radical substitution reaction. Let's go over the mechanism for this process. So the first thing that happens is you have two bromine molecules, I mean two bromine atoms in a molecule, and they absorb energy from an ultraviolet photon, and they're going to break apart.

So this bromine bond is going to cleave homolytically. So each bromine atom will gain one electron from those two electrons in that bond. So we have two bromine radicals.

So this step, going from a neutral molecule to two radicals, is known as initiation. And then the alkane will react with the bromine radical in a propagation step. Now, one of the electrons in this carbon-hydrogen bond will pair up with the bromine radical, forming hydrobromic acid as one of the products. And the other electron in this carbon-hydrogen bond is going to go on the carbon, so we have a carbon radical.

So, whenever you have a radical on the left side and on the right side, it's going to be a propagation step. Now, this radical will react with... the bromine molecule.

And so one of the electrons in the bond between the bromine atoms will form a bond with the carbon atom. So we're going to get a carbon-bromine bond and then this other electron will go to the other bromine atom generating a new bromine radical. So this is another propagation step.

But that's how we can get our alkyl halide. So the end result is we started with this and we replaced the hydrogen with a bromine by means of radicals. So it's called a free radical substitution reaction. Now let's say if we have benzene, what's going to happen if we react it with nitric acid and sulfuric acid? What's going to be the major product in this example?

So this is nitration and the end result is that we're going to replace the hydrogen. with a nitro group known as an NO2 group. And so this is going to give us nitrobenzene. The nitro group is an electrophile. It looks like carbon dioxide, but we have a nitrogen atom at the center with a positive charge.

It has the same linear geometry as carbon dioxide. So notice that we've replaced a hydrogen atom with an electrophile. And so this is...

substitution reaction because we substituted this with NO2, but it's an electrophilic substitution reaction. And we're also dealing with an aromatic ring, so collectively this is called an electrophilic aromatic substitution reaction. Now let's go over the mechanism for a typical EAS reaction.

So we're going to start with benzene. Now the first thing that's going to happen is that the benzene ring is going to react with a generic electrophile. And so this first step is an addition step because we're adding an electrophile to the benzene ring. And notice that the double bond will be converted to a single bond.

So that is an addition step. In the second step, a base will come in, remove the proton, and regenerate the aromatic ring. So the second step is an elimination step. And so whenever you have addition and elimination combined, it doesn't matter the order, overall it's a substitution reaction. Because we started with a double bond and we ended with a double bond.

So overall it's not addition or elimination, it's substitution. But if you take it step by step, the first step is addition and the second step is elimination. But overall it's substitution.

So that's the basic mechanism for an electrophilic aromatic substitution reaction. Now what about the next example? So let's say if we have a benzene ring with a nitro group and a chlorine atom attached to it as well. What's going to happen if we react it with, let's say, hydroxide and if we heat the solution?

In this reaction, the chloride leaving group will be replaced with an OH group. And so we're going to get a phenol derivative with a nitro group attached to it. So this is called paranitrophenol.

Now, what type of reaction is this? So we're replacing a chloride group with a nucleophile. So this is going to be a nucleophilic substitution. And we have an aromatic ring, so collectively, it's called a nucleophilic aromatic substitution reaction. So now let's go over the mechanism for this reaction.

So I need to draw out the nitro group, which looks like this. The nitrogen atom has a positive formal charge. And so what's going to happen is... The nucleophile is going to attack the carbon that has the leading group.

And so this double bond is going to move here. This one is going to move to the nitrogen. And the double bond between the nitrogen and the oxygen will break, putting a negative charge on the oxygen. And so the nitro group, what it does is it basically stores the negative charge that you place on the ring when adding a nucleophile with a negative charge.

And so right now we have two groups. Attached to this carbon we have a chlorine group and an OH group So we have a double bond here here and here now the negative charge that was stored on the nitro group will be released back Into the ring causing the chloride group to leave and so we're going to get our final product which will look like this. So here is our OH group.

We now have a double bond here, here, and here. And we have a single bond between the carbon and the nitrogen group. We have a double bond O and a single bond O, and a positive charge on the nitrogen.

And so this is going to be the product for this reaction. So this reaction overall is a substitution reaction. But in the first step, where we added the OH group and we still had the chloride group, In that step, that was an addition step.

The second step was the removal of the chloride group, and so that was an elimination step. So this reaction is also known as an addition-elimination reaction because we added something first and then we removed something later. But overall, it's still a substitution reaction, but specifically a nucleophilic aromatic substitution reaction. Now I'm going to go over one more example. So let's say we have bromobenzene and let's react it with sodium amide, NaNH2.

Now the sodium ion is a spected ion, so we could drop that off. So basically we're just reacting this with NH2-. The NH2-is a very strong base.

What do you think is going to happen? So this is going to be another aromatic reaction. And what's going to happen is we're going to replace the leaving group with an NH2 group.

So this is going to produce aniline. So we're replacing the leaving group with a nucleophile. So this is another nucleophilic aromatic substitution reaction, but it's different from the last one. In the last example, we had an addition-elimination reaction, but this one is going to be an elimination-addition reaction. Let's go over the mechanism.

So the strong base is going to remove the hydrogen. We're going to get a triple bond kicking out the leaving group. And so this intermediate is called the benzyne intermediate. Some textbooks will say that this involves more like a diradical than a triple bond.

But notice that this triple bond, it's not linear, and so it's very reactive. It's in a high energy state. In the next step, another NH2-ion will attack from either side of the triple bond, putting a negative charge on this carbon. So notice the first step was an elimination step.

We eliminated... H and Br. Now the second step is an addition step because we're going to add NH2 and also we're going to add a hydrogen as well. You'll see it.

So now we have this. So we regenerated the aromatic ring, but now we have an NH2 group in this position. And we have a lone pair on this carbon with a negative charge.

So now this is going to react with ammonia. By the way, this reaction, the reactants are NaNH2, but it's to dissolve in liquid ammonia, which is around negative 33 degrees Celsius. Now, the carbon with the negative charge is going to grab the hydrogen from ammonia, regenerating the NH2-ion.

And so the end result is that we get aniline. And so here is the hydrogen atom. So overall, this is a substitution reaction because we replaced a bromine atom with an NH2 group.

But the first step was elimination because we removed or eliminated a hydrogen atom and a bromine atom. And the second step was an addition reaction because we added hydrogen and we added the NH2 group. So these two can be cancelled. And as you can see that the net result is that we replace a bromine atom with the NH2 group. So overall, it's a substitution reaction, but specifically a nucleophilic aromatic substitution reaction.

And so that's it for this video. Hopefully it gave you a good understanding of addition reactions, elimination reactions, substitution reactions, and rearrangement reactions. So those are the main four reactions that you need to know.

But there's different variations of those four reactions, like electrophilic addition reactions or nucleophilic substitution reactions. So there's a lot of different types of those four categories. But make sure you know those main four, addition, elimination, substitution, and rearrangements.

Transcript for:Overview of Organic Chemistry Reactions

Transcript for:
Overview of Organic Chemistry Reactions