Overview of Organic Chemistry Mechanisms

Leah here from leah4sci.com and in this video we're going to look at two simple organic chemistry mechanisms to identify the types of arrow patterns that happen within every step of the mechanism. This is part 3 in this series where part 1 and 2 looked at the four different common pattern types, nucleophilic attack, loss of a leaving group, proton transfer and rearrangement. If you haven't seen it yet, make sure to catch the series on my website leah4sci.com slash mechanism. The first reaction will be an SN1 reaction where 2-butanol is reacting with hydrochloric acid. To start this mechanism, ideally we want OH to leave. The first reaction we'll look at is an SN1 reaction of 2-butanol reacting with hydrochloric acid. I cover this in detail in the Substitution Elimination Series. But if you haven't studied that yet, forget about why the mechanism happens until you get to that chapter and for now only focus on the arrow patterns so they understand where the electrons are moving. This reaction starts out with the protonating or bribing of OH to make it into a good leaving group. Oxygen uses a lone pair of electrons to reach out for and grab that partially positive hydrogen atom. Oxygen can only have one bond. Which means as soon as the hydrogen is grabbed by oxygen, the bond between itself and chlorine has to break so that it only has a net one bond. As a result, we get our first intermediate which is oxygen bound to the green hydrogen from the starting molecule also bound to the purple hydrogen. It now has just one lone pair of electrons and a new formal charge of plus one. Chlorine started out with three lone pairs but now has those initial lone pairs plus the extra blue electrons that collapsed onto chlorine when it broke away from hydrogen. This gives chlorine a charge of minus one. The mechanism isn't over so this is just our first intermediate. But to get to this intermediate, what type of reaction happened? We had oxygen reaching for the hydrogen atom. and chlorine breaking away. When oxygen grabbed hydrogen, hydrogen, the proton, was transferred from being on the chlorine to being on the oxygen. This makes step one of the mechanism a proton transfer. But we're not done. Oxygen was a poor leaving group but now as an OH2+, it's a pretty good leaving group. And what does a good leaving group do? It leaves. We show this. With an arrow starting at the electrons binding oxygen to carbon, these electrons are pulled and pulled and pulled to the point where they collapse onto the oxygen atom breaking the bond between itself and carbon, giving me a carbon chain and solution that is no longer bound to oxygen and now has a formal charge of plus one. The oxygen that broke away still has the green and purple hydrogens attached, it still has the green lone pair. But now it also has a black lone pair from the electrons that bounded the carbon but collapsed onto the oxygen atom. Since oxygen now has two bonds and two lone pairs, the formal charge is gone, it's neutral in solution. Always look for your net charge, we started neutral, we had plus and minus which is neutral. In the next step it looks like our charge is off balance but don't forget that Cl-is still floating around somewhere in solution. So the net charge for this portion of the molecule, positive to positive, hasn't changed. What happened in this reaction? When oxygen yanked at the electrons binding at the carbon and broke that bond, it left as a leaving group making the second step loss of leaving group. And last but not least, we still have that chlorine in solution so I'll redraw it here as a Cl-with four lone pairs of electrons. Chloride will use one of those lone pairs to attack the carbocation forming a bond between itself and carbon giving me the final product that has chlorine sitting on that secondary carbon. Chlorine started out with four lone pairs, three of which were shown in purple and haven't changed. The fourth lone pair was in green but these electrons are now sitting as a bond between itself and carbon. We have Positive reacting with negative net neutral, our final product is neutral and the mechanism, the attack that gave us the final product. We had chlorine which is negative and therefore positive seeking acting as the nucleophile attacking a carbocation which is positive and therefore negative seeking electrophile. When the nucleophile attacks the electrophile this step is simply a nucleophilic attack to give us our final product. Let's look at another reaction. In this mechanism we're given an alkene reacting with sulfuric acid dissolved in a solution of methanol. This is a twist on the acid catalyzed hydration of an alkene but instead of water we're using an alcohol to give us a different product. Whenever you have an acid in solution, for example H2SO4, it's going to dissociate. That would be our first step or pre-mechanism but most professors will not ask you to show this. In this step, We want to show the sulfuric acid dissolving in solution to give us acidic protons to catalyze the reaction. But protons don't float around freely in solution, instead they're picked up by the solvent molecules. In this case, we have an oxygen atom with two lone pairs of electrons reaching for and grabbing the hydrogen off the sulfuric acid. Hydrogen remember, can only have one bond. So the bond between oxygen and hydrogen will break and collapse onto the oxygen atom, freeing that hydrogen to attach to methanol. As a result, we have an oxygen now with three lone pairs and no hydrogen. Two lone pairs from the initial oxygen and the third lone pair as the bond that collapsed from the hydrogen atom. Our solvent molecule methanol still has the CH3 and blue hydrogen attached but it now has just one lone pair because the second lone pair was used as the electrons to grab the hydrogen and now sits as a bond between oxygen and the new purple hydrogen atom. You can think of this as the activation step where the former sulfuric acid now has a charge of negative one and the methanol is activated or protonated with a charge of positive one. That should be your clue what type of reaction happened here. We protonated the methanol, we moved that proton from one molecule to the other making this step a proton transfer. Now the real mechanism begins. We have a methanol solvent and solution but we'll show a protonated methanol so that we have the acid catalyst. How did it get protonated? It's what we looked at earlier. This has a charge of plus one. What is most negative in solution here to attack? We have an alkene. Alkenes have pi bonds and pi bonds have extra electrons. These electrons are looking for something positive, they're positive seeking making them good nucleophiles. The protonated methanol is positive and therefore negative seeking making it the electrophile and so the first step is a nucleophile attacking hydrogen. Stealing one of the hydrogens off of methanol, so the bond between hydrogen and oxygen is broken and collapses back onto the oxygen atom, it collapses back onto methanol. As a result, the pi bond is broken, the hydrogen goes to the less substituted carbon giving us a carbocation on the more substituted carbon following Markovnikov's rule. If you're not familiar with this, see the tutorial link below. And then we have methanol just floating around freely in solution with two lone pairs. The green one that it had initially and now the black one from the electrons that collapsed from the hydrogen bond. We started with a positive molecule, we ended with a positive intermediate and methanol is once again neutral. The attack in this case can be classified as two different ways. First you can consider it as a proton transfer. because the proton was taken out of solution and onto the molecule or you can think of it as a nucleophilic attack because the nucleophile grabbed the hydrogen. If your professor is very specific, find out which one they want. Most professors will want a proton transfer as the definition for this step. In the next step, we have a carbocation sitting on a secondary carbon next to a tertiary carbon. Don't forget the trick from the last video, a secondary carbocation near a tertiary carbon will undergo a carbocation rearrangement specifically a hydride shift. We'll show the invisible hydrogen for the sake of this mechanism and show an arrow from the bond between hydrogen and carbon going towards that carbocation where the hydrogen is going to move. This now gives us a hydrogen at the secondary position. We don't have to show this hydrogen anymore because hydrogens in skeletal structure are invisible. We're only showing this one to emphasize that it moved from the tertiary to secondary carbon. The tertiary carbon now has a positive charge giving us a more stable carbocation and we're ready for the next step. But what happened to give us this step? We had a rearrangement. We had a carbocation rearrangement to give us a more stable intermediate. Don't forget, we still have methanol floating around in solution. And methanol has oxygen, which is electronegative and two lone electron pairs. These electronegative electrons are seeking positive, making them a nucleophile. And the carbocation is positive, therefore seeking negative, making it an electrophile. As any good nucleophile will do, we show an arrow from the nucleophilic electrons attacking the carbocation, specifically the carbon atom to form a new bond. As a result, we now have a new bond from the tertiary carbon to the oxygen of methanol which still has both a CH3 and a hydrogen attached. The purple lone pair is still sitting on the oxygen but the red lone pair is now sitting as a bond. between carbon and oxygen. We started with something positive, now oxygen having 3 bonds and 1 lone pair is our new positive atom and not very happy. But before we see what happens, what happened in the previous step? The nucleophile attacked the electrophile making this step a nucleophilic attack. And last but not least, we have an unhappy positive molecule and ideally we want our product to be neutral. So we'll bring in yet another methanol from solution, one that has two lone pairs on the oxygen making it very greedy for something positive and we show oxygen reaching for the hydrogen atom to form a new bond and as a result breaking the bond between hydrogen and the initial oxygen atom. What does that look like? On the tertiary carbon we still have an oxygen but now the only thing bound to it is a CH3. Oxygen had a purple lone pair and now gets a second lone pair which is green giving us a neutral product but in solution we have the methanol that did the attack as a CH3OH. It has just one lone pair because the second one was used to form a bond between Oxygen and the purple hydrogen from the molecule here. We had a positive molecule, we now have a positive molecule. But our final product which is an ether is neutral. How did we get to that final product? We had a proton on the molecule, we transferred it to a molecule in solution making this a proton transfer. This concludes the mechanism series. If you're not very confident, go back and watch it from the beginning. You can find it on my website leah4sci.com slash mechanism. For even more practice in identifying the steps and mechanisms and so much more, come join me in the Organic Chemistry Study Hall where you'll find lots of practice in this and every other topic in Organic Chemistry 1 and 2. Full details are on my website leah4sci.com slash join.

The first reaction we'll look at is an SN1 reaction of 2-butanol reacting with hydrochloric acid. I cover this in detail in the Substitution Elimination Series. But if you haven't studied that yet, forget about why the mechanism happens until you get to that chapter and for now only focus on the arrow patterns so they understand where the electrons are moving. This reaction starts out with the protonating or bribing of OH to make it into a good leaving group. Oxygen uses a lone pair of electrons to reach out for and grab that partially positive hydrogen atom.

Oxygen can only have one bond. Which means as soon as the hydrogen is grabbed by oxygen, the bond between itself and chlorine has to break so that it only has a net one bond. As a result, we get our first intermediate which is oxygen bound to the green hydrogen from the starting molecule also bound to the purple hydrogen.

It now has just one lone pair of electrons and a new formal charge of plus one. Chlorine started out with three lone pairs but now has those initial lone pairs plus the extra blue electrons that collapsed onto chlorine when it broke away from hydrogen. This gives chlorine a charge of minus one. The mechanism isn't over so this is just our first intermediate. But to get to this intermediate, what type of reaction happened?

We had oxygen reaching for the hydrogen atom. and chlorine breaking away. When oxygen grabbed hydrogen, hydrogen, the proton, was transferred from being on the chlorine to being on the oxygen. This makes step one of the mechanism a proton transfer. But we're not done.

Oxygen was a poor leaving group but now as an OH2+, it's a pretty good leaving group. And what does a good leaving group do? It leaves. We show this.

With an arrow starting at the electrons binding oxygen to carbon, these electrons are pulled and pulled and pulled to the point where they collapse onto the oxygen atom breaking the bond between itself and carbon, giving me a carbon chain and solution that is no longer bound to oxygen and now has a formal charge of plus one. The oxygen that broke away still has the green and purple hydrogens attached, it still has the green lone pair. But now it also has a black lone pair from the electrons that bounded the carbon but collapsed onto the oxygen atom.

Since oxygen now has two bonds and two lone pairs, the formal charge is gone, it's neutral in solution. Always look for your net charge, we started neutral, we had plus and minus which is neutral. In the next step it looks like our charge is off balance but don't forget that Cl-is still floating around somewhere in solution. So the net charge for this portion of the molecule, positive to positive, hasn't changed. What happened in this reaction?

When oxygen yanked at the electrons binding at the carbon and broke that bond, it left as a leaving group making the second step loss of leaving group. And last but not least, we still have that chlorine in solution so I'll redraw it here as a Cl-with four lone pairs of electrons. Chloride will use one of those lone pairs to attack the carbocation forming a bond between itself and carbon giving me the final product that has chlorine sitting on that secondary carbon. Chlorine started out with four lone pairs, three of which were shown in purple and haven't changed.

The fourth lone pair was in green but these electrons are now sitting as a bond between itself and carbon. We have Positive reacting with negative net neutral, our final product is neutral and the mechanism, the attack that gave us the final product. We had chlorine which is negative and therefore positive seeking acting as the nucleophile attacking a carbocation which is positive and therefore negative seeking electrophile. When the nucleophile attacks the electrophile this step is simply a nucleophilic attack to give us our final product. Let's look at another reaction.

In this mechanism we're given an alkene reacting with sulfuric acid dissolved in a solution of methanol. This is a twist on the acid catalyzed hydration of an alkene but instead of water we're using an alcohol to give us a different product. Whenever you have an acid in solution, for example H2SO4, it's going to dissociate. That would be our first step or pre-mechanism but most professors will not ask you to show this.

In this step, We want to show the sulfuric acid dissolving in solution to give us acidic protons to catalyze the reaction. But protons don't float around freely in solution, instead they're picked up by the solvent molecules. In this case, we have an oxygen atom with two lone pairs of electrons reaching for and grabbing the hydrogen off the sulfuric acid.

Hydrogen remember, can only have one bond. So the bond between oxygen and hydrogen will break and collapse onto the oxygen atom, freeing that hydrogen to attach to methanol. As a result, we have an oxygen now with three lone pairs and no hydrogen. Two lone pairs from the initial oxygen and the third lone pair as the bond that collapsed from the hydrogen atom. Our solvent molecule methanol still has the CH3 and blue hydrogen attached but it now has just one lone pair because the second lone pair was used as the electrons to grab the hydrogen and now sits as a bond between oxygen and the new purple hydrogen atom.

You can think of this as the activation step where the former sulfuric acid now has a charge of negative one and the methanol is activated or protonated with a charge of positive one. That should be your clue what type of reaction happened here. We protonated the methanol, we moved that proton from one molecule to the other making this step a proton transfer.

Now the real mechanism begins. We have a methanol solvent and solution but we'll show a protonated methanol so that we have the acid catalyst. How did it get protonated? It's what we looked at earlier.

This has a charge of plus one. What is most negative in solution here to attack? We have an alkene.

Alkenes have pi bonds and pi bonds have extra electrons. These electrons are looking for something positive, they're positive seeking making them good nucleophiles. The protonated methanol is positive and therefore negative seeking making it the electrophile and so the first step is a nucleophile attacking hydrogen. Stealing one of the hydrogens off of methanol, so the bond between hydrogen and oxygen is broken and collapses back onto the oxygen atom, it collapses back onto methanol. As a result, the pi bond is broken, the hydrogen goes to the less substituted carbon giving us a carbocation on the more substituted carbon following Markovnikov's rule.

If you're not familiar with this, see the tutorial link below. And then we have methanol just floating around freely in solution with two lone pairs. The green one that it had initially and now the black one from the electrons that collapsed from the hydrogen bond. We started with a positive molecule, we ended with a positive intermediate and methanol is once again neutral.

The attack in this case can be classified as two different ways. First you can consider it as a proton transfer. because the proton was taken out of solution and onto the molecule or you can think of it as a nucleophilic attack because the nucleophile grabbed the hydrogen.

If your professor is very specific, find out which one they want. Most professors will want a proton transfer as the definition for this step. In the next step, we have a carbocation sitting on a secondary carbon next to a tertiary carbon. Don't forget the trick from the last video, a secondary carbocation near a tertiary carbon will undergo a carbocation rearrangement specifically a hydride shift.

We'll show the invisible hydrogen for the sake of this mechanism and show an arrow from the bond between hydrogen and carbon going towards that carbocation where the hydrogen is going to move. This now gives us a hydrogen at the secondary position. We don't have to show this hydrogen anymore because hydrogens in skeletal structure are invisible. We're only showing this one to emphasize that it moved from the tertiary to secondary carbon. The tertiary carbon now has a positive charge giving us a more stable carbocation and we're ready for the next step.

But what happened to give us this step? We had a rearrangement. We had a carbocation rearrangement to give us a more stable intermediate. Don't forget, we still have methanol floating around in solution.

And methanol has oxygen, which is electronegative and two lone electron pairs. These electronegative electrons are seeking positive, making them a nucleophile. And the carbocation is positive, therefore seeking negative, making it an electrophile.

As any good nucleophile will do, we show an arrow from the nucleophilic electrons attacking the carbocation, specifically the carbon atom to form a new bond. As a result, we now have a new bond from the tertiary carbon to the oxygen of methanol which still has both a CH3 and a hydrogen attached. The purple lone pair is still sitting on the oxygen but the red lone pair is now sitting as a bond.

between carbon and oxygen. We started with something positive, now oxygen having 3 bonds and 1 lone pair is our new positive atom and not very happy. But before we see what happens, what happened in the previous step?

The nucleophile attacked the electrophile making this step a nucleophilic attack. And last but not least, we have an unhappy positive molecule and ideally we want our product to be neutral. So we'll bring in yet another methanol from solution, one that has two lone pairs on the oxygen making it very greedy for something positive and we show oxygen reaching for the hydrogen atom to form a new bond and as a result breaking the bond between hydrogen and the initial oxygen atom.

What does that look like? On the tertiary carbon we still have an oxygen but now the only thing bound to it is a CH3. Oxygen had a purple lone pair and now gets a second lone pair which is green giving us a neutral product but in solution we have the methanol that did the attack as a CH3OH.

It has just one lone pair because the second one was used to form a bond between Oxygen and the purple hydrogen from the molecule here. We had a positive molecule, we now have a positive molecule. But our final product which is an ether is neutral. How did we get to that final product?

We had a proton on the molecule, we transferred it to a molecule in solution making this a proton transfer. This concludes the mechanism series. If you're not very confident, go back and watch it from the beginning.

You can find it on my website leah4sci.com slash mechanism. For even more practice in identifying the steps and mechanisms and so much more, come join me in the Organic Chemistry Study Hall where you'll find lots of practice in this and every other topic in Organic Chemistry 1 and 2. Full details are on my website leah4sci.com slash join.

Transcript for:Overview of Organic Chemistry Mechanisms

Transcript for:
Overview of Organic Chemistry Mechanisms