Anionic Polymerization - Online Tutor, Practice Problems & Exam Prep

Anionic polymerization involves alkenes with electron withdrawing groups, initiated by strong nucleophiles like sodium amide or butyl lithium. The process consists of three steps: initiation, where the nucleophile attacks the double bond; propagation, where the monomer anion reacts with another monomer to form a new anion; and termination, which requires a proton donor, such as water, to neutralize the carbanion. This method allows for living polymers, meaning the reaction can continue as long as monomers are available.

Anionic Polymerization Concept 1

Hey, everyone. So let's take a look at anionic polymerization. Here, we're going to say that alkenes with electron withdrawing groups undergo anionic polymerization. Here, we're going to say that a strong nucleophile, and this is typically in the form of sodium amide or the very strong butyl lithium, initiates the reaction. So if we take a look here, we have our alkene with its electron-withdrawing group.

And then here, we have our nucleophile through anionic polarization. We create a polymer with this portion representing our repeating unit. Again, we don't specify the number. It's just a long chain of them. So just remember, when it comes to anionic polymerization, it's alkenes with electron withdrawing groups that are basically going to be initiating this process.

Alright. So look out for that electron withdrawing group on our respective alkenes.

Anionic Polymerization Concept 2

So, we're going to say the reaction mechanism has 3 steps. We have step 1 which is the initiation step, step 2 which is propagation, and step 3 which is termination. Here, let's take a look at step 1, which is the initiation step. We're going to say, the strong nucleophile attacks the double bonded or the double bond similar to conjugate addition. And we're going to say the electron withdrawing group stabilizes the negative charge.

So, if we take a look here, we have butyllithium. It's a very strong nucleophile, very basic as well. We're going to say that this butyl could come in here. It's going to hit this carbon, which is going to cause this pi bond to come here to this carbon. This carbon now has a pi bond, so it's negatively charged.

And here is the butyl that we've attached. Now, that negative carbon, it's okay because, again, the electron withdrawing group in the form of the Nitrile group is going to stabilize that negative charge. Right. So, again, this is step 1, which is initiation, and this is the process that it takes similar to a conjugate addition reaction.

Anionic Polymerization Concept 3

In step 2, we have the propagation step. Here, we're going to say the monomer anion reacts with the monomer molecule and forms a new anion through a head to tail addition. So, here is our head, which is negatively charged carbon. It'll come in and attack the tail of this monomer molecule, causing this pi bond to move here to this carbon. So at the end, what do we have?

This lone pair is used to make this bond now to our CH₂. That CH₂ is still connected to CH, which is connected to our nitrile. The pi bond moved to this carbon, so now it's negatively charged. And remember, that nitrile group is an electron withdrawing group. It'll help to stabilize the carbanion that we just created.

Now that we've seen the propagation step, click on the next video and let's take a look at the termination step.

Anionic Polymerization Concept 4

We now take a look at the termination step. Here, we're going to say, unlike radical and cationic polymerization, anionic polymerization does not self-terminate. Here, we're going to say we have living polymers. That means the chain remains active after consuming all monomers. And that means if we introduce another monomer, it would just keep going.

It wouldn't just stop. So how do we terminate it? Well, we terminate the chain by adding a proton donor to the mixture. Typically, this proton donor can be in the form of water. So here goes our carbanion.

It spots that water, so it's going to deprotonate it. Oxygen holds on to the electrons. So now we've neutralized the carbanion. It's no longer negative. It's neutral.

So we don't have to worry about polymerization continuing. And then we've also created this hydroxide ion. So again, unlike radical and cationic polymerization, anionic polymerization does not self-terminate. We have to introduce a proton donor in order to stop the reaction altogether.

Anionic Polymerization Example 1

In this example, we're told that a catalytic amount of strong base, such as sodium hydroxide, can catalyze the polymerization of 4-vinylpyridine. It says, draw the structure of the anion from the initiation step and the repeating unit of the polymer. Alright. So for vinylpyridines, remember, pyridine here is going to be a benzene ring with a nitrogen in it. We're going to say where the nitrogen is at carbon number 1 or position 1, 2, 3, 4.

Vinyl just means we have double-bonded carbons, 2 of them. This is going to react with our hydroxide ion by way of a conjugate addition mechanism. So this would come in here, hitting the tail causing this pi bond to break and go to this carbon here. So what we get as a result of this is we'd have this OH attached here is the carbanion that formed. So we have this structure here.

And then if we want to show the repeating unit that is possible here, we just show it like this. K. And then in brackets, we'd have this portion here, this portion here with an n subscript to show that it's a chain of them linked together. Alright. So this is how we approach this particular example problem.

Which one of the following compounds would undergo anionic polymerization at the fastest rate?

• A.
• B.
• C.
• D.

A student was preparing nitroethene from the reaction of nitromethane with formaldehyde in an acidic solution. In the workup step, they accidentally added a small excess of NaOH, resulting in a white coating on the inside wall of the reaction flask. What is the structure of the compound in the white coating?

In this practice question, we're told a student was preparing nitroethane from the reaction of nitromethane with formaldehyde in an acidic solution. In the workup step, they accidentally added an excess, a small excess, of sodium hydroxide resulting in a white coating on the inside wall of the reaction flask. What is the structure of the compound in the white coating? Alright. So here we have our Nitroethane. So, ethene is our 2 carbons double bonded and Nitro is an \( \text{NO}_2 \) group. Remember, Nitro is a strong electron-withdrawing group, which means that this is going to want to do anionic polymerization. We've added a small excess of hydroxide, so it's floating around. This will want to do a conjugate addition mechanism where this hydroxide attacks the tail causing this pi bond to break and move here. We will get as a result of this, here is the \( \text{OH} \) that attached. Attached there, comes down. Here's an \( \text{O}_2 \). And then this carbanion that we have right here. Now here, if we wanted to show the polymer that forms, we could. The repeating unit itself would resemble this. It would look like such. And then here we put it in brackets. The subscript \( n \) showing that it's a string of them with an undesignated number strung together to form our polymer. Right? So this is how we'd approach this particular practice question.