Radical Synthesis - Online Tutor, Practice Problems & Exam Prep

Radical bromination initiates the transformation of alkanes into functionalized products, with bromine preferentially attaching to the tertiary position. Utilizing LDA leads to a Hofmann E2 elimination, yielding the least substituted alkene. The addition of HBr in the presence of peroxides results in anti-Markovnikov products. A strong nucleophile like NaOEt facilitates an SN2 reaction with a primary alkyl halide, culminating in ether formation. This process exemplifies organic synthesis, showcasing the conversion of simple alkanes into more complex functional groups through a series of strategic reactions.

Predict the MAJOR product.

All right. So the first reaction would be what? This is what we call a radical halogenation or a radical bromination. And remember that this is actually the only starting point I have for an alkane. I'm starting with an alkane, so I really need to start from there. So let's go ahead and attach the bromine. Where would it go? To the tertiary position. Now I have LDA. LDA with a tertiary alkyl halide is going to do what? This is a flowchart question. When I say flowchart, that means it's my flowchart that has to do with substitution elimination. Look in those chapters if you want to know more about that flowchart. But what that would be is that would be a Hoffman E2. And that should be something that's pretty comfortable to you guys at this point because we've done it a lot. So a Hoffman E2, basically any bulky base would do a Hoffman E2. That means that where should I make my double bond? I actually have 2 different double bonds available. I could use the alphas I'm sorry, the beta h

Predict the MAJOR product.

Alright. So what was this first reaction? HCl on a double bond. That would just be a regular hydrohalogenation. What that's going to do is it's going to give me a Markovnikov alkyl halide. So, my first step, I just expect to get a Cl here. Now usually, we draw the mechanism, full mechanism for these, but notice that this carbocation wouldn't shift because it's already going to be tertiary, so I don't have to worry too much about the mechanism. The mechanism's really important when a carbocation could shift, but in this case, it's not going to.

Now I have ωt butoxide. What's that going to do to a tertiary alkyl halide? This is a flowchart question again, so I would say, okay, is this a negatively charged nucleophile? Yes. Is it a bulky base? Yes. ωt butoxide is one of the bulky bases, so this is going to be a Hoffman E2. Hoffman E2 means that I'm going to prefer the less substituted double bond, so my product should actually look like this. So, I'm going for the less substituted. If I made it go down or up, that would have been more substituted. That would have had more R groups. So, that would have been a bad choice.

Now we've got Cl₂ over water. Do you guys remember what that is? That is going to be a halohydrin formation. This is one of the addition reactions you need to know. So, the way that a halohydrin formation works is that you're going to get a chlorine with a positive charge. That's going to be your intermediate. Remember that then water comes in and attacks the Markovnikov location. So, we wind up getting at the end of that mechanism, I was just kind of trying to draw it shorthand for you guys so you'd remember, is you get obviously like an alcohol on one side and you'd get an anti facing halogen on the other. So, notice that they have opposite stereochemistries.

Okay? So, that's from the whole halohydrin. Now the last step is NaOH. What does NaOH do to a halohydrin? Do you guys remember? This is a very, very particular reaction. Okay? And if you don't remember it, I'm sorry. I'm not trying to be mean, but it is an important reaction for you guys. This is actually going to be a form of epoxidation. How? Well, remember epoxidation means I'm making epoxide. Remember there are 2 ways to make epoxides. We could use peroxyacids, which we would just add directly to the double bond. So if I wanted to use a peroxyacid right here, I could.

But once we have the halohydrin, I can also use a base to do what we call an intramolecular SN₂. Okay. So just to remind you of this mechanism, what winds up happening is that my O⁻ pulls off an H and gives me something that looks like this. Cl here and an O⁻ here. So, notice that now I have a nucleophile and a leaving group right next to each other. So these wind up doing a backside attack on each other and what I wind up getting for my final, final product is an epoxide that looks like this. Plus my leaving group Cl. Okay? And that is actually my product here. My product is this epoxide. I could have even made this harder because remember that epoxides open with an acid and a base-catalyzed method. I could have made you do that as well, but we did it. I took it easy on you guys. We just made the epoxide. But just letting you know, these are all common synthetic pathways that you need to know. Okay?

Cool, guys. So I hope that made sense. Let me know if you have any questions. If not, let's wrap this one up.

Here’s what students ask on this topic:

What is radical bromination and how does it work?

Radical bromination is a chemical reaction where a bromine atom is introduced into an alkane molecule. The process involves the formation of bromine radicals through the homolytic cleavage of Br₂ under UV light or heat. These bromine radicals then abstract a hydrogen atom from the alkane, creating an alkyl radical. The alkyl radical reacts with another Br₂ molecule to form the alkyl bromide and regenerate the bromine radical, propagating the chain reaction. This reaction preferentially occurs at the tertiary carbon due to the stability of the resulting radical.

What is the significance of using LDA in radical synthesis?

Lithium diisopropylamide (LDA) is a strong, non-nucleophilic base used in organic synthesis. In the context of radical synthesis, LDA is used to perform Hofmann E2 eliminations. This reaction involves the removal of a hydrogen atom and a leaving group from adjacent carbon atoms, resulting in the formation of a double bond. LDA favors the formation of the least substituted alkene, known as the Hofmann product, due to its bulky nature, which hinders the formation of more substituted alkenes.

How does the addition of HBr in the presence of peroxides lead to anti-Markovnikov products?

The addition of HBr in the presence of peroxides (such as ROOR) leads to anti-Markovnikov products through a radical mechanism. Peroxides decompose to form radicals, which initiate the reaction by abstracting a hydrogen atom from HBr, generating a bromine radical. This bromine radical adds to the less substituted carbon of the alkene, forming a more stable radical intermediate. The intermediate then abstracts a hydrogen atom from another HBr molecule, resulting in the anti-Markovnikov product, where the bromine is attached to the less substituted carbon.

What is the role of NaOEt in radical synthesis?

Sodium ethoxide (NaOEt) is a strong nucleophile used in radical synthesis to facilitate SN2 reactions. When reacting with a primary alkyl halide, NaOEt performs a backside attack, displacing the leaving group and forming an ether. This reaction is highly efficient for primary alkyl halides due to minimal steric hindrance, allowing the nucleophile to approach the carbon center easily. The result is the formation of an ether, showcasing the transformation of simple alkanes into more complex functional groups.

How can radical synthesis be used to convert alkanes into functionalized products?

Radical synthesis can convert alkanes into functionalized products through a series of strategic reactions. Starting with radical bromination, a bromine atom is introduced at the tertiary position of the alkane. Subsequent reactions, such as Hofmann E2 elimination using LDA, form the least substituted alkene. The addition of HBr in the presence of peroxides results in anti-Markovnikov products. Finally, a strong nucleophile like NaOEt facilitates an SN2 reaction with a primary alkyl halide, forming an ether. This sequence of reactions demonstrates the transformation of simple alkanes into more complex and functionalized molecules.