Dehydrohalogenation - Online Tutor, Practice Problems & Exam Prep

Dehydrohalogenation is an E2 elimination mechanism where a hydrogen and a halogen are removed from an alkyl halide, typically favoring more substituted alkyl halides (secondary or tertiary). The process involves beta hydrogen elimination in an anti-coplanar orientation, leading to the formation of a double bond. The choice of base influences the product, adhering to Zaitsev's and Hofmann's rules. Understanding the geometry and the role of the leaving group is crucial for predicting the outcome of the reaction.

The dehydrohalogenation mechanism.

Hey guys. So now we're going to talk about a named reaction called dehydrohalogenation. I know the name sounds tricky, but actually, it turns out that you already know all the parts of this mechanism already. So it's actually pretty easy. Let's go ahead and check it out. As you can see, the name is pretty long, but all this really is is an E2 mechanism because if you think about the name, it's saying dehydro, we're taking away 1 hydrogen and we're taking away 1 halogen. Well, that's exactly what happens with a typical E2 mechanism. Remember that we always break those 2 sigma bonds and make a pi bond at the end. And that's exactly what we're going to do. So let's go ahead and check it out. Basically, you would have an alkyl halide and in this case, do we prefer that alkyl halide to be like primary or tertiary? What do you think is better? We just said this is an E2 reaction, so that means that we're going to prefer the more substituted alkyl halide that's going to favor elimination more. Okay? So that means that hopefully, we have like a secondary or tertiary alkyl halide and we're reacting that with some kind of base. Now notice here I'm just using the word base in general, but remember the type of base could lead to a different type of product. Okay? Because we had Zaitsev and we had Hoffman and the type of base that you use could prefer one product over another. Let's just go ahead and just draw the general E2 elimination product right now. I would take my base and where would those arrows go to? Do you remember? Remember that you'd always take off a Beta hydrogen. This is actually called Beta hydrogen elimination. So I'll take my minus, grab a Beta hydrogen. Now notice that the geometry of that beta hydrogen is in a special position and it's in the anticoplanar position. Remember that that's important because if you were to make a Newman projection out of this guy, you would want to make sure that your groups are facing opposite directions or in the anti position so that they can be in the most favorable orientation to eliminate. Okay? So I would take that, but remember that elimination always has 3 arrows. So I would take the electrons from here and make a double bond and finally I would kick out my X and what I'm going to get at the end is just a new double bond where basically these 2 methyl groups here are now located here and these 2 methyl groups here are now located there. Plus I would get, obviously, my base with the new hydrogen on it, so that would be the conjugate acid And I would also get the leaving group X-. Okay? So that was really easy. But now you guys just understand that that's the name associated with this type of reaction. Whenever we're using a strong base to eliminate an alkyl halide through an E2 mechanism, that's called dehydrohalogenation. And you have to think of all those things in terms of anticoplanar, in terms of Zaitsev and Hoffman, all of that is fair game. Okay? So now I have a practice problem for you guys. I want you guys to take your time trying to draw the products based on exactly what reagents you see and then I'll give you the answer. So anyway, go for it.

Dehydrohalogenation mechanism and products

So right when we look at this problem, we notice that we have a nucleophile, which would be in this case, terbutoxide. And we have a leaving group, which in this case is an alkyl fluoride. So what that means is that this is a perfect situation to use the flow chart. Okay? The flow chart that I use for substitution and elimination reactions because right away, we don't know exactly what this is. We're going to need to identify it first. So let's go through the flowchart. The first question is my nucleophile negatively charged or neutral? And in this case, because I have a potassium, that's going to leave as a spectator ion, so it is negatively charged. So that means I'm going to go down the left side of my flowchart and I'm going to go to 2. 2, do I have a bulky base? Yes, I do. In this case, this is terbutoxide and terbutoxide is one of my bulky bases. So what that means is that I'm going to say yes here and that's going to indicate that I have a certain type of E2. I have an E2 and it's going to be Hoffman. Why is that? Well, my flowchart tells you that, so in case you just wanted to use the flowchart, you could. But on top of that, you know that it's Hoffman because of the fact that we have a bulky base and bulky bases prefer that kinetic product or the one that's easiest to form. Okay. The one that's fastest to form. So what that means is that if I have more than one option possible, I'm going to go with the less substituted option. Alright? So now we have to go ahead and identify beta carbons and we have to see how many different ones there are. So this is a beta carbon here, I'm going to call that beta and this is a beta carbon here.

So now my next question is do both of these beta carbons have at least 1 beta hydrogen on them? Yes, they do. Both of them do. So then my last question is, do they have hydrogens in the anticoplanar position? And now it turns out that I don't need to ask that question in this case. Do you remember why? Because I have not been given stereochemistry of the alkyl fluoride. So it says no chirality given. K? And since there's no chirality given, I don't really have to worry about if the hydrogen is in the anti position or not because I don't even know what position the fluoride is in. Alright? So it turns out that that last question I can ignore. I can say that I'm going to get both of these products. Now it's saying that I provide the major and minor products and the mechanism for this reaction, so I'm going to go ahead and draw the mechanism for what I think is going to be the major product and then I'll draw the other one as well.

The major product is going to go along in the less substituted direction, so it's probably going to be this H right here. Okay? So let's go ahead and draw our arrows. It's going to go basically terbutoxide looks like this. Okay? And I'm going to do the following. I'm going to grab the beta hydrogen, make a double bond, kick out the fluoride and what I'm going to get for that blue product is this. Okay? Cool. But now we also have another product that's possible. If it would have attacked the red position, then I would have gotten a double bond that looked like this. Okay? Now I just have to figure out which is going to be major, which one is going to be minor. So I look at how substituted each double bond is. This one is disubstituted. This one is trisubstituted. How did I know that? Actually, wow, okay, I messed up. This one is not disubstituted. This double bond only has one chain coming off of it, so it's actually only monosubstituted. Sorry about that. And then the red one is trisubstituted because it's got 3 different branches coming off of it. So one is way less substituted than the other. This is going to be my Hoffmann product and this is going to be my Zaitsev product. And when I'm using this base, which one do I prefer? I actually prefer the Hoffmann, so this is going to be my major. Okay? And then obviously that means that this is my minor. Does that make sense, guys? So really we haven't changed anything from the E2 mechanism, it's just that now we have a name for it. When you do an E2 with just an alkyl halide and a base, called dehydrohalogenation. Alright? Cool. So I hope that made sense. Let's go ahead and move on to the next topic.

Here’s what students ask on this topic:

What is dehydrohalogenation in organic chemistry?

Dehydrohalogenation is an E2 elimination mechanism in organic chemistry where a hydrogen atom and a halogen atom are removed from an alkyl halide. This process typically favors more substituted alkyl halides, such as secondary or tertiary ones. The reaction involves the elimination of a beta hydrogen in an anti-coplanar orientation, leading to the formation of a double bond. The choice of base can influence the product, following Zaitsev's or Hofmann's rules. Understanding the geometry and the role of the leaving group is crucial for predicting the reaction outcome.

How does the choice of base affect the product in dehydrohalogenation?

The choice of base in dehydrohalogenation significantly affects the product. A strong, bulky base tends to favor the formation of the less substituted alkene, following Hofmann's rule. In contrast, a smaller, more nucleophilic base usually leads to the more substituted alkene, adhering to Zaitsev's rule. This is because the steric hindrance of the bulky base makes it difficult to abstract the more hindered beta hydrogen, leading to the formation of the less substituted product.

What is the role of anti-coplanar orientation in dehydrohalogenation?

In dehydrohalogenation, the anti-coplanar orientation is crucial for the elimination process. This orientation ensures that the beta hydrogen and the leaving group (halogen) are positioned opposite each other in the same plane. This geometric arrangement allows for the most favorable overlap of orbitals, facilitating the formation of the double bond. Without this anti-coplanar arrangement, the reaction would be less efficient or might not occur at all.

What is the difference between Zaitsev's and Hofmann's rules in dehydrohalogenation?

Zaitsev's and Hofmann's rules predict the major product in dehydrohalogenation reactions. Zaitsev's rule states that the more substituted alkene will be the major product when a smaller, more nucleophilic base is used. This is because the more substituted alkene is generally more stable. Hofmann's rule, on the other hand, predicts that the less substituted alkene will be the major product when a bulky base is used. The steric hindrance of the bulky base makes it difficult to abstract the more hindered beta hydrogen, leading to the formation of the less substituted product.

Why are secondary and tertiary alkyl halides preferred in dehydrohalogenation?

Secondary and tertiary alkyl halides are preferred in dehydrohalogenation because they are more likely to undergo elimination rather than substitution. These more substituted alkyl halides stabilize the transition state better during the E2 elimination process. Additionally, the resulting alkenes from secondary and tertiary alkyl halides are generally more stable due to hyperconjugation and alkyl group electron-donating effects, making the elimination process more favorable.