The elimination reaction is exactly what it sounds like. Use a base to take away (de-) one hydrogen and one halogen. Voila! We’ve got a double bond.
Concept: The dehydrohalogenation mechanism.4m
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 it's an E2 mechanism. Because if you think abut the name it's saying, dehydro, we're taking away one hydrogen and we're talking away one halogen. That's exactly what happens with a typical E2 mechanism. Remember that we always break those two 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 the case do we prefer that an alkyl halide to be primary or tertiary? What do you think is better? Well, we just said this is an E2 reaction so that means we're going to prefer the more substituted alkyl halide that's going to favor elimination more. That means that hopefully we have a secondary or a 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 because we had Zaitsev and we had Hoffman and the type of base that you use could prefer one product over another.
But let's just go ahead and 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'd 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 anti-coplanar position. Remember that that's important because if you were to make a Newman projection out of this guy, you'd want to make sure that your groups were facing opposite directions or in the anti position so that they can be in the most favorable orientation to eliminate.
So I would take that. But remember that elimination always has three 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 two methyl groups here are now located here. And these two methyl groups here, are now located there. Plus, I would get obviously my base with the new hydrogen on it, so that would be a conjugate acid and I would also get the leaving group X-.
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 anti-coplanar, in terms of Zaitsev and Hofmann, all of that is fair game.
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 answers. So anyway, go for it.
Even more simply put, this is simply the name given to an E2 mechanism with a base an alkyl halide.
Example: Supply the mechanism and major/minor products for the following dehydrohalogenation reaction.5m
Predict the product for the reaction below.
Identify possible product(s) of dehydrohalogenation of trans–1–bromo–2–methylcyclohexane.
What would be the major product of dehydrohalogenation of 2-chloropentane by potassium hydroxide?
Identify possible product(s) of dehydrohalogenation of cis–1–bromo–2–methylcyclohexane.
What would be the major product of the dehydrohalogenation of 3-chloropentane by KOH?
What would be the major product of the dehydrohalogenation of 2-chloropentane by KOH?