Zaitsev and Hofmann Elimination Products - Legendas Bilíngues para YouTube

Zaitsev and Hofmann Elimination Products - Legendas Bilíngues

Hey guys, Professor Dave here.

Let's talk a little bit about Zeitzif and Hoffman elimination products.

Sometimes, in an elimination reaction, there can be more than one elimination product depending on which beta proton is extracted by the base.

So let's take a look at this example here.

There are two different kinds of protons that can be extracted that will yield two different products.

So, for example, let's see if a base takes this proton here and then the electrons there

go to form the double bond,

kick off the bromine,

we're going to get the Okay,

that is a sp2 trigonal planar,

so this methyl group flattens out,

or we can extract a proton over on this side, form the double bond here, leaving this side of the molecule untouched.

So, we would get that, this would remain sp3, that would remain a wedge bond for the methyl group.

So, these are different molecules, they are different products, and both are possible, and most likely we're going to get a mixture.

So, we want to be able to discuss these two products.

And, as it happens, happens, we have these terms zitzif and Hofmann to refer to which one is the more highly substituted alkene product.

So this alkene is tri-substituted.

Now the way we want to understand that is there are two carbons per in that carbon-carbon double bond,

and we want to be able to count the number of alkyl groups projecting from the carbons, participating in the double bond.

So this carbon has one alkyl group, right?

It's connected to one carbon atom over here and then there's the implied hydrogen.

This carbon over here has two alkyl groups, has that methyl group and the rest of the ring over here.

So that's three alkyl groups, so this is tri-substituted.

Whereas over here, So, this has this side of the ring and the implied hydrogen.

This carbon has this side of the ring and the implied hydrogen, so that's two alkyl groups.

So, it is disubstituted.

So, you can have an unsubstituted, a-substituted, disubstituted, tri The maximum would be tetra-substituted.

And, and we are going to label the mono-substituted.

substituted alkene as the Zaitsef product.

The less substituted will be Hoffman.

So it is not the case that any tri-substituted alkene elimination product would be labeled Zaitsef.

It is simply that the more substituted of the two is a Zaitsef product.

So it's certainly a relative to Zaitsef.

term because a tri-substituted alkene could be the Hoffman product if the zites of product were tetra-substituted or more substituted.

We need to understand which product will dominate and when.

So in order to understand that we need to know something about alkene stability.

And it is the case that the more substituted an alkene is, the more stable it is.

Now this is similar for reasons of carbocation stability with hyperconjugation.

It's kind of like that.

So a tetra-substituted alkene is going to be the most stable alkene, then tri-substituted, then a e-di-substituted alkene.

Z disubstituted and then the mono-substituted and then the unsubstituted will be the least stable alkene.

That means that the Zaitsef product is more stable and therefore thermodynamically favored.

It is the thermodynamically favored product and the kinetically favored product will be the Hoffman product.

So it is less stable, but it could happen more frequently if the base is very sterically hindered.

So let's take a look at two possible bases.

If it was high...

that was acting as the base.

We would probably expect more of the zits of product than the Hoffman product.

This is because the base is very sterically unhindered.

It has relatively equal access to any proton on the molecule,

so it's not going to have a very hard time getting to that proton and because it can approach any portion of the molecule with equal

ease, then it will have a tendency to the lowest energy product, the most thermodynamically stable product.

So that would be the zits of product.

However, let's say we're looking at something a little bit more sterically hindered.

Let's say we're looking at ethoxide.

Now, That's not quite as sterically hindered as, say, turt butoxide, but that's still a fair amount of steric hindrance because we have an ethyl group and there's protons, you know, there's five different protons there.

So this is not going to have as easy a time approaching every proton in the molecule.

There is going to be a lower activation energy associated with approaching that hydrogen.

So if we were looking at an energy We could say that there was a lower activation energy.

approaching one of these protons as opposed to a higher activation energy approaching that proton.

And so this is the kinetically favored product because more collisions on this side of the molecule will result in a reaction because

more collisions on this side will be able to surpass the smaller activation energy.

So that's why the Hoffman product is kinetically favored and will be

favored in the product mixture if the base is quite sterically hindered whereas the

zites of product and will probably be favored in the product mixture if the base is sterically unhindered.

So let's look at a couple of examples to see what's going on.

on here.

So let's look at this substrate.

We're reacting with methoxide so just to go back to the mechanism stuff we know that this is a tertiary substrate so we cannot do SN2.

This is a strong base that is relatively unhindered so we could do SN2 or E2 however we just said SN2 is impossible

so we know we're going to do E2.

We know we're going to do an elimination here.

However there are multiple protons that could be extracted that would yield different elimination products.

So let's draw some of the implied hydrogens in.

Let's say if we got this one grab that double bond there lose the bromine we're going to get this product right the double bond will be here

leaving group leaves so that's what we would get.

However we could also extract a proton from this methyl group up here.

So if we did that We would get this product here.

So What is the degree of substitution of these products?

Well here the left carbon has two alkyl groups,

the one on the right has one and the implied hydrogen, so that is a tri-substituted alkene.

Over here this carbon has two alkyl groups, the two sides of the ring and this has none, those are the two implied hydrogens.

So that is a dis-substituted alkene.

Tri-substituted versus dis-substituted means this will be the zites of product and this will be the Hoffman product product.

And, in this case, the Zaitsef is going to be favored because it is more thermodynamically favorable and it's a relatively unhindered base.

So it's going to have roughly equal access to this secondary pro proton as it would to that primary proton.

So in this case, the Zaitsef product is going to dominate.

We would still get some Hoffman.

at the majority to be zeitze.

Now, if the base was significantly, sterically hindered, such as, say, terbutoxide, then we might expect the Hoffman product to be able to dominate,

even though it is less thermodynamically favorable, because kinetics would then drive that reaction.

A would be able to access this proton with much more ease because there would be a lower activation energy

associated with that acid-base reaction.

Now let's look at something like this here.

Let's say we're using methoxide again, now the thing with the cyclic systems is we're going to have to draw the chair.

So I went ahead and drew that.

So here's the chair, there's that isopropyl group extending upwards, there is the chloro group there, there is the methyl group going down there.

So now the reason we had to draw the chair is because we have to be able to see which protons are available for elimination.

Now we know that we're going to have to extract a proton from a carbon adjacent to the carbon that bears the leaving group.

So that means a proton from here or a proton from here.

But as we learned before,

we have to have that proton that's eliminating anti to And on this carbon it's the isopropyl group that's anti to the leaving group.

So so this proton is not available for elimination That's anti to the rest of the ring.

So this one's no good.

We're not going to get elimination there over here We do have two protons.

So even though this one's not going to work out this one will do just fine because if we drew the Newman projection we would see that because they're

both axial, those are anti to one another and that's where the elimination is going to occur.

So let's see that.

So, in this case, we're only going to get one product, and we don't have to

analyze CITSA versus Hoffman, whereas you might think that you would have if you just went based on the top-down view.

It would be very possible to draw in this implied hydrogen there.

Assume that that was available for abate elimination just as one of the ones on this carpet But when we draw the chair,

we see it is not the case We do not get a mixture of Zaitzev and Hoffman products.

We get only one product and that is the dye substituted alkene there Thanks for watching guys subscribe to my channel for more tutorials and as always feel free to email me of questions

Professor Dave explains at gmail.com You

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