Elimination Reactions With Rearrangments

One last (weird) reaction to show you with respect to elimination reactions. Can you see what’s weird about it?

How did that double bond get over?there??Normally when elimination occurs, we remove a hydrogen from the carbon adjacent to the leaving group. But here, something extra has taken place.

Let’s look at all the bonds that form and the bonds that break so we can track down?exactly what happens:

Notice how it differs from a typical elimination reaction? Sure, we’re forming C-C (π), and breaking C-H and C-OH, but we have an extra C-H that forms and an extra C-H that breaks.

This is a sure sign of a rearrangement step!

So what’s going on here?

Well, we start by protonating the alcohol. This allows for water to leave in the next step, which is going to form a carbocation.?Here’s the thing:?the carbocation is?secondary, and we’re adjacent to a?tertiary carbon. So if the hydrogen (and its pair of electrons) were to migrate from C3 in our example to C-2, we’d now have a tertiary carbocation, which is?more stable. Then, a base (water in this example) could remove C-H, forming the more substituted alkene (the Zaitsev product in this case). And that’s how the alkene ends up there.

OK. So that’s one mystery solved.

You might remember that these types of rearrangements can occur in SN1 reactions too. And if you read that post, you might recall that in addition to shifts of hydrogen (“hydride”, because there’s a pair of electrons attached) we can also have alkyl shifts. Here’s a final example. Note – I’ve also made a video of this, you can watch it here.

This pretty much does it for elimination reactions.

In the next series of posts, let’s go though one of the biggest questions students struggle with. Okay, now that we’ve gone through substitution and elimination reactions, HOW DO WE DECIDE WHICH ONE IS GOING TO OCCUR IN EACH SITUATION?

Great question. That’s next.

Next Series, post 1: SN1/SN2/E1/E2 Decision (1) – The Substrate

Tagged as: alkyl shift, e1, eliminations, hydride shift, rearrangement, sn1

View the original article here

Comparing the E1 and SN1 Reactions

Imagine you’re starting with the alcohol on the left and you’d like to get to the alkene on the right.

What bonds are formed and broken here? We’re forming C-C (π), we’re breaking C-H, and we’re breaking C-OH. It’s an elimination reaction.

Notice a problem here? We need to have HO(-) as a leaving group. If you’ll recall, strong bases [like HO(-) ] are terrible leaving groups – which makes the E1 pathway unlikely. So what if we tried to use a strong base, maybe trying to promote an E2 reaction? Well, that would be even worse – we’d likely deprotonate OH before the C-H, and you can imagine that we’d have to have O(2-) as a leaving group here. Not good!

That means that the reaction, as written, is very unlikely to happen.

Yet, there is something very simple that we can do to make this reaction work. We’d need to have a better leaving group (a weaker base). How can we do this?

Add acid!

If we add a strong acid, we turn OH into H2O+, the conjugate acid is a better leaving group. Now, water can leave, forming a carbocation; and then a base can break the C-H bond, forming the alkene.
Notice that this is now a classic E1 reaction. The rate is going to be dependent on the stability of the carbocation. This one is tertiary, so it should proceed at a reasonably high rate.

A question arises here. What’s going to act as the base? As it stands, a C-H bond adjacent to a carbocation has an extremely high acidity (at least below -2, if you follow pKa). That means that just about any weak base (water, or the conjugate base of the acid) is sufficient to deprotonate the carbon. It’s possible that more than one species can act as a base here. I’ve shown water removing the proton, but it’s not unreasonable to show the conjugate base of the acid removing the proton in most circumstances.

Now comes one of the things about organic chemistry that often causes trouble for students. For one of the first times in our discussions here, we’re dealing with a situation where we can have?competing reactions.

Let’s back up. The E1 reaction goes through a carbocation, correct? Well, if you’ll recall, so does the SN1 reaction. We’ve already seen examples where a carbocation was formed from an alcohol by adding a strong acid like HCl, HBr, or HI, and we ended up with the alkyl halide. Why???The halide ions are decent nucleophiles under the reaction conditions.

So how can we stack the deck in favor of the E1 process? Use a strong acid with a conjugate base that is a poor nucleophile.?The usual choice is H2SO4. The HSO4(-) ion is a relatively poor nucleophile due to the negative charge of the oxygen being distributed throughout the molecule (resonance). Two other acids you might see for this purpose are p-toluenesulfonic acid (p-TsOH), which is essentially a cousin of H2SO4, and phosphoric acid (H3PO4).

Also, don’t forget that elimination reactions are favored by heat.

In summary, if you’d like E1 to predominate over SN1: choose an acid with a weakly nucleophilic counterion [H2SO4, TsOH, or H3PO4], and heat.

If you’d like SN1 to predominate over E1, choose an acid like HCl, HBr, or HI.

We’re almost done talking about elimination reactions. Next post – we’ll talk about rearrangements.

Next Post: The E1 Reaction With Rearrangements

Tagged as: base, conjugate acid, e1, e2, elimination, leaving groups, nucleophile

View the original article here

THE HYDROGENATION OF ALKENES

THE HYDROGENATION OF ALKENES

This page looks at the reaction of the carbon-carbon double bond in alkenes with hydrogen in the presence of a metal catalyst. This is called hydrogenation. It includes the manufacture of margarine from animal or vegetable fats and oils.