The Cope Elimination
Last updated: June 27th, 2019 |
The “Cope Elimination” is a reaction where an amine is oxidized to an intermediate called an “N-oxide”, which, when heated, acts as the base in an intramolecular elimination reaction.
Table of Contents
- The Cope Elimination, Part One: Formation Of The N-Oxide
- The Cope Elimination, Part Two: Heating Induces An Intramolecular Elimination Reaction
- The Cope Elimination vs. The “E2” Elimination: Stereochemistry
- The Cope Elimination: Mechanism
- Summary: The Cope Elimination
The Cope Elimination starts with formation of a species called an “N-oxide”. Treating a tertiary amine (which itself can be created through reductive amination, for instance) with hydrogen peroxide and base results in an “N-oxide”, which has a positive charge on nitrogen and a negative charge on oxygen:
In the first part of this process, the amine attacks hydrogen peroxide (or the OH of a peroxyacid, such as m-CPBA), forming N-OH and breaking the weak O-O bond. The amine oxide O-H bond is then deprotonated by the strong hydroxide base, resulting in formation of the N-oxide.
(Note that there are opposite charges on adjacent atoms in the N-oxide; it is a “zwitterion“. Also, the starting amine must be tertiary for this process – using a secondary or primary amine will lead down other reaction pathways that we need not concern ourselves with here)
The O– in N-oxides is a weak base (pKaH = 4.5 about as strong as pyridine). When heated sufficiently [e.g. 160°C], an elimination reaction occurs, leading to the formation of an alkene, along with a substituted hydroxylamine:
The Cope Elimination differs from the familiar E2 elimination mechanism in several important ways.
- The E2 always proceeds such that the two sigma bonds that break [the C-H bond and the C-LG (leaving group) bond] are oriented anti to each other (dihedral angle of 180°, like the hour-hand and minute-hand of a clock at 6:00)
- In the Cope elimination, the breaking C-H bond and the breaking C-N bond are oriented syn to each other (dihedral angle of 0°, like the hour-hand and minute-hand of a clock at midnight).
Secondly, the E2 is almost always [Note] an intermolecular process, where the rate depends both on the concentration of the substrate and the concentration of the base.
The Cope is an intramolecular elimination reaction, which is to say that the base (the N-oxide) and the acid (the C-H bond) are always contained within the same molecule.
Note how the proton marked in green remains in the Cope elimination (above) but is lost in the E2 (below):
So how does it work?
The Cope proceeds through a concerted syn-elimination mechanism. The oxygen from the N-oxide acts a base, forming an O-H bond, while the C-H and C-N bonds break to form the new C-C pi bond:
Here’s how it would look in the example we’ve been working with:
The Cope rearrangement is an intramolecular, syn-selective elimination reaction that converts a tertiary amine to an alkene.
Why might one use the Cope elimination instead of a traditional elimination like the E2 ? Well, sometimes it’s much easier to perform an elimination reaction when the base is literally next-door (i.e. an intramolecular process) as opposed to heating an alkyl halide with a strong base. This especially has an advantage in situations where many side reactions can occur or the base cannot readily access the desired C-H bond due to steric hindrance [Note] .
Being a concerted, six-electron process, the Cope is also a member of the family of sigmatropic reactions which includes the Diels-Alder, the Claisen rearrangement, the Cope rearrangement (different reaction, same Cope!) and many others.
What About Zaitsev’s Rule?
The Hofmann elimination of ammonium salts tends to give the less substituted alkene [see: The Hofmann Elimination ].
So what about the Cope elimination? It’s got a bulky leaving group too, after all. Does it undergo elimination to give the less substituted alkene, just like in the Hofmann?
My apologies, but I’m going to go all “Choose Your Own Adventure” on this.
Click on the answer you want to hear:
(In short: for reasons beyond the scope of our discussion there’s contradictory evidence, and for now, if you are led to believe that it undergoes Hofmann elimination, that’s probably OK unless you take advanced level studies in organic chemistry).
Note 1. “almost always”, because it’s possible to come up with examples of molecules that, once deprotonated, perform anti-selective elimination reactions that are formally an intramolecular process – a rare exception.
Note 3. An example of the Cope Elimination in action – this one required heating to 160°C.