The Cope Elimination
Last updated: December 16th, 2020 |
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
- (Advanced) References and Further Reading
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.
(Advanced) References and Further Reading
The Cope Elimination is a thermal syn elimination; they are thermally activated, unimolecular, require no acid/base catalysis, and because the intramolecular hydrogen transfer is concerted with the formation of the C=C double bond, this reaction proceed via a cyclic transition state. The cyclic transition state also dictates that the elimination occur with syn stereochemistry.
- Zur Einwirkung von Wasserstoffsuperoxyd auf Fettamine
Mamlock R. Wolffenstein
Chem. Ber. 1900, 33 (1), 159-161
The thermolysis of a tertiary amine N-oxide to give an olefin and an N,N-disubstituted hydroxylamine was first described by Mamlock and Wolffenstein in this paper at the turn of the century, but the reaction remained unused until Cope and co-workers, starting in 1947, explored the scope and mechanism of this alternative to the Hofmann elimination reaction for the conversion of tertiary amines into olefins.
- METHYLENECYCLOHEXANE AND N,N-DIMETHYLHYDROXYLAMINE HYDROCHLORIDE
Arthur C. Cope and Engelbert Ciganek
Org. Synth. 1959, 39, 40
An Organic Syntheses prep by Prof. Cope himself demonstrating the utility in this reaction in forming exocyclic double bonds, as an alternative to Wittig chemistry. Additionally, as Prof. Cope mentions in the notes, “This route from amines to olefins in many cases yields pure olefins where the alternative method, the Hofmann exhaustive methylation reaction, is accompanied by some rearrangement to more stable isomeric olefins.”
- Amine Oxides. VIII. Medium-sized Cyclic Olefins from Amine Oxides and Quaternary Ammonium Hydroxides
Arthur C. Cope, Engelbert Ciganek, Charles F. Howell, and Edward E. Schweizer
Journal of the American Chemical Society 1960, 82 (17), 4663-4669
This paper has illustrations of the cyclic transition state of this reaction, which is what the authors propose as the mechanism to rationalize the observed syn
- Room Temperature Wolff-Kishner Reduction and Cope Elimination Reactions
Donald J. Cram, Melville R. V. Sahyun, and Graham R. Knox
Journal of the American Chemical Society 1962, 84 (9), 1734-1735
This paper by Prof. D. J. Cram (UCLA), who later received the Nobel Prize in Chemistry for his contributions to supramolecular chemistry, shows that dry THF or DMSO can be used as solvents for the Cope Elimination. In these solvents, elimination occurs at room temperature!
- Mechanism of the Cope elimination
Robert D. Bach, Denis Andrzejewski, and Laurence R. Dusold
The Journal of Organic Chemistry 1973, 38 (9), 1742-1743
In the words of this paper, it “report[s] unequivocal deuterium-labeling evidence that establishes the mechanism of the Cope elimination as a 100% syn elimination”.
- Reverse Cope elimination reactions. 1. Mechanism and scope
Engelbert Ciganek, John M. Read Jr., and Joseph C. Calabrese
The Journal of Organic Chemistry 1995, 60 (18), 5795-5802
Reverse Cope Eliminations are also possible. This is the formation of a tertiary amine-N-oxide from an olefin and an N,N-disubstituted hydroxylamine.