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Heterocyclic Chemistry I

Introduction

In 1965 an undergraduate student by the name of Cooper at Bucknell University was working on an independent project investigating the chemistry of aziridines, 3-membered rings in which one of the atoms is a nitrogen. Late one evening Mr. Cooper completed the preparation of a sample of 2-benzoyl-3-(4-nitropheny)aziridine, the starting material he needed for his project. Rumor has it that while cleaning some glassware, an errant spray of acetone from Mr. Cooper's squeeze bottle was misinterpreted by a nearby co-worker as an act of aggression. In the battle that ensued some acetone contaminated Mr. Cooper's reaction. As serendipity would have it, and as later investigation revealed, the accidental contamination led to the formation of a compound called 2,2-dimethyl-4-phenyl-6-(4-nitrophenyl)-1,3-diazabicyclo[3.1.0]hex-3-ene. Since no one wanted to be saying 2,2-dimethyl-4-phenyl-6-(4-nitrophenyl)-1,3-diazabicyclo[3.1.0]hex-3-ene all the time, and since the compound was bright blue, it quickly became known as Cooper Blue:

Synthesis

The synthesis of Cooper Blue provides an excellent example of the application of many of the topics we have considered

The first step in the synthesis involves the crossed aldol condensation of acetophenone with 4-nitrobenzaldehyde as shown in Equation 1 where Ar represents the 4-nitrophenyl ring. The product of reaction 1 is a trans-chalcone.

Four points about reaction 1 deserve comment.

  1. Since 4-nitrobenzaldehyde does not have any a hydrogens, it can act only as an electrophilic component in the crossed aldol reaction.
  2. As an electrophile, 4-nitrobenzaldehyde is more reactive than acetophenone. Therefore the crossed aldol condensation occurs faster than the self-condensation of acetophenone.
  3. Dehydration of the initial addition product occurs spontaneously under the reaction conditions.
  4. The E stereoisomer of the chalcone is formed preferentially. It is the thermodynamically preferred isomer.

Exercise 1 Write an equation for the self-condensation of acetophenone. Draw the structure of the b-hydroxyketone and the a,b-unsaturated ketone.
Equation 2 outlines the electrophilic addition of dibromine to the double bond of the trans-chalcone. Presumably this reaction involves the formation of a bromonium ion intermediate, in which case the stereochemistry of the addition is anti.

Exercise 2 Draw the structure of the bromonium ion intermediate in reaction 2. Note that the product of reaction 2 contains two chiral centers. Would this material be optically active? Yes No
Treatment of the dibromide formed in reaction 2 with a saturated solution of ammonia in ethanol initiates a series of three reactions which ultimately lead to the formation of an aziridine. These reactions are shown in Equations 3, 4, and 5.

The initial reaction between the chalcone dibromide and ammonia is a 1,2-elimination of HBr. Ammonia acts as a base, deprotonating the carbon a to the carbonyl group. The mechanism of the reaction is most likely E2.

Exercise 3 Why does ammonia deprotonate the carbon a to the carbonyl group rather than the one b to it?

Exercise 4 If reaction 3 proceeds by an E2 mechanism, what is the dihedral angle between the H and the Br in the reactive conformation?

The product of reaction 3 undergoes further reaction with ammonia:

Reaction 4 is an example of a Michael addition. Conjugation of the double bond to the carbonyl group reduces the electron density at the b carbon, making it susceptible to nucleophilic attack by a molecule of ammonia. The net result is that the components of ammonia, H and NH2, add across the double bond.

Exercise 5 Use curved arrows to show the resonance interaction between the pi electrons of the double bond and those of the carbonyl group. Draw the structure of a resonance contributor that indicates clearly the electron deficient character of the b carbon.
The 3-membered aziridine ring is then formed by an intramolecular nucleophilic aliphatic substitution reaction:

In Equation 5 the reactant is shown in that conformation where the dihedral angle between the nucleophile and the leaving group is 180o. This geometry allows for the lone pair of electrons on the nitrogen atom to approach the a carbon from the back side of the C-Br bond. Recall that this is the orientation required for the Sn2 mechanism.

The product of reaction 5 is the compound that Mr. Cooper needed as the starting material for his research project. The work-up of the reaction normally involves cooling the solution overnight and filtering the crystals that precipitate. However, if acetone is added to the solution before the work-up, two additional reactions occur. Each of them involves nucleophilic addition to a carbonyl group. The order in which these reactions occur is not certain. In Equation 6 the addition of NH3 to the carbonyl group is arbitrarily shown first.

Exercise 6 Devise a mechanism depicting all the steps in the conversion of the keto group to an imine in reaction 6. Use curved arrows to show the movement of electrons. Your mechanism should clearly account for the formation of the water as well as the aziridine.
In the last step of the sequence the two nitrogen atoms in the product of reaction 6 bond to the carbonyl carbon of a molecule of acetone. This reaction, Equation 7, is analogous to the formation of a ketal.

Exercise 7 Devise a mechanism depicting the formation of Cooper Blue. Use curved arrows to show the movement of electrons.
Cooper Blue is a photochromic compound, which is to say that it is white when stored in the dark, but turns blue on exposure to light. When Mr. Cooper returned to the laboratory the next morning, he knew immediately that something had gone awry. When he removed his sample from the refigerator the white crystals turned blue!

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