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Neutral Diborene Is A First

Unexpected compound's boron-boron double bond confirmed by calculations

by Ron Dagani
October 1, 2007 | A version of this story appeared in Volume 85, Issue 40

A team led by Gregory H. Robinson was trying to make a compound with a boron-boron triple bond using solution-phase chemistry. That didn't quite work out. But they got something else that will shed some light on multiple bonding: the first stable, neutral compound containing a boron-boron double bond (J. Am. Chem. Soc., DOI: 10.1021/ja075932i).

To make this "diborene," Robinson, a chemistry professor at the University of Georgia, Athens, and his coworkers synthesized a new starting material: RBBr3, in which R is a bulky N-heterocyclic carbene ligand. The chemists treated this compound with a reducing agent (potassium graphite, KC8) in diethyl ether, hoping to make RB=BR. But instead, the potassium liberated hydrogens from the solvent, and the researchers isolated two crystalline hydrogen-containing products: the diborene RHB=BHR and the diborane RH2B–BH2R.

Although dianions containing a B=B bond have been reported by other groups, no one previously had been able to get two boron atoms to form a double bond in a stable, neutral molecule.

It's "a major development in the chemistry of the main-group elements," comments inorganic chemist Alan H. Cowley of the University of Texas, Austin. Chemist Jerry L. Atwood of the University of Missouri, Columbia, agrees, noting that "this is the type of fundamental discovery that will find its way very quickly into textbooks."

The neutral diborene's B=B bond distance is 1.560, considerably shorter than the single boron-boron bond in the corresponding diborane RH2B, BH2R and also shorter than the B=B distance in some dianions that purportedly contain a strong B=B π bond, the Georgia researchers note.

The doubly bonded nature of the diborene's B=B bond is supported by theoretical studies carried out by Robinson's chemistry department colleagues R. Bruce King, Henry F. Schaefer III, and Paul v. R. Schleyer.

The carbene ligands are key to the diborene's B=B bond, Robinson points out. The divalent carbon atom of each carbene donates its two free electrons to form a carbon-boron bond, allowing the boron's three valence electrons to form a bond to hydrogen and a σ and a π bond to the other boron.

As far as making a stable boron-boron triple bond is concerned, that's still on Robinson's "to do" list. After all, the only molecule known with a bond of this type had to be prepared at cryogenic temperatures in a frozen argon matrix.


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