Question

Frustrated Lewis Pairs easily hydrogenate imines, ketones with more difficulty  but are bad at hydrogenating alkenes. Provide an explanation for this.

Answer 1

Frustrated Lewis pair react with hydrogen to form a salt or adduct. Hydrogenation takes place by means of protonation and hydride transfer to the unsaturated functional group that is being reduced.

The basic or nucleophilic atom of the unsaturated functional group being reduced is first protonated. The carbon bonded to the protonated atom becomes much more electrophilic and is then attacked by hydride. This implies that rate of reaction is higher when the unsaturated functional group contains more basic atom.

R₂C=X + H⁺ --------> [R₂C=XH]⁺ + H^- -------> R₂HC-XH    .........................(1)

(representative scheme; not actual meachanism)

When X is R¹N, R₂C=X is R₂C=NR₁, an imine.

When X is O, R₂C = X is R₂C=O, a ketone.

When X is CR₂, R₂C = X is R₂C=CR₂, an alkene.

H⁺ and H^- are supplied by salt formed upon reaction of frustrated lewis pair with hydrogen.

[Example: PCy₃ and B(C₆F₅)₃ are a frustrated lewis pair that can split hydrogen and form salt.

PCy₃ + B(C₆F₅)₃ + H₂ -------------> [HPCy₃]⁺[HB((C₆F₅)₃]^-

During hydrogenation, proton transfer occurs from [HPCy₃]⁺ and hydride transfer occurs from [HB((C₆F₅)₃]^- ​.]

From representative scheme (1), it is clear that, the functional group which contains more basic atom, is protonated faster and hence is reduced faster. We know that nitrogen is more basic than oxygen which is much more basic than carbon, in neutral molecules. Hence, imine is protonated (at nitrogen) more readily than ketone (at oxygen) which is protonated much more readily than alkene (at one of the doubly bonded carbons).

Therefore, frustrated lewis pairs hydrogenate imines easily, ketones with more difficulty but are bad at hydrogenating alkenes.