Stoichiometry and Stereochemistry of Reductions with

Mechanism, Stoichiometry and Stereochemistry of Reductions with Hydrides 

The reaction of complex hydrides with carbonyl compounds can be exemplified by the reduction of an aldehyde with lithium aluminum hydride. The reduction is assumed to involve a hydride transfer from a nucleophile -tetrahydroaluminate ion onto the carbonyl carbon as a place of the lowest electron density. The alkoxide ion thus generated complexes the remaining aluminum hydride and forms an alkoxytrihydroaluminate ion. This intermediate reacts with a second molecule of the aldehyde and forms a dialkoxy-dihydroaluminate ion which reacts with the third molecule of the aldehyde and forms a trialkoxyhydroaluminate ion. Finally the fourth molecule of the aldehyde converts the aluminate to the ultimate stage of tetraalkoxyaluminate ion that on contact with water liberates four molecules of an alcohol, aluminum hydroxide and lithium hydroxide. Four molecules of water are needed to hydrolyze the tetraalkoxyaluminate. The individual intermediates really exist and can also be prepared by a reaction of lithium aluminum hydride 

From the equation showing the mechanism it is evident that 1 mol of lithium aluminum hydride can reduce as many as four molecules of a carbonyl compound, aldehyde or ketone. The stoichiometric equivalent of lithium aluminum hydride is therefore one fourth of its molecule, i.e. 9.5 g/mol, as much as 2 g or 22.4 liters of hydrogen. Decomposition of 1 mol of lithium aluminum hydride with water generates four molecules of hydrogen, four hydrogens from the hydride and four from water. 

The stoichiometry determines the ratios of lithium aluminum hydride to other compounds to be reduced. Esters or tertiary amides treated with one hydride equivalent (one fourth of a molecule) of lithium aluminum hydride are reduced to the stage of aldehydes (or their nitrogen analogs). In order to reduce an ester to the corresponding alcohol, two hydride equivalents, i.e. 0.5 mol of lithium aluminum hydride, is needed since, after the reduction of the carbonyl, hydrogenolysis requires one more hydride equivalent. 

Free acids require still an additional hydride equivalent because their acidic hydrogens combine with one hydride ion of lithium aluminum hydride forming acyloxy trihydroaluminate ion. Complete reduction of free carboxylic acids to alcohols requires 0.75 mol of lithium aluminum hydride. The same amount is needed for reduction of monosubstituted amides to secondary amines. Unsubstituted amides require one full mole of lithium aluminum hydride since one half reacts with two acidic hydrogens while the second half achieves the reduction. 

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