# How is sodium borohydride similar to lithium aluminum hydride as a reducing agent? Could they be used interchangeably?

May 16, 2017

The molar mass of $N a B {H}_{4}$ is $37.83 \cdot g \cdot m o {l}^{-} 1$

#### Explanation:

Whereas the molar mass of $L i A l {H}_{4}$ is $37.95 \cdot g \cdot m o {l}^{-} 1$.

Given that the molar masses are almost identical, equal masses of lithium salt and sodium salt would be required. Depending on the quality of the hydride salt, each reagent would be assumed to deliver at least 3 equiv of hydride ion.

"Moles of lithal"=(0.5*g)/(37.95*g*mol^-1)=??*mol.

"Moles of sodium borohydride"=(0.5*g)/(37.83*g*mol^-1)=??*mol.

Note that lithal is a FAR more fierce reducing agent than sodium borohydride. Organic chemists often use lithal to DELIVER the one equiv of hydride...and water workup becomes pretty hairy. I always used lithal to deliver THREE equiv of hydride...and the workups of reaction based on 1/2 molar quantities were moderate.