Reducing Agents in Organic Chemistry: LiAlH4 vs. NaBH4

When students start getting to the chapters on alcohols and carbonyl groups (aldehydes, ketones, etc.), they come across the important process of oxidation and reduction. Many of us remember traces of oxidation and reduction from general chemistry, tossing around electrons on balanced equations, LEO GER (or OIL RIG) to help us remember which is which.

In organic chemistry, oxidation is usually the addition of oxygen-carbon bonds and a removal of hydrogens, while reduction is the addition of hydrogens and removal of oxygen -carbon bonds. Reduction in organic chemistry is usually accomplished by a very strong base/nucleophile called a hydride. When most people imagine hydrogen they think either hydrogen gas (elemental hydrogen) or acidic solution (H+). However hydrogen can exist with a negative charge (H-) and when it does, it becomes a hydride.

In organic chemistry we normally learn about two important reducing reagents, sodium borohydride (NaBH4) and lithium aluminum hydride (LiAlH4 or LAH). We learn that NaBH4 is a “weak reducing agent” and can only take aldehydes and ketones to alcohols easily. NaBH4 can handle esters, but it is very slow at converting them and thus not preferable. We then learn that if you want to turn esters and carboxylic acids into alcohols, we better use LiAlH4 because it is a strong reducing agent. LiAlH4 can convert aldehydes, ketones, esters, and carboxylic acids all to alcohols in the blink of an eye, but why is LiAlH4 stronger then NaBH4? 

The answer has to do with electronegativity. When considering each compound you must see it in the Lewis form which is basically a metal cation and a hydride complex anion. This is pictured below:



The atom holding on to the hydrides should be examined here. In the case of NaBH4 we have boron with an electronegativity of around 2.0 where as with LiAlH4 we have aluminum with an electronegativity around 1.6. Remember, the more electronegative an atom is, the less likely it will be to share or give up electrons. A hydride (H-) is just that, extra electrons (one to be exact) on a hydrogen. So if the hydride is bound to boron, it is less likely to be released for organic transformation when compared to aluminum (because boron will hold on to electrons, and thus the hydride, better).

An argument can also be made that lithium will better stabilize the negative oxygen intermediate that results from the hydride attacking the carbonyl group.

So why use NaBH4 at all? Because it is stable compared to LiAlH4, it can be used in aqueous solutions and measured in open air where as LiAlH4 will react violently to moisture and water. Hope this helps to clarify the difference!



3 thoughts on “Reducing Agents in Organic Chemistry: LiAlH4 vs. NaBH4

    • In order to solve this you must know Avagadro’s number which states there is 6.022 x 10^23 atoms per mole so lets set up a conversion equation:

      1.1 X 10^22 atoms of silver x 1 mole of silver/6.022 x 10^23 atoms of silver = 0.0183 moles of silver.

      Hope this helps!

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