Several students wanted to know why lithium aluminium hydride is such a stronger reducing agent than sodium borohydride. Rather than having to keep saying the answer I thought I would show you here the reasons behind the answer.
We need to consider a series of four reducing agents, if you want to know how to make dot and cross diagrams of the anions then please click here otherwise keep on reading.
LiAlH4, here is a picture of the anion and the cation.
Reduces esters, ketones, aldehydes, alkyl halides, epoxides and amides
LiBH4, here is a picture of the anion and cation.
Reduces esters, ketones and aldehydes
NaBH4, here is a picture of the anion and the cation.
Reduces ketones and aldehydes
NaB(CN)H3, here is a pciture of the anion and the cation.
Weaker version of NaBH4
So looking at this list we have a set of four reagents, in terms of strength the order is
LiAlH4 > LiBH4 > NaBH4 > Na(BH3CN)
We now need to consider why the order exists.
1. The cation.
The lithium is better able to act as a lewis acid than the sodium, the lewis acid bonds to the carbonyl oxygen. This in turn increases the amount of positive charge density on the carbonyl carbon. As the reduction is favoured by an increase in the positive character of the carbonyl carbon a change from sodium to lithium will make the reducing agent better. For more details please read here.
2. The electronegativity of the atom at the centre of the EH4 anion.
The more electronegative the atom is the less electron density will be on the hydrides, the less electron density of the hydrides the less able they are to act as nucleophiles to reduce the carbonyl.
Pauling electronegativity of Al is 1.61
Pauling electronegativity of B is 2.04
The two factors combined to make LiAlH4 a stronger reducing agent than NaBH4.
If we were to make a LiGaH4 then it would be an even stronger reducing agent than LiAlH4, this is becuase the gallium would be less electronegative than the aluminium.
3. I would also like to point out that as we go down the groups in the p block the hydrides become less and less stable. This is because the hydrogen to whatever atom bond becomes weaker. For example phosphine (PH3) is more stable than arsine (AsH3) which in turn is more stable than SbH3. This general reduction in the strength of the bond to the hydrogen may also help to increase the reactivity of LiAlH4 when it is compared with LiBH4.
The weakening of the bonding to hydrogen is not an isolated example, the carbon whatever bonds also become weaker as you go down a group in the p block. For example consider the alkyls
CEt4, SiEt4, GeEt4, SnEt4 and PbEt4.
While CEt4 (3,3-diethylpentane) is a very stable saturated hydrocarbon (alkane) the lead compound (tetraethyl lead) is unstable when it is heated. It breaks down with ease to form lead atoms and ethyl radicals. It used to be used in motor fuel (petrol) as an additive increase the octane rating but for public health reasons it has been withdrawn from use in petrol. The other ethyl compounds will have stabilities which will be partway between the alkane and the tetraalkyl lead. I reason that as the central atom becomes more heavy the bonds between the whatever and the ethyl will become weaker.
I have noticed that this blog entry seems to have taken on a life of its own, what started with a student asking “why” has resulted in my most popular work on my organic chemistry teaching blog. If you like my explanation of why one reducing agent is stronger than the other then please spread the good news by giving people the URL to this blog post.
Now before I go I would like to point out that lots of fun chemistry other than the reduction of aldehydes and ketones can be done using sodium borohydride. For example the reaction of nickel chloride with sodium borohydride is a good way to make nanoparticles of nickel which are embedded in a boron oxide matrix. This makes one of my favourite hydrogenation cataylsts.
If you like this post maybe you should check out the “joy of organic chemistry” where I have put up some more stuff.