Understanding Reducing Agents: LiAlH4, NaBH4, H2/Nickel

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SUMMARY

This discussion clarifies the differences in reactivity between lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4) as reducing agents. LiAlH4 effectively reduces nitro groups due to lithium's higher electropositivity compared to sodium, while NaBH4 does not, primarily because it operates under slightly basic conditions that stabilize the reducing agent. Additionally, hydrogen gas in the presence of Raney nickel can reduce C=C bonds, but LiAlH4 does not due to the lack of a catalyst and the nature of its reactivity. The discussion emphasizes that NaBH4 generally does not reduce carboxylic acids under standard conditions.

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  • Understanding of reducing agents in organic chemistry
  • Familiarity with lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4)
  • Knowledge of hydrogenation processes and catalysts like Raney nickel
  • Basic concepts of radical mechanisms in chemical reactions
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  • Research the mechanisms of reduction using lithium aluminum hydride (LiAlH4)
  • Explore the reactivity of sodium borohydride (NaBH4) in various solvents
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Chemistry students, organic chemists, and researchers interested in the mechanisms of reduction reactions and the application of reducing agents in synthetic pathways.

garytse86
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I have got some problems on reducing agents, can someone explain why:

1) LiAlH4 and NaBH4 both reduce carboxylic acids / ketones, but only LiAlH4 reduces a nitro group? Why doesn't NaBH4 reduce the Nitro group?

2) How come H2/Nickel reduces C=C bond but LiAlH4 doesnt?

3) Similarly, why doesn't H2 reduce a carboxylic acid / ketone?

Thanks
 
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Well, these are tough indeed, but let's try something; I hope they will be useful for your brainstorming...

Okay, the first one is probably about the improved reactivity of lithium over sodium. Lithium loves oxygen more than sodium does; this is because nitro groups can be reduced with lithium aluminum hydride; you know that slightly basic sodium borohydride in alcohol or water is used for reduction; lithium aluminium hydride is not. That may be why the different reactivities are.

Hydrogen gas in the presence of raney nickel as catalyst is very efficient, yet not very selective in reductions. However, alkali metal hydrides don't reduce C=C bond, since there is no catalyst, I mean, any species reactive to both C=C bond and hydride. However, it is not correct to say that alkali metal hydrides never reduce C=C bonds; but the yield and reaction time may not be practical, that is all.

Hydrogen cannot very easily leave the other hydrogen and cleave its bond to be free and bind to oxygen; however, alkali metal hydrides can do so, since the alkali metals used in these compounds are far more electropositive than hydrogen, so much more reactive than hydrogen to facilitate this kind of reduction.
 
ok, it sounds reasonable
so the difference in reducing power is due to electropositive nature of Li.

Is the reason LiAlH4 has to be in dry ether is because its much more reactive than sodium borohydride?
 
Lithium is a much better Lewis acid than sodium. This accounts for much of the reactivity difference between LiAlH4 and NaBH4. However, in the NaBH4 reduction the active reducing agent is probably an alkoxyborohydride, where one or more of the hydrides from the BH4- has reacted with the alcohol solvent. The alkoxyborohydride is less reactive then because of the electronegativity of the oxygen atoms. They serve to stabilize the negative charge on boron and in turn make the delivery of the hydride less favorable. Since LiAlH4 reductions are performed under aprotic conditions, the AlH4- is the active reducing species (at first, at least). Also note that in general NaBH4 will not reduce carboxylic acids, at least under standard conditions.

Reduction of C=C bonds with H2 and a metal catalyst is often thought to proceed through a radical mechanism. Such radical mechanisms are more accessible with transition metals such as Ni and Pd than with metals like aluminum.

H2 can be used to reduce carbonyls to alcohols, but it is much, much more difficult. You can think of this in terms of radicals too; if you first delivered a hydrogen atom (a proton with an electron) to the carbon of the carbonyl group you would form an oxygen radical, which is an unstable species because the octet isn't filled around an electronegative atom. Since the NaBH4 and LiAlH4 deliver a proton with two electrons, you never get an unfilled octet around oxygen.
 
Thanks movies, you are a great expert :approve:
 
chem_tr said:
Thanks movies, you are a great expert :approve:

Heh, thanks, but I think it's more that I have a lot of experience writing and grading these types of mechanisms!
 

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