Reducing Agents

[MedPR]

I remember reading somewhere in TBR that there are specific (somewhat) uses for the following reducing agents. The point of the explanation was to help you distinguish between the two pairs of reducing agents if they were both answer choices. For instance, LiAlH4 is more for reducing esters, carboxylic acids, and amides. While NaBH4 is more for reducing ketones and aldehydes.

I'm sure there are more ways/tricks to differentiate these on the MCAT. Help?

NADH
FADH2

LiAlH4
NaBH4

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[chiddler]

My professor use to compare LiAlH4 to an explosive bomb compared to NaBH4 in strength. He says is very strong and annoying to use so avoid if possible.

Given that FADH2 gives electrons in lower energy levels during ETC, i'd expect it to be a weaker reducing agent just like NaBH4 is to LiAlH4.

 

[MedPR]

My professor use to compare LiAlH4 to an explosive bomb compared to NaBH4 in strength. He says is very strong and annoying to use so avoid if possible.

Given that FADH2 gives electrons in lower energy levels during ETC, i'd expect it to be a weaker reducing agent just like NaBH4 is to LiAlH4.

Hmm, can you explain a little bit more about why/how we kow that LiAlH4 is a more powerful reducing agent than NaBH4? Why are esters/carboxylic acids/amides more difficult to reduce than ketones/aldehyes?

 

[SaintJude]

Have you seen NADH & FADH2 discussed outside of cellular respiration? As laboratory reactants for instance ?!

 

[chiddler]

Hmm, can you explain a little bit more about why/how we kow that LiAlH4 is a more powerful reducing agent than NaBH4? Why are esters/carboxylic acids/amides more difficult to reduce than ketones/aldehyes?

Notice that aluminum and boron are in the same column. Aluminum is below so it is of course less electronegative. 4 H's bind the aluminum in a polar covalent bond. Since Al is less electronegative, the bonds are more strongly polar towards the hydrogens. So when this molecule reacts, it offers a hydrogen anion, right? It is more reactive than boron because the aluminum-hydrogen bonds are weaker and more polar than the boron-hydrogen bonds.

Also, wiki says that "LiAlH4 is more reactive than NaAlH4 because the smaller lithium cation is a better Lewis acid." Not sure what this implies...do you?

I think carboxylic acids and its derivatives are more difficult to reduce because first you need to break a carbon-oxygen bond then replace it with a hydrogen, then reduce the carbonyl carbon. More energetic, especially replacing the non-carbonyl C-O bond or C-N in case of amide.

On the other hand, reduction of a carbonyl is adding an anion to the already polar C=O bond so the reaction is less energetic.

 

[chiddler]

Have you seen NADH & FADH2 discussed outside of cellular respiration? As laboratory reactants for instance ?!

ummm i think so? vaguely remember a passage....maybe...

 

[MedPR]

Have you seen NADH & FADH2 discussed outside of cellular respiration? As laboratory reactants for instance ?!

Probably, but didn't think much of it. One of those types of questions where you just need to know that NADH is a reducing agent, and when you see NADH->NAD+, you know that something got reduced.

Notice that aluminum and boron are in the same column. Aluminum is below so it is of course less electronegative. 4 H's bind the aluminum in a polar covalent bond. Since Al is less electronegative, the bonds are more strongly polar towards the hydrogens. So when this molecule reacts, it offers a hydrogen anion, right? It is more reactive than boron because the aluminum-hydrogen bonds are weaker and more polar than the boron-hydrogen bonds.

Also, wiki says that "LiAlH4 is more reactive than NaAlH4 because the smaller lithium cation is a better Lewis acid." Not sure what this implies...do you?

I think carboxylic acids and its derivatives are more difficult to reduce because first you need to break a carbon-oxygen bond then replace it with a hydrogen, then reduce the carbonyl carbon. More energetic, especially replacing the non-carbonyl C-O bond or C-N in case of amide.

On the other hand, reduction of a carbonyl is adding an anion to the already polar C=O bond so the reaction is less energetic.

Not sure about the Lithium being smaller thing, but I guess it is implying that Lithium cation is less stable than Sodium cation because Sodium has a larger electron cloud so the electrons are more delocalized? Smaller Lithium also means that the valence electrons are closer to the nucleus, so there is more of an attraction = higher ionization energy = weaker lewis base = stronger lewis acid?

Thanks for the info about LiAlH4 vs NaBH4. With that in mind, is there a reason not to alawys use LiAlH4? If it's stronger, you shouldn't have a problem getting it to react in the lab. So what reasons are there for using NaBH4? If NaBH4 reduces ketones to secondary alcohols, would LiAlH4 reduce it to an alkane? An alkene?

 

[chiddler]

If it's a compound with a single functional group, then I don't think it makes a difference.

But suppose you have a carboxylic acid and a ketone on the same molecule. Now you can see how NaBH4 can be useful.

The most they can reduce is alcohol. I'm not sure why, nor do I know why NaH is never used for this purpose but it inspired me to ask the pro's earlier today. No compelling answers yet, though.

"I think part of the answer is that sodium hydride is a much better base than it is a reducing agent, so molecules tend to condense."

maybe.

 

[MedPR]

If it's a compound with a single functional group, then I don't think it makes a difference.

But suppose you have a carboxylic acid and a ketone on the same molecule. Now you can see how NaBH4 can be useful.

The most they can reduce is alcohol. I'm not sure why, nor do I know why NaH is never used for this purpose but it inspired me to ask the pro's earlier today. No compelling answers yet, though.

"I think part of the answer is that sodium hydride is a much better base than it is a reducing agent, so molecules tend to condense."

maybe.

Ah, I see. That example about multiple functional groups makes sense, thank you.

There are some replies about the s orbital not being able to overlap. I guess that makes sense too, in regard to NaH.