[Ben Elgar]

The reducing agent LiAlH4 is never mentioned in the text book but is accepted nearly all the time in the mark scheme. I normally pick that as I have real difficulty remembering all the alternative reducing agents. I just wanted to ask when should I NOT use it. I know there are occasions when it is not accepted. Any ideas?

[TiTo20]

(Original post by Ben Elgar)
The reducing agent LiAlH4 is never mentioned in the text book but is accepted nearly all the time in the mark scheme. I normally pick that as I have real difficulty remembering all the alternative reducing agents. I just wanted to ask when should I NOT use it. I know there are occasions when it is not accepted. Any ideas?

LiAlH4 is only used in the reduction of either carboxylic acids to aldehydes to primary alcohols, or from ketones to secondary alcohols if I remember correctly. So you should only use it then. =]


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[JMaydom]

LiAlH4 is one of the strongest reducing agents available and will reduce all the carbonyl functional groups down to alcohols. It is unselective and will reduce all the carbonyls in the molecule

[thegodofgod]

(Original post by TiTo20)
LiAlH4 is only used in the reduction of either carboxylic acids to aldehydes to primary alcohols, or from ketones to secondary alcohols if I remember correctly. So you should only use it then. =]


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Are there any cases where NaBH₄ can't be used - I remember my teacher telling me that one of them couldn't be used for complete reduction from carboxylic acid to primary alcohol, but I can't remember right now?

Cheers

[illusionz]

(Original post by thegodofgod)
Are there any cases where NaBH₄ can't be used - I remember my teacher telling me that one of them couldn't be used for complete reduction from carboxylic acid to primary alcohol, but I can't remember right now?

Cheers

Reduction of amides, esters and carboxyllic acids requires LiAlH4 (or something else, but stronger than NaBH4). Technically NaBH4 will do esters but it's very slow (LiAlH4 or other stuff you haven't met is normally used.)

[thegodofgod]

(Original post by illusionz)
Reduction of amides, esters and carboxyllic acids requires LiAlH4 (or something else, but stronger than NaBH4). Technically NaBH4 will do esters but it's very slow (LiAlH4 or other stuff you haven't met is normally used.)

Never come across the reduction of amides / esters - would they be reduced to amines / ethers respectively?

Also, you know how you have to distill off the product for an aldehyde but reflux for a carboxylic acid when oxidising a primary alcohol, which of those do you need for oxidising a secondary alcohol to a ketone?

Cheers!

[TiTo20]

(Original post by thegodofgod)
Never come across the reduction of amides / esters - would they be reduced to amines / ethers respectively?

Also, you know how you have to distill off the product for an aldehyde but reflux for a carboxylic acid when oxidising a primary alcohol, which of those do you need for oxidising a secondary alcohol to a ketone?

Cheers!

Amides become amines and a carboxylic acid (I think? Haven't done chemistry for over 4 months =[) and esters become a carboxylic acid and an alcohol.

You reflux for oxidising a secondary alcohol to a ketone =]

[illusionz]

(Original post by thegodofgod)
Never come across the reduction of amides / esters - would they be reduced to amines / ethers respectively?

Also, you know how you have to distill off the product for an aldehyde but reflux for a carboxylic acid when oxidising a primary alcohol, which of those do you need for oxidising a secondary alcohol to a ketone?

Cheers!

Amides are reduced to the amine of the same order the amide was. Esters are reduced to primary alcohols. The person above me is thinking of hydrolysis rather than reduction by the way.

As for your oxidation question, I'm not sure. In reality no-one oxidises alcohols to aldehydes the way A levels tell you it's done - there are reagents which will stop at the aldehyde with no worry of over oxidation to the acid. I would presume that oxidation to a ketone does not require reflux (analogous to the oxidation to an aldehyde), but you could probably reflux it if you wanted to make it go faster, although the A level mark scheme is pretty good at disagreeing with that chemists would actually do so check that with something official.

[thegodofgod]

(Original post by TiTo20)
Amides become amines and a carboxylic acid (I think? Haven't done chemistry for over 4 months =[) and esters become a carboxylic acid and an alcohol.

You reflux for oxidising a secondary alcohol to a ketone =]

Cheers, although you're thinking of hydrolysis of amides / esters rather than reduction of the carbonyl group

Yeah, I thought reflux too, but just wanted to be sure

(Original post by illusionz)
Amides are reduced to the amine of the same order the amide was. Esters are reduced to primary alcohols. The person above me is thinking of hydrolysis rather than reduction by the way.

As for your oxidation question, I'm not sure. In reality no-one oxidises alcohols to aldehydes the way A levels tell you it's done - there are reagents which will stop at the aldehyde with no worry of over oxidation to the acid. I would presume that oxidation to a ketone does not require reflux (analogous to the oxidation to an aldehyde), but you could probably reflux it if you wanted to make it go faster, although the A level mark scheme is pretty good at disagreeing with that chemists would actually do so check that with something official.

Yeah, my teacher keeps on saying that it's a lot different in theory (A level) and in practice (industry).

Amazing diagram by the way, all of the reduction mechanisms for the aldehydes / ketones / esters seem to be nucleophilic addition?

[illusionz]

(Original post by thegodofgod)
Amazing diagram by the way, all of the reduction mechanisms for the aldehydes / ketones / esters seem to be nucleophilic addition?

They are. LiAlH4 is a source of nucleophilic hydrogen, can be thought of as H^-.

The mechanism for reduction of aldehydes and ketones is very simple, but other carbonyls are a bit more complicated and perhaps not what you would guess.

[thegodofgod]

(Original post by illusionz)
They are. LiAlH4 is a source of nucleophilic hydrogen, can be thought of as H^-.

The mechanism for reduction of aldehydes and ketones is very simple, but other carbonyls are a bit more complicated and perhaps not what you would guess.

Also, (last thing ) would you need to acidify the LiAlH₄ in order to provide protons for the last stage of the mechanism?

[illusionz]

(Original post by thegodofgod)
Also, (last thing ) would you need to acidify the LiAlH₄ in order to provide protons for the last stage of the mechanism?

Not quite. It's done as two steps. First you use the LiAlH4 and generate the O- coordinated to the Li+ ion, then addition of water (slightly acidified) gives the desired product.

Can you think what could happen if you added acid to the LiAlH4? (Or water for that matter)

[thegodofgod]

(Original post by illusionz)
Not quite. It's done as two steps. First you use the LiAlH4 and generate the O- coordinated to the Li+ ion, then addition of water (slightly acidified) gives the desired product.

Can you think what could happen if you added acid to the LiAlH4? (Or water for that matter)

Yeah, that makes more sense - the Li^+-O bond must be ionic, so water would break the lattice and separate the ions, and the one of the lone pairs of electrons would attack the extra proton on the hydronium ion, forming R-OH and H₂O?

Ahh - if you acidified the LiAlH₄, would you get some LiOH / H₂ being formed?

[illusionz]

(Original post by thegodofgod)
Yeah, that makes more sense - the Li^+-O bond must be ionic, so water would break the lattice and separate the ions, and the one of the lone pairs of electrons would attack the extra proton on the hydronium ion, forming R-OH and H₂O?

Ahh - if you acidified the LiAlH₄, would you get some LiOH / H₂ being formed?[/QUOTE]

The Lithium ion isn't really that relevant (well apart from stabilising the O- formed during reduction). As I said previously, reducing agents are sources of nucleophilic H^-, so that would react with H+ to form hydrogen gas, as you guessed. So you'll end up destroying your reducing agent!!

There are some (weaker) reducing agents which can be used in water (or even weak acid), as this can help speed up reactions by protonating the group which is to be reduced.

[thegodofgod]

(Original post by illusionz)


Ahh - if you acidified the LiAlH₄, would you get some LiOH / H₂ being formed?

Final question

What's the deal with tertiary alcohols - can they be oxidised (but would need a stronger oxidising agent than K₂Cr₂O₇?), or just not be oxidised at all, as it seems that the Carbon atom which has the -OH group attached to it would otherwise have 5 bonds (when there's a C=O group), which is not possible?

My chemistry teacher said it's possible, but didn't delve into it - would he have meant complete oxidation, i.e. forming CO₂?

Cheers!

[TiTo20]

(Original post by thegodofgod)
Final question

What's the deal with tertiary alcohols - can they be oxidised (but would need a stronger oxidising agent than K₂Cr₂O₇?), or just not be oxidised at all, as it seems that the Carbon atom which has the -OH group attached to it would otherwise have 5 bonds (when there's a C=O group), which is not possible?

My chemistry teacher said it's possible, but didn't delve into it - would he have meant complete oxidation, i.e. forming CO₂?

Cheers!

Your chemistry teacher is correct, you would use potassium permanganate and rigorous heating if I remember in order to oxidise a tertiary alcohol. I don't know about making carbon dioxide, I would have thought it would be creating alkyl chains and creating water in the process?

Good lord I'm rusty on my chemistry

[thegodofgod]

(Original post by TiTo20)
Your chemistry teacher is correct, you would use potassium permanganate and rigorous heating if I remember in order to oxidise a tertiary alcohol. I don't know about making carbon dioxide, I would have thought it would be creating alkyl chains and creating water in the process?

Good lord I'm rusty on my chemistry

Thanks - I thought of KMnO₄ too, but thought that might not be strong enough either

By complete oxidation I meant complete combustion, forming CO₂ + H₂O

[illusionz]

(Original post by TiTo20)
Your chemistry teacher is correct, you would use potassium permanganate and rigorous heating if I remember in order to oxidise a tertiary alcohol. I don't know about making carbon dioxide, I would have thought it would be creating alkyl chains and creating water in the process?

Good lord I'm rusty on my chemistry

(Original post by thegodofgod)
Final question

What's the deal with tertiary alcohols - can they be oxidised (but would need a stronger oxidising agent than K₂Cr₂O₇?), or just not be oxidised at all, as it seems that the Carbon atom which has the -OH group attached to it would otherwise have 5 bonds (when there's a C=O group), which is not possible?

My chemistry teacher said it's possible, but didn't delve into it - would he have meant complete oxidation, i.e. forming CO₂?

Cheers!

Formation of an alkyl chain and water from a tertiary alcohol is reduction. You are going from the alcohol oxidation level to the alkyl oxidation level. Oxidation would be going from the alcohol oxidation level to the carbonyl one. This is only possible if you break a C-C bond. The only way I am aware of this being possible is if you have a good leaving group on a carbon next to the alcohol so you can get a C-C bond migration as you form the carbonyl.

MnO4- is a very powerful oxidant, but I am not sure it would sucessfully oxidise a tertiary alcohol with any sort of control as to what products you product. It has a tendancy to create some rather annoying side reactions and as such, is rarely used in the lab.

An example of going from a tertiary alcohol to a ketone

[NutterFrutter]

Also, used for the reduction of nitriles to amines.