Quick O-chem question: reducing agents and carboxylic acids

[JFK90787]

Hey guys I'm having trouble finding an answer to this. There are 2 common reducing agents we need to know, LiAlH4 and NaBH4. I know that LiAlH4 will always turn a carboxylic acid to a primary alcohol. Does NaBH4 do the same or will it turn a carboxylic acid into an aldehyde first?

TY

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[PiBond]

Hey guys I'm having trouble finding an answer to this. There are 2 common reducing agents we need to know, LiAlH4 and NaBH4. I know that LiAlH4 will always turn a carboxylic acid to a primary alcohol. Does NaBH4 do the same or will it turn a carboxylic acid into an aldehyde first?

TY

LiAlH4 reduces all carbonyl compounds, NaBH4 reduces only ketones and aldehydes

 

[JFK90787]

LiAlH4 reduces all carbonyl compounds, NaBH4 reduces only ketones and aldehydes

Perfect, thank you 😀

 

[TurkSurg]

Hey guys I'm having trouble finding an answer to this. There are 2 common reducing agents we need to know, LiAlH4 and NaBH4. I know that LiAlH4 will always turn a carboxylic acid to a primary alcohol. Does NaBH4 do the same or will it turn a carboxylic acid into an aldehyde first?

TY

NaBH4 is NOT added to carboxylic acids. Just remember NaBH4 is for Family A (aldehydes and ketones) and LiAlH4 is for Family B (any other carbonyl groups besides aldehydes and ketones). However, sometimes you can use LiAlH4 on Family A but NaBH4 is preferred.

 

[wanderer]

Hey guys I'm having trouble finding an answer to this. There are 2 common reducing agents we need to know, LiAlH4 and NaBH4. I know that LiAlH4 will always turn a carboxylic acid to a primary alcohol. Does NaBH4 do the same or will it turn a carboxylic acid into an aldehyde first?

TY

I believe that there is an aldehyde as an intermediate, but the aldehyde is actually more reactive than the carboxylic acid, so that using standard LiAlH4 always leads to reduction to primary alcohol.
(As has been mentioned before NaBH4 will not work on a carboxylic acid.)

 

[JFK90787]

Heys guys, I have a related question, this time about oxidizing agents. I found out through a wrong answer today that the oxidizing agent Cr2O7 causes a primary alcohol to become a carboxylic acid, and for MCAT purposes it 'skips' the aldehyde step. As far as I know the other common oxidizing agents we need to know are CrO4, CrO3, PCC, and MnO4. Anyone know if one of these does something similar or do they all go to aldehyde form?

Thanks

 

[BerkReviewTeach]

Heys guys, I have a related question, this time about oxidizing agents. I found out through a wrong answer today that the oxidizing agent Cr2O7 causes a primary alcohol to become a carboxylic acid, and for MCAT purposes it 'skips' the aldehyde step. As far as I know the other common oxidizing agents we need to know are CrO4, CrO3, PCC, and MnO4. Anyone know if one of these does something similar or do they all go to aldehyde form?

Thanks

The oxidation of a primary alcohol (RCH2OH) involves the loss of two hydrides from carbon (the two designated in red). The loss of the first hydride forms either an aldehyde or a geminal diol (where both OH groups are on the terminal carbon), which are in an equilibrium with one another that depends on the concentration of water. A simplified rule is that if the oxidizing agent can be applied in the absence of water (such as is the case with PCC--pyridine, HCl, CrO3), then the aldehyde can be isolated. If the reagents require the presence of water (which is true with the salt forms of the oxidizing agents, such as Na2Cr2O7), then it will proceed past the diol/aldehyde equilibrium and on to the carboxylic acid.

 

[JFK90787]

A simplified rule is that if the oxidizing agent can be applied in the absence of water (such as is the case with PCC--pyridine, HCl, CrO3), then the aldehyde can be isolated. If the reagents require the presence of water (which is true with the salt forms of the oxidizing agents, such as Na2Cr2O7), then it will proceed past the diol/aldehyde equilibrium and on to the carboxylic acid.

Nice, thanks 👍

 

[TurkSurg]

Heys guys, I have a related question, this time about oxidizing agents. I found out through a wrong answer today that the oxidizing agent Cr2O7 causes a primary alcohol to become a carboxylic acid, and for MCAT purposes it 'skips' the aldehyde step. As far as I know the other common oxidizing agents we need to know are CrO4, CrO3, PCC, and MnO4. Anyone know if one of these does something similar or do they all go to aldehyde form?

Thanks

Well, take the example of chromic acid. You get the oxidizing agent CrO4 from the chromic acid (H2CrO4), right? The reason why you don't stop at the aldehyde product when using CrO4 as a reagent is because aldehydes react very quickly with the chromic acid to produce carboxylic acids.
The main use of PCC is basically to stop the oxidation at the aldehydic stage. Let me give you an example. Lets say you want to synthesize a compound by reacting a carboxylic acid with a Grignard reagent. Well, Grignard reagents cannot be added to carboxylic acids, so there's no reaction. So you have to reduce the carboxylic acid to do anything with it. Since carboxylic acid is a Family B member, you reduce it with LiAlH4 and you end up with an alcohol. But this time you cannot add the Grignard reagent to the alcohol either, so you must oxidize it first. But any oxidizing agent will oxidize the alcohol back to carboxylic acid. However, we do not want this since NO REACTION occured between a carboxylic acid and a Grignard reagent. Thats why you choose to oxidize it with PCC to stop the oxidation at aldehyde stage so in that case you will add the Grignard reagent to the aldehyde and proceed the reaction.

😉

 

[Uisa]

Well, take the example of chromic acid. You get the oxidizing agent CrO4 from the chromic acid (H2CrO4), right? The reason why you don't stop at the aldehyde product when using CrO4 as a reagent is because aldehydes react very quickly with the chromic acid to produce carboxylic acids.
The main use of PCC is basically to stop the oxidation at the aldehydic stage. Let me give you an example. Lets say you want to synthesize a compound by reacting a carboxylic acid with a Grignard reagent. Well, Grignard reagents cannot be added to carboxylic acids, so there's no reaction. So you have to reduce the carboxylic acid to do anything with it. Since carboxylic acid is a Family B member, you reduce it with LiAlH4 and you end up with an alcohol. But this time you cannot add the Grignard reagent to the alcohol either, so you must oxidize it first. But any oxidizing agent will oxidize the alcohol back to carboxylic acid. However, we do not want this since NO REACTION occured between a carboxylic acid and a Grignard reagent. Thats why you choose to oxidize it with PCC to stop the oxidation at aldehyde stage so in that case you will add the Grignard reagent to the aldehyde and proceed the reaction.

😉

Perfect explanation 👍

 

[BerkReviewTeach]

Well, Grignard reagents cannot be added to carboxylic acids, so there's no reaction.

Be careful following rote memorized pathways when studying for the MCAT.

There is indeed a reaction between a Grignard reagent and a carboxylic acid, just not the addition reaction you generally desire with a Grignard reagent and a carbonyl. The Grignard reagent is basic, which means it can deprotonate the carboxylic acid, resulting in an alkane (R-H, where R comes from R-MgBr) and a carboxylate. This is the reason you cannot carry out a Grignard reaction in a protic solvent.

 

[TurkSurg]

Be careful following rote memorized pathways when studying for the MCAT.

There is indeed a reaction between a Grignard reagent and a carboxylic acid, just not the addition reaction you generally desire with a Grignard reagent and a carbonyl. The Grignard reagent is basic, which means it can deprotonate the carboxylic acid, resulting in an alkane (R-H, where R comes from R-MgBr) and a carboxylate. This is the reason you cannot carry out a Grignard reaction in a protic solvent.

Yeah in this case I was talking about the addition reaction since since the OP was talking about the addition of oxidizing agents.