Organic Chemistry Help (Sodium Borohydride Reduction)

In summary, The driving force for the B-H bond to break and attack the ketone in a sodium borohydride reaction is not fully understood. It is believed that the O-B bond is harder to break than the B-H bond due to differences in electronegativity. The solvent system for borohydride reductions is typically an alcohol, which solvates the sodium borohydride. The oxygen from the ketone does not usually bond to the borane, but rather the solvent does. The reduction process ends with the protonation of the alkoxide generated from the ketone to produce an alcohol. It takes one molecule of BH4- to reduce each ketone, and all four hydrogens can participate in the reduction
  • #1

twt

2
0

Homework Statement



Hi all,

In a sodium borohydride reaction, one of the hydride in BH4- will attack the ketone, and the O- in the ketone will bond with the BH3. This would continue until all the hydride ions are used up.

My question is..what would be the driving force that makes the B - H break and attack the ketone.

Also, after you have one or more ketone bonded with the Boron, can't the O - B bond break and attack the ketone instead of the B - H bond?



The Attempt at a Solution



My group have no idea for the first question...the most possible answer we came up with is: "it just happen".

The second question...we just take a guess that the O - B is much harder to break compared with B - H, and maybe it has to do with the electronegativity.

Thank you very much.

I will add some more questions later when we finish figuring out the exact mechanism for NaBH4
 
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  • #2
The solvent system for most borohydride reductions is an alcohol. It must be a solvent that solvates the sodium borohydride. The oxygen from the ketone does not usually bond to the borane but rather the solvent does. The proton from the solvent protonates the alkoxide generated from the ketone to produce an alcohol. That is where the reduction ends for that particular borohydride anion... a new molecule of BH4- is required for each ketone reduced.
 
  • #3
chemisttree said:
...That is where the reduction ends for that particular borohydride anion... a new molecule of BH4- is required for each ketone reduced.

Whoops! Wrong again! All 4 hydrogens can reduce the carbonyl and the stoichiometry is 4 carbonyls to 1 borohydride...
:redface:
 

1. What is Sodium Borohydride Reduction?

Sodium Borohydride Reduction is a chemical reaction that is commonly used in organic chemistry to reduce aldehydes and ketones to their corresponding alcohols.

2. What is the mechanism of Sodium Borohydride Reduction?

The mechanism of Sodium Borohydride Reduction involves the transfer of a hydride ion (H-) from sodium borohydride to the carbonyl group of the aldehyde or ketone, resulting in the formation of an alkoxide ion. The alkoxide ion then reacts with the solvent, typically an alcohol, to form the final product.

3. What are the advantages of using Sodium Borohydride Reduction?

Sodium Borohydride Reduction is a mild and selective reducing agent, meaning that it only reacts with aldehydes and ketones and does not affect other functional groups. It is also relatively inexpensive and easy to handle, making it a popular choice in organic chemistry laboratories.

4. Are there any safety precautions when using Sodium Borohydride Reduction?

Sodium Borohydride is a flammable and reactive chemical, so it is important to handle it with caution. It should be stored in a cool, dry place and should not be allowed to come into contact with water or acids. Protective equipment, such as gloves and goggles, should be worn when handling this compound.

5. Can Sodium Borohydride Reduction be used on all types of carbonyl compounds?

Sodium Borohydride Reduction is most commonly used to reduce aldehydes and ketones, but it can also be used to reduce esters, acid chlorides, and lactones. However, it is not effective in reducing carboxylic acids or amides.

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