[Meter]

Hello.

I had a lab where I had to reduce benzophenone with NaBH₄. After a solution of benzophenone in ethanol (96%) was left to react with NaBH₄ (dissolved in ~20 drops of water) for 10-15 minutes, I had to add 6 M NaOH.

What role does the NaOH play? The general mechanism for reduction of a ketone with NaBH₄ utilizes an acidic workup to turn the alkoxide intermediate into the final alcohol product.

Any help would be appreciated.

[Orcio_87]

In my opinion:

4 (C₆H₅)₂CO + NaBH₄ + 4 H₂O ---> 4 (C₆H₅)₂CHOH + Na(B(OH)₄)

NaBH₄ is given in excess and NaOH is needed to convert all H₃BO₃ to Na(B(OH)₄).

Maybe Na(B(OH)₄) is better than H₃BO₃, as H₃BO₃ can dehydrate during distillation:

H₃BO₃ ---> HBO₂ + H₂O

[Meter]

In my opinion:

4 (C₆H₅)2CO + NaBH₄ + 4 H₂O ---> 4 (C₆H₅)2CHOH + Na(B(OH)₄)

NaBH₄ is given in excess and NaOH is needed to convert all H₃BO₃ to Na(B(OH)₄).

Maybe Na(B(OH)₄) is better than H₃BO3, as H₃BO₃ can dehydrate during distillation:

H₃BO₃ ---> HBO₂ + H₂O

Hmm, this seems strange.

It seems that NaBH₄ decomposes in solvents such as water and ethanol to form hydrogen gas. Are hydroxide ions formed in that process? That would make it easy to explain then.

[Orcio_87]

It seems that NaBH4 decomposes in solvents such as water and ethanol to form hydrogen gas. Are hydroxide ions formed in that process? That would make it easy to explain then.

Hydroxide ions doesn't come from hydrolisis of NaBH₄ itself, but rather of hydrolisis Na(B(OH)₄) as it is salt of strong base (NaOH) and weak acid (H₃BO₃):

NaBH₄ + 4 H₂O ---> Na(B(OH)₄) + 4 H₂

Na(B(OH)₄)  NaOH + H₃BO₃

[Meter]

It seems that NaBH4 decomposes in solvents such as water and ethanol to form hydrogen gas. Are hydroxide ions formed in that process? That would make it easy to explain then.

Hydroxide ions doesn't come from hydrolisis of NaBH₄ itself, but rather of hydrolisis Na(B(OH)₄) as it is salt of strong base (NaOH) and weak acid (H₃BO₃):

NaBH₄ + 4 H₂O ---> Na(B(OH)₄) + 4 H₂

Na(B(OH)₄)  NaOH + H₃BO₃

Ah, that makes sense. Thanks.