when NaBH4 is used as a reducing agent in organic chemistry, is the B or the H being oxidized?
H has an oxidation state of -1 in NaBH4 and it is going to +1. That is a loss of electrons and that makes H^-1 oxidized.
When NaBH4 (sodium borohydride) is used as a reducing agent in organic chemistry, the boron (B) is being oxidized.
To understand this, let's look at the reaction of NaBH4 as a reducing agent. NaBH4 can be represented as [BH4]-, where the negative charge is balanced by sodium ions (Na+). In the presence of a reducing species, such as an aldehyde or ketone, NaBH4 donates a hydride ion (H-) to the carbonyl carbon of the organic compound.
The reaction can be represented as follows:
[R-C(=O)-H] + [BH4]- → [R-CH(OH)-H] + B(OH)3
In this reaction, the boron (B) in NaBH4 is being oxidized. Initially, it has a formal oxidation state of -1. However, after the reaction, the boron (B) in the borohydride ion ([BH4]-) is oxidized to boron trihydroxide (B(OH)3), which has a formal oxidation state of +3.
Therefore, in the process of reducing the aldehyde or ketone, the boron (B) in NaBH4 is oxidized from -1 to +3.