You need a negative charge to balance the positive charge
But importantly - you do not lose a ''naked'' H- from NaBH4. Typically mechanisms are drawn with the borohydride transferring the hydride to the site of reduction like this.
You need a negative charge to balance the positive charge
But importantly - you do not lose a ''naked'' H- from NaBH4. Typically mechanisms are drawn with the borohydride transferring the hydride to the site of reduction like this.
Oh because in the books it just has the hydride ions on its own with a lone pair attacking the carbonyl carbon. I think i might lose marks if i draw an arrow from the H-B bond to the carbonyl carbon.
Oh because in the books it just has the hydride ions on its own with a lone pair attacking the carbonyl carbon. I think i might lose marks if i draw an arrow from the H-B bond to the carbonyl carbon.