By considering the mechanism of the two step reaction of butanone and NaBH4 followed by dilute acid, explain why the product has no effect on plane polarised light.

CH₃-CO-CH₂-CH₃ + 2[H] --> CH₃-CHOH-CH₂-CH₃ NaBH₄ is a reducing agent that effectively produces an H^- ion to act as the reducing agent. The H^- ion attacks the electrophilic carbon centre (electron deficient due to the polar C=O bond), breaking one of the bonds between carbon and oxygen (electron pair move onto oxygen). Dilute acid is added, forming H₃O⁺ ions in solution. A proton is removed from H₃O⁺ by O^-, forming a hydroxyl group. The overall product is butan-2-ol, which is optically active due to the chiral centre. However, the product of the reaction does not rotate plane polarised light (this suggests that there is a racemic mixture in the product). The mechanism shows that the H^- ion can attack either the top or the bottom of the trigonal planar carbonyl group, and each is equally likely. The reaction produces two enantiomers which are of equal concentrations. Enantiomers rotate plane polarised light in opposite directions, therefore the effects of this rotation cancel when the enantiomers are of equal concentrations.