Selective reduction introducing a chiral center

Steps:

• Identify starting compound as pyruvic acid, CH3C(O)COOH (likely intended, achiral ketone).
• NaBH4 reduces only the ketone group to -CH(OH)-, yielding CH3CH(OH)COOH.
• The reduced carbon bears CH3, H, OH, COOH—four different groups.
• Confirm chirality: no plane of symmetry in the product.

Why B is correct:

• NaBH4 reduces ketones to alcohols per standard organic reduction rules, creating a tetrahedral carbon with four distinct substituents, defining a chiral center.

Why the others are wrong:

• A: HCN forms cyanohydrin CH3C(OH)(CN)COOH (chiral), but reagent typically pairs with base for ketones; alone, reaction incomplete.
• C: LiAlH4 reduces both ketone and COOH to CH3CH(OH)CH2OH (chiral diol), but over-reduces, altering functional groups beyond chiral introduction.
• D: NaHCO3 forms salt CH3C(O)COO⁻Na⁺, retaining achiral ketone.

Final answer: B