NAD⁺ hydride transfer (alcohol dehydrogenase) template

In liver alcohol dehydrogenase ethanol binds to the catalytic Zn²⁺, loses its proton (relayed to His51), and the alkoxide transfers a hydride ion to C4 of NAD⁺, giving acetaldehyde and NADH.

Mechanism step by step

A Zn²⁺-bound alkoxide gives a hydride to C4 of the nicotinamide ring: ethanol → acetaldehyde, NAD⁺ → NADH

1. Alkoxide formation – Zn²⁺ lowers the pKa of bound ethanol; the proton is relayed (Ser48, ribose 2′-OH) to His51
2. Hydride transfer – C=O forms as H⁻ moves to C4 of the nicotinamide; the ring N⁺ takes the electrons

Key points

• The hydride (H⁻: a proton and two electrons) goes to C4, para to the ring nitrogen; the positive charge of NAD⁺ is the electron sink.
• Transfer is stereospecific: ADH is an A-side (pro-R) dehydrogenase, so NADH carries the transferred H on a defined face of C4.
• Zn²⁺ (bound to Cys46, His67 and Cys174 in horse liver ADH) is a Lewis acid that holds and polarises the substrate oxygen.
• In yeast fermentation the reaction runs backwards (acetaldehyde + NADH → ethanol), regenerating NAD⁺ for glycolysis.

Drawn with 73 atoms, 9 arrows, 6 shapes – every one editable in MolDraw.

Sources

• Nelson & Cox, Lehninger Principles of Biochemistry, 8th ed., 2021, ch. 14 Textbook
• Berg, Tymoczko, Gatto & Stryer, Biochemistry, 9th ed., 2019, ch. 16 Textbook