Chem 27 Section 7 Handout: Redox Chemistry: We are now usually talking about H-, which is nucleophilic, not H+, which is electrophilic. Oxidation: a loss of electrons NADH Æ NAD+ + H- FADH2 Æ FAD +H2 Reduction: a gain of electrons NAD+ + H- Æ NADH FAD + H2 Æ FADH2 Oxidizing Agent: is reduced (NAD+ and FAD) while the substrate is oxidized. Reducing Agent: is oxidized (NADH and FADH2) while the substrate is reduced. NAD+/NADH: (often abbreviated as nicotinamide with an R group on N) • NADPH is NADH with a phosphate group attached to the adenosine. The P aids molecular recognition. • H- attacks NAD+ at the para position • Loss of aromaticity is countered by N l.p. delocalization • Reactions are stereospecific (Re/Si, prochiral) • In alcohol dehydrogenase, H- is only delivered to the Re face of NAD+ (and only lost) • NADH only adds into Re face of acetaldehyde • NADH is a cofactor and is bound to an enzyme FAD/FADH2: • R groups are used for recognition by specific enzymes • Azadiene is the reactive part of the molecule Modes of Reactivity: I. NAD+/FAD coupling Makes FADH2 II. O2/FADH2 coupling Makes a superoxide • Breaks disulfide bonds (glutathione reductase) • Dehydrogenates (succinate dehydrogenase) • Baeyer-Villiger reaction (ketone monooxygenase) • Decarboxylation/oxidation (lactate oxidase) Things to consider in your mechanisms: • What bonds need to be broken/made to get from the reactants to products • Can you think of an enzyme/cofactor discussed in lecture which can do any of these or similar reactions? • If you get lost, try working from both directions. • Write down the steps you know and then try to link them together. • ***Remember to consider stereochemistry in reactant, products, enzymes, and especially when you are asked to draw a transition state and asked why something reacts in a given way. Problems: K04, L01, LO3, L05 Answers to R/S handout: F C H H O 3 OH S R CHO Cl H R R O H H H Cl R S S S S Cl Cl R R H CH3 CHO CH3 H H NH S R S R 2 HS CH=CH CH2CH3 CH2CH3 CH3 (H CO) HC 2 HO Br CO2H 3 2 H R CH(CH3)2 CH3 H CH3 .