Nicotinamide Adenine Dinucleotide (NAD+) and Nicotinamide Adenine Dinucleotide Hydride (NADH) make up another pair of nature's redox cofactors. Another mechanistically identical redox pair is Nicotimamide Adenine Dinucleotide monoPhosphate (NADP+) and Nicotinamide Adenine Dinucleotide monoPhosphate Hydride (NADPH). Now that you have seen their full names once, you don't have to worry about remembering them ever again (at least in this course). So instead just remember the acronym pairs: NAD+/NADH and NADP+/NADPH. In fact, the situation can be further simplified. Take a look at the diagram below. It might take you a moment to notice what is different between these two molecules. The only difference is in the lower righthand corner, where NAD+ has a hydroxyl group attached and NADP+ has a phosphate group. These groups are for recognition and enzyme binding purposes only; neither group actually participates in the mechanistic chemistry of the enzyme catalyzed redox reactions.
|NADH Structure EJB.cdx|
Since the mechanisms of these pairs of redox cofactors are identical, this course will almost always use NAD+/NADH, and you can file away NADP+/NADPH in an accessible, albeit seldom used, part of your brain. If you are worried about remembering the complex structure of NAD+, things are about to be simplified one more time. The phosphate backbone, nitrogen lone pairs, hydroxyl groups and anything else below top-most pyridinium ring is used for recognition and binding purposes only, not for the chemistry of the enzyme catalyzed reactions. Therefore NAD+/NADH will be written in the form seen in the next section. This doesn't mean forget that the binding groups exist (after all salt bridges and "hydrophobic" specificity pockets keep us alive), but just file the information away with NADP+/NADPH.
|NADH Redox EJB.cdx|
As it is an oxidizing agent, NAD+ serves as a hydride acceptor. When an NAD+ molecule is reduced to NADH by the addition of a substrate bound hydride, it simultaneously oxidizes the substrate. The "reduction" and "oxidation" above the arrows in the reaction shown above refer to the NAD+/NADH, which is REDUCED as you go to the right and OXIDIZED as you go to the left. The substrate experiences the REVERSE! Remember the memory trick: whatever the "hydride" (real or imaginary) is getting added to, that is the substance that is being reduced. Your Chem-120 intuition may be saying that hydride is the worst leaving group possible, so there is no way a hydride could mechanistically jump from one molecule to another. Remember however, that this reaction is not taking place in solution (where a hydride would never be a viable leaving group), but within an enzyme where the precise alignment and spacing of molecules makes this possible. If you notice some striking similarities between the reduction of NAD+ and the mechanism of PLP, you are correct. Both PLP and NAD+ feature pyridinium rings that serve as temporary electron storage sinks, so the flow of electrons through the conjugated pi-system closely corralates.
Both coenzymes lose aromaticity when the conjugated pi-system is changed, and this is energetically unfavorable. However, the lone pair on the nitrogen is being stabilized by the extended pi-system, helping to lower the energy cost. In fact, the amide group attached to NAD+ and NADH serves to extend the conjugated pi-system of the coenzyme, providing additional stability (electron density can be resonated onto the the oxygen). Don't worry though, aromaticity is soon restored when NADH is oxidized back into NAD+ in a subsequent reaction. NAD+/NADH is a good coenzyme because it essentially serves as a good mobile carrier of energy. PLP normally stays within the enzyme, but NAD+/NADH is free to carry out one half of a redox reaction in one place and the other half in another. In fact, the main function of NADH in the body is to carry energy between cells. NAD+ is important for portable redox chemistry, and can easily reduce or oxidize numerous moities in many locations throughout the body.
Here is one example of a reaction NAD+ facilitates within the body. This reaction oxidizes ethanol to form acetaldehyde (which is the cause of a hangover after a night of heavy drinking). Remember that the oxidation of ethanol and reduction of NAD+ occur at the same time due to their close proximity within the enzyme. A hydride never fully forms.
When chemists started studying the mechanism of enzyme catalyzed NAD+/NADH reactions in detail, they noticed the reactions were stereospecific. But before we go into more detail about this principle, some background information is needed. Let’s look at the stereochemistry of ethanol:
Ethanol is achiral, but it is also prochiral, meaning that if one of its hydrogens were replaced by another atom, such as deuterium (D), it would become chiral. As seen in the diagram, one hydrogen is labeled “pro-R” and the other is labeled “pro-S.” If the pro-R hydrogen were substituted with a deuterium, the resulting molecule would have (R) stereochemistry. If ethanol were deuterated at the pro-S position, it would have (S) stereochemistry. The pro-R and pro-S deuterated ethanol molecules are enantiomers.
When ethanol is oxidized to acetaldehyde, we need another labeling system to show what lobe of the p orbital on the sp2 carbon will be attacked if the molecule is once again reduced to ethanol:
If the substituents around the sp2 carbon are arranged so that decreasing IUPAC priority is observed in a clockwise direction (the substituents get smaller in a clockwise direction), the lobe facing outward is known as the Re lobe. The opposite applies for the Si lobe. We find that this system of nomenclature is also applicable to the sp2 carbon where reduction occurs on an NAD+ molecule:
Another way to visualize the Re side verse the Si face is to think about looking from above NAD+ verses looking at from the bottom. In the diagram below, Eye 1 is looking down at the molecule from above. From above, the priority around the sp2 carbon goes clockwise, the direction of typical (R) chirality. This means the face that Eye 1 is looking at is the Re face of the molecule. Eye 2 is looking up at the molecule from the bottom of the NAD+. From that angle, the priority is counterclockwise, the direction of typical (S) chirality. This makes the face that Eye 2 is looking at the Si face.
Dr. Goess doesn't require you to memorize the Si/Re method of naming the faces of pi systems. However, at the very least, it is necessary that you know what will happen to the stereochemistry of NADH after it has been reduced from NAD+. If the reduction occurs at the Re lobe of NAD+, the added hydrogen (or deuterium, in the diagram) will be facing forward, towards the viewer in NADH (when NADH is viewed from the standard angle we use in Bioorganic class). If the reduction occurs on the Si lobe, the added hydrogen/deuterium will face backwards, away from the viewer.
We are now ready to discuss the stereospecificity of NAD+/NADH redox reactions mentioned earlier. Again, we will focus on the oxidation of ethanol via Alcohol Dehydrogenase. Chemists observed that the pro-R hydrogen of ethanol is always picked up by NAD+, and that the hydrogen always attacks the Re-face of NAD+. The opposite never happens. The stereospecificity of this redox reaction would have been unheard of in Chem-120. However, this reaction is happening in an enzyme's active site and not in solution. therefore, the substrate and cofactor can be held in a specific orientation relative to one another.
To hold the substrate in place (CH3CH2OH) in place, there is a hydrophobic interaction with the methyl group in the enzyme. There are two possible hydrogens that may be removed. The terminal Hydrogen attached to the Oxygen is removed using the B--ENZ the electron density is used to eventually create a pi sistem, and then the Hydrogen that is pro-R (located directly above the NAD+ in the model) is removed and leaves as a hydride with attacks the Re-Face of the NAD+. This causes a flow of electrons down the "sink". Remeber, Enzymes are Chiral and have the ability to impose chirality on their substrate. This would not be possible if chirality was not around in the environment itself, and nonracemic products are formed. In addition, this reaction occurs more readily in an enzyme because it is able to hold its substrate in a specific orientation, allowing for a stereospecific reaction to occur. This would be very very difficult to do under lab conditions. First of all a hydride is being removed which has bad kinetics, and the carbanion is not stabilized. Second of all, laboratory settings depend on a random orientation and collision. Another note-worthy feature of the Yeast Dehydrogenase is that inititally a water molecule is chelated on a Zn2+. A metal cation is found within the enzyme active site, which can hold more positive charge than a hydrogen. So this can easily handle oxidation state, (lewis metals are best activators). As for the chelation, it is drawn as a covalent bond between the zinc 2+ and the water, but its not. The formal charge is not positive nor negative, but there is some electron density taken away.
Kinetic Isoptope Effect
From this reaction we see that H does in fact react 4 times faster that D does. This 4-fold difference is relatively nothing in most situations, but here we need to know if this is signficant enough to cause the difference in stereospecificity of NADH/NAD+.
To test whether the kinetic isotope effect was causing the difference or whether it was due to stereochemistry the following experiments were conducted:
Since the reaction goes in the exact same way whether the deuterium is in the pro-R or pro-S position, this shows that the reaction is not affected by different isotopes of hydrogen. Ethanol always loses the hydrogen in the pro-R position regardless of the difference in weight of D and H. The reaction is truly stereospecific.
It is also important to note that reversibility can take place and stereospecificity is important in the reverse reaction as well. We see this in the reaction with alcohol dehydrogenase.
Just as the forward reaction above takes place, the backwards reaction occurs in the same way. In the case of alcohol dehydrogenase, where stereoselectivity is shown, if an atom is in the upward facing Re position on NADH after the forward reaction, this must mean that in the reverse reaction, that same atom will attach to the pro-R position on the product substituent, as shown in the mechanisms below. The first mechanism shows that reversibility takes place and that it is the same backwards as forward. The second mechanism depicts that there is stereoselectivity in the reverse reaction as well; the Re-face tritium is shown to attach only to the pro-R position on the substituent.
h1. Extra Problems
h4. Drill Problems
D15-1. A hypothetical alcohol dehydrogenase with propanol as its substrate has been isolated. It is in turn associated with a novel ketoreductase to yield the coupled scheme below. If the dehydrogenase removes the pro-R hydrogen and the reductase delivers to the re face of the ketone, predict the product (Y) and define its stereochemistry by the R/S designation.D15-1.cdxD15-3. A newly discovered alcohol dehydrogenase (AD) catalyzes the following reaction:
AD removes the pro-(S) hydrogen at C-1 of ethanol. Predict the products of the following reactions. Indicate the absolute stereochemistry of the chiral products formed, neglecting the NADX, or indicate if the product is achiral.
D15-5. What products would be observed from the following reaction were it not stereospecific?
D15-2. The following reaction is catalyzed by an NAD+-dependent decarboxylase enzyme. Propose a reasonable mechanism for this transformation. Note: NAD+-is both consumed and then reproduced during this sequence of transformations and the reaction does not occur in the absence of NAD+.
D15-4. Different substitutions of the NAD ring can change its redox potential. For each of the modifications below, which form of the cofactor (NAD+ or NADH) would be more stabilized with respect to the natural compound? Would the modified NAD+ be a better or worse oxidizing agent? Explain. Do parts (b) and (c) only.