# What are high-energy electrons?

I read that (in cellular respiration) the transported electrons in NADH have a higher energy than those in FADH2.

I can't find a (simple or otherwise) explanation of what a "high-energy" electron is.

I came across a post on Reddit which asks the same question:

I am not a chemist but I am trying to understand what my biochemistry book means by "high energy electrons" that are obtained from glycolysis and the citric acid cycle. I am assuming that this "high energy" is not a property of the electrons themselves, but it is not clear to me in what way they are "high energy". Do they mean that when these electrons are transferred to the electron carriers they are part of molecular structures in which they sit in high energy positions?

(There were no satisfactory answers to the question).

Could anyone come up with a definition understandable by this non-chemist (i.e. me)?

• Interesting question. It appears that in the context of biochemistry the definition of "high-energy electrons" is not the same as used for, say, HEED. – andselisk Dec 3 '20 at 8:43
• I would assume that in analogues reactions the energy gain is higher when NADH in involved instead of FADH2. Or that NADH can undergo a step that FADH2 can't. So what actually has high(er) energy is NADH. Without something more satisfactory consider Biology SE. – Alchimista Dec 3 '20 at 11:37
• @Alchimista Saying that there will be a better answer about biochemistry questions on Biology SE is a trigger for me, compelling me to write an answer right away. – Karsten Theis Dec 3 '20 at 13:33
• That's funny! I knew there was a chemist/biologist rivalry going on. Great answer, thanks. – Naj Dec 3 '20 at 13:35
• @KarstenTheis meaning was "without something more satisfactory than my comment THEN look at...." but my comment was already in tune. :) – Alchimista Dec 3 '20 at 19:24

transported electrons in $$\ce{NADH}$$ have a higher energy than those in $$\ce{FADH2}.$$

This is jargon describing the redox potential of the electron carrier $$\ce{NADH}/\ce{NAD+}$$ vs the electron carrier $$\ce{FADH2}/\ce{FAD}.$$ So if you compare the two reduction half reactions

\begin{align} \ce{NAD+ + H+ + 2e- &-> NADH} &\quad E^{\circ '}_\mathrm{red} &= \pu{-0.320 V}\tag{1}\\ \ce{FAD + 2H+ + 2e- &-> FADH2} &\quad E^{\circ '}_\mathrm{red} &= \pu{0.031 V}\tag{2} \end{align}

$$\ce{NADH}$$ is the stronger reducing agent.

Could anyone come up with a definition understandable by this non-chemist (i.e. me)?

Another way of saying this is that the reaction of $$\ce{NADH}$$ with dioxygen is more exergonic (the equilibrium lies further on the side of the products, more free energy is available from it) than the reaction of $$\ce{FADH2}$$ with dioxygen. For completeness, the reduction half reaction for dioxygen/water is:

$$\ce{1/2 O2 + 2 e- + 2H+ -> H2O} \quad E^{\circ '}_\mathrm{red} = \pu{0.816 V}\tag{3}$$

Do they mean that when these electrons are transferred to the electron carriers they are part of molecular structures in which they sit in high energy positions?

It has to do with how tightly the electrons are bound in one and the other electron carrier. The OP is correct to place the term "high-energy" in quotes when referring to electrons; electrons are indistinguishable, so their energy is with respect to the bound state that they are in.

• @KarstenTheis So the electrons are less tightly bound in NADH than in FADH2? – Naj Dec 3 '20 at 14:01
• @Naj Yes, and most tightly bound in water. – Karsten Theis Dec 3 '20 at 15:27
• I understand the 'Yes' (electrons are less tightly bound in NADH than in FADH2). I'm not sure I understand 'and most tightly bound in water'. Does this mean 'Electrons are the most tightly bound when they are within an H2O molecule'? – Naj Dec 3 '20 at 17:32
• @Naj Of the three molecules, electrons are mostly tighly bound in water, which is why oxygen is the "terminal electron acceptor" and the final product is water. – Karsten Theis Dec 3 '20 at 18:27

@KarstenTheis made it clear by using E0 values of each half reaction. Here I describe what they actually mean by using the phrase "high energy electrons" in biochemistry texts. Consider Ethyne and Ethane. Therefore Ethynide has a high energy electron compared to Ethanide

Another Example : H- , OH- , SH- . H- is stronger than OH- and OH- is stronger than HS- in the terms of basicity. We can say that H- has more active electrons compared to those 2. Therefore in the opinion of biology systems , its electron has more energy.(discussed more in the upcoming paragraphs)

The more thermodynamically unfavorable reaction gives out the more instable product. In turn, the instable product participates in other reactions , in a fast rate.

High Energy Electrons do mean the unstable electrons present in the biochemicals. These biochemicals are unstable because they've missed aromaticity ,their previous octet configuration or became radicals etc. therefore they have high Gibbs energy.

NAD+ loses its entire aromaticity by the reduction. This makes the reduction reaction ,to some extents, thermodynamically unfavorable for NAD+ (compared to FAD). This is why E0 of FAD is bigger than E0 of NAD+.

As I said Reduction of NAD+ is less favorable than reduction of FAD. Therefore NADH is more unstable/has high energy electrons than FADH2 and you can say that electrons in NADH are more accessible because they want to gain their aromaticity back.(not because of the radius). So NADH is more reactant than FADH2.

EDIT: This just came to my mind. In biochemistry texts , usually they say that "covalent bond between carbon and carbon in sugars, has a lot of energy that eventually gives the energy to NADH,FADH2 and later for ATP."

This highness of energy comes from the oxidation number of carbons in sugars(e.g. glucose) . See this

Each of them eventually participate in a CO2 by-product. The oxidation number in CO2 is +4. This results in huge movement of electrons from glucose to 6 CO2 molecules, hence the term "high energy covalent bonds in glucose". In one instance i.e. reduction of NAD+ to NADH in Step 6 of glycolysis , there is a nucleophilic substitution , HPO4(-2) acting as nucleophile and interestingly H- as leaving group. Here, the attack of Pi and the exit of H-(carrying electrons) , oxidizes the aldehyde carbon located at the end of the molecule.

Step 6

this unstable product containing high energy electrons (H-) , reacts with NAD+ at an extremely high rate.(the more unstable the reagent , the faster the reaction). {H- = H+ + 2e-)

Hydrid acting as a leaving group , is an interesting concept. look at Cannizzaro reaction . In the Step 6 of glycolysis, Aldehyde becomes oxidized into phosphonic acid and later carboxylic acid (reducing another particle- NAD+) ,While in Cannizzaro reaction , an aldehyde becomes oxidized into a carboxylic acid ( reducing another particle- another aldehyde).