In many ways organic chemistry is very different from general chemistry. However, there are things that you learned in general chemistry that are essential for grasping organic chemistry concepts. In particular, we will be reviewing equilibrium and how it applies to acid-base reactions.
Often these concepts are reviewed during the first semester of organic chemistry and heavily used second semester. If you feel like you could use a quick review then keep reading.
The theme for this post: Lower energy is favorable.
All reactions are reversible. (In some reactions the product is so heavily favored that we say the reaction is irreversible and that it goes to completion.) This means that the reaction can go from starting materials to product and vice versa. In fact, the mechanism for going from products to starting materials is just the reverse of the mechanism for the forward reaction. For example the mechanism for the protection of the ketone in Aldehyde and Ketone 14 is reversed to accomplish the deprotection in Aldehyde and Ketone 16.
Let’s consider this reversible nature of reactions further using a simple generic reaction
Both the forward and reverse reactions never stop occurring but at some point there is no longer an observable change in the concentration of reactants and products. When this has happened we say the reaction has reached a state of equilibrium.
The extent to which starting material is converted to product is governed by thermodynamics (changes in energy). Remember, it is favorable for a system to go from high energy to low energy. Therefore, the side that is lower in energy is favored at equilibrium.
Quantitatively, we use the equilibrium constant (Keq) to describe the reaction at equilibrium. Using the concentrations at equilibrium we can calculate Keq using the following equation:
Based on this equation we can conclude that at equilibrium:
If Keq > 1 then [B] > [A]
If Keq < 1 then [B] < [A]
If Keq = 1 then [B] = [A]
Let’s take this a step further, since the extent of conversion is a function of the relative energies we can restate the first two above statements as follows:
The larger the value of Keq, the lower in energy the products are relatively to starting materials.
The smaller the value of Keq, the higher in energy the products are relatively to starting material
I’m from Minnesota, and one of my favorite winter activities is sledding. We can use the analogy of sledding and climbing back up the hill to visualize equilibrium and the associated energetics changes.
When Keq > 1 (the greater than symbol points to the right towards, or products) energetically the reaction is sliding down the hill to get to products. When Keq < 1 (the less than symbol points to the left, or toward starting materials) energetically the reaction is climbing up the hill to get to products.
No doubt about it, going down the hill is much more favorable!!!
No matter how you think about it, knowing which side of the equilibrium is favored informs you whether the starting materials or products are energetically favorable. When is this useful? A good example can be seen by looking at acid-base chemistry.
Acid and Base Chemistry
The strength of an acid and base are measured by equilibrium constants. Let’s explore this using a generic acid-base reaction:
From this we can define the equilibrium constant:
Generally, we can assume the concentration of water to be nearly constant since acid-base reactions are usually done in dilute aqueous solutions. As a result, we can use a new term, Ka in which the concentration of water has been removed from the expression by multiplying both sides of the above equation by [H2O]. The result is
Similar to equilibrium constants, Ka measures the extent to which products are formed. In this case, we are measuring the extent to which the acid dissociates. The stronger the acid the more it dissociates. Therefore, the larger the Ka value the stronger the acid. The value of Ka is usually a very large or a very small number that require use of scientific notation. Therefore, chemists prefer to use pKa values instead. Remember, pKa is related to Ka using the following equation:
If you want, you can review logarithmic math functions elsewhere, but the key thing to remember is that the smaller the pKa value the stronger the acid. This is reverse of the trend we saw for Ka because we looking at the -log. Also keep in mind that a change in 1 pKa unit corresponds to an order of magnitude change in Ka.
Now that we have a better idea of equilibrium and pKa let’s take a look at how we can use pKa values to predict which side of the equilibrium will be favored in acid-base reactions.
Let’s use Acid Base 4 as an example
Remember, the products of an acid base reaction are the conjugate acid and conjugate base. Therefore, we can compare pKa values of the acid and conjugate acid to determine which side of the equilibrium is favored. In this example the pKa on the left is approximately 4 and on the right is 15.7. Because acids and bases are paired as conjugates, the stronger acid is on the same side as the stronger base, so the reaction that is energetically favored is the one where stronger/stronger → weaker/weaker.
Key things to remember for acid-base reactions:
- The stronger the acid the lower the pKa value
- Equilibrium favors the side of the weaker acid
Therefore, pKa values can be used to determine which side of the equilibrium will be favored.
It is helpful to get a good feeling of the relative pKa of common acids and/or functional groups. Your instructor and your textbook will highlight the ones you need to know.
Remember: Acidity increases as the stability of the conjugate base increases. To review how resonance and inductive effects can impact the stability of the conjugate base watch the following videos from the app Mechanisms:
- Resonance: Acid Base 4-7, 10, 14-17
- Inductive Effects: Acid Base 9,12
Practice determining which side of an acid-base equilibrium is favored can be found following this link. Read through the answers to learn about examples of common equilibria you will likely encounter during the second semester of organic chemistry.