Now that we’ve described how to figure out where the electrons are in a molecule, the factors that stabilize negative charge and positive charge, and gone into the curved arrow formalism for reactions, we’re ready to start going into more detail on some of the key reactions in organic chemistry.
Let’s start with acid base reactions.
In the first post I made introducing acid base reactions, I just said that the first order of business is just to figure out the bonds that form and the bonds that break, and recognize the four components. Like this.
This is the pattern of an acid-base reaction: we’re swapping a hydrogen between two atoms. Actually, since the hydrogen doesn’t take any electrons with it when this occurs, we’re technically swapping H+ (a proton). We have names for each of the species involved:
- The molecule which loses H+ is called the acid
- The molecule which gains H+ is called the base
- After gaining H+, the base becomes the conjugate acid
- After losing H+, the acid becomes the conjugate base.
Although this helps us recognize the pattern, it doesn’t really help us answer “why?”. For instance, why does the above reaction proceed well, but the reaction below does not? After all, it’s also technically an acid-base reaction.
Now we can circle back and start addressing this important question.
Let’s start with some simple examples. Here are four sample acid-base reactions. If you read about the stability of negative charge earlier, you should be able to gauge which of these will be more favorable and which are less favorable. (Feel free to ignore the backwards arrow for now)
- In each reaction, the molecule on the far left is the acid, which is donating a proton to water (the base), forming H3O(+) [the conjugate acid] and an anion (the conjugate base).
- Notice how the charges change: in each case, the conjugate base is more negative than the acid, and the conjugate acid is more positive than the base.This is always true for acid-base reactions.
- Since every reaction here involves H2O and H3O(+), the only difference between each reaction is the identity of the acid and the conjugate base.
- All else being equal, a neutral molecule is more stable than its conjugate base (nature seeks to minimize charges). The differences in stability between the neutral molecules (on the left) is trivial compared to the difference in stability between the charged molecules (on the right).
- Therefore, understanding the factors that stabilize negative charge is going to be key to understanding which of these are most favorable and which are least favorable.
If you think back to the factors that stabilized negative charge, as we went from left to right across the periodic table we got this trend:
Of these four species, CH3(-) is the least stable. Therefore of all the four species listed we should expect H-CH3 to be the least likely to give up its proton to form its conjugate base, CH3(-). In other words, H-CH3 is the least acidic.
Let’s repeat this, in different form. THE ACIDITY OF A SPECIES IS DIRECTLY RELATED TO THE STABILITY OF ITS CONJUGATE BASE.
Sorry for yelling but this is important! I’m so excited I have to yell this out.
For a good time, you could go back to the seven factors that stabilize negative charge (charge, electronegativity, polarizability, resonance, inductive effects, orbitals, and aromaticity), and look at them through this lens. Every trend that leads to a stabilization of negative charge is going to have a direct impact on acidity. More specifically, stabilizing negative charge will make the conjugate acid more acidic.
So for the reaction at the very top, we’re going from a less stable anion (H2N–) to a more stable anion (F-). This is energetically favorable, like water flowing downhill. So this reaction proceeds.
The reverse reaction would involve going from a more stable anion (F-) to a less stable anion (H2N-). This is energetically unfavorable, and does not proceed satisfactorily.
Of course this is still pretty vague at the moment – we can actually be a lot more exact about this, as we’ll see. In the next post, though, we’ll look at these exact same reactions from a slightly different angle.