Following up on the 4 major classes of reactions you encounter in Org 1, let’s look at the first of those four: acid base reactions. The focus here is not initially to understand why things happen the way they do – that comes later. The initial focus is to be able to recognize what is happening.
Just like when you read a novel or watch a movie for the first time, your primary focus is probably just following the plot. Once you understand the plot of the movie, then you can start a deeper level of analysis with respect to the characters, setting, the meaning of the movie and so on.
Kurt Vonnegut’s great speech about plots in novels could also be applied to organic chemistry.[See below]
The “plot” of a chemical reaction is basically analyzing which bonds break and which bonds form. So let’s look the following four reactions. Before you do so it’s assumed you’ll be familiar with bond line diagrams and “hidden” hydrogens – if you need a refresher, you might find this video helpful. It’s also assumed that you know that reactants are the structures to the left of (or above) the reaction arrow, and that products are the structures to the right of the reaction arrow.
You’ll notice that they might look very different on the surface – all those different structures! – but the basic plot of each reaction is the same. We’re breaking an H-(atom) bond and forming an H-(atom) bond. At the same time we’re also connecting the two “leftover” partners to form a salt, composed of two oppositely-charged ions.
Let’s look at that last reaction in more detail. Here, we’re breaking a C-H bond and an (ionic) Na-NH2 bond, and forming an N-H bond as well as an (ionic) C-Na bond.
There are four “actors” in this reaction – as there are in every acid-base reaction – and we have names for all of them.
The reactant where the bond to H is breaking is the acid.
The reactant where the bond to H is forming is the base
The product formed when the bond to H is broken is called the conjugate base.
The product formed when the bond to H is formed is called the conjugate acid.
We can also draw the reverse of the previous reaction. Look at this carefully. We’re still breaking a bond to H and forming a bond to H, but we’ve swapped everything. We’re breaking N-H and C-Na, and forming N-Na and C-H. It’s still an acid-base reaction.
There’s just one thing – experiment tells us that this reaction doesn’t happen to any appreciable extent.
One way of combining both the “forward” and “reverse” reactions is to use an equilibrium arrow. If we draw it this way, another way of making the same point is that equilibrium favors the product on the right.
Interesting! So not all acid-base reactions are equally likely. That brings up some questions.
- Why do some acid base reactions work, but not others?
- What factors determine the strengths of acids and bases?
- How can we determine the equilibrium lies for an acid-base reaction?