Brønsted Acids and Bases

Proton transfer is a fundamental reaction across all of chemistry. If you can understand this both qualitatively and quantitatively, you will have mastered a large chunk of organic chemistry! Core concept #1: the bare proton (H⁺) is good for bookkeeping but never exists in reality in chemical systems. Protons can exist by themselves in interstellar space, and in the Sun, but when we we refer to "H⁺" we are using a formalism. The positive charge on the proton is attracted to anything with electrons and will bind to whatever it can find. In the gas phase, generation of bare protons (by electric discharge) in methane results in the CH₅⁺ cation! However, we can (and do, often) illustrate proton transfer by showing "H⁺"--recognizing that it is in reality coming from some strongly acidic proton carrier. Core concept #2: the Brønsted acid concept revolves around the relative abilities of species to give up or accept protons. We can define a proton transfer equilibrium reaction as follows (recognize that this is a formalism--since we are using "H⁺") The equilibrium constant will reflect how easy it is for H-A to give up a proton: We are using the construct of a bare proton as the basis for a "ruler" to compare acidity of any H-A, and conversely, the basicity of any A^-. Since many of these values are very large or very small, it is more convenient to use a logarithmic scale called pK_a, defined as: pK_a = -log K_a Core concept #3: the location of equilibrium for any proton transfer is determined by the K_a values Let's take an example: dissolve HCl in water and describe the equilibrium that occurs. We'll analyze this in several steps: Identify every acid and its conjugate base. Choose (arbitrarily) which acid will be the proton donor. (If the equilibrium is not favored, we can turn everything around.) Write the equation for K_a for the proton donor. Look up and list K_a. For the proton acceptor, write the reverse reaction for K_a for its conjugate acid. Look up K_a for the conjugate acid; K_eq for the reaction as you wrote it is 1/K_a. Add the two reactions. You will see that the H⁺ term cancels out--if it does not, you made a mistake. The equilibrium constant is the product of the two K_eq values, or more simply K_a(donor)/K_a(acceptor-conjugate acid). For HCl dissolving in water: H-Cl is the proton donor, and H₂O is the proton acceptor. The conjugate base of HCl is Cl^-, and the conjugate acid of water is H₃O⁺. H-Cl ⇋ H⁺ + Cl^- K_a = 10⁸ (pK_a = -8) H⁺ + H₂O ⇋ H₃O⁺ K_eq = 1/50 (pK_a = -1.7) H-Cl + H₂O ⇋ H₃O⁺ + Cl^- K_eq = 10⁸/50 = 2 x 10⁶ We demonstrate that HCl is completely dissociated in water with this. If we know K_a (or pK_a) values--or can estimate them--we can quickly predict proton transfer equilibria. You should look at Table 2.2 carefully; I strongly recommend memorizing the following 6 pK_a values. In any proton transfer analysis you then have merely to estimate which of your memorized compounds any of the proton donors or conjugate acids most closely resembles.
Compound	pK_a	Comments
H-Cl	-8.0	Strong acids all have very negative pK_as.
H₂O⁺-H Hydronium	-1.7	The strongest acid possible in water.
CH₃C(=O)O-H Acetic acid	4.75	A typical weak organic acid
H₃N⁺-H Ammonium ion	9.3	Ammonia is a base
HO-H Water	15.74	Hydroxide is the strongest base possible in water.
HC≡C-H Acetylene	25	Acetylenes are among the most acidic hydrocarbons. The high s character of the sp orbital holds the lectron close to the nucleus.
H₂N-H Ammonia	36	Amide (NH₂^-) is a good example of a very strong base.
H₂C=CH-H Ethylene	43	An sp² carbon is intermediate in its electronegativity.
H₃C-H Methane	50	C-H bonds are usually not acidic, so their conjugate bases are strong bases.
Final notes: As you learn more about the reactivity of organic compounds, you will need to add to this list a small number of examples with pK_a between 16 and 50. Also, these numbers are all scaled to reactivity in water as a solvent; this creates difficulty in providing meaningful numbers for either very strong acids (and their conjugate bases) or very strong bases (and their conjugate acids). For an exhaustive comparison that you can refer to, see this table from a research group at Harvard.
Core concept #4: Stablilizing charge and/or electrons on either side of the equation will shift the equilibrium toward that side. We will be looking at a variety of ways that a structure will stabilize charge and electrons. In general you should look at the following: Electronegative atoms or groups will stabilize electrons and negative charge (in the conjugate base), and destabilize positive charge (in the conjugate acid). Makes a stronger acid. This applies to both the atom losing a proton, and when that atom is close to other electronegative atoms. Increasing atom size will stabilize electrons and negative charge (in the conjugate base). Makes a stronger acid. Electron delocalization (illustrated by multiple resonance forms) stabilizes electrons and negative charge (in the conjugate base). Makes a stronger acid. Resonance stabilization of a positive charge is also possible and will make for a weaker conjugate acid. Electron-donating groups will also weaken any conjugate acid.