Pauling's rules

These are two empirical rules which predict the strength of mononuclear oxoacids. I'm not sure why they are so commonly taught to undergrads - since I presume modern wavefunction software could make more advanced predictions.

1. For the oxoacid O_pE(OH)_q, pK_a ≈ 8 – 5p
2. The successive pK_a values of polyprotic acids (those with q > 1), increase by 5 units for
each successive proton transfer.

It should be obvious why oxo groups increase acidity, they can delocalize charge on the anion by resonance:

This allows us to rationalize why the acidities of chloro oxoacids decrease in the order HOCl₄ > HClO₃ > HClO₂ > HClO. Likewise for H₂SO₄ > H₂SO₃ and HNO₃ > HNO₂.

One good use of these rules is to notice structural anomalies. For example, if you dissolve carbon dioxide in water to make carbonic acid, you will find an empirical pK_a of 6.4 - but the rules predict a pK_a of 3.

The mistake is in assuming that all of the CO₂ dissolved in a sample of water exists as carbonic acid. In reality, only 1-2% of dissolved carbon dioxide exists in carbonic acid form. The pure pK_a is 3.6 when this difference is taken into account.

Many non-metal oxides, and some metal oxides, do not fully become their acidic form when dissolved in water. Deviation from Pauling's rules can help us detect that.

Bronsted acid classes

1. Aqua acid

This is an acidic proton located on a water molecule which is co-ordinated to a metal ion. Example:

2. Hydroxoacid

This is an acidic proton on a hydroxyl (OH) group, which does not have a neighboring oxo group (=O). An example is telluric acid:

Which is a white solid that dissolves in water.

3. Oxoacid

This is an acidic proton on a hydroxyl group, which has a neighboring oxo group. The most common ones are mononuclear, examples:

Often one or more of the OH groups can be replaced with an electron-withdrawing group such as F or CF₃, or an electron-donating group such as NH₂ - which can donate electrons by pi-resonance. Charge is more stable when spread out, so we can increase acidity by making an anion more stable by delocalizing charge (ie. pulling it off the oxygen atom).

Not all oxoacids follow the simple pattern of one atom surrounded by =O and -OH. For example, phosphonic acid (H₃PO₃) is only diprotic. Its easy to lose a mark in an exam by forgetting that one of the hydrogens isn't attached to an OH:

Solvent-system definition of acids and bases

This recognizes autoionisation of some aprotic solvents, such as boron trifluoride:

Notice the similarity with autoionisation (ie. autoprotolysis) of water.

In the solvent-system definition, a solute which increases the concentration of the cation generated by autoionisation of the solvent is called an acid, and a solute which increases the concentration of the anion generated by autoionisation of the solvent is called a base.

BrF₂AsF₆, called Difluorobromine(III) hexafluoroantimonate(V), is a salt without any obvious uses. But it is soluble in BrF₃. It disassociates into BrF₂⁺ and AsF₆^-, hence we would call it an acid.

Another example is dissolving sodium amide in ammonia. The autoionisation equation of ammonia is:

Since sodium amide splits into Na⁺ and NH₂^-, we would call it a base.

Solvent levelling

It is very hard to discriminate between strong acids in water since they are fully deprotonated, so a mol of HI and a mol of HBr each act like one mol of H₃O⁺. A solvent which is a weaker proton acceptor is needed to tell the difference.

The same logic applies to strong bases, one mole of any strong base can be treated as one mole of hydroxyl ions.

This is called the leveling effect - the limitation by the solvent of how strong acids or bases can be.

If two weak acids are put into a solvent which is a strong proton acceptor, such as ammonia, it is likewise hard to tell the difference between the two acids, since they will both act like a mol of ammonium ions. So the leveling effect changes with the solvent.

So a solvent HSol has a range for the allowed pH of a dissolved compound. Any acid stronger than the range will act like H₂Sol⁺, and any base stronger than this range will act like Sol^-. This is called the acid–base discrimination window:

The width of each bar is actually proportional to the pK_w of the solvents, we can see why below.

For a solute acting as an acid:

Notice that pK_{a depends on the solvent.}

An acid is considered strong at pK_a < 0, which corresponds to K_a > 1

So an acid with pK_a < 0 in a particular solvent has an acidity at the limit allowed by the leveling effect of the solvent, which is the same acidity as [H2Sol⁺]

For a solute acting as a base:

A base is considered strong at pK_b < 0, which corresponds to K_b > 1

So a base with pK_b < 0 has a basicity at the limit allowed by the leveling effect of the solvent, which is the same basicity as [Sol^-]

Now for the clever bit. The relationship between pK_a and pK_b is:

Therefore all bases with pK_a > pK_sol give a negative value of pK_b thus they behave like pure [Sol^-]

So the upper limit (strongest base) of pK_a is pK_sol

The lower limit (strongest acid) of pK_a is 0

Hence the windows of strengths that are not leveled in the solvent are is from pK_a = 0 to pK_a = pK_sol

Which explains why the window size is proportional to pK_w.

You may have difficulty grasping intuitively why a substance bad at protonating itself (high pK_w, so a small autopyrolysis constant) is also good at discriminating between acids. I have never been able to grasp it intuitively either. This explanation just shows you from the equations.

Rate of bond rotations

t_1/2 refers to the time taken for half the molecules in a sample to rotate.

I've always been told "double bonds don't rotate" so it is very interesting to see that some have half-lifes measured in days. I don't know why there is such a big difference in half-life for the bottom two molecules.

Arrhenius equation

This is a relationship between reaction rate, activation energy, and temperature. It is an empirical relationship, but it works remarkably well:

k = Rate constant of reaction
A = Constant, depending on the reaction
R = Molar gas constant
T = Temperature
E_a = activation energy for the reaction

Notice that this shows rate increasing with temperature, and decreasing with activation energy.

Examples will often give you some variables and have you calculate another, usually requiring you to rearrange the equation by making use of logarithm rules.

Isooctane

Isooctane is a major additive to petrol, it defines the octane rating - pure isooctane has an octane rating of 100. While pure n-octane has a rating of 0.

The strict meaning of isooctane would refer to 2-methylheptane, but 2,2,4-trimethylpentane ended up taking the name historically due to its far higher importance.

Combustion of isooctane has a ∆G° of -1000 kJ / mol, yet can exist comfortably at room temperature if there are no spark sources about. It is a good example of a kinetically stable but thermodynamically unstable molecule.

Variation of equilibrium constant with temperature

The relationship can be shown by putting together two previously-mentioned equations:

This has the form of y = mx + c. So plotting lnK against (1/T) will give a slope with a gradient of (-ΔH° / R) and an intercept of (ΔS° / R).

Notice that whether the gradient is positive or negative depends on the sign of -ΔH°. So an exothermic reaction has K decrease with temperature, and an endothermic reaction has K increase with temperature.

Le Chatelier didn't know about equilibrium constants or these equations, but now we do, we are able to explain why he observed his principle consistently.

Entropy and acids

We should already expect that adding electronegative atoms to a carboxylic acid will increase acidity, example:

The obvious reason is that electronegative atoms will pull electrons away from the oxygen atoms, delocalizing the negative charge in the carboxylate, hence stabilizing the molecule.

It turns out that this obvious reason is a minor one - the main reason is entropy.

You might guess that entropy increases when the acid disassociates, but this is only true in the gaseous phase. In a solvent such as water, charged species have a "shell" of solvent molecules, which is more ordered, hence total entropy change is negative.

When the charge is spread out over the carboxylate, there are more ways of arranging this shell of solvent molecules, so the entropy change is not as negative.

Diagram:

Entropy and hemiacetal formation

Entropy explains why intramolecular hemiacetals are more stable.

The bonds formed at the same in both cases, but the former reaction has a more negative entropy change. Hence we can expect a more positive ΔG from:

Driving reactions with equilibrium constants

In the reaction above, K₁ is about equal to K₂, and they are both in favor of the carbonyls. So we would expect a 50:50 mixture.

The reason that doesn't happen is due to an additional equilibrium - the deprotonation of the acid:

This drives it over to the right.

The process can also be represented as an energy diagram:

Below is another example of a reaction which can be driven left or right, depending on pH:

How the equilibrium constant varies with the energy change of a reaction

The equilibrium constant of a reaction is related to the difference in energy between reactants and products by this equation:

ΔG° = The standard Gibbs energy of the reaction, the difference between two states in kJ / mol. The ° represents standard conditions.

T = Temperature in kelvin
R = Molar gas constant
K = Equilibrium constant of the reaction

For example, this equation can be used to work out the energy change in the addition of water to isobutyraldehyde:

The equilibrium concentrations of hydrate and water can be measure by comparing UV adsorptions of the compound dissolved in water and the compound dissolved in hexane. As usual, approximately constant concentrations such as H₂O are set to 1.

These experiments reveal K_eq at 25° to be around 0.5.

So ∆G° = –8.314 × 298 × ln(0.5) = +1.7 kJmol^–1

The sign of G tells us about the direction the equilibrium favors. At K = 1, there is a 50:50 mixture of products:reactants, and ln(1) = 0, so ∆G° = 0.

At K > 1, ln(K) gives a positive number. At K < 1, ln(K) is negative.

In other words, equilibrium concentration is shifted in favor of the side with the lowest energy - in favor of the reaction with a negative ∆G. This should be pretty intuitive even without working it out from the equation. 

A small change in ∆G makes a big difference in K, which you might notice by considering the log term. Also have a look at this table:

Every reaction is theoretically at equilibrium. But a typical C-C bond is 350 kJ mol^-1. You might want to input that and see what percentage of products you get... it gives you an appreciation for why we consider these reactions to only operate in one direction.

The point of using Gibbs energy instead of enthalpy is that the Gibbs energy takes entropy into account.

Rotation and energy profile diagrams

An amide bond has double-bond character from conjugation. This tends to prevent rotation, but rotation does happen slowly, and can be measure by NMR.

Depending on the relative sizes of attached R groups, we can expect one form to be more stable than another. This can be shown on an energy profile diagram:

The maximum at 90 degrees is from the lone pair on nitrogen not having the right symmetry to overlap with the pi* of the C=O.

When both substituents on N are the same, we can expect equal energies:

Energy profile diagrams can also be used with alkenes:

The energy at 90 degrees going off the scale makes sense, since cis and trans alkenes are not observed to intercovert.

One way to measure these energies is to use the heat of hydrogenation:

We expect movement to a lower energy state to be spontaneous, ignoring entropy effects. Hence we can often consider the higher energy state to be a reactant, and the lower energy state to be a product.

Making an aldehyde from a carbanion by attacking a carbonyl

This is a problem. Attacking a carbonyl which doesn't have a good leaving group will create an alcohol:

While attacking a carbonyl with a good leaving group will result in it being attacked twice:

What we need is something which forms a stable tetrahedral intermediate, yet which has a group which leaves during the acid workup. By leaving at the same time as the acid workup, the organometallic is destroyed at the same time, preventing another attack.

A solution is DMF, or dimethylformamide:

In otherwords, adding DMF to an organometallic will replace the metal with an aldehyde group. Another example:

You might wonder what makes the tetrahedral intermediate stable, since amine anions can occasionally act as leaving groups. I don't know the answer.

Making ketones from esters

When attacking esters with carbanions, the product ketone is more reactive than the starting ester, so another alcohol attacks to produce a secondary alcohol. There are two solutions to this:

1. Making the starting materials more reactive

This can be done by converting to ester to an acid chloride using HCl. The acid chloride can then be attacked by the carbanion, example:

This can be helped by transmetallating a grignard reagent with a copper salt, making the carbanion less reactive too, hence the reaction is more likely to only go halfway.

2. Making the product less reactive

One example is a carboxylate salt. These are bad electrophiles, but an organolithium is just about sufficiently nucleophillic to attack one:

Obviously the dianion has no good leaving group, so it sits stable until the acid workup.

The water/acid quench also destroys any remaining organolithium, preventing further attack.

Another way of stabilizing the tetrahedral intermediate is to use a Weinreb amide, which is a N-methoxy-N-methyl amide:

The stabilization is from chelation from the methoxy group:

Quenching mechanism:

Overall reaction:

Carbonyl tetrahedral intermediate shorthand

This can save some time when drawing:

Making alcohols from esters

A carbanion will attack an ester, kicking off RO^-:

Another carbanion will then attack the ketone to produce a stable alkoxide, which becomes an alcohol on acid workup. :

Assuming there is a single type of carbanion used, notice that this results in at least two identical substituents on the product. 

Using this, you should be able to tell what type of organolithium is used for this reaction:

We can also use hydrides as the twice-attacking nucleophile, such as LiAlH₄.

You might reasonably predict that simply using a 2:1 carbanion:ester ratio will give aldehydes or ketones. But this doesn't work - because the aldehyde/ketone is more reactive to nucleophiles than an ester is. So a 2:1 ratio would just cause half of the esters to stay as esters, and half to be converted to alcohols.

But stopping the reaction halfway is achievable by other methods.