IR of alkanes

C-H stretch: For saturated carbons, this always occurs under 3000, with the rare exception of cyclopropane due to its ring strain. It can go higher than 3000 in unsaturated carbons.

As a general rule, ring strain and s character increases the frequency of a vibration, moving the peak to the left.

CH₂ bending: Occurs around 1465

CH₃ bending; Occurs around 1375 

Long-chain bending: This is a rocking motion which occurs with four or more CH₂ in an open chain. It is around 720

C-C stretch: This has many weak peaks, so it is not considered useful.

Below is the spectra of cyclohexane, notice how the CH₃ bend and the long-chain bend disappear:

IR spectroscopy 1

In A-level chemistry, questions requiring you to interpret IR spectra have a list of wavenumbers and the bonds they represent. Many undergraduate exams will not provide this, and will expect you to recognize the main absorptions from memory.

The most important absorption to memorise is the carbonyl stretch around 1715 cm^-1. You should also recognize the C-H stretch, which is just below 3000 cm^-1.

Unfortunately the carbonyl stretch can dip into the 1600s, and the 1600s is where the C=C stretch is located. But there is a difference - the carbonyl stretch is much more intense:

Another important stretch is the broad O-H peak at 3400:

The broadness is caused by hydrogen bonds. So the O-H stretch becomes a sharp peak if the spectrum is done on gaseous alcohol.

The N-H stretch can also be found in this region, at similar intensity. However, the N-H stretch is shaper, and for primary amines it is split into two peaks:

Index of hydrogen deficiency

Each time a ring or π bond is added to an alkane, the number of hydrogens decrease by two.

The index of hydrogen deficiency (IHD), sometimes called the unsaturation index, is the amount of rings or π bonds which a molecule contains. When working out a molecule's structure, this is generally calculated before considering other spectral data. It is calculated by taking the difference between a molecular formula and the general formula for an acyclic alkane, then dividing that difference by two.

Take C₄H₆ for example. Compared to the general formula C_nH_2n+2, there is a difference of four hydrogens. Dividing four by two gives us an unsaturation index of 2.

If other elements are contained in the molecular formula, there are three simple rules to account for them:

1. For group V atom in the molecular formula (N, P, As, Sb, Bi), add one hydrogen to the general formula.

2. For group VI atoms in the molecular formula (O, S, Se, Te), leave the general formula unchanged.

3. For group VII atoms in the molecular formula (F, Cl, Br, I), subtract one hydrogen from the general formula.

These rules are also simple enough to visualize intuitively. Remember that double bonds on these elements count in the unsaturation index. Below are two examples where the rules are applied.

C₇H₁₄O₂:

1. Compared to the general formula C_nH_2n+2 (where n = 7), there is a difference of two hydrogens.

2. The oxygen atoms have no effect on the general formula.

3. The unsaturation index therefore equals one.

The molecule could, for example, be an ester:

C₁₀H₁₄N₂:

1. The general formula for n = 10 is C₁₀H₂₂

2. Since there are two group V atoms, we add two hydrogens to the general formula to give C₁₀H₂₄

3. There is a difference of 10 hydrogens, giving an unsaturation index of 5.

So the structure has 5 rings, 5 double bonds, or a mixture of both. One possibility is nicotine:

C₅H₇O₂Cl

1. The general formula for C₅ is C₅H₁₂

2. Oxygen atoms have no effect

3. There is one halogen atom, so we subtract one hydrogen from the general formular to give C₅ is C₅H₁₁

There is a difference of 4 hydrogens, so there is a IHD of 2.

One possibility is 2-Chloro-1-methylcyclopropanecarboxylic acid:

Determination of molecular mass

In the example in the previous post, we were able to work out the moles of H, O, and C in the sample. But that doesn't tell us the moles of the sample. For example, 14 moles of C could be contained in one mole of C₁₄H₂₈O₄ or could be contained in two moles of C₇H₁₄O₂. In this example we can only derive the empirical formula, while we need the molecular mass to derive the molecular formula.

In a modern lab, the molecular mass is determined using a mass spectrometer. But the oldschool methods are also good to know:

Titration: This can be used if the sample is an acid. Since we know the moles used in the titrant, and we know the ratio of the reaction, we therefore know the number of moles of the sample. Dividing the sample mass by the moles gives us the molecular weight.

Vapor density method: This can be used if the sample is a gas. One mole of an ideal gas will occupy 24000 cm³ at standard pressure/temperature, so we can use the volume to determine the number of moles in a given mass of gas.

Many gases deviate from ideal, but determining the molecular weight can be done with rough numbers. For example, if we had the empirical formula H₂O, we know that the molecular weight has to be 18, 36, 54, 72... So an inaccurate experimental value of 22 is still good enough to tell us that the molecular weight is 18 and the molecular formula is therefore H₂O.

Cryoscopic method: This measures the freezing point depression when a known quantity of solid is dissolved in a liquid - which is a function of the solid's molecular weight.

Vapor pressure osmometry: This measures the change in the vapour pressure of a liquid when a known amount of solid is dissolved it - which is a function of the solid's molecular weight.

Calculating percentage composition from combustion data

I hope this subject isn't too obvious to include. Combustion analysis is a way to determine the percentage composition of elements in an organic molecule. The process is time-consuming and hence rarely done in research labs, unless it is a specialist lab which the sample has been sent to. Non-specialist labs typically just put a sample straight into modern spectrometers.

The logic used here is that moles of the product CO₂ equals the moles of C in the sample, while moles of product H₂O equals half the moles of H in the sample.

The moles of oxygen in the sample is what is left - assuming there are no other elements to take into account. In the above example, you can work out the moles of C and H in the sample, convert to weight, and subtract their weights from 9.83 mg. Convert the remaining weight to moles to get the moles of oxygen. 

With the moles of all three elements, you can convert into percentage composition.

The original process involved trapping the produced water in a hygroscopic agent, and the carbon dioxide in a strong base. Since carbon dioxide produces carbonic acid in water, the formation of an insoluble carbonate salt will, via Le Chatelier's principle, encourage the solvation of CO₂.

Structure of carbonic acid

Modern methods will separate the products via gas chromatography. Gas chromatography is a modern and complex way of separating and analysing compounds which can be vaporised without decomposition. I look forward to learning how they work.

Autoprotolysis constant

If you consider the equilibrium constant of water protonating itself, and once again assume the solution is dilute enough to ignore [H₂O], you get the autoprotolysis constant K_w.

The experiment value of K_w at room temperature is 1 x 10^-14. We know that pure water has a ph of 7; a pH of 7 corresponds to a H₃O⁺ concentration of 1 x 10^-7. Since with pure water [H₃O⁺] = [OH^-], then we can get K_w by multiplying [H⁺] and [OH^-], which is 1 x 10-7 x 1x-10-7 = 1 x10^-14.

What makes the autoprotolysis constant really useful is that it remains constant even when adding acid or base in dilute quantities, and nearly all acid or base used in a lab can be called dilute. So because K_w = [H⁺][OH^-] , doubling the concentration of hydrogen ions will half the concentration of OH^-. Unfortunately I have never been able to grasp why intuitively. But the equation works.

We can also express [H⁺] and [OH^-] as an acidity and basicity constant respectively, and substitute them into the autoprotolysis equation:

Or in log format:

The K_a is the strength of the acid (ability to give a proton to water), the K_b is the strength of the conjugate base (the ability to receive a proton from water). So the stronger the acid, the weaker its conjugate base. The stronger a base, the weaker its conjugate acid. We already knew that of course, but the equations both confirm this fact and gives a more precise relationship.

pK_w = 14. Similar expressions apply to other solvents, with pK_w replaced by the autoprotolysis constant of the solvent, pK_sol.

Basicity constant

The basicity constant (K_b) is similar to the acidity constant. It is the equilibrium constant for a base absorbing a proton from water to produce OH^-, while assuming enough dilution to ignore [H₂O].

Like the K_a, the basicity constant is often written in -log form: