Category Archives: Chem 2

Organic acids and bases

Knowing how to determine the relative strength of organic acids and bases is the key to understanding, (not just memorising) the organic reactions that make up the rest of this organic course.

You already know that a strong acid is a molecule that will readily donate a proton, and a strong base will gain one. You should already know some strong inorganic acids and bases, but you may not know many organic ones. That’s okay – this isn’t a place to memorise a great big list of compounds, instead you just need to know three rules, (and the reasons behind them) and remember one important thing: it all comes down the the stability of the conjugate acid or base.

The key to a strong acid/base is that the conjugate base/acid must be stable in solution. A strong acid has a weak conjugate base. A strong base has a weak conjugate acid. Confused? Prove it to yourself using some inorganic acids or bases. This means that when you’re looking at a list of acids or bases, sometimes it is easiest to look at their conjugate bases or acids and determine their stability and relative strength.

There are three things that will influence the conjugate base/acid’s stability:

Let’s start by looking at how the acidity is affected, then deal with bases later

  1. Electronegativity

Electronegativity is the measure of how strongly an atom attracts electrons. It increases across a period and decreases down a group – i.e. fluorine is the most electronegative element, and francium is the least. Highly electronegative atoms are happy to hold a negative charge and exist as their negative ions.

There are two parts to how electronegativity dictates acidity:

a) the atom bearing the charge in the conjugate ion:

b) inductive effect

The atom bearing the charge in the conjugate ion is the more significant of the two effects. If the charge-bearing atom is highly electronegative, (fluorine, chlorine etc) the anion (conjugate base) will be more stable than if you had an electropositive atom in its place.

The inductive effect is a little more subtle, and involves having electronegative atoms that are not directly attached to the acidic proton. When you lose that proton, an electronegative atom further down the molecule, (provided it isn’t too far, the inductive effect falls away very quickly with distance) will help stabilise the negative charge.

2. Hybridisation:

The acidity of a compound changes with the hybridisation of the carbon. Ethyne (sp) is more acidic than ethene (sp2) or ethane (sp3), for example. Orbitals with higher s character, (a higher proportion of s orbital to p orbital; sp orbitals have 50% s character, sp2 33% and sp3 25%) are more stable, as the electrons in s orbitals are close to the nucleus. It follows that in an anion, electrons would rather be in a high s character orbital, as they will be closer to the nucleus and therefore more stable.

This is actually an extension of electronegativity, as an sp carbon is more electronegative than an sp3 carbon.

3. Resonance stabilisation:

An anion that has a resonance form is going to be far more stable than one where the charge is localised to one atom. Spreading out the charge through resonance will increase acidity. We will talk more about resonance stabilisation in a few weeks.

To summarise, having electronegative atoms, carbons with high s character and resonance structures for the anion will all make a compound more acidic. Of course, they will also make something a weaker base. A strong base will have electropositive atoms, (to hold the positive charge), low s character and will not be able to form resonance structures.

For practice, place the following in order of decreasing acidity:

Some links that you might find useful:

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Substitution reactions

A nucleophilic substitution reaction occurs when a nucleophile (link to nucleophile post) reacts with the substrate to replace the leaving group. This reaction can occur via two pathways – Sn1 or Sn2. We’ll start with Sn1.

The Sn1 reaction occurs in two steps, the first is the removal of the leaving group from the substrate, and the second is the addition of the nucleophile to the carbocation.

Formation of the carbocation

Nucleophilic attack

The formation of the carbocation is the rate-determining step, (the slow step that determines the overall rate of the reaction), and as it relies only on the concentration of the substrate, the kinetics are first order – hence Sn1.

In order for an Sn1 reaction to occur, the carbocation formed in step one must be stable enough for the leaving group to leave without causing a reverse reaction. Carbocation stability depends on the number of alkyl side groups attached to the positively charged carbon – the more substituents, the more stable the carbocation is:

In order to create a carbocation at all, the leaving group must be good enough to detach from the substrate on its own.

By going through a planar intermediate, Sn1 reactions lose any stereochemistry, as the nucleophile is free to attack from either face of the carbocation. The product of an Sn1 reaction will always be a racemic mixture.

The Sn2 reaction occurs in a single, concerted step, the rate of which depends on both the nucleophile and substrate concentrations.

The nucleophile attacks the substrate from the opposite side to the leaving group, (backside attack), pushing the leaving group away. This results in an inversion of stereochemistry, known as an umbrella inversion or Walden inversion. Of course, if you start with a racemic mixture, you’ll end up with one.

Because it is like an umbrella being blown inside out in the wind

Sn2 reactions must go through a 5 membered transition state, which is crowded and unpleasant for the central carbon. This transition state can only form provided the substrate is not bulky, and the nucleophile is also small. If there are too many atoms around the central carbon, the nucleophile simply won’t fit.

Now you know all you need to in order to distinguish between Sn1 and Sn2 reactions.

  1. Is the leaving group a very good leaving group? If no, it cannot be Sn1
  2. Is the nucleophile a good nucleophile? If no, it cannot be Sn2.
  3. Will the removal of the leaving group from a stable carbocation? If no, it cannot be Sn1.
  4. Is the substrate sterically hindered? If yes, it cannot be Sn2.
  5. Has there been inversion of stereochemistry? Yes? Sn2.

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Elimination reactions

Read the post on substitution reactions first, as many of the concepts follow on…

In an elimination reaction, fragments of a molecule are removed from adjacent atoms and replaced with a double bond. There are two mechanisms for an elimination, E1 and E2.


The E1 reaction starts in the same way as an Sn1 reaction: the removal of the leaving group to form a carbocation. The nucleophile can now either attach to the carbocation and reaction becomes a substitution, or it may act as a base and take one of the hydrogens attached to an adjacent carbon. If this occurs, the elimination product is formed.




E1 reactions will occur under similar conditions to Sn1: they must form a stable carbocation and have a good leaving group.






E2 reactions are bimolecular eliminations, with the elimination occurring in a single step



An E2 reaction requires the nucleophile to act as a base and taken a proton from a carbon. Alkyl hydrogens are not very acidic, so this reaction requires a strong base to occur.


In order to distinguish E1 from E2, you need to ask the same questions for Sn1 and Sn2, the only difference is that a strong base is required for E2.


Zaitzev’s rule:


There are situations where it is possible to get two elimination products, just as in addition reactions it is possible to place the hydrogen on two difference carbons. For elimination reactions, the most substituted product will be the major product, as the more substituted an alkene is, the most stable it is.


Image from wikipedia



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Quantum theory 2: The Bohr model of the atom

At the beginning of the 20th century, the atom was explained using Rutherford’s model of a small, central nucleus with orbiting electrons like planets around a star. You still see this type of atom used in symbols today:

This model doesn’t work – orbiting electrons would lose energy and eventually (in about 10-10s) fall into the nucleus. Niels Bohr came up with the solution to this problem by considering the absorption and emission spectra of hydrogen. When you shine light through hydrogen gas, you don’t see a continuous spectrum, but a line spectrum (as shown below).

Bohr was able to explain that by postulating two things.
1) An electron can only exist in certain energy levels within the atom and 2) the electron can undergo a transition from one energy level to another. It is these transitions that we see on the line spectrum.

The energy of each energy level can be calculated using the following:

En = -2.18 x 10-18 (Z2/n2)

where Z is the atomic number and n is the energy level. This equation can then be used to find the energy of the transitions.

The transitions are grouped together in “series”. Each series corresponds to all the transitions to/from a certain energy level.

The Bohr model accurately describes the energy levels of single electron atoms, but fails for anything larger. For that, we need to look at exactly how the electrons behave when they are near the nucleus.


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Quantum theory 1: The dual nature of light

The wave nature of light:

One way of looking at light is as an electromagnetic wave, consisting of perpendicular, oscillating electric and magnetic fields.




We characterise it by is wavelength, (and therefore its frequency). The wavelength, (λ) is the distance between two waves, and the frequency (ν) is the number of waves that pass through a point per unit time.

The wavelength and frequency are related to one another by c, the speed of light:

c = λν

The range of electromagnetic radiation is called the electromagnetic spectrum, (shown below) and ranges from wavelengths smaller than atoms (gamma rays) to kilometres (radio waves). Visible light makes up a very small section, ranging from about 400 nm (violet) to 700 nm (red).

There is significant evidence for the wave nature of light – we see diffraction effects, it can be polarised and it can interfere with itself.

The particle nature of light:

Isaac Newton was a big proponent of the particle, (or, in his words “corpuscular”) theory of light, but this view was largely abandoned in favour of the wave model until the beginning of the 20th century. Einstein won his Nobel prize for his paper on the photoelectric effect which showed the energy in light was “quantised”, based on the work of Max Planck.

Planck showed that there was a close relationship between the light emitted by a solid and its temperature. A material will glow “white hot” at 1200°C, but red at 750°C. He found that, when using Boltzmann’s statistical version of the second law of thermodynamics, the energy emitted by a solid could only be E = nhν, where n is an integer value – ie, the energy is quantised. The value h is Planck’s constant and equal to 6.63 x 10-34 J.s.

Einstein took this idea and showed that it applied to the photoelectric effect. This was an experiment where light is shone on the clean surface of a metal and the electrons released are measured. Electrons are ejected from an atom when enough energy is absorbed; however this only occurs when light above a certain frequency is used. This threshold frequency is characteristic of each metal.

To explain this, the electron must be ejected using the energy from a single photon, and that photon must have greater energy than the ionisation energy of the atom. If the photon doesn’t have enough energy, no matter how many photons hit the metal, the electron won’t be ejected. So, light must be quantised and behave as a particle!

How do we reconcile these two behaviours? Not in any really satisfying way. We say that light behaves as both a wave and a particle, (known as the wave-particle duality).






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Kinetics 2: Collision theory

It is well established that increasing the temperature, increases the rate of a reaction. That dependence can be explained by looking at what is happening on a molecular level during a reaction. You might want to go back and read up on the kinetic theory of gases before we go on.

According to collision theory, for a reaction to occur, reactant molecule must collide with sufficient energy (activation energy) and the correct orientation. Activation energy is the potential barrier that the molecule has to overcome for bonds to break and reform.

The frequency of “correct” collisions is related to the temperature, as molecules will have higher kinetic energy at higher temperatures as well as the activation energy and an orientation factor. Combining these gives us the Arrhenius equation:


 We can make this equation more useful by taking the log:


We can now measure the rate at two temperatures and use this to find Ea:


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Kinetics 1: Rates of reaction

Chemical kinetics is the study of reaction rates, which includes how fast a reaction occurs, how the conditions change the rate and what molecular interactions are involved in a reaction step.
We define the reaction rate as the amount of reactant used, or amount of product used per unit time, and always has the units M/s. As rates are dependent on concentration, rate laws are typically written like so:

For that imaginary reaction, we have defined the rate as the decrease in concentration of A (which is why it is negative) over change in time. The rate is proportional to the concentrations raised to some power, (which must be determined experimentally, it is not the stoichiometric value!) and k, the rate (proportionality) constant.


The exponent values are referred to as the order of the reaction with respect to whichever species it is associated with, (in the previous example, the rate is nth order with respect to the concentration of B). The overall order of the reaction is the sum of the exponents.


Determining the rate law (initial rates method):


To determine the rate law, you need to know the order of reaction with respect to each reactant, and find the rate law. Typically, the experiment measures the initial rate at different concentrations, so the order can be calculated using simultaneous equations. Once the orders have been determined, it is easy to find k. An example (taken from General Chemistry, Ebbing, 5th Edition) is shown below:


Iodide ions are oxidised in acidic solution to form triiodide ion (I3), by hydrogen peroxide.

 H2O2(aq) + 3I(aq) + 2H+(aq) → I3(aq) + 2H2O(l)

 The reaction was run at a variety of concentrations and the rate recorded:

We give the rate law the form:


Rate = k[H2O2]m[I]n[H+]p


To calculate the order with respect to H2O2, we look at experiments 1 and 2 where the concentration of H2O2 doubles, but the other two stay the same. We divide one by the other:

We substitute the values in and see that the I, H+ and k values cancel leaving:


2 = 2m therefore m = 1.

This procedure is repeated for [I] and [H+], giving the rate law

 Rate = k[H2O2][I]

 The values for one of the experiments can be substituted in to find k, giving us:

 Rate = 1.2×10-2[H2O2][I]


Integrated rate laws:

The initial rate law is all well and good, but it doesn’t tell you what the concentration of a species will be at a given time during the reaction. To find this, we integrate the rate law and get the following:

For a first order rate law, (eg. Rate = k[A]a) we get:

ln[A]t = -kt + ln[A]0

and a half life, (the time taken for half of your reactant to react) of:

For a second order reaction, we get the following:


Collision theory and reaction mechanisms to come…

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Crystal field theory

We have been discussing metal complexes using the valence bond theory, but while it is a useful method for discussing simple bonds, it cannot account for the colours and magnetic properties of metal complexes. To understand why metal complexes are such beautiful colours, we have to use crystal field theory.

Crystal field theory gives us information on the electronic structure of the metal atom, considering how the d orbitals will be affected by the ligands.

For an octahedral complex, we imagine the ligands are point charges that sit on the Cartesian (x,y,z) axes. The d orbitals are arranged around the nucleus as shown below, with two orbitals pointing directly towards the ligands, and the other three in-between.

The d(x2-y2) and d(z2), which point at the ligands, experience electrostatic repulsion and are therefore at higher energy than the d(xy), d(yz) and d(xz) orbitals. This means the d orbitals are no longer degenerate, and the orbital diagram is split:

The energy difference between the high and low energy orbital is ∆O (the ligand field splitting parameter, if you’re feeling wordy), and it is the size of ∆O that dictates the spin, colour and magnetic properties. The size of the splitting parameter is determined field strength of the ligand. A strong field ligand, such as CN, will give a large splitting parameter and a weak field ligand, I for example, will have a small value for ∆. The ligands are arranged into the spectrochemical series, indicating the strength of the ligand field.

Now that we know about splitting, the next question is how do the d electrons distribute themselves between the non-degenerate orbitals? For example, we have two Fe2+ complexes, [Fe(CN)­6]4- and [Fe(H2O)6]2+. We get two different splitting diagrams:

When we go to fill the orbitals according to Hund’s rule, we should half fill each orbital before going back and pairing the electrons, as it requires energy to put two electrons in an orbital. This is fine, if ∆ is low, as with [Fe(H2O)6]2+, and we get a high spin complex, with a higher number of unpaired electrons. However, if ∆ is large, as with [Fe(CN)­6]4-­, the pairing energy is lower than the energy required to get an electron into the high energy orbitals, so the complex is low spin.

High spin complexes are paramagnetic, as the unpaired electrons give the complex an inherent magnetism. The magnetic moment can be calculated according to the following equation: μs.o. = √{4S(S+1)}. This is the spin-only formula, where S is the number of unpaired electrons.

The colour of transition metal complexes also depends on the magnitude of ∆, as the energy of absorbed light will correspond to the energy difference between the low and high energy orbitals. If ∆ is small, the complex will absorb lower energy light, (red) and appear blue,  the opposite is true for complexes with high ∆.

Some links: (warning – animated gifs)







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Transition metal complexes

Transition metal atoms and ions can act as Lewis acids, accepting electron pairs from molecules or ions with electrons to spare, (Lewis bases).


A complex ion is a metal ion with lewis bases attached through covalent bonds, a metal complex or coordination compound is the same thing, but neutral.

Ligands are the lewis bases attached the metal ion. They may be small molecules or ions

The coordination number is the number of ligands attached to the metal ion.

 Denticity refers to how many bonding sites exist on the ligand – or how many times a single ligand can bond to the central metal ion.


The rules for naming metal complexes are much the same as for other nomenclatures:

  1. Always name the cation before the anion.
  2. The ligands are named first, then the metal atom but all in one word.
  3. The ligands are preceded by a Greek prefix telling you how many of them there are. If you have more than one type of ligand, they are listed in alphabetical order, (ignoring the prefix).
    1. The prefixes are the standard di-, tri-, tetra-… for simple ligands but become bis-, tris-, tetrakis-…  for more complex ligands. As a general rule, if a ligand name already as a prefix in it, (for example ethylenediamine) it is “complex”.
  4. Anionic ligands end in –o, neutral ligands have the same name as the compound, (there are a few exceptions: NH3 = ammine; H2O = aqua).
  5. The metal atom generally has the normal name, but will end with –ate if it is an anion. Some atoms, however, use the old-fashioned names for the anion, (eg. Cu = cuprate; Au = aurate etc). If the symbol on the periodic table doesn’t match the modern name – give it its ye-olde name. You always give the oxidation state of the metal in roman numerals.

Some examples:

[Pt(NH3)Br2]Cl2: tetraamminedichloroplatinum(IV) chloride

Fe(CN)64-: hexacyanoferrate (II)

[Co(en)3]Cl3: tris(ethylenediamine)cobalt (III) chloride


Just like your favourite part of organic chemistry, metal complexes have isomers too! And they are nearly as much fun to name.

There are, as before, two main types of isomers.

Structural isomers deal with differences in the way the atoms are bonded together.

Ionisation isomers: where the ligands and the counterions are exchanged.

[Pt(NH3)4Cl2]Br2 vs [Pt(NH3)4Br2]Cl2

Coordination isomers: are where compounds containing complex anions and cations differ by the distribution of ligands.

[Co(en)3][Cr(CN)­6] vs [Cr(en)3][Co(CN)6]

Linkage isomers: are isomers where a ligand attaches via different atoms.

[Mn(CO)5(SCN)] vs [Mn(CO)5(NCS)]

Stereoisomers have the same bonds, but a different arrangement of them in space.

Geometric isomers: differ in the relative positions of the ligands in space.

If you have a pair of the same ligands, you can have cis- and trans- isomers:

Images courtesy of  Doc Brown (

If you have three, you can get fac- and mer- isomers:

Just to make your life difficult, there are also enantiomers, non-superimposable mirror images:


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Introduction to transition metals

Transition metals are the d-block of the periodic table and therefore have unfilled d-orbitals in common oxidation states. Technically, zinc, cadmium and mercury aren’t transition metals, as they have full d-orbitals, they are generally included anyway.

Trends of the transition metals:

 Electronic configuration:

The electronic configuration of the transition metals is important, as it is the partially filled d-orbitals that are at the root of the interesting chemistry. For the most part, the electronic configuration is predicted by the aufbau principle (which we will cover in greater detail in week 10, but you should know from school. If you’ve forgotten, here’s a worksheet for you.)

Scandium, the first of the transition metals has the configuration [Ar]4s23d1, (remember that the 4s fills before the 3d), and you continue along the period, adding in one electron to the d shell as you go. There are, however, two exceptions. Chromium: [Ar]4s13d5 and copper: [Ar]4s13d10, as the half filled and fully filled d shells are more stable than a full 4s shell.

Chromium - breaking all the rules


Melting and boiling points:

Transition metals all have high melting and boiling points, though some are higher than others.


The melting/boiling point is related to the strength of the metallic bonding, which is determined by the number of unpaired d electrons. The first 5 elements have increasing numbers of unpaired bonds, but they start to pair up after that.


Oxidation states:

The first electrons to be knocked out of a transition metal tend to be the s orbital ones, so the 2+ oxidation state is common across the board, (with a few exceptions). The maximum oxidation state possible is equal to the number of s and d electrons present in the valence shell up to iron, and then decreases after that. Most transition metals have many oxidation states, due to all the different arrangements the d electrons can take.


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