Chemistry olympiad roadmap.

EDTA is a hexadentate ligand: one ion wraps around a metal ion using two N and four O donor atoms. It is usually written $\text{H}_4\text{Y}$, and the fully deprotonated form $\text{Y}^{4-}$ is the one that binds.

Key facts:

• EDTA reacts with almost every metal ion in a 1 : 1 ratio, whatever the charge on the metal. This makes the arithmetic easy.
• The formation constants are large, but only the $\text{Y}^{4-}$ fraction binds. At low pH most EDTA is protonated, so titrations are buffered, often at pH 10 with ammonia. The effective (conditional) constant is $K' = \alpha_{\text{Y}^{4-}} K_f$.
• A metal indicator such as Eriochrome Black T is a weaker ligand with a different colour when bound. At the end point EDTA has taken all the metal and the free indicator colour appears (wine red to blue for Eriochrome Black T).
• Back titration: add a known excess of EDTA, then titrate the leftover EDTA with a standard $\text{Mg}^{2+}$ or $\text{Zn}^{2+}$ solution. This is used when the metal reacts slowly or has no good indicator, for example aluminium.
• Masking: a second ligand, such as cyanide or fluoride, ties up an interfering ion so it does not react with EDTA.

Water hardness is reported as mg of $\text{CaCO}_3$ per dm³, as if all the $\text{Ca}^{2+}$ and $\text{Mg}^{2+}$ were calcium carbonate.

For a one-electron atom or ion (H, $\text{He}^+$, $\text{Li}^{2+}$) with nuclear charge $Z$, the allowed energies are

$$E_n = -\frac{R_H Z^2}{n^2}, \qquad R_H = 2.18 \times 10^{-18}\ \text{J} = 13.6\ \text{eV}$$

where $n = 1, 2, 3, \dots$ When the electron drops from $n_2$ to $n_1$, a photon carries the difference:

$$\frac{1}{\lambda} = R_\infty Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right), \qquad R_\infty = 1.097 \times 10^{7}\ \text{m}^{-1}$$

• Lines ending on $n_1 = 1$ are in the ultraviolet (Lyman series); lines ending on $n_1 = 2$ are visible (Balmer series).
• Ionisation from the ground state is $n = 1 \rightarrow \infty$: $13.6 Z^2$ eV per atom, or $1312 Z^2$ kJ mol^-1.

Each orbital is labelled by $n$, $l$ (0 to $n-1$; s, p, d, f) and $m_l$ ($-l$ to $+l$), and each electron also has $m_s = \pm\tfrac{1}{2}$. Shell $n$ holds $n^2$ orbitals. An orbital has $n-1$ nodes in total: $l$ angular nodes and $n-l-1$ radial nodes. So 3p has one radial node and one angular node.

For many-electron atoms the formula is only a guide: inner electrons shield the nucleus, so use an effective $Z$.

An electron trapped in a one-dimensional box of length $L$ can only have certain energies:

$$E_n = \frac{n^2 h^2}{8 m_e L^2}, \qquad n = 1, 2, 3, \dots$$

Three things to notice. The lowest energy is not zero. Levels spread out as $n^2$. A longer box gives closer levels.

Conjugated molecules. Treat the delocalised $\pi$ electrons as moving freely along the chain of length $L$. Fill the levels two electrons at a time. With $N$ $\pi$ electrons ($N$ even), the highest occupied level is $n = N/2$ and the lowest empty one is $n = N/2 + 1$. The absorption that moves one electron up costs

$$\Delta E = \frac{h^2}{8 m_e L^2}\left[\left(\tfrac{N}{2}+1\right)^2 - \left(\tfrac{N}{2}\right)^2\right] = \frac{h^2 (N+1)}{8 m_e L^2}$$

and the absorbed wavelength is $\lambda = hc/\Delta E$. As the chain gets longer, $\lambda$ grows roughly like $L$, so long polyenes absorb visible light and look coloured (carotene is orange).

Useful constants: $h = 6.626 \times 10^{-34}$ J s, $m_e = 9.109 \times 10^{-31}$ kg, $c = 2.998 \times 10^{8}$ m s^-1. Keep $L$ in metres.

When two atomic orbitals overlap they make two molecular orbitals: a bonding one, lower in energy, with electron density between the nuclei, and an antibonding one (marked *), higher in energy, with a node between them.

For diatomics of the second period, the valence orbitals fill in this order:

• B₂, C₂, N₂: $\sigma_{2s},\ \sigma^*_{2s},\ \pi_{2p}$ (two orbitals), $\sigma_{2p},\ \pi^*_{2p}$ (two), $\sigma^*_{2p}$
• O₂, F₂: the same, except $\sigma_{2p}$ comes below $\pi_{2p}$

The swap happens because in the lighter atoms the 2s and 2p levels are close and mix.

Then use

$$\text{bond order} = \tfrac{1}{2}\left(\text{bonding electrons} - \text{antibonding electrons}\right)$$

A higher bond order means a shorter, stronger bond. Fill degenerate orbitals singly first (Hund's rule). Any unpaired electrons make the species paramagnetic. O₂ has two unpaired electrons in $\pi^*_{2p}$, which a Lewis structure with a double bond does not show.

Removing an electron from an antibonding orbital strengthens the bond; removing one from a bonding orbital weakens it. So the bond order of $\text{O}_2^+$ is 2.5, while that of $\text{O}_2^{2-}$ is 1.

The first law: the internal energy of a system changes by the heat put in plus the work done on it.

$$\Delta U = q + w$$

Signs: $q \gt 0$ when heat flows in, $w \gt 0$ when work is done on the system. A gas expanding against a constant outside pressure does work on the surroundings, so $w = -p_{\text{ext}}\Delta V$.

• At constant volume (a sealed bomb calorimeter) no expansion work is done, so $q_V = \Delta U$.
• At constant pressure (an open beaker) $q_p = \Delta H$, where $H = U + pV$.
• For a reaction involving gases: $\Delta H = \Delta U + \Delta n_{\text{gas}} RT$, where $\Delta n_{\text{gas}}$ is moles of gaseous products minus gaseous reactants. Liquids and solids hardly change volume, so ignore them.
• Reversible isothermal expansion of an ideal gas: $w = -nRT \ln(V_2/V_1)$, and $\Delta U = 0$, so $q = -w$.
• Heat capacity $C_p$ tells you how $H$ changes with temperature. For a reaction, Kirchhoff's law gives $\Delta H(T_2) = \Delta H(T_1) + \Delta C_p (T_2 - T_1)$.

Watch units: $R = 8.314$ J K^-1 mol^-1, so convert to kJ before adding to a value in kJ.

A phase diagram plots pressure against temperature. The lines show where two phases coexist; the triple point is where all three do; beyond the critical point liquid and gas cannot be told apart.

Vapour pressure rises steeply with temperature. The Clausius-Clapeyron equation gives the change:

$$\ln\frac{p_2}{p_1} = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$

Raoult's law: in an ideal solution the vapour pressure of component A is $p_A = x_A p_A^*$, where $x_A$ is its mole fraction and $p_A^*$ the pure value.

Colligative properties depend only on the number of solute particles, not their identity:

• Freezing point depression: $\Delta T_f = i K_f b$
• Boiling point elevation: $\Delta T_b = i K_b b$
• Osmotic pressure: $\Pi = i c R T$

Here $b$ is molality (mol of solute per kg of solvent), $c$ is concentration (mol m^-3 in SI units), and $i$ is the van 't Hoff factor: about 2 for NaCl, 3 for $\text{CaCl}_2$, 1 for sugar, and less than 1 for a solute that pairs up (dimerises). For water, $K_f = 1.86$ K kg mol^-1.

A mechanism lists elementary steps. For each elementary step the rate law follows from the equation itself: $\text{A} + \text{B} \rightarrow$ products has rate $k[\text{A}][\text{B}]$. Intermediates must not appear in the final rate law, so you eliminate them.

Pre-equilibrium. If a fast reversible step comes before a slow one, treat the fast step as being at equilibrium. For $\text{A} + \text{B} \rightleftharpoons \text{I}$ (constant $K$) then $\text{I} \rightarrow \text{P}$ (slow, $k_2$):

$$[\text{I}] = K[\text{A}][\text{B}], \qquad \text{rate} = k_2 K [\text{A}][\text{B}]$$

Steady state. A reactive intermediate is used up as fast as it forms, so its concentration stays small and roughly constant: set $d[\text{I}]/dt = 0$. For $\text{A} + \text{B} \rightleftharpoons \text{I}$ ($k_1$, $k_{-1}$) and $\text{I} + \text{C} \rightarrow \text{P}$ ($k_2$):

$$k_1[\text{A}][\text{B}] - k_{-1}[\text{I}] - k_2[\text{I}][\text{C}] = 0 \ \Rightarrow\ [\text{I}] = \frac{k_1[\text{A}][\text{B}]}{k_{-1} + k_2[\text{C}]}$$ $$\text{rate} = \frac{k_1 k_2 [\text{A}][\text{B}][\text{C}]}{k_{-1} + k_2[\text{C}]}$$

Always look at the limits. If $k_2[\text{C}] \gg k_{-1}$, the rate becomes $k_1[\text{A}][\text{B}]$ (the first step is rate-determining). If $k_2[\text{C}] \ll k_{-1}$, it becomes $(k_1/k_{-1})k_2[\text{A}][\text{B}][\text{C}]$, the pre-equilibrium result. A rate law with a sum in the denominator is a clue that a steady state is involved.

An enzyme E binds its substrate S, then turns it into product:

$$\text{E} + \text{S} \rightleftharpoons \text{ES} \rightarrow \text{E} + \text{P}$$

with rate constants $k_1$, $k_{-1}$ and $k_2$. The steady-state approximation on ES gives the Michaelis-Menten equation:

$$v = \frac{V_{\max}[\text{S}]}{K_M + [\text{S}]}, \qquad K_M = \frac{k_{-1} + k_2}{k_1}, \qquad V_{\max} = k_2[\text{E}]_0$$

• At low $[\text{S}]$ the rate is first order in S; at high $[\text{S}]$ the enzyme is saturated and the rate levels off at $V_{\max}$ (zero order in S).
• When $[\text{S}] = K_M$, $v = V_{\max}/2$. A small $K_M$ means the enzyme works well at low substrate concentration.
• $k_2$ is also called the turnover number $k_{\text{cat}}$: substrate molecules converted per enzyme per second.
• Taking reciprocals gives a straight line (Lineweaver-Burk): $\dfrac{1}{v} = \dfrac{K_M}{V_{\max}}\cdot\dfrac{1}{[\text{S}]} + \dfrac{1}{V_{\max}}$. The intercept gives $V_{\max}$ and the gradient gives $K_M/V_{\max}$.
• A competitive inhibitor competes for the active site: $V_{\max}$ is unchanged but the apparent $K_M$ grows. A non-competitive inhibitor lowers $V_{\max}$.

A crystal is a repeating 3D pattern. The smallest repeating box is the unit cell. For a cube of edge $a$, count the atoms that belong to one cell:

• corner atom: shared by 8 cells, counts $\tfrac{1}{8}$; edge: $\tfrac{1}{4}$; face: $\tfrac{1}{2}$; body centre: 1.
• Simple cubic: $Z = 1$. Body-centred cubic (bcc): $Z = 2$. Face-centred cubic (fcc, also called cubic close packed): $Z = 4$.

Atoms touch along the line where they are closest: the edge in simple cubic ($a = 2r$), the body diagonal in bcc ($\sqrt{3}a = 4r$), and the face diagonal in fcc ($\sqrt{2}a = 4r$).

The density follows from the mass in one cell divided by its volume:

$$\rho = \frac{Z M}{N_A a^3}$$

Use $a$ in cm to get g cm^-3 (1 nm = $10^{-7}$ cm).

Ionic solids are often a close-packed array of anions with cations in the holes. An fcc array of 4 anions has 4 octahedral holes and 8 tetrahedral holes. NaCl: $\text{Na}^+$ in all octahedral holes ($Z = 4$ formula units). Zinc blende ZnS: $\text{Zn}^{2+}$ in half the tetrahedral holes. CsCl is different: simple cubic $\text{Cl}^-$ with $\text{Cs}^+$ at the body centre. The formula always follows from the ratio of ions in one cell.

You know that nucleophiles attack the $\delta+$ carbon of C=O. The olympiad adds two ideas: carbon nucleophiles, and the carbon next to C=O (the alpha carbon) acting as a nucleophile itself.

Grignard reagents $\text{RMgBr}$, made from a halogenoalkane and magnesium in dry ether, behave like $\text{R}^-$. Followed by dilute acid:

• methanal gives a primary alcohol; other aldehydes give secondary alcohols; ketones give tertiary alcohols;
• carbon dioxide gives a carboxylic acid with one more carbon;
• esters react twice, giving a tertiary alcohol with two identical R groups.

Enolates. An H on the alpha carbon is weakly acidic (p$K_a$ about 20 for a ketone, about 11 between two C=O groups) because the anion is shared between C and O. The enolate is a carbon nucleophile.

• Aldol addition: the enolate of one aldehyde or ketone attacks the C=O of another, giving a $\beta$-hydroxy carbonyl compound. Heating removes water to give an $\alpha,\beta$-unsaturated carbonyl (aldol condensation).
• Claisen condensation: an ester enolate attacks another ester; alkoxide leaves; the product is a $\beta$-keto ester.
• Michael addition: an enolate adds to the end carbon of an $\alpha,\beta$-unsaturated carbonyl. A Michael addition followed by an intramolecular aldol makes a new six-membered ring (the Robinson annulation).

In a pericyclic reaction the electrons move round a closed loop in one step, with no intermediate. The most important one for you is the Diels-Alder reaction, a [4+2] cycloaddition:

• A conjugated diene (4 $\pi$ electrons) reacts with a dienophile, an alkene or alkyne (2 $\pi$ electrons), to give a cyclohexene.
• Two new C-C $\sigma$ bonds form; the new C=C sits between what were C2 and C3 of the diene.
• The diene must be able to take the s-cis shape (both double bonds on the same side of the central bond). Cyclopentadiene is locked s-cis and is very reactive.
• It goes fastest when the dienophile carries electron-withdrawing groups (CHO, COOR, CN) and the diene carries electron-donating ones.
• Stereospecific: groups that are cis on the dienophile stay cis in the ring.
• Endo rule: with cyclic dienes, the electron-withdrawing group usually ends up under the new ring (endo) in the main product.

To work backwards (retrosynthesis), find the C=C in the cyclohexene ring, count two bonds away from each end of it, and break those two ring bonds. The four-carbon piece holding the C=C was the diene; the other two carbons were the dienophile.

Other pericyclic reactions you may meet: electrocyclic ring closures and [3,3] sigmatropic rearrangements (Cope and Claisen).

A saturated, acyclic hydrocarbon has the formula $\text{C}_n\text{H}_{2n+2}$. Every ring or $\pi$ bond removes two hydrogens. The degree of unsaturation (also called double bond equivalent, DBE) counts how many:

$$\text{DBE} = C - \frac{H}{2} - \frac{X}{2} + \frac{N}{2} + 1$$

where $C$, $H$, $X$ (halogens) and $N$ are the numbers of each atom. Oxygen and sulfur do not change the count, so ignore them.

How to read the number:

• each ring = 1; each C=C or C=O = 1; each triple bond (C≡C or C≡N) = 2;
• a benzene ring = 4 (one ring and three C=C), so a DBE of 4 or more with 6 or more carbons suggests an aromatic ring;
• DBE = 0 means no rings and no multiple bonds: alcohols, ethers, amines, alkanes.

The answer must be a whole number of 0 or more; if not, recheck the formula. Combine the DBE with the IR (is there a C=O?) and NMR (are there aromatic H near $\delta$ 7?) to decide where the unsaturation is.

When light passes through a solution, some is absorbed. If $I_0$ goes in and $I$ comes out:

$$T = \frac{I}{I_0}, \qquad A = \log_{10}\frac{I_0}{I} = -\log_{10} T$$

The Beer-Lambert law says absorbance is proportional to concentration and path length:

$$A = \varepsilon c l$$

Here $c$ is in mol dm^-3, $l$ is the cell length in cm (usually 1.00 cm) and $\varepsilon$ is the molar absorption coefficient in dm³ mol^-1 cm^-1, a property of the substance at that wavelength.

• A calibration line of $A$ against $c$ for standard solutions lets you read off an unknown. The law works best for $A$ below about 1.
• Absorbances add: for a mixture, $A = (\varepsilon_1 c_1 + \varepsilon_2 c_2) l$. Measuring at two wavelengths gives two equations for two unknowns.
• A compound looks the colour complementary to the one it absorbs: absorbing orange-red light makes a solution look blue-green.
• Longer conjugated systems absorb at longer wavelength (see the particle in a box), and d-d transitions in complexes give the colours of transition-metal compounds.

Vibration. Treat a bond as a spring of force constant $k$ joining two masses. It vibrates at

$$\nu = \frac{1}{2\pi}\sqrt{\frac{k}{\mu}}, \qquad \mu = \frac{m_1 m_2}{m_1 + m_2}$$

where $\mu$ is the reduced mass. Spectra quote the wavenumber $\tilde{\nu} = \nu / c$ in cm^-1. Stiffer bonds (triple bonds) and lighter atoms (H) give higher wavenumbers. Changing an isotope leaves $k$ the same, so $\tilde{\nu} \propto 1/\sqrt{\mu}$: swapping H for D lowers a C-H stretch by a factor of about 0.73.

Rotation. A diatomic molecule turning end over end has energies

$$E_J = hcB\,J(J+1), \qquad B = \frac{h}{8\pi^2 c I}, \qquad I = \mu r^2$$

with $J = 0, 1, 2, \dots$ and $B$ in cm^-1 when $c$ is in cm s^-1. Allowed jumps are $\Delta J = \pm 1$, so the pure rotational (microwave) spectrum is a row of lines spaced $2B$ apart. Measure the spacing, find $B$, then $I$, then the bond length $r$.

In a gas-phase IR spectrum, each vibrational jump comes with a rotational jump, giving two branches of lines on either side of the band centre, also spaced about $2B$ apart.

Use masses in kg: 1 u = $1.6605 \times 10^{-27}$ kg.