2026 paper questions

An increase in oxidation number means the substance loses electrons, so it is oxidized. A substance that is oxidized gives its electrons to another species and so acts as the reducing agent. An oxidizing agent does the opposite: its oxidation number decreases. A catalyst is not changed by the reaction.

The textbook defines $K_{pc}$ with the concentration in the organic (non-polar) solvent on top and the concentration in water (the polar solvent) at the bottom, $K_{pc}=[\text{solute}{]}_{\text{organic}}/[\text{solute}{]}_{\text{water}}$. A solute that prefers the polar solvent has the larger concentration in the denominator, so $K_{pc}<1$.

It cannot be zero, because some solute always enters each layer; it would equal 1 only if the solute were equally soluble in both. (If the ratio were taken the other way, polar over non-polar, the value would be greater than 1, so the answer depends on the book's convention.)

In a weak acid and strong base titration the salt formed (for example $\mathrm{CH}{}_3\mathrm{COONa}$) contains the anion of a weak acid. It reacts with water and releases $\mathrm{OH}{}^{-}$ ($\mathrm{CH}{}_3\mathrm{COO}{}^{-}+\mathrm{H}{}_2\mathrm{O}\rightleftharpoons \mathrm{CH}{}_3\mathrm{COOH}+\mathrm{OH}{}^{-}$), so the equivalence point is at about pH 8.7 to 9.

Strong acid + strong base gives pH 7, strong acid + weak base gives about 5.3, and weak acid + weak base depends on $K_a$ and $K_b$ (about 7 for ethanoic acid and ammonia).

Group 2 sulphates become less soluble down the group, because the hydration enthalpy of the cation falls more than the lattice energy (the large sulphate ion keeps the lattice energy nearly constant). Barium is lowest, so $\mathrm{BaSO}{}_4$ is the least soluble (about $10^{-5}\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$).

A ligand has a lone pair of electrons and donates it into a vacant orbital of the metal ion, forming a dative (coordinate) bond. The ligand is a Lewis base.

It does not accept electrons (A), the bonds are not ionic (B), and losing protons (C) describes an acid, not a ligand.

Stereoisomers have the same structural formula (same atoms joined in the same order) but a different arrangement of groups in space. Its two types are geometrical (cis-trans) and optical isomerism. Chain, position and functional group isomerism change the order in which atoms are joined, so they are types of structural isomerism.

Benzene has six π electrons delocalised over a closed ring; its resonance energy is $152\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. 1,3-Cyclohexadiene has only slight conjugation (about $8\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$), cyclohexene has one isolated C=C bond and hexane has no π electrons at all, so none of them is stabilised by resonance like benzene.

Aqueous hydroxide at a lower temperature acts as a nucleophile and substitutes the chlorine: $\mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{Cl}+\mathrm{OH}{}^{-}\to \mathrm{CH}{}_3\mathrm{CH}{}_2\mathrm{OH}+\mathrm{Cl}{}^{-}$. Ethene forms by elimination with hot alcoholic KOH. Benzene and phenol cannot form from a two-carbon compound.

Diazonium salts are unstable. Above about $10^{\circ }\mathrm{C}$ the benzenediazonium ion reacts with water, losing $\mathrm{N}{}_2$ gas and forming phenol: $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{N}{}_2{}^{+}+\mathrm{H}{}_2\mathrm{O}\to \mathrm{C}{}_6\mathrm{H}{}_5\mathrm{OH}+\mathrm{N}{}_2+\mathrm{H}{}^{+}$. That is why diazotisation is carried out at $0$ to $5^{\circ }\mathrm{C}$.

Hot alkaline $\mathrm{KMnO}{}_4$ followed by acid oxidises any alkyl side chain that has a benzylic H atom to a carboxyl group, whatever its length: methylbenzene and ethylbenzene both give benzoic acid, $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{COOH}$. The ring itself is not oxidised.

The cyanide ion is a nucleophile. It replaces the halogen: $\mathrm{R}-\mathrm{X}+\mathrm{CN}{}^{-}\to \mathrm{R}-\mathrm{CN}+\mathrm{X}{}^{-}$. No double bond is formed (not elimination) and nothing adds across a multiple bond (not addition).

The segment repeats the two-carbon unit $-\mathrm{CH}{}_2-\mathrm{CH}{}_2-$. Putting back the double bond between these two carbons gives ethene, $\mathrm{CH}{}_2=\mathrm{CH}{}_2$, so the polymer is polythene: $-[\mathrm{CH}{}_2-\mathrm{CH}{}_2]{}_n-$.

Propene would give a $\mathrm{CH}{}_3$ branch on every second carbon, and butene would give an ethyl branch. Methane has no double bond, so it cannot polymerise.

Enzymes are proteins whose shape, and so activity, depends on temperature: the rate rises up to the optimum temperature and falls when the enzyme is denatured. Density, colour and pressure are not factors that control enzyme activity.

The molecular ion is the whole molecule minus one electron and carries a charge of $+1$, so the $m/z$ value of its peak equals the relative molecular mass of the compound. It says nothing about colour, and functional groups are found from the fragment peaks and from other kinds of spectra.

The absorption moves to longer wavelength and becomes more intense as conjugation increases. Hexa-1,3,5-triene has three conjugated $\mathrm{C}=\mathrm{C}$ bonds (${\lambda }_{\max }\approx 258\mathrm{nm}$), buta-1,3-diene has two ($217\mathrm{nm}$), and ethene and propene have one isolated $\mathrm{C}=\mathrm{C}$ each (below $200\mathrm{nm}$, so no absorption in the normal UV-visible range).

The two $\mathrm{CH}{}_3$ carbons are equivalent (the molecule is symmetrical about $\mathrm{C}=\mathrm{O}$) and give one signal at $30.8\mathrm{ppm}$; the carbonyl carbon gives the second at $206.6\mathrm{ppm}$. Three carbon atoms, but only two environments.

A compound that is held strongly by the stationary phase spends most of its time there and moves forward only while it is in the mobile phase, so it moves slower and has a low $R_f$. $R_f=1$ is for a compound that does not interact at all and moves with the solvent front; colour has nothing to do with the interaction.

Split the reaction into half-reactions. Iron loses electrons (oxidation, so it is the anode) and $\mathrm{Pb}{}^{2+}$ gains electrons (reduction, so it is the cathode).

$E_{\text{cell}}^{\circ }$ is positive, so the reaction is feasible (spontaneous). Iron is above lead in the electrochemical series (more negative $E^{\circ }$), so it is the stronger reducing agent and can displace lead from a solution of its ions.

The organic layer holds 2.5 times the concentration of the aqueous layer, as expected for $K_{pc}>1$: the solute is more soluble in the organic solvent.

Oxidation number of N in $\mathrm{NH}{}_3$ (1 mark)

Oxidation number of N in $\mathrm{NO}{}_2$ (1 mark)

Conclusion (1 mark)

The oxidation number of nitrogen increases from $-3$ to $+4$, a change of $+7$ (7 electrons lost per N atom). An increase in oxidation number is oxidation, so $\mathrm{NH}{}_3$ is oxidized in this conversion (for example by oxygen in the Ostwald process).

HCl (strong acid) is in the flask and $\mathrm{NH}{}_3$ (weak base) is added from the burette. Take $25\mathrm{c}{\mathrm{m}}^3$ of $0.1\mathrm{M}$ HCl and $0.1\mathrm{M}$ $\mathrm{NH}{}_3$, so the equivalence point is at $25\mathrm{c}{\mathrm{m}}^3$.

Curve (1.5 marks)

Labels (1.5 marks)

• Initial pH about 1: only the strong acid is present, $[\mathrm{H}{}^{+}]=0.1\mathrm{M}$.
• **Equivalence point at $25\mathrm{c}{\mathrm{m}}^3$, pH about 5.3 (below 7):** the flask contains $\mathrm{NH}{}_4\mathrm{Cl}$, and the $\mathrm{NH}{}_4{}^{+}$ ion is acidic: $\mathrm{NH}{}_4{}^{+}+\mathrm{H}{}_2\mathrm{O}\rightleftharpoons \mathrm{NH}{}_3+\mathrm{H}{}_3\mathrm{O}{}^{+}$. The steep section is short, from about pH 3 to 7.5.
• Final pH about 9 to 10: excess ammonia is a weak base (it forms an $\mathrm{NH}{}_3$/$\mathrm{NH}{}_4{}^{+}$ buffer), so the pH never reaches the 12 to 13 of a strong alkali.

Molten NaCl contains only \ce{Na+} and \ce{Cl-} ions, so these are the ions discharged.

Anode (+): chlorine gas (1.5 marks)

Chloride ions move to the positive anode and are oxidized; greenish-yellow chlorine gas is given off.

Cathode (−): sodium metal (1.5 marks)

Sodium ions move to the negative cathode and are reduced; molten sodium metal collects at the cathode.

Overall: \ce{2NaCl(l) -> 2Na(l) + Cl2(g)}.

Direct method, as in the book: K_a=\dfrac{[\ce{CH3COO-}][\ce{H+}]}{[\ce{CH3COOH}]} gives 1.8\times10^{-5}=\dfrac{(1)[\ce{H+}]}{1}, so [\ce{H+}]=1.8\times10^{-5}\,\mathrm{M} and \mathrm{pH}=-\log(1.8\times10^{-5})=4.74. When salt and acid are equal, \mathrm{pH}=\mathrm{p}K_a.

Type of ions (1 mark)

Group 2 elements are metals with the valence configuration ns^2. They lose both valence electrons and form positive ions (cations) with a noble gas configuration: \ce{M -> M^2+ + 2e-}, for example \ce{Mg -> Mg^2+ + 2e-}. Their compounds are ionic (except those of Be).

Charge (1 mark)

The charge on each ion is \mathbf{2+} (\ce{Be^2+}, \ce{Mg^2+}, \ce{Ca^2+}, \ce{Sr^2+}, \ce{Ba^2+}), so their oxidation state is +2.

Reactivity compared with Group I (1 mark)

Group 2 elements are less reactive than the Group 1 elements of the same period. Their atoms are smaller and have a greater nuclear charge, so their ionisation energies are higher (Mg 738, Na 496\,\mathrm{kJ\,mol^{-1}}), and they must lose two electrons instead of one. For example, sodium reacts vigorously with cold water, \ce{2Na + 2H2O -> 2NaOH + H2}, but magnesium reacts only very slowly.

• Shape (1 mark): linear. The two \ce{NH3} ligands lie on opposite sides of the silver ion. Two bond pairs repel each other least when they are as far apart as possible (VSEPR).
• Bond angle (1 mark): \ce{H3N-Ag-NH3} angle =180^\circ.
• Coordination number (1 mark): 2, because two monodentate \ce{NH3} ligands each form one dative bond to \ce{Ag+}.

Silver is +1 ([\mathrm{Kr}]\,4d^{10}) and ammonia is neutral, so the ion carries a charge of +1. It is the active species in Tollens' reagent.

Lithium with oxygen (1 mark)

Lithium (Group 1) burns readily when heated in air or oxygen, with a crimson flame, forming the white normal oxide \ce{Li2O}:

Beryllium with oxygen (1 mark)

Beryllium (Group 2) is much less reactive. The massive metal is protected by a thin layer of \ce{BeO}; it burns only when powdered and strongly heated, forming beryllium oxide:

Comparison and reason (1 mark)

Lithium is more reactive with oxygen than beryllium. Li loses one electron (first IE 520\,\mathrm{kJ\,mol^{-1}}) while Be must lose two electrons from a smaller atom (first IE 900, second IE 1757\,\mathrm{kJ\,mol^{-1}}). Both form only normal oxides (no peroxides), because their small ions cannot stabilise the large peroxide ion; \ce{Li2O} is basic while \ce{BeO} is amphoteric.

The colour of a transition metal compound comes from a d-d transition. Ligands split the 3d orbitals into a lower and a higher set separated by \Delta E. An electron in the lower set absorbs a photon of visible light with energy \Delta E=h\nu and jumps into a vacant place in the higher set; we see the complementary colour.

In a d^{10} ion, such as \ce{Zn^2+}, \ce{Cu+} or \ce{Ag+}, all five d orbitals are completely filled. There is no vacancy in the upper set, so no electron can be promoted and no visible light is absorbed. All the light is transmitted or reflected, so the compounds are white or colourless (for example \ce{ZnSO4}, \ce{CuI}).

Answer (1 mark)

Yes. \ce{CH3-CH(OH)-Br} (1-bromoethan-1-ol) is chiral, so each of its pure enantiomers is optically active.

Justification (2 marks)

Look at the middle carbon, \ce{CH3-\overset{*}{C}H(OH)-Br}. It is bonded to four different groups: \ce{-H}, \ce{-OH}, \ce{-Br} and \ce{-CH3}. So it is a chiral carbon (asymmetric centre). The molecule has no plane of symmetry, and its mirror image cannot be superimposed on it. The two mirror images are a pair of enantiomers: one turns plane-polarized light clockwise (+), the other turns it anticlockwise (−) by the same angle.

Comparison of the enthalpies of hydrogenation (2 marks)

The enthalpy of hydrogenation of cyclohexene, which has one C=C bond in a six-membered ring, is -120\,\mathrm{kJ\,mol^{-1}}. The hypothetical 1,3,5-cyclohexatriene would have three localised C=C bonds, so its expected value is three times as large:

The measured value for benzene is much smaller:

Both reactions give the same product, cyclohexane. Benzene therefore releases 360-208=152\,\mathrm{kJ\,mol^{-1}} less heat, which means it starts 152\,\mathrm{kJ\,mol^{-1}} lower in energy than the cyclohexatriene structure.

Significance of the difference (1 mark)

The difference, 152\,\mathrm{kJ\,mol^{-1}}, is the resonance (delocalisation) energy of benzene. It shows that benzene is not cyclohexatriene: its six π electrons are delocalised over all six carbon atoms, and this makes it much more stable. Because addition would destroy this stable π system, benzene resists addition reactions and prefers electrophilic substitution, which keeps the aromatic ring.

Retrosynthesis works backwards from the target. The target, propan-2-ol, is a secondary alcohol with the \ce{-OH} group on the middle carbon (C2). It can be made by nucleophilic substitution of a halogenoalkane by \ce{OH-}.

Product (1 mark)

Phenol decolourises bromine water at once and gives a white precipitate of 2,4,6-tribromophenol:

Reason (2 marks)

A lone pair on the oxygen of the \ce{-OH} group is delocalised into the benzene ring, which greatly increases the electron density of the ring, especially at positions 2, 4 and 6. This activated ring polarises the \ce{Br2} molecule by itself and attacks it, so no catalyst or heating is needed. Benzene has no such electron-donating group; its ring is less electron-rich, so it needs liquid bromine with an \ce{FeBr3} catalyst to form the \ce{Br+} electrophile.

Comparison (1 mark)

Chloroacetic acid, \ce{CH2ClCOOH} (\mathrm{p}K_a=2.9), is a stronger acid than acetic acid, \ce{CH3COOH} (\mathrm{p}K_a=4.7). It is roughly 100 times stronger (the K_a ratio is 10^{1.8}, about 60 to 80).

Explanation (2 marks)

Chlorine is more electronegative than carbon, so the \ce{C-Cl} bond pulls electron density towards chlorine along the chain (-I inductive effect). In the chloroacetate ion this pulls part of the negative charge away from the \ce{COO-} group, spreading it over more atoms. The anion is therefore more stable and has less tendency to recombine with \ce{H+}, so the acid ionises more. The pull also makes the \ce{O-H} bond more polar and easier to break.

Acetic acid has a methyl group instead, which pushes electrons towards \ce{COO-} (+I effect). This concentrates the negative charge, makes the acetate ion less stable, and so acetic acid is the weaker acid.

Equations (2 marks)

Nitrous acid is prepared in the mixture from sodium nitrite and dilute hydrochloric acid, and reacts with phenylamine in the cold (diazotisation):

Reason for the low temperature (1 mark)

Benzenediazonium chloride is unstable. If the mixture warms up (above about 10^\circ\mathrm{C}) it decomposes, giving off nitrogen gas and forming phenol, so the product is lost. Nitrous acid also decomposes on warming. The mixture is therefore kept ice-cold and the salt is used at once.

Zwitterionic form (1 mark)

In an amino acid, \ce{H2N-CHR-COOH}, the \ce{-COOH} group gives a proton to the \ce{-NH2} group of the same molecule. The result is the zwitterion \ce{H3N+-CHR-COO-}: it has a positive and a negative part but no overall charge. For glycine (R = H):

Acidic site (1 mark)

The \ce{-NH3+} group is the acidic site: it can donate \ce{H+} to a base, \ce{H3N+CH2COO- + OH- -> H2NCH2COO- + H2O}.

Basic site (1 mark)

The \ce{-COO-} group is the basic site: it can accept \ce{H+} from an acid, \ce{H3N+CH2COO- + H+ -> H3N+CH2COOH}.

Equation (2 marks)

Vinyl chloride (chloroethene) is heated to about 40 to 60\,^\circ\mathrm{C} under about 13\,\mathrm{atm} with a small amount of a peroxide initiator. The \pi bond of each \ce{C=C} opens and the molecules join end to end:

Type of polymerisation (1 mark)

It is addition (free-radical) polymerisation. The monomer has a \ce{C=C} double bond, the monomers simply add together, and no small molecule is eliminated, so the repeat unit \ce{C2H3Cl} has the same formula as the monomer.

Addition polymerisation is the joining of unsaturated monomers (with a \ce{C=C} bond) without the loss of any atom. Condensation polymerisation is the joining of monomers that have two functional groups each, with the elimination of a small molecule such as \ce{H2O} or \ce{HCl} at every link.

Any three of the following, each with one line of explanation:

1. Temperature: the rate increases up to the optimum temperature (37^\circ\mathrm{C} for human enzymes); above it the enzyme is denatured and the rate falls sharply.
2. pH: each enzyme works best at its optimum pH (pepsin about 2, trypsin about 8); a change in pH alters the charges on the active site and extreme pH denatures the enzyme.
3. Substrate concentration: the rate increases with substrate concentration until all active sites are occupied (saturation), then stays constant at V_{\max}.
4. Enzyme concentration: with excess substrate the rate is directly proportional to the enzyme concentration.
5. Inhibitors: substances such as malonic acid or \ce{Hg^2+} ions reduce or stop enzyme activity.

On the basis of chemical structure, lipids are classified as follows.

Definition (1 mark)

The molecular ion peak is the peak in a mass spectrum produced by the molecular ion, the ion formed when a molecule loses one electron. It appears at the highest m/z value in the spectrum, apart from the small M+1 and M+2 isotope peaks.

Symbol (1 mark)

It is written \ce{M+} (more exactly \ce{M^{+\bullet}}, a radical cation), formed by electron impact: \ce{M + e- -> M^{+\bullet} + 2e-}.

Importance (1 mark)

Its m/z value equals the relative molecular mass of the compound, because the charge on the ion is +1. For example, hexane gives \ce{M+} at m/z=86, so M_r=86. It is also the reference point for the M+1 peak (number of carbon atoms), the M+2 peak (chlorine or bromine) and the mass losses that identify fragments.

The molecule contains 5 carbon atoms. With the SLO form of the formula, n=\dfrac{5.5}{0.01107\times100}=4.97\approx5, the same answer.

Principle (2 marks)

A covalent bond behaves like a spring and vibrates (stretches and bends) at a natural frequency that depends on the bond strength and the masses of the bonded atoms. When infrared radiation (4000 to 625\,\mathrm{cm^{-1}}) passes through a sample, each bond absorbs the radiation whose frequency matches its natural vibrational frequency, if the vibration changes its dipole moment. The bond then vibrates with a larger amplitude. Each functional group absorbs at a characteristic wavenumber (for example \ce{C=O} near 1715\,\mathrm{cm^{-1}}, \ce{O-H} at 3200\text{–}3500\,\mathrm{cm^{-1}}), so the spectrum of % transmittance against wavenumber reveals the functional groups.

One application (1 mark)

Identification of the functional groups in an unknown organic compound, for example a broad band at 3300\,\mathrm{cm^{-1}} shows an alcohol \ce{O-H} group. (Other acceptable answers: checking purity, following a reaction such as the oxidation of ethanol, confirming identity from the fingerprint region, breathalysers.)

An NMR sample is dissolved in a solvent, and the solvent molecules greatly outnumber the sample molecules. A solvent with ordinary hydrogen (\ce{^1H}) would give a huge signal that would swamp the small signals of the sample and hide them.

In deuterated solvents, such as \ce{CDCl3}, \ce{D2O} and \ce{(CD3)2CO}, hydrogen is replaced by deuterium (\ce{^2H} or D). Deuterium absorbs at a very different frequency from \ce{^1H}, so the solvent gives no peak in the proton spectrum and does not interfere with the analysis.

Deuterated solvents have almost the same chemical properties as the normal solvents, so they dissolve the sample in the same way. (Solvents without hydrogen, such as \ce{CCl4}, serve the same purpose.)

a. Stationary phase (1.5 marks)

The stationary phase is the phase that does not move during chromatography. It is a solid such as silica gel, alumina or paper, or a liquid coated on the surface of a solid. The components of the mixture are held back by it (by adsorption or by dissolving in it) to different extents: a component held strongly moves slowly. Example: the silica gel layer on a TLC plate, or the water in the fibres of chromatography paper.

b. Solvent front (1.5 marks)

The solvent front is the highest level reached by the solvent (mobile phase) as it moves up the paper or TLC plate. It is marked with a pencil line as soon as the chromatogram is removed from the tank, before the solvent evaporates. The distance from the baseline to the solvent front is the denominator of the R_f value:

Catalyst (1 mark): a substance that increases the rate of a chemical reaction without being used up; it is chemically unchanged at the end of the reaction.

Action (2 marks): a catalyst provides an alternative pathway (mechanism) with a lower activation energy E_a. At the same temperature many more colliding particles now have energy equal to or greater than E_a, so more collisions are successful and the reaction is faster. The energies of reactants and products do not change, so \Delta H is the same, and the equilibrium position is not changed (it is only reached sooner).

Transition metals are good catalysts because they have variable oxidation states and vacant d orbitals that form temporary bonds with reactants, for example iron in the Haber process.

Definition (1 mark)

Pesticides are chemical substances used in agriculture to control or kill pests, including insects, fungi, rodents and weeds, which damage crops and reduce yields. Examples are insecticides, herbicides, fungicides and rodenticides.

Two negative aspects (2 marks)

1. Harm to human health: pesticides can be carcinogenic and neurotoxic, so they can cause adverse health effects in farmers, co-workers and consumers. Residues left on fruits and vegetables also affect consumers.
2. Harm to non-target organisms and the environment: beneficial insects, pollinators such as bees, birds and aquatic organisms are also killed. Runoff water from fields contaminates water sources, and the loss of beneficial insects causes loss of biodiversity.

Mechanism and the carbocation intermediate (3 marks)

tert-Butyl bromide (2-bromo-2-methylpropane) is a tertiary halogenoalkane, so it reacts by the \mathrm{S_N1} mechanism (substitution, nucleophilic, unimolecular). Water is the nucleophile, and the overall reaction is hydrolysis:

The carbocation is sp^2 hybridised and planar, with an empty 2p orbital above and below the plane, so water can attack from either side.

Role of the inductive effect of alkyl groups (2 marks)

Methyl groups are electron-releasing (+I effect). The three methyl groups push σ-bond electron density towards the positive carbon. This partly neutralises and spreads out (disperses) the positive charge, and a dispersed charge is more stable. Hyperconjugation with the 9 neighbouring C–H bonds adds to this stabilisation.

Because the tertiary carbocation is the most stable kind (3^\circ > 2^\circ > 1^\circ > \text{methyl}), step 1 has a relatively low activation energy, so the ionisation can happen and the \mathrm{S_N1} route is followed. The three bulky methyl groups also block attack from the back of the carbon, which rules out \mathrm{S_N2}.

Rate equation (2 marks)

Only tert-butyl bromide takes part in the slow step, so the rate depends only on its concentration. The reaction is first order; water does not appear in the rate equation because it reacts in a fast step after the slow one (and it is also the solvent). k is the rate constant, with unit \mathrm{s^{-1}}.

Transition elements (1 mark)

Transition elements are d-block elements which form one or more stable ions with an incomplete (partly filled) d sub-shell. In period 4 they are Sc to Cu (Ti to Cu in the book's study). Zinc is a d-block element but not a transition element, because \ce{Zn^2+} is 3d^{10}.

Trend in melting point, Ti to Cu (3 marks)

• Melting points are high because of strong metallic bonding: both 3d and 4s electrons are delocalised into the metallic bond.
• Ti to V (and Cr): the melting point rises (1668\to1910^\circ\mathrm{C}) because the number of unpaired 3d electrons available for bonding increases.
• Mn: a sharp dip (1246^\circ\mathrm{C}). Its stable half-filled 3d^{5} and full 4s^{2} sub-shells hold their electrons firmly, so fewer electrons take part in metallic bonding.
• Fe to Cu: the melting point falls gradually (1538\to1085^\circ\mathrm{C}) because the 3d electrons pair up and become less available for bonding.

Trend in atomic radius, Ti to Cu (3 marks)

• Ti to Cr: decreases (147 to 128 pm). The nuclear charge rises by one at each step and outweighs the extra shielding, pulling the electrons closer.
• Cr to Ni: almost constant (128 to 124 pm). Each new electron enters the inner 3d sub-shell; its shielding of the outer 4s electrons and electron-electron repulsion balance the increase in nuclear charge.
• Ni to Cu: slight increase (124 to 128 pm). The 3d sub-shell becomes full; the full 3d^{10} shields strongly and repels the 4s electron, pushing it outwards.

Two factors that affect the basic strength of amines

1. Availability of the lone pair on nitrogen. A base accepts \ce{H+} with the nitrogen lone pair. Electron-donating alkyl groups (+I effect) increase the electron density on nitrogen and make the base stronger; delocalisation of the lone pair into a benzene ring (resonance) makes it less available and the base weaker.
2. Stability of the cation formed. The more the positive charge of \ce{RNH3+} is spread out (by alkyl groups) and stabilised by hydrogen bonding with water (solvation), the more readily the amine accepts \ce{H+}.

Correct order and explanation

All three accept a proton: \ce{RNH2 + H2O <=> RNH3+ + OH-}. A lower \mathrm{p}K_b means a stronger base.

Ethylamine > ammonia. The ethyl group is electron-donating. It pushes electron density onto nitrogen, so the lone pair attracts \ce{H+} more strongly (factor 1). It also spreads the positive charge in \ce{CH3CH2NH3+}, making this ion more stable than \ce{NH4+} (factor 2).

Ammonia > phenylamine. In phenylamine the nitrogen lone pair overlaps with the \pi electrons of the benzene ring and is delocalised over the ring, so it is far less available for \ce{H+} (factor 1). On protonation this delocalisation is lost, so the phenylammonium ion gains no extra stability (factor 2).

The resonance structures show negative charge at positions 2, 4 and 6 of the ring and a positive charge on nitrogen: the lone pair is shared with the ring. This is also why phenylamine reacts so easily with bromine water.

So electron donation makes ethylamine the strongest base and delocalisation makes phenylamine the weakest.

Reaction

Friedel–Crafts alkylation introduces an alkyl group into the benzene ring. Benzene is warmed with chloromethane in the presence of anhydrous aluminium chloride; methylbenzene (toluene) and hydrogen chloride are formed.

Role of the Lewis acid (2 marks)

• A Lewis acid is an electron-pair acceptor. The Al atom in \ce{AlCl3} has only six valence electrons (incomplete octet) and an empty 3p orbital, so it accepts a lone pair from the Cl atom of \ce{CH3Cl}.
• This polarises and breaks the C–Cl bond heterolytically, so the Lewis acid generates the electrophile, the methyl carbocation \ce{CH3+}. Chloromethane on its own is far too weak an electrophile to attack the stable benzene ring.
• \ce{AlCl3} is regenerated in the last step (\ce{AlCl4- + H+ -> AlCl3 + HCl}), so it is a catalyst (a halogen carrier). It must be anhydrous, because water hydrolyses it and destroys the Lewis acid.

Mechanism (4 marks)

Trend (2 marks)

Group 2 carbonates and nitrates decompose on heating to the metal oxide. Thermal stability increases down the group: a higher temperature is needed to decompose the compound as the group is descended.

For example, \ce{MgCO3} decomposes at about 540^\circ\mathrm{C}, \ce{CaCO3} at 900^\circ\mathrm{C}, \ce{SrCO3} at 1280^\circ\mathrm{C} and \ce{BaCO3} at 1360^\circ\mathrm{C}. Nitrates give off brown \ce{NO2} gas and oxygen; \ce{Mg(NO3)2} decomposes most easily and \ce{Ba(NO3)2} least easily.

Reasons (3 marks)

• The carbonate ion \ce{CO3^2-} and nitrate ion \ce{NO3-} are large anions whose electron clouds are easily polarised (distorted).
• All Group 2 cations have a 2+ charge, but their size increases down the group (\ce{Mg^2+ < Ca^2+ < Sr^2+ < Ba^2+}), so their charge density and polarising power decrease.
• A small cation like \ce{Mg^2+} pulls electron density towards itself and weakens a C-O (or N-O) bond in the anion, so the anion breaks up at a low temperature, releasing \ce{CO2} (or \ce{NO2} and \ce{O2}).
• A large cation like \ce{Ba^2+} hardly distorts the anion, so its bonds remain strong and much more heat is needed. The enthalpy of decomposition also becomes more positive down the group (+117 for \ce{MgCO3} to +268\,\mathrm{kJ\,mol^{-1}} for \ce{BaCO3}).

Order of stability (1 mark)

Principle

The strength of an acid depends on the stability of its conjugate base. The more the negative charge of the conjugate base is spread out (delocalised), the more stable it is and the stronger the acid.

Phenol: resonance structures of phenoxide (3 marks)

In the phenoxide ion a lone pair of the negative oxygen is delocalised into the ring. Four resonance structures can be drawn; the charge is spread over O and carbons 2, 4 and 6:

The phenoxide ion is therefore resonance-stabilised, and phenol is the strongest acid of the three (K_a=1.3\times10^{-10}).

Water and ethanol (3 marks)

Hydroxide ion, \ce{H-O^-}: only one structure can be drawn; the charge is localised on one oxygen atom and there is no resonance. \ce{OH-} is less stable than phenoxide, so water (K_a=1.8\times10^{-16}) is weaker than phenol.

Ethoxide ion, \ce{CH3-CH2-O^-}: again only one structure (no resonance). In addition the ethyl group is electron-releasing (+I effect) and pushes electron density onto the oxygen, intensifying the negative charge:

So ethoxide is less stable than hydroxide, and ethanol (K_a\approx10^{-16}) is the weakest acid.

The ease of hydrolysis depends on how electron-deficient the carbon bonded to chlorine is, on the mechanism available, and on the strength of the \ce{C-Cl} bond.

Acyl chloride: hydrolysed most easily (2 marks)

In \ce{CH3COCl} the carbonyl carbon is attached to two electronegative atoms, O and Cl, which both pull electrons away. The carbon is highly \delta+. It is also planar (sp^2) with a \ce{C=O} \pi bond that can take the nucleophile's electron pair. Even water attacks it at room temperature by nucleophilic addition-elimination, giving steamy fumes of \ce{HCl}.

Alkyl chloride: slower (2 marks)

In \ce{CH3CH2Cl} the carbon is attached to only one electronegative atom, so it is less electron-deficient. It is sp^3 with no \pi bond, so the \ce{C-Cl} bond must break during the substitution. Water is too weak a nucleophile; the reaction needs \ce{OH-} and heating under reflux.

Aryl chloride: not hydrolysed (2 marks)

In chlorobenzene a lone pair of chlorine is delocalised into the benzene ring by resonance. The \ce{C-Cl} bond gains partial double-bond character, so it is shorter and stronger and hard to break even by a strong nucleophile. The ring's \pi cloud also repels nucleophiles. So hydrolysis does not occur under ordinary conditions.

Hence: acyl chloride > alkyl chloride > aryl chloride.

The highest level the solvent reaches on the paper or plate is the solvent front. The starting point is the baseline, R_f is a ratio of distances, and the mobile phase is the solvent itself, not a distance.

A catalyst provides an alternative pathway with a lower activation energy, so more collisions are successful. It does not change the enthalpy change (B) or the concentrations (C), and it works without raising the temperature (A).

Zinc goes from 0 to +2: it loses two electrons, so Zn is oxidized. \ce{Cu^2+} is reduced (it is the oxidizing agent); \ce{Zn^2+} and Cu are products, not the species being oxidized.

A strong base such as NaOH is completely ionised in water, so its degree of ionisation is high (\alpha\approx1). Weak bases ionise only partly, which rules out B, and they have low K_b values, which rules out D. Option C contradicts the meaning of 'strong'.

Down the group the atoms get larger and the ionisation energies decrease, so the two outer electrons are lost more easily. Reactivity with oxygen, water and acids therefore increases from Be to Ba.

Zinc ([\mathrm{Ar}]\,3d^{10}4s^{2}) and its only ion \ce{Zn^2+} (3d^{10}) have completely filled d orbitals, so zinc has no stable ion with an incomplete d sub-shell.

Zinc is a metal (not B), it does have d orbitals (not C), and it is not highly reactive (not D).

Optical activity is defined by what the compound does to light: it rotates the plane of plane-polarized light, clockwise (dextrorotatory, +) or anticlockwise (laevorotatory, −). It needs a chiral molecule; double bonds, colour or smell have nothing to do with it.

Alkenes have a localised, exposed π bond and react easily with electrophiles. Benzene has π electrons too, but they are delocalised and stabilised by 152\,\mathrm{kJ\,mol^{-1}}, so it reacts only with strong electrophiles made with a catalyst. Alkanes have only strong, non-polar σ bonds and react only with free radicals, so they are the least reactive.

\ce{CH3CH2Cl + KCN -> CH3CH2CN + KCl}. The cyanide carbon joins the chain, so the two-carbon chloroethane gives a three-carbon nitrile, propanenitrile. Ethanol would need \ce{OH-} as the nucleophile.

On warming with water the \ce{-N2+} group leaves as nitrogen gas and phenol is formed: \ce{C6H5N2+Cl- + H2O -> C6H5OH + N2 ^ + HCl}.

Dichloroacetic acid has two chlorine atoms on the carbon next to \ce{COOH}. Their combined -I effect stabilises the carboxylate ion the most, so it is the strongest acid here: \mathrm{p}K_a values are \ce{Cl2CHCOOH} 1.3, \ce{ClCH2COOH} 2.9, \ce{CH3COOH} 4.7, \ce{CH3CH2COOH} 4.9.

Methylamine (\mathrm{p}K_b 3.36) is the strongest base here. Trimethylamine has more electron-donating groups, but in water its cation \ce{(CH3)3NH+} has only one \ce{N-H} bond to hydrogen-bond with water and is crowded by three methyl groups, so it is poorly solvated (\mathrm{p}K_b 4.20). Ammonia (4.75) and phenylamine (9.36) are weaker still.

Trimethylamine would be the answer only in the gas phase, where solvation plays no part.

An amino acid has two different functional groups, \ce{-NH2} and \ce{-COOH}. The \ce{-NH2} of one molecule reacts with the \ce{-COOH} of the next to form a peptide (amide) link and eliminate \ce{H2O}, giving a polypeptide:

Vinyl chloride, ethene and propene all have a \ce{C=C} bond and form addition polymers (PVC, polythene, polypropene).

Haemoglobin in red blood cells binds \ce{O2} at the \ce{Fe^2+} ion of each of its four haem groups. Keratin, elastin and collagen are fibrous structural proteins.

\ce{CH3CH2CH2+} contains 3 carbon and 7 hydrogen atoms: 3(12)+7(1)=43. Its charge is +1, so m/z=43. It is the propyl cation, the base peak in the spectrum of n-pentane.

The base peak is the tallest peak, caused by the most abundant ion, and the most abundant ion is the most stable one (for n-pentane, \ce{C3H7+} at 43). The molecular ion (A) is the base peak only in a few cases such as ethene, and the heaviest ion (C) is usually the molecular ion, which is often small.

Ethanol has two carbon atoms in different environments: C1 bonded to O (58\,\mathrm{ppm}) and the \ce{CH3} carbon (18\,\mathrm{ppm}). So there are 2 environments and 2 peaks.

The half-reaction with the more positive reduction potential occurs as reduction; the one with the more negative potential is reversed and occurs as oxidation.

Oxidation reaction (1 mark)

Reduction reaction (1 mark)

Overall: \ce{Zn + 2Fe^3+ -> Zn^2+ + 2Fe^2+}. The iron half-equation is doubled to balance electrons, but its E^\circ is not doubled.

Cell potential (1 mark)

The positive value shows that zinc reduces \ce{Fe^3+} ions spontaneously.

K_{pc} has no units because it is a ratio of two concentrations. Since K_{pc}>1, the solute is twice as concentrated in ether: it is more soluble in ether than in water.

Definition (1 mark)

The standard hydrogen electrode (SHE) is the primary reference electrode. It is a platinum foil coated with platinum black, dipped in 1\,\mathrm{M} \ce{H+} solution (1 M HCl), over which pure hydrogen gas at 1\,\mathrm{atm} is bubbled at 298\,\mathrm{K}. Its potential is taken as exactly 0.00\,\mathrm{V}.

Working (2 marks)

Platinum black adsorbs hydrogen and sets up an equilibrium between \ce{H2} gas and \ce{H+} ions; platinum itself does not react and only carries electrons. The SHE is joined through a salt bridge to the half-cell under test, and the voltmeter reading gives that half-cell's standard electrode potential. The SHE can act either way:

With zinc the SHE is the cathode and the cell reads 0.76\,\mathrm{V}, so E^\circ(\ce{Zn^2+}/\ce{Zn})=-0.76\,\mathrm{V}. With copper the SHE is the anode and the cell reads 0.34\,\mathrm{V}, so E^\circ(\ce{Cu^2+}/\ce{Cu})=+0.34\,\mathrm{V}.

Blood pH is kept at 7.35 to 7.45 by the bicarbonate buffer. Carbon dioxide dissolves in blood to form carbonic acid, which ionizes to bicarbonate; carbonic anhydrase in red blood cells catalyses these steps:

Removal of excess acid (1.5 marks)

When acid (\ce{H+}) enters the blood, bicarbonate acts as a base and combines with it to form carbonic acid, a weak acid. The carbonic acid breaks down to \ce{CO2} and water, and the lungs breathe out the \ce{CO2}. So [\ce{H+}] does not rise.

Removal of excess base (1.5 marks)

When a base (\ce{OH-}) enters the blood, carbonic acid releases \ce{H+}, which neutralises it to water and forms more bicarbonate. So [\ce{H+}] does not fall.

The lungs (control of \ce{CO2}) and kidneys (control of \ce{HCO3-}) support these reactions, so blood pH stays nearly constant.

Stronger base (1 mark)

Base B (K_b=1.0\times10^{-2}) is the stronger base.

Justification (2 marks)

K_b measures the extent of ionisation of a base. A larger K_b means the equilibrium lies further to the right: more base molecules accept protons from water and more \ce{OH-} ions are formed. K_b of B is about 550 times that of A (\mathrm{p}K_b = 2.00 for B and 4.74 for A).

So B is about 20 times more ionised than A and gives a much higher [\ce{OH-}] (pH 12.4 against 11.1 for 0.1\,\mathrm{M} solutions). Greater ionisation means a stronger base.

Take 25\,\mathrm{cm^3} of 0.1\,\mathrm{M} ammonia (weak base) in the flask and add 0.1\,\mathrm{M} HCl (strong acid) from the burette.

Sketch (2 marks)

• The curve starts at about pH 11 (weak base, only partly ionised).
• It falls slowly through a buffer region (\ce{NH3}/\ce{NH4+}); half-way, \mathrm{pH}=9.25.
• It then drops steeply from about 7.5 to 3 and levels off near pH 1.5 in excess acid.

Equivalence point (1 mark)

The equivalence point is at 25\,\mathrm{cm^3} of acid, in the middle of the steep drop, at about pH 5.3 (below 7). At this point the flask holds \ce{NH4Cl}, whose \ce{NH4+} ion is acidic: \ce{NH4+ + H2O <=> NH3 + H3O+}. Methyl orange is a suitable indicator.

Comparison (1 mark)

The first ionisation energies of Group 2 elements are higher than those of the Group 1 elements in the same period. Examples: Li 520 vs Be 900\,\mathrm{kJ\,mol^{-1}}; Na 496 vs Mg 738\,\mathrm{kJ\,mol^{-1}}; K 419 vs Ca 590\,\mathrm{kJ\,mol^{-1}}. (The second IE of Group 1 is, however, far higher than that of Group 2, because it removes an electron from a noble gas core.)

Reason (2 marks)

• Greater nuclear charge: a Group 2 atom has one more proton than the Group 1 atom before it, while the added electron enters the same shell and shields poorly. The valence electrons feel a larger effective nuclear charge.
• Smaller atomic size: the stronger attraction pulls the electrons closer, so Group 2 atoms are smaller (Mg 0.160\,\mathrm{nm}, Na 0.186\,\mathrm{nm}) and the outer electrons are held more tightly.
• Stable configuration: Group 2 atoms have a completely filled ns^2 subshell, which is more stable than the half-filled ns^1 subshell of Group 1.

So more energy is needed to remove an electron from a Group 2 atom. This is why Group 2 metals are less reactive and weaker reducing agents than Group 1 metals.

(a) 1.5 marks

(b) 1.5 marks

• Industry (1 mark): limestone, \ce{CaCO3}, is used as a building material and is heated in a lime kiln to make quicklime, which is roasted with clay to make cement: \ce{CaCO3 ->[heat] CaO + CO2}. Magnesium is used in flares and in light alloys for aircraft.
• Agriculture (1 mark): slaked lime, \ce{Ca(OH)2}, is spread on acidic soils to neutralise the acid and raise the pH: \ce{Ca(OH)2 + 2H+ -> Ca^2+ + 2H2O}.
• Medicine (1 mark): barium sulphate, \ce{BaSO4}, is given as a 'barium meal' for X-ray pictures of the stomach and intestines; it absorbs X-rays and is too insoluble to be absorbed into the blood. Magnesium hydroxide (Milk of Magnesia) and calcium carbonate are antacids: \ce{Mg(OH)2 + 2HCl -> MgCl2 + 2H2O}.

1. High melting and boiling points (for example Fe 1538^\circ\mathrm{C}), because strong metallic bonding uses both 3d and 4s electrons.
2. High density (for example Cu 8.96\,\mathrm{g\,cm^{-3}}), because the atoms are heavy, small and closely packed in the lattice.
3. Good conductors of heat and electricity, because they have delocalised (free) electrons.

Other acceptable properties: hard and strong; shiny metallic lustre; malleable and ductile; paramagnetic or ferromagnetic (Fe, Co, Ni); form alloys because of similar atomic sizes.

Structures (2 marks)

But-2-ene is \ce{CH3-CH=CH-CH3}. Each carbon of the double bond carries two different groups (\ce{H} and \ce{CH3}), and the \pi bond stops rotation, so two geometric isomers exist:

More stable isomer (1 mark)

trans-But-2-ene is more stable. In the cis isomer the two bulky \ce{CH3} groups are on the same side and close together, so their electron clouds repel (steric strain). In the trans isomer they are far apart, so there is no such strain. Evidence: the enthalpy of hydrogenation of cis-but-2-ene (-120\,\mathrm{kJ\,mol^{-1}}) is more negative than that of trans-but-2-ene (-116\,\mathrm{kJ\,mol^{-1}}), so the cis isomer has more energy to start with.

Benzene is warmed at 50–55^\circ\mathrm{C} with a nitrating mixture (concentrated \ce{HNO3} and concentrated \ce{H2SO4}). The electrophile is the nitronium ion, \ce{NO2+}, and the reaction is an electrophilic substitution in three steps (one mark each).

Rule: a molecule with one carbon bonded to four different atoms or groups is chiral. Check every carbon of each molecule.

Chiral centre (1 mark)

2-Butanol (butan-2-ol) is \ce{CH3-\overset{*}{C}H(OH)-CH2-CH3}. Carbon 2 (marked *) is the chiral centre: it is bonded to four different groups, \ce{-H}, \ce{-OH}, \ce{-CH3} and \ce{-C2H5}.

Pair of enantiomers (2 marks)

Draw the chiral carbon as a tetrahedron (plain lines in the page, a wedge coming towards you, a dashed bond going away), then draw its mirror image. The two cannot be superimposed, so they are enantiomers.

The two enantiomers have the same boiling point and density, but they rotate plane-polarized light by equal angles in opposite directions (+13.5^\circ and -13.5^\circ).

The target is butan-2-ol (2-butanol), \ce{CH3CH(OH)CH2CH3}, a secondary alcohol with \ce{-OH} on C2. Alcohols are made from halogenoalkanes by substitution with \ce{OH-}, so we work backwards to the halogenoalkane.

Comparison. A carbocation is a carbon with three bonds and a positive charge. In a primary carbocation the positive carbon carries one alkyl group, in a secondary two and in a tertiary three. The more alkyl groups, the more stable the ion.

Order of stability.

Reason. Alkyl groups are electron-releasing (+I inductive effect). They push electron density towards the positive carbon, which reduces and spreads out the positive charge; a dispersed charge is more stable. Three alkyl groups do this most. Hyperconjugation also helps: the C–H bonds next to \ce{C+} (9 in a tertiary, 6 in a secondary, 3 in a primary cation) share electron density with the empty 2p orbital.

Order (1 mark)

Reason (2 marks)

• Phenol is the most acidic because its conjugate base, the phenoxide ion, is stabilised by resonance: the negative charge is delocalised from oxygen into the benzene ring (carbons 2, 4 and 6).
• Water gives \ce{OH-}, in which the charge stays on one oxygen with no resonance, so it is less stable than phenoxide.
• Ethanol gives \ce{C2H5O-}; the ethyl group releases electrons (+I effect) and intensifies the charge on oxygen, so ethoxide is the least stable and ethanol is the weakest acid.

The more stable the conjugate base, the stronger the acid.

Step 1: diazotisation (1.5 marks)

Aniline (phenylamine) in dilute hydrochloric acid is treated with nitrous acid (made from \ce{NaNO2 + HCl}) at 0 to 5\,^\circ\mathrm{C} (below 10\,^\circ\mathrm{C}). Benzenediazonium chloride forms:

Step 2: hydrolysis (1.5 marks)

The diazonium salt solution is warmed above 10\,^\circ\mathrm{C}. Nitrogen gas is given off and phenol is formed:

In an amide, \ce{RCONH2}, the nitrogen is attached to a carbonyl group. The lone pair on nitrogen is delocalised onto the \ce{C=O} group and the electronegative oxygen atom (resonance), so it is not free to bond to a proton.

The carbonyl group is also strongly electron-withdrawing, which lowers the electron density on nitrogen further. In amines the lone pair is localised on nitrogen and alkyl groups push electrons towards it (+I), so amines accept protons readily. As a result amide solutions are neutral while amine solutions are alkaline.

An addition polymer is formed when unsaturated monomers join without losing any atoms, for example polythene and PVC. A condensation polymer is formed when monomers with two functional groups join and a small molecule (\ce{H2O} or \ce{HCl}) is eliminated at each link, for example nylon-6,6 and terylene.

Monosaccharides are the simplest carbohydrates, which cannot be hydrolysed into smaller sugars; general formula (\ce{CH2O})_n, n = 3 to 6. They are sweet, crystalline and soluble in water. Examples: glucose and fructose (\ce{C6H12O6}), ribose (\ce{C5H10O5}).

Disaccharides give two monosaccharide molecules on hydrolysis; the two units are joined by a glycosidic bond. They are sweet and soluble. Examples (\ce{C12H22O11}): sucrose (glucose + fructose), maltose (glucose + glucose), lactose (glucose + galactose).

Polysaccharides give hundreds to thousands of monosaccharide units on hydrolysis, (\ce{C6H10O5})_n. They are tasteless, amorphous and insoluble in water. Examples: starch and glycogen (energy stores), cellulose (plant cell walls).

Molecular ion peak (1.5 marks)

The peak produced by the molecular ion \ce{M+}, formed when the whole molecule loses one electron (\ce{M + e- -> M^{+\bullet} + 2e-}). It lies at the highest m/z (apart from the small M+1 and M+2 peaks) and its m/z value gives the relative molecular mass. It may be small, or even missing, if the molecular ion breaks up easily.

Base peak (1.5 marks)

The tallest peak in the spectrum, produced by the most abundant, most stable ion. Its height is set to 100 and all other peaks are measured relative to it. It can be at any m/z and is usually a fragment ion.

One mark for each correct bond. Together, these three bands are typical of compounds such as ethanol (\ce{O-H} and \ce{C-O}) and propanone (\ce{C=O}), the examples of the book (Figs 18.2 and 18.3).

Number of peaks (1 mark)

One peak (a singlet), at about \delta=3.3\,\mathrm{ppm}, with an area corresponding to 6 H.

Justification (2 marks)

Dimethyl ether, \ce{CH3-O-CH3}, is symmetrical about the oxygen atom. Both \ce{CH3} groups are bonded to the same O atom, so all six hydrogen atoms are chemically equivalent (one environment), and equivalent protons give one signal. The atom next to each \ce{CH3} is oxygen, which carries no hydrogen (n=0), so by the n+1 rule the peak is not split. Because the \ce{CH3} groups are bonded to electronegative oxygen, they are deshielded and appear downfield at about 3.3\,\mathrm{ppm}.

a. Mobile phase (1.5 marks)

The mobile phase is the phase that moves over or through the stationary phase and carries the components of the mixture with it. It is a liquid (solvent or eluent) in paper, thin layer and column chromatography, and an inert gas such as helium or nitrogen (the carrier gas) in gas chromatography. A component that is more soluble in the mobile phase moves faster.

b. R_f value (1.5 marks)

The R_f value (retardation factor) is the ratio of the distance travelled by a solute from the baseline to the distance travelled by the solvent front from the baseline. It has no unit and lies between 0 and 1.

Under the same conditions each compound has a fixed R_f, so comparing R_f values with standards identifies the components of a mixture. A low R_f shows a polar compound held strongly by the stationary phase.

An alloy is a homogeneous mixture of a metal with other metals or non-metals. Three properties that make alloys important (1 mark each):

1. Harder and stronger than the pure metal. Atoms of the added element have a different size, so they distort the regular layers of metal atoms and stop them sliding over each other. Example: steel (iron + carbon) is much harder than pure iron and is used for buildings and tools.
2. More resistant to corrosion. Stainless steel (iron, chromium, nickel) does not rust, because chromium forms a thin protective oxide layer; bronze statues are not affected by sun and rain.
3. Lower melting point than the constituent metals. This makes melting and casting easier; solder (tin and lead) melts easily, so it is used to join wires in electric circuits.

Other acceptable points: alloys have lower electrical conductivity (useful for heating elements), attractive colours (brass is golden), and remain shiny, malleable and ductile.

Fertilizers are natural or artificial substances containing elements that improve plant growth and productivity. The three main types are named after the macronutrient they supply.

1. Nitrogen fertilizers (1 mark)

Examples: urea \ce{CO(NH2)2} (46% N), ammonium nitrate \ce{NH4NO3} and calcium ammonium nitrate. Importance: nitrogen is part of proteins and chlorophyll, so it promotes green leafy growth. N-fertilizers are especially useful in the middle phases of the plant's life cycle, when it needs the most support to keep growing and leafing out.

2. Phosphorus fertilizers (1 mark)

Examples: single superphosphate \ce{Ca(H2PO4)2 + CaSO4}, triple superphosphate and diammonium phosphate \ce{(NH4)2HPO4}. Importance: they reinforce roots and stems and boost flowering, seed production and fruit; they are crucial at all stages of growth.

3. Potassium fertilizers (1 mark)

Examples: potassium chloride \ce{KCl}, potassium sulphate \ce{K2SO4} and potassium nitrate \ce{KNO3}. Importance: they promote root development, help plants maintain healthy photosynthesis and limit the spread of diseases. Without potassium, leaf edges turn yellow or brown and leaves wither and fall.

Trend (2 marks)

Group 2 carbonates and nitrates decompose on heating to the metal oxide. Thermal stability increases down the group: a higher temperature is needed to decompose the compound as the group is descended.

For example, \ce{MgCO3} decomposes at about 540^\circ\mathrm{C}, \ce{CaCO3} at 900^\circ\mathrm{C}, \ce{SrCO3} at 1280^\circ\mathrm{C} and \ce{BaCO3} at 1360^\circ\mathrm{C}. Nitrates give off brown \ce{NO2} gas and oxygen; \ce{Mg(NO3)2} decomposes most easily and \ce{Ba(NO3)2} least easily.

Reasons (3 marks)

• The carbonate ion \ce{CO3^2-} and nitrate ion \ce{NO3-} are large anions whose electron clouds are easily polarised (distorted).
• All Group 2 cations have a 2+ charge, but their size increases down the group (\ce{Mg^2+ < Ca^2+ < Sr^2+ < Ba^2+}), so their charge density and polarising power decrease.
• A small cation like \ce{Mg^2+} pulls electron density towards itself and weakens a C-O (or N-O) bond in the anion, so the anion breaks up at a low temperature, releasing \ce{CO2} (or \ce{NO2} and \ce{O2}).
• A large cation like \ce{Ba^2+} hardly distorts the anion, so its bonds remain strong and much more heat is needed. The enthalpy of decomposition also becomes more positive down the group (+117 for \ce{MgCO3} to +268\,\mathrm{kJ\,mol^{-1}} for \ce{BaCO3}).

Order of stability (2 marks)

In a free metal ion the five 3d orbitals are degenerate (same energy). When ligands approach, their lone pairs repel the d electrons. Orbitals whose lobes point towards the ligands are repelled more and are raised more in energy, so the d orbitals split into two non-degenerate sets. The pattern depends on where the ligands are.

Octahedral complexes (3.5 marks)

• Six ligands approach along the x, y and z axes.
• d_{x^2-y^2} and d_{z^2} point directly at the ligands, are repelled more and form the higher e_g set (2 orbitals).
• d_{xy}, d_{yz}, d_{xz} point between the ligands, are repelled less and form the lower t_{2g} set (3 orbitals).
• The gap \Delta_o is large. Example: \ce{[Ti(H2O)6]^3+}, \ce{[Cu(H2O)6]^2+}.

Tetrahedral complexes (3.5 marks)

• Four ligands sit at alternate corners of a cube, between the axes.
• d_{xy}, d_{yz}, d_{xz} point nearer to the ligands and form the higher t_2 set (3 orbitals).
• d_{x^2-y^2} and d_{z^2} point away from the ligands and form the lower e set (2 orbitals).
• The pattern is the reverse of octahedral, and the gap is smaller, \Delta_t\approx\tfrac49\Delta_o, because there are fewer ligands and none points straight at an orbital. Example: \ce{[CuCl4]^2-}, \ce{[CoCl4]^2-}.

Because \Delta E differs, octahedral and tetrahedral complexes of the same ion absorb different frequencies of light and have different colours, for example pink \ce{[Co(H2O)6]^2+} and blue \ce{[CoCl4]^2-}.

Nylon-6,6 (3 marks)

Monomers: hexanedioic acid (adipic acid), \ce{HOOC-(CH2)4-COOH}, and hexane-1,6-diamine (hexamethylenediamine), \ce{H2N-(CH2)6-NH2}. Each has 6 carbon atoms, which gives the name 6,6.

On heating (about 200 to 280\,^\circ\mathrm{C}), the \ce{-COOH} of the acid reacts with the \ce{-NH2} of the diamine to form an amide link \ce{\bond{-}CO-NH\bond{-}} and a molecule of water. Both monomers react at both ends, so the chain keeps growing: this is condensation polymerisation.

Kevlar (4 marks)

Monomers: benzene-1,4-dicarboxylic acid (terephthalic acid), \ce{HOOC-C6H4-COOH}, and benzene-1,4-diamine (p-phenylenediamine), \ce{H2N-C6H4-NH2}. In both, the two groups are on opposite (1,4 or para) positions of the benzene ring.

The acid and amine groups condense to amide links with loss of water, giving an aromatic polyamide:

In industry the more reactive acid chloride, benzene-1,4-dioyl chloride, is used; the reaction then takes place at room temperature and \ce{HCl} is eliminated:

The straight, rigid chains of Kevlar lie parallel and are held together by many hydrogen bonds between \ce{C=O} and \ce{N-H} groups. This makes Kevlar very strong, light and heat resistant, so it is used in bullet-proof vests, helmets, ropes and tyres.

An acid is stronger when its conjugate base is more stable. A negative charge is most stable when it is delocalised, and best of all when it is shared by electronegative atoms such as oxygen.

Ethanoic acid: ethanoate ion (2 marks)

The ethanoate ion has two resonance structures that are identical in energy. The negative charge is shared equally by two oxygen atoms, and both C–O bonds are the same length (between single and double). This is very effective stabilisation, so ethanoic acid is the strongest (K_a=1.8\times10^{-5}).

Phenol: phenoxide ion (2 marks)

The phenoxide ion is also resonance-stabilised: the charge is delocalised over O and ring carbons 2, 4 and 6.

But in structures II to IV the charge sits on carbon, which is less electronegative than oxygen. These structures are less stable and contribute less, so most of the charge stays on one oxygen. Phenoxide is less stable than ethanoate, and phenol (K_a=1.3\times10^{-10}) is weaker than ethanoic acid.

Ethanol: ethoxide ion (2 marks)

The ethoxide ion has no resonance structures: the charge is localised on one oxygen. The ethyl group releases electrons (+I effect), which intensifies the charge and destabilises the ion:

So ethanol (K_a\approx10^{-16}) is the weakest acid.

Two factors that affect the basic strength of amines

1. Availability of the lone pair on nitrogen. A base accepts \ce{H+} with the nitrogen lone pair. Electron-donating alkyl groups (+I effect) increase the electron density on nitrogen and make the base stronger; delocalisation of the lone pair into a benzene ring (resonance) makes it less available and the base weaker.
2. Stability of the cation formed. The more the positive charge of \ce{RNH3+} is spread out (by alkyl groups) and stabilised by hydrogen bonding with water (solvation), the more readily the amine accepts \ce{H+}.

Correct order and explanation

All three accept a proton: \ce{RNH2 + H2O <=> RNH3+ + OH-}. A lower \mathrm{p}K_b means a stronger base.

Ethylamine > ammonia. The ethyl group is electron-donating. It pushes electron density onto nitrogen, so the lone pair attracts \ce{H+} more strongly (factor 1). It also spreads the positive charge in \ce{CH3CH2NH3+}, making this ion more stable than \ce{NH4+} (factor 2).

Ammonia > phenylamine. In phenylamine the nitrogen lone pair overlaps with the \pi electrons of the benzene ring and is delocalised over the ring, so it is far less available for \ce{H+} (factor 1). On protonation this delocalisation is lost, so the phenylammonium ion gains no extra stability (factor 2).

So electron donation makes ethylamine the strongest base and delocalisation makes phenylamine the weakest.

Balanced equation (2 marks)

The white precipitate is 2,4,6-tribromophenol; the orange colour of bromine water disappears at once.

Comparison with the bromination of benzene (4 marks)

Reason. In phenol a lone pair on the oxygen of \ce{-OH} is delocalised into the ring. The ring becomes strongly activated, with extra electron density at positions 2, 4 and 6. It polarises \ce{Br2} by itself, so no catalyst is needed, and all three activated positions are substituted. Benzene has no activating group; its π electrons are less available, so \ce{FeBr3} must polarise \ce{Br2} to produce the electrophile, and substitution stops after one bromine.

(a) TMS as the standard (3 marks)

Tetramethylsilane, \ce{Si(CH3)4}, is added in small amount, and its signal is taken as \delta=0\,\mathrm{ppm}. Chemical shifts of all other protons are measured from it: \delta=\dfrac{\nu_{\text{sample}}-\nu_{\text{TMS}}}{\nu_{\text{spectrometer}}}\times10^{6}. It is chosen because:

• One sharp signal: the molecule is tetrahedral and symmetrical, so all 12 protons are equivalent and give a single strong singlet; only a little TMS is needed.
• Highly shielded: silicon (electronegativity 1.8) is less electronegative than carbon (2.5), so electron density is pushed onto the methyl groups. The TMS protons are more shielded than those of almost all organic compounds, so its peak lies upfield, at one end of the spectrum, and does not overlap the sample peaks.
• Inert, soluble and volatile: it does not react with the sample, dissolves in most organic solvents, and boils at about 27^\circ\mathrm{C}, so it is easily removed to recover the sample.
• It serves as the reference for both \ce{^1H} and \ce{^{13}C} NMR.

(b) Deuterated solvents (3 marks)

The sample is dissolved in a solvent that is present in much greater amount than the sample. If the solvent contained ordinary \ce{^1H} atoms, its very large peak would swamp and hide the small peaks of the sample.

In deuterated solvents, such as \ce{CDCl3}, \ce{D2O} and \ce{(CD3)2CO}, the hydrogen atoms are replaced by deuterium. Deuterium resonates at a very different frequency from \ce{^1H}, so the solvent gives no signal in the proton spectrum and does not interfere. Chemically, the deuterated solvent behaves like the ordinary solvent and dissolves the sample equally well. (A small peak at 7.26\,\mathrm{ppm} from traces of \ce{CHCl3} in \ce{CDCl3} is ignored.)

Thermal stability of Group 2 nitrates increases down the group. \ce{Mg^2+} is the smallest and most polarising of the four cations, so it weakens the N-O bonds of the nitrate ion most, and \ce{Mg(NO3)2} decomposes at the lowest temperature.

Six \ce{NH3} ligands each form one dative bond to cobalt, so the coordination number is 6. The three \ce{Cl^-} ions are counter ions outside the coordination sphere, so 9 is wrong; 3 is the oxidation state of cobalt, not its coordination number.

\ce{Fe} is [\mathrm{Ar}]\,3d^{6}4s^{2}. The two 4s electrons are lost first, giving \ce{Fe^2+} [\mathrm{Ar}]\,3d^{6}. Options B and C wrongly remove 3d electrons before 4s; D has one electron too many.

Methane, \ce{CH4}, is a hydrocarbon, the simplest organic compound. \ce{CaCO3} contains carbon, but carbonates (like \ce{CO}, \ce{CO2} and cyanides) are treated as inorganic. \ce{NaCl} and \ce{H2SO4} contain no carbon.

Ethane and propane both fit the alkane formula \ce{C_{n}H_{2n+2}} and differ by one \ce{CH2} unit. In B, \ce{C2H4} is an alkene. In C, one is an alcohol and the other an acid (different functional groups). In D, an alkyne and an aldehyde.

Hydrocarbons that contain one or more benzene rings are called aromatic hydrocarbons (arenes). Alkanes, cycloalkanes and other open-chain or non-benzene ring compounds are aliphatic.

Ethanol is oxidised (the carbon's oxidation number rises from -1 to +3) by acidified dichromate, which is reduced at the same time (\ce{Cr}: +6 \to +3). Oxidation and reduction together make a redox reaction.

The middle carbon of propan-2-ol carries \ce{-H}, \ce{-OH} and two identical \ce{-CH3} groups, so it is not chiral. The end carbons carry three H atoms each. With no chiral centre the molecule is superimposable on its mirror image and shows no optical isomerism.

In \ce{(CH3)3C-Br} the carbon carrying Br is bonded to three other carbon atoms, so it is tertiary. Options A and D are primary; in \ce{CH3CHCl2} the chlorine-bearing carbon is bonded to only one carbon.

\mathrm{S_N2} needs attack from the back of the carbon. Chloromethane has only small hydrogen atoms around its carbon, so it is the least hindered. The secondary halide is slower, the tertiary halide is too crowded, and bromobenzene does not undergo \mathrm{S_N2} (partial double bond).

Alcohols contain the hydroxyl group \ce{-OH} on a saturated carbon. \ce{-COOH} is the carboxyl group of acids, \ce{-CHO} the aldehyde group and \ce{-O-} the ether link.

The carbonyl carbon is \delta+ and accepts the electron pair of the attacking nucleophile, so the carbonyl group acts as an electrophile. It is not a nucleophile, a radical or a leaving group.

Like other acids, carboxylic acids react with reactive metals to give a salt and hydrogen gas: \ce{2CH3COOH + 2Na -> 2CH3COONa + H2}.

Simple carbohydrates are polyhydroxy aldehydes or ketones: several hydroxyl (\ce{-OH}) groups plus one aldehyde (\ce{-CHO}, aldoses such as glucose) or keto (\ce{-CO-}, ketoses such as fructose) group. \ce{-NH2} with \ce{-COOH} describes amino acids.

Trace elements are present at the level of parts per million or less. Spectroscopic methods (such as atomic absorption spectroscopy and mass spectrometry) can detect such tiny amounts. Gravimetric analysis, titration and precipitation need weighable masses or measurable volumes, so they are not sensitive enough.

Comparison (1.5 marks)

\ce{MgSO4} is soluble in water (1.83\,\mathrm{mol\,dm^{-3}}), while \ce{BaSO4} is practically insoluble (9.43\times10^{-6}\,\mathrm{mol\,dm^{-3}}). Solubility of Group 2 sulphates decreases down the group: \ce{MgSO4 > CaSO4 > SrSO4 > BaSO4}.

Reason (1.5 marks)

The sulphate ion is very large, so the lattice energies of \ce{MgSO4} and \ce{BaSO4} are nearly the same (r_++r_- hardly changes). But the small \ce{Mg^2+} ion has a much larger hydration enthalpy (about -1920\,\mathrm{kJ\,mol^{-1}}) than the large \ce{Ba^2+} ion (about -1360\,\mathrm{kJ\,mol^{-1}}). For \ce{MgSO4} the hydration energy overcomes the lattice energy and \Delta H_{sol} is exothermic; for \ce{BaSO4} it does not, so \Delta H_{sol} is endothermic and it does not dissolve.

In dilute sulphuric acid, orange dichromate(VI) is reduced to green chromium(III) (Cr: +6 to +3, 6 electrons per \ce{Cr2O7^2-}).

(a) With Mohr's salt, \ce{FeSO4.(NH4)2SO4.6H2O}

Iron(II) is oxidised to iron(III) (1 electron each), so 6 \ce{Fe^2+} are needed per \ce{Cr2O7^2-}:

(b) With oxalic acid, \ce{H2C2O4}

Oxalic acid is oxidised to carbon dioxide (C: +3 to +4, 2 electrons per molecule), so 3 molecules are needed per \ce{Cr2O7^2-}:

\ce{[Fe(CN)6]^3-} (1.5 marks)

Oxidation state: x+6(-1)=-3, so iron is +3. The complex is an anion, so iron is named ferrate. Six cyano ligands give 'hexacyano'.

\ce{[Co(NH3)4Cl2]Cl} (1.5 marks)

The sphere \ce{[Co(NH3)4Cl2]+} is the cation (one \ce{Cl^-} counter ion). Oxidation state: x+4(0)+2(-1)=+1, so cobalt is +3. Ligands in alphabetical order: ammine, then chloro. The cation is named first, then the chloride anion.

Melting points (1 mark)

Beryllium has a much higher melting point (about 1280^\circ\mathrm{C}) than magnesium (650^\circ\mathrm{C}).

Boiling points (1 mark)

Beryllium also has a much higher boiling point (about 2470^\circ\mathrm{C}) than magnesium (about 1090^\circ\mathrm{C}).

Reason (1 mark)

Both are held together by metallic bonds, each atom giving two electrons to the sea of delocalised electrons. The beryllium atom is much smaller (0.122\,\mathrm{nm} vs 0.160\,\mathrm{nm}), so its nucleus is closer to the delocalised electrons and attracts them more strongly. The metallic bond in Be is therefore much stronger, and more energy is needed to melt and boil it.

Comparison of the enthalpies of hydrogenation (2 marks)

The enthalpy of hydrogenation of cyclohexene, which has one C=C bond in a six-membered ring, is -120\,\mathrm{kJ\,mol^{-1}}. The hypothetical 1,3,5-cyclohexatriene would have three localised C=C bonds, so its expected value is three times as large:

The measured value for benzene is much smaller:

Both reactions give the same product, cyclohexane. Benzene therefore releases 360-208=152\,\mathrm{kJ\,mol^{-1}} less heat, which means it starts 152\,\mathrm{kJ\,mol^{-1}} lower in energy than the cyclohexatriene structure.

Significance of the difference (1 mark)

The difference, 152\,\mathrm{kJ\,mol^{-1}}, is the resonance (delocalisation) energy of benzene. It shows that benzene is not cyclohexatriene: its six π electrons are delocalised over all six carbon atoms, and this makes it much more stable. Because addition would destroy this stable π system, benzene resists addition reactions and prefers electrophilic substitution, which keeps the aromatic ring.

Carbon forms more compounds (about 20 million) than any other element because of the following properties of the carbon atom.

1. Catenation (1.5 marks)

Carbon atoms join to one another to form long straight chains, branched chains and rings. This is possible because the \ce{C-C} bond is very strong (about 348\,\mathrm{kJ\,mol^{-1}}), much stronger than the \ce{Si-Si} bond (226\,\mathrm{kJ\,mol^{-1}}). Example: hexane, 2-methylpentane and cyclohexane all have six carbons.

2. Tetravalency (1.5 marks)

Carbon has four valence electrons (\mathrm{2s^2\,2p^2}), so it forms four covalent bonds. Four bonds per atom allow branching and let each carbon carry different atoms or groups, which gives a huge number of structures and isomers.

Other reasons (any can be used)

• It forms stable multiple bonds: \ce{C=C}, \ce{C#C}, \ce{C=O}, \ce{C#N}.
• It bonds strongly with H, O, N, S and halogens, giving many functional groups.
• Isomerism: one molecular formula can have many structures, for example \ce{C10H22} has 75 structural isomers.

Alkanes are saturated hydrocarbons (paraffins, 'little affinity'). Acids, alkalis, oxidising agents and bromine water do not attack them in ordinary conditions, while alkenes, alkynes and benzene react with electrophiles. Two reasons:

1. Strong σ bonds only, no π electrons (1.5 marks). All C–C (348\,\mathrm{kJ\,mol^{-1}}) and C–H (413\,\mathrm{kJ\,mol^{-1}}) bonds are strong σ bonds. There are no loosely held π electrons and no lone pairs, so there is no electron-rich site for an electrophile to attack, and a lot of energy is needed to break a bond. Other hydrocarbons have π electrons that attract electrophiles.
2. Non-polar bonds (1.5 marks). The electronegativities of C (2.5) and H (2.1) are almost the same, so the C–H bond is almost non-polar. The molecule has no partial positive or negative centre to attract a nucleophile or an electrophile.

Therefore alkanes react only under harsh conditions (UV light or high temperature) by free-radical substitution, for example \ce{CH4 + Cl2 ->[\text{UV light}] CH3Cl + HCl}.

Stable compound (1 mark)

1,3-Butadiene (\ce{CH2=CH-CH=CH2}) is the more stable, when compared per C=C bond, because its two double bonds are conjugated.

Reason (2 marks)

In 1,3-butadiene the two C=C bonds are separated by one single bond. All four carbons are sp^2 hybridised, so their four unhybridised 2p orbitals lie parallel and overlap side by side. The four π electrons are delocalised over all four carbons, and the central C–C bond gets partial double-bond character (1.47 Å instead of 1.54 Å). Delocalisation lowers the energy. In 1-butene the π electrons stay between two carbons only.

Enthalpies of hydrogenation prove it: 1-butene releases 127\,\mathrm{kJ\,mol^{-1}} for one C=C, so two isolated C=C bonds should release 2\times127=254\,\mathrm{kJ\,mol^{-1}}. 1,3-Butadiene releases only 239\,\mathrm{kJ\,mol^{-1}}, so it is about 15\,\mathrm{kJ\,mol^{-1}} more stable than expected (its delocalisation energy).

An amphoteric oxide reacts with both an acid and a base. \ce{BeO} is amphoteric because the small, highly polarising \ce{Be^2+} ion gives the Be-O bond considerable covalent character.

With an acid: BeO acts as a base (1.5 marks)

With a base: BeO acts as an acid (1.5 marks)

sodium beryllate; in water it is the complex \ce{Na2[Be(OH)4]}: \ce{BeO + 2NaOH + H2O -> Na2[Be(OH)4]}

Stronger base (1 mark)

Methylamine, \ce{CH3NH2}, is the stronger base (\mathrm{p}K_b 3.36, against 4.75 for ammonia).

Reason (2 marks)

Both bases accept a proton with the lone pair on nitrogen: \ce{CH3NH2 + H2O <=> CH3NH3+ + OH-}. The methyl group is electron-donating (positive inductive effect). It increases the electron density on the nitrogen atom, so the lone pair is more available and attracts \ce{H+} more strongly than in ammonia. The methyl group also spreads the positive charge of the \ce{CH3NH3+} ion, so this ion is more stable than \ce{NH4+}, which also favours protonation.

Tertiary alkyl halides favour \mathrm{S_N1} (1.5 marks)

• Stable carbocation. Ionisation gives a tertiary carbocation, stabilised by the +I effect and hyperconjugation of three alkyl groups, so the slow ionisation step (\mathrm{S_N1}) happens easily.
• Steric hindrance. Three bulky alkyl groups crowd the back of the C–X carbon, so a nucleophile cannot attack from the back; the one-step \mathrm{S_N2} route is blocked.

Primary alkyl halides favour \mathrm{S_N2} (1.5 marks)

• Little steric hindrance. Only one alkyl group is on the C–X carbon, so the nucleophile easily attacks from the back and the substitution takes place in one step.
• Unstable carbocation. A primary carbocation has only one electron-releasing group, so it is very unstable and the \mathrm{S_N1} ionisation is too slow.

Reaction (2 marks)

Alcohols react with sodium metal at room temperature. The hydrogen of the \ce{O-H} group is replaced by sodium, an alkoxide is formed and hydrogen gas is evolved (effervescence). For ethanol:

general equation; the product is a sodium alkoxide

What it shows (1 mark)

The reaction shows that alcohols are weakly acidic: the H of the \ce{O-H} group can be released as \ce{H+} and replaced by a reactive metal. They are weaker acids than water (the reaction is gentler than sodium with water), so they do not affect litmus and do not react with \ce{NaOH}.

The \ce{-OH} group of phenol is a strongly activating group. One lone pair on its oxygen atom overlaps with the π electrons of the ring and is delocalised into it. This raises the electron density of the ring, especially at the ortho (2, 6) and para (4) positions.

The electron-rich ring polarises an approaching \ce{Br2} molecule (\ce{Br^{\delta+}-Br^{\delta-}}) and attacks its \delta+ end directly, and the intermediate is stabilised by the O lone pair. So no Lewis acid catalyst is needed to make \ce{Br+}, unlike benzene.

Since all three positions 2, 4 and 6 are activated, three bromine atoms are substituted at once and insoluble 2,4,6-tribromophenol separates as a white precipitate.

Comparison (1 mark)

**Formaldehyde (\ce{HCHO}) is more reactive** towards nucleophilic addition than acetone (\ce{CH3COCH3}).

Reasons (2 marks)

Inductive effect: the two methyl groups of acetone push electrons towards the carbonyl carbon (+I effect) and reduce its \delta+ charge, so it attracts nucleophiles less. Formaldehyde has only hydrogen atoms, so its carbonyl carbon keeps a larger \delta+ charge.

Steric effect: the bulky methyl groups crowd the carbonyl carbon of acetone and hinder the approach of the nucleophile; they also crowd the tetrahedral product. The small H atoms of formaldehyde give almost no hindrance. For example, formaldehyde adds water almost completely in solution, while acetone hardly does.

\ce{C4H10O} fits \ce{C_{n}H_{2n+2}O}, so it is saturated and can be an alcohol (\ce{R-OH}) or an ether (\ce{R-O-R'}). Any three of the seven isomers earn full marks (1 mark each). Three alcohols:

The other four isomers are listed below. Butan-1-ol and butan-2-ol are position isomers; butan-1-ol and 2-methylpropan-1-ol are chain isomers; an alcohol and an ether are functional group isomers.

Structural difference (2 marks)

D-glucose and L-glucose are enantiomers (non-superimposable mirror images) with the same formula \ce{C6H12O6}. In the Fischer projection with \ce{CHO} at the top, the \ce{-OH} on C-5 (the chiral carbon farthest from the \ce{CHO} group) is on the right in D-glucose and on the left in L-glucose, compared with D- and L-glyceraldehyde. Being mirror images, L-glucose has the arrangement reversed at all four chiral carbons (C-2 to C-5).

Effect on biological function (1 mark)

Enzymes and glucose-transport proteins are chiral, and their active sites fit only D-glucose, like a right hand in a right glove. So D-glucose is absorbed and respired for energy, while L-glucose does not fit, is not absorbed or metabolised, and gives no energy, although it tastes sweet. That is why natural sugars are D sugars.

esterification: methyl acetate (methyl ethanoate) is formed

The \ce{-OH} removed as water comes from the acid and the \ce{H} from methanol. The reaction is reversible; concentrated \ce{H2SO4} also absorbs water and increases the yield of ester.

Reason (2 marks)

An irreversible inhibitor reacts with a functional group of an amino acid in (or near) the active site and forms a strong covalent bond with it. This bond does not break under body conditions, so the inhibitor cannot leave the enzyme, unlike reversible inhibitors held by weak bonds. The active site is permanently blocked or its shape permanently changed, so no enzyme–substrate complex can form, and adding more substrate cannot displace the inhibitor. The enzyme molecule is inactivated for good; activity returns only when the cell makes new enzyme.

Example (1 mark)

Medical: aspirin transfers an acetyl group covalently to a serine in the active site of cyclooxygenase (COX), permanently stopping prostaglandin formation, which relieves pain, fever and inflammation. (Other examples: penicillin permanently inhibits the bacterial transpeptidase that builds cell walls; nerve gases and organophosphate insecticides bond to acetylcholinesterase.)

4-Nitrobenzoic acid (\mathrm{p}K_a=3.44) is about 6 times stronger than benzoic acid (\mathrm{p}K_a=4.20). Acid strength depends on the stability of the anion (conjugate base) formed after losing \ce{H+}.

The nitro group is strongly electron-withdrawing. It pulls electrons through the \sigma bonds (inductive effect, -I) and also out of the \pi system of the ring by resonance (-R effect), because its N atom carries a positive charge. This draws electron density away from the \ce{COO-} group and spreads its negative charge, so the 4-nitrobenzoate ion is more stable than the benzoate ion. A more stable conjugate base means a stronger acid. Benzoic acid has only H at position 4 and gets no such stabilisation.

Comparison: cyclohexene has one isolated \ce{C=C} bond. Its \pi\to\pi^* absorption lies at about 182\,\mathrm{nm}, below 200\,\mathrm{nm}, so it shows practically no absorption in the normal UV region. Buta-1,3-diene has two conjugated \ce{C=C} bonds (\ce{C=C-C=C}) and absorbs at a longer wavelength, \lambda_{\max}=217\,\mathrm{nm}, with a more intense band.

Explanation: in buta-1,3-diene the p orbitals of the two double bonds overlap side by side, so the \pi electrons are delocalised over four carbon atoms. This raises the highest filled \pi orbital and lowers the lowest empty \pi^* orbital, so the energy gap \Delta E becomes smaller. Since \lambda=hc/\Delta E, a smaller gap means that light of longer wavelength (lower energy) is absorbed. Cyclohexene has no conjugation, so its gap is larger and it absorbs at a shorter wavelength.

Mass spectrometry (2 marks)

Mass spectrometry is an analytical technique in which a sample is changed into gaseous positive ions, the ions are separated according to their mass-to-charge ratio (m/z), and the relative abundance of each kind of ion is recorded as a mass spectrum. It is used to find the isotopes of an element and their abundances, relative atomic and molecular masses, and the structures of organic compounds.

Instrumentation (labelled diagram)

Working (5 marks)

For example, the mass spectrum of boron shows two peaks, at m/z=10 (19.90\%) and m/z=11 (80.10\%), from which A_r(\ce{B})=10.80.

Nitration of benzene (3 marks)

Benzene is warmed at 50–55^\circ\mathrm{C} with a mixture of concentrated nitric acid and concentrated sulphuric acid. The electrophile is the nitronium ion, \ce{NO2+}.

Sulphonation of benzene (4 marks)

Benzene is warmed at about 40^\circ\mathrm{C} with fuming sulphuric acid (oleum, \ce{SO3} in conc. \ce{H2SO4}), or refluxed with concentrated \ce{H2SO4} for several hours. The product is benzenesulphonic acid. The electrophile is sulphur trioxide: it is neutral, but three electronegative O atoms leave the S atom strongly \delta+.

Addition and condensation polymers (4 marks)

Addition polymer

A polymer formed by the repeated addition of unsaturated monomers (with a \ce{C=C} bond) without the loss of any atom, for example polythene and PVC.

Condensation polymer

A polymer formed when monomers that each have two functional groups join with the elimination of a small molecule such as \ce{H2O} or \ce{HCl} at every link, for example nylon-6,6 and terylene.

Example of an addition polymer

Example of a condensation polymer

Preparation of nylon-6,6 (3 marks)

Nylon-6,6 is made by heating hexanedioic acid (adipic acid) with hexane-1,6-diamine (hexamethylenediamine). The \ce{-COOH} and \ce{-NH2} groups condense to amide links, and one molecule of water is lost for each link:

Nylon-6,6 is strong, elastic, resistant to abrasion and has a high melting point, so it is used for clothing, stockings, ropes, carpets, fishing nets, gears and bearings.

Reaction and conditions

Two molecules of acetaldehyde (ethanal) join in the presence of dilute NaOH in the cold to give 3-hydroxybutanal (aldol). On heating, the aldol loses water to give but-2-enal (crotonaldehyde). Acetaldehyde can do this because it has \alpha-hydrogens (on the \ce{CH3} next to \ce{C=O}).

Step 1: formation of the carbanion

The hydroxide ion removes an acidic \alpha-hydrogen from one acetaldehyde molecule. The carbanion (enolate ion) formed is stabilised by resonance with the carbonyl group.

Step 2: nucleophilic attack

The carbanion acts as a nucleophile and attacks the \delta+ carbonyl carbon of a second acetaldehyde molecule. The \ce{C=O} \pi electrons move onto oxygen, giving an alkoxide ion and a new \ce{C-C} bond.

Step 3: protonation (aldol formed)

The alkoxide ion takes a proton from water, giving 3-hydroxybutanal and regenerating the hydroxide ion (catalyst).

Step 4: dehydration (condensation)

On heating, \ce{OH-} removes an \alpha-hydrogen from the aldol and the \ce{-OH} group leaves as water, forming a \ce{C=C} bond conjugated with \ce{C=O}.

The overall process is a condensation because two molecules combine and a small molecule (water) is lost.