Class 12 Chemistry — Important Board Questions with Answers

Everything the Class 12 Chemistry (Plus Two) board paper tends to ask, in one place — 100 most-asked questions across 12 chapters, each with a model answer and the exact marking-scheme points examiners reward. Revise chapter by chapter, and walk in sure of yourself.

100 questions+2 · Plus Two12 chaptersModel answersCBSE · ISC · State boards

The Solid State6 questions

2 markseasyCrystalline and amorphous solids

Distinguish between crystalline and amorphous solids with one example each.

Reveal model answer + marking points

Crystalline solids have a regular, long-range ordered arrangement of particles, sharp melting points, definite geometrical shapes and are anisotropic (properties differ with direction); e.g. sodium chloride and diamond. Amorphous solids have only a short-range order (irregular arrangement), no sharp melting point (they soften over a range), no definite shape and are isotropic (same properties in all directions); e.g. glass and rubber.

Marking-scheme points

  • Crystalline: long-range order, sharp melting point, anisotropic (NaCl)
  • Amorphous: short-range order, no sharp melting point, isotropic (glass)
  • Amorphous solids soften over a range of temperature
2 markseasyUnit cell

What is a unit cell? Distinguish between a primitive and a centred unit cell.

Reveal model answer + marking points

A unit cell is the smallest repeating three-dimensional portion of a crystal lattice which, when repeated in different directions, generates the entire crystal. In a primitive (simple) unit cell, the constituent particles are present only at the corners of the unit cell. In a centred unit cell, particles are present at positions other than the corners as well, such as body-centred (one at the centre of the body), face-centred (one at the centre of each face) or end-centred.

Marking-scheme points

  • Unit cell = smallest repeating unit of a crystal lattice
  • Primitive: particles only at the corners
  • Centred: extra particles (body-, face- or end-centred)
3 marksmediumAtoms per unit cell

Calculate the number of atoms present per unit cell in simple cubic, body-centred cubic (bcc) and face-centred cubic (fcc) structures.

Reveal model answer + marking points

A corner atom is shared by 8 unit cells (contributes 1/8), a face atom by 2 (contributes 1/2) and a body-centre atom belongs entirely to one cell (contributes 1). Simple cubic: 8 corners x 1/8 = 1 atom. Body-centred cubic (bcc): (8 x 1/8) + 1 (centre) = 2 atoms. Face-centred cubic (fcc): (8 x 1/8) + (6 x 1/2) = 1 + 3 = 4 atoms.

Marking-scheme points

  • Corner atom contributes 1/8, face atom 1/2, body-centre 1
  • Simple cubic: 1 atom; bcc: 2 atoms
  • fcc: 1 + 3 = 4 atoms
3 markshardDensity of a unit cell

An element with atomic mass 56 g/mol has a bcc structure with edge length 288 pm. Calculate its density. (NA = 6.022 x 10^23)

Reveal model answer + marking points

For bcc, Z = 2. Edge a = 288 pm = 2.88 x 10^-8 cm, so a^3 = (2.88 x 10^-8)^3 = 2.39 x 10^-23 cm^3. Density d = Z M/(a^3 NA) = (2 x 56)/(2.39 x 10^-23 x 6.022 x 10^23) = 112/14.39 = 7.79 g/cm^3.

d = Z M/(a^3 NA)

Marking-scheme points

  • d = Z M/(a^3 NA); bcc Z = 2
  • a^3 = (2.88e-8)^3 = 2.39 x 10^-23 cm^3
  • d = 112/14.39 = 7.79 g/cm^3
2 marksmediumPoint defects

Distinguish between Schottky and Frenkel defects.

Reveal model answer + marking points

Schottky defect is a vacancy defect in which an equal number of cations and anions are missing from their lattice sites, so the density of the solid decreases; it occurs in ionic solids with high coordination number and similar cation and anion sizes (e.g. NaCl, KCl). Frenkel defect is a dislocation defect in which an ion (usually the smaller cation) leaves its lattice site and occupies an interstitial site, so the density remains unchanged; it occurs in solids with a large difference in the sizes of the ions (e.g. AgCl, ZnS).

Marking-scheme points

  • Schottky: equal cations and anions missing; density decreases (NaCl)
  • Frenkel: smaller ion shifts to an interstitial site; density unchanged (AgCl)
  • Both are point defects in ionic solids
2 marksmediumPacking efficiency

State the packing efficiency and coordination number of simple cubic, bcc and fcc structures.

Reveal model answer + marking points

Packing efficiency is the percentage of the total space occupied by the particles. Simple cubic: packing efficiency 52.4 percent, coordination number 6. Body-centred cubic (bcc): packing efficiency 68 percent, coordination number 8. Face-centred cubic (fcc, also ccp/hcp): packing efficiency 74 percent (the highest), coordination number 12.

Marking-scheme points

  • Simple cubic: 52.4 percent, coordination number 6
  • bcc: 68 percent, coordination number 8
  • fcc/ccp/hcp: 74 percent (highest), coordination number 12

Solutions9 questions

2 markseasyConcentration terms

Define molarity, molality and mole fraction.

Reveal model answer + marking points

Molarity (M) is the number of moles of solute per litre of solution (mol/L). Molality (m) is the number of moles of solute per kilogram of solvent (mol/kg); it is independent of temperature. Mole fraction of a component is the ratio of the number of moles of that component to the total number of moles of all components in the solution; it is dimensionless and the sum of the mole fractions equals 1.

M = n/V(L); m = n/mass of solvent(kg)

Marking-scheme points

  • Molarity = moles of solute/litre of solution (mol/L)
  • Molality = moles of solute/kg of solvent (mol/kg), temperature independent
  • Mole fraction = moles of component/total moles (dimensionless)
2 marksmediumMolality calculation

Calculate the molality of a solution containing 18 g of glucose (molar mass 180 g/mol) dissolved in 500 g of water.

Reveal model answer + marking points

Moles of glucose = mass/molar mass = 18/180 = 0.1 mol. Mass of solvent (water) = 500 g = 0.5 kg. Molality = moles of solute/mass of solvent in kg = 0.1/0.5 = 0.2 mol/kg (0.2 m).

molality = moles of solute/mass of solvent (kg)

Marking-scheme points

  • Moles of glucose = 18/180 = 0.1 mol
  • Mass of water = 0.5 kg
  • Molality = 0.1/0.5 = 0.2 m
2 marksmediumHenry's law

State Henry's law. Give one application.

Reveal model answer + marking points

Henry's law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas over the liquid. Mathematically, p = KH x (mole fraction of gas), where KH is Henry's law constant. Applications: it explains why soft drinks (aerated under high pressure) fizz when opened, and why deep-sea divers can suffer from bends (nitrogen dissolving in blood under high pressure).

p = KH x (mole fraction)

Marking-scheme points

  • Solubility of a gas is proportional to its partial pressure
  • p = KH x (mole fraction of gas)
  • Application: aerated drinks, deep-sea diving (bends)
2 marksmediumRaoult's law

State Raoult's law for a solution of two volatile liquids.

Reveal model answer + marking points

Raoult's law states that for a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction in the solution. For components A and B, pA = pA(pure) x xA and pB = pB(pure) x xB, and the total vapour pressure = pA + pB. For a solution of a non-volatile solute, the relative lowering of vapour pressure equals the mole fraction of the solute.

pA = pA(pure) x xA

Marking-scheme points

  • Partial pressure of each component is proportional to its mole fraction
  • pA = pA(pure) x xA
  • Total pressure = pA + pB
3 marksmediumColligative properties

What are colligative properties? Name the four colligative properties.

Reveal model answer + marking points

Colligative properties are those properties of a solution that depend only on the number of solute particles present and not on their chemical nature. The four colligative properties are: (1) relative lowering of vapour pressure; (2) elevation of boiling point; (3) depression of freezing point; and (4) osmotic pressure. They are used to determine the molar mass of a solute.

Marking-scheme points

  • Depend only on the number of solute particles, not their nature
  • Relative lowering of vapour pressure and elevation of boiling point
  • Depression of freezing point and osmotic pressure
3 markshardDepression of freezing point

When 3 g of a non-volatile solute is dissolved in 100 g of water, the freezing point is depressed by 0.93 K. Calculate the molar mass of the solute. (Kf for water = 1.86 K kg/mol)

Reveal model answer + marking points

The depression of freezing point is given by delta Tf = Kf x molality = Kf x (w2 x 1000)/(M2 x w1), where w2 = mass of solute, w1 = mass of solvent, M2 = molar mass of solute. So 0.93 = 1.86 x (3 x 1000)/(M2 x 100) = 55.8/M2. Therefore M2 = 55.8/0.93 = 60 g/mol.

delta Tf = Kf x (w2 x 1000)/(M2 x w1)

Marking-scheme points

  • delta Tf = Kf x (w2 x 1000)/(M2 x w1)
  • 0.93 = 1.86 x (3 x 1000)/(M2 x 100) = 55.8/M2
  • M2 = 60 g/mol
3 marksmediumOsmotic pressure

Calculate the osmotic pressure of a 0.1 M glucose solution at 300 K. (R = 0.0821 L atm K^-1 mol^-1)

Reveal model answer + marking points

Osmotic pressure is given by pi = C R T, where C is the molar concentration. Here C = 0.1 mol/L, R = 0.0821 L atm K^-1 mol^-1, T = 300 K. So pi = 0.1 x 0.0821 x 300 = 2.463 atm. Osmotic pressure is a colligative property used to find the molar mass of macromolecules.

pi = C R T

Marking-scheme points

  • pi = C R T
  • = 0.1 x 0.0821 x 300
  • pi = 2.463 atm
2 marksmediumVan't Hoff factor

What is the Van't Hoff factor? What does its value indicate for association and dissociation?

Reveal model answer + marking points

The Van't Hoff factor (i) is the ratio of the observed (experimental) value of a colligative property to the calculated (theoretical) value assuming no association or dissociation; equivalently, i = (actual number of particles after association/dissociation)/(number of particles before). For a solute that dissociates (like NaCl), i is greater than 1; for a solute that associates (like acetic acid in benzene), i is less than 1; and for a solute that neither associates nor dissociates, i = 1.

i = observed value/calculated value

Marking-scheme points

  • i = observed colligative property/calculated value
  • Dissociation: i greater than 1 (e.g. NaCl)
  • Association: i less than 1 (e.g. acetic acid in benzene)
2 marksmediumIdeal and non-ideal solutions

Distinguish between ideal and non-ideal solutions.

Reveal model answer + marking points

An ideal solution is one that obeys Raoult's law over the entire range of concentration; the enthalpy of mixing and the volume change on mixing are zero (e.g. benzene and toluene). A non-ideal solution does not obey Raoult's law; it shows either positive deviation (when A-B interactions are weaker than A-A and B-B, giving higher vapour pressure, e.g. ethanol and water) or negative deviation (when A-B interactions are stronger, giving lower vapour pressure, e.g. chloroform and acetone).

Marking-scheme points

  • Ideal: obeys Raoult's law; delta H(mix) = 0 and delta V(mix) = 0 (benzene-toluene)
  • Non-ideal positive deviation: higher vapour pressure (ethanol-water)
  • Non-ideal negative deviation: lower vapour pressure (chloroform-acetone)

Electrochemistry9 questions

2 markseasyGalvanic cell

What is a galvanic (voltaic) cell? Name its two electrodes and the reaction at each.

Reveal model answer + marking points

A galvanic cell is an electrochemical device that converts chemical energy into electrical energy through a spontaneous redox reaction (e.g. the Daniell cell). It has two electrodes: the anode, where oxidation (loss of electrons) takes place and which is the negative terminal; and the cathode, where reduction (gain of electrons) takes place and which is the positive terminal. A salt bridge connects the two half-cells and maintains electrical neutrality.

Marking-scheme points

  • Converts chemical energy into electrical energy (spontaneous redox)
  • Anode: oxidation, negative terminal
  • Cathode: reduction, positive terminal; salt bridge maintains neutrality
2 marksmediumStandard electrode potential

What is standard electrode potential? What is taken as the reference electrode?

Reveal model answer + marking points

The standard electrode potential is the potential difference developed between an electrode and its solution of unit concentration (1 M) at 298 K and 1 bar pressure, measured with respect to a reference electrode. The standard hydrogen electrode (SHE) is taken as the reference electrode and is assigned a potential of exactly zero volt. Electrodes are compared with the SHE to obtain their standard potentials.

Marking-scheme points

  • Electrode potential under standard conditions (1 M, 298 K, 1 bar)
  • Reference: standard hydrogen electrode (SHE)
  • SHE assigned a potential of zero volt
3 marksmediumEMF of a cell

Calculate the standard EMF of a Daniell cell given the standard electrode potentials E(Cu2+/Cu) = +0.34 V and E(Zn2+/Zn) = -0.76 V.

Reveal model answer + marking points

In the Daniell cell, zinc (lower/more negative potential) is the anode and copper is the cathode. The standard EMF = E(cathode) - E(anode) = E(Cu2+/Cu) - E(Zn2+/Zn) = (+0.34) - (-0.76) = 0.34 + 0.76 = 1.10 V. Since the EMF is positive, the cell reaction is spontaneous.

E(cell) = E(cathode) - E(anode)

Marking-scheme points

  • Zn = anode (more negative), Cu = cathode
  • EMF = E(cathode) - E(anode)
  • = 0.34 - (-0.76) = 1.10 V (spontaneous)
3 marksmediumNernst equation

Write the Nernst equation for a general electrode and explain the terms.

Reveal model answer + marking points

The Nernst equation gives the electrode (or cell) potential under non-standard conditions. For the reaction M(n+) + n e- -> M, the electrode potential E = E(standard) - (2.303 RT/nF) log(1/[M(n+)]). At 298 K, substituting the constants, E = E(standard) - (0.0591/n) log(1/[M(n+)]), where E(standard) is the standard electrode potential, n is the number of electrons transferred, F is the Faraday constant, and the term in brackets is the reaction quotient. It shows how potential varies with concentration.

E = E(standard) - (0.0591/n) log Q

Marking-scheme points

  • Gives potential under non-standard conditions
  • E = E(standard) - (0.0591/n) log Q at 298 K
  • n = electrons transferred; depends on ion concentration
2 marksmediumGibbs energy and EMF

Write the relation between the standard Gibbs energy change and the EMF of a cell.

Reveal model answer + marking points

The standard Gibbs energy change of a cell reaction is related to the standard EMF by delta G(standard) = -n F E(standard), where n is the number of electrons transferred, F is the Faraday constant (96500 C/mol) and E(standard) is the standard cell EMF. A positive EMF gives a negative delta G, meaning the cell reaction is spontaneous. This relation also links electrochemistry with thermodynamics.

delta G(standard) = -n F E(standard)

Marking-scheme points

  • delta G(standard) = -n F E(standard)
  • n = electrons transferred; F = 96500 C/mol
  • Positive EMF gives negative delta G (spontaneous)
3 marksmediumKohlrausch's law

State Kohlrausch's law of independent migration of ions and give one application.

Reveal model answer + marking points

Kohlrausch's law states that at infinite dilution, the molar conductivity of an electrolyte is the sum of the individual contributions of its cations and anions, each migrating independently. Mathematically, the limiting molar conductivity = (number of cations x limiting molar conductivity of cation) + (number of anions x limiting molar conductivity of anion). Applications: it is used to calculate the limiting molar conductivity of weak electrolytes (which cannot be found by extrapolation) and the degree of dissociation of a weak electrolyte.

Marking-scheme points

  • At infinite dilution, molar conductivity = sum of ionic contributions
  • Ions migrate independently
  • Used to find limiting molar conductivity of weak electrolytes and degree of dissociation
2 marksmediumFaraday's laws of electrolysis

State Faraday's two laws of electrolysis.

Reveal model answer + marking points

Faraday's first law states that the mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte (m is proportional to Q = I t). Faraday's second law states that when the same quantity of charge is passed through different electrolytes, the masses of substances deposited or liberated are directly proportional to their chemical equivalent weights.

m proportional to Q = I t

Marking-scheme points

  • First law: mass deposited is proportional to charge passed (m proportional to It)
  • Second law: for same charge, mass is proportional to equivalent weight
  • One Faraday (96500 C) deposits one gram equivalent
2 marksmediumElectrolysis calculation

A current of 5 A is passed through a silver nitrate solution for 30 minutes. Calculate the mass of silver deposited. (Atomic mass of Ag = 108, F = 96500 C/mol)

Reveal model answer + marking points

Charge passed Q = I t = 5 x (30 x 60) = 5 x 1800 = 9000 C. Silver is deposited by Ag+ + e- -> Ag, so 1 mole of electrons (96500 C) deposits 108 g of silver. Mass of Ag = (108/96500) x 9000 = (108 x 9000)/96500 = 972000/96500 = 10.07 g.

mass = (equivalent mass x Q)/96500

Marking-scheme points

  • Q = I t = 5 x 1800 = 9000 C
  • 96500 C deposits 108 g of Ag (Ag+ + e- -> Ag)
  • Mass = (108 x 9000)/96500 = 10.07 g
2 markseasyCells and batteries

Distinguish between primary and secondary cells with one example each.

Reveal model answer + marking points

A primary cell is one in which the redox reaction occurs only once and cannot be reversed, so it cannot be recharged and is discarded after use; e.g. the dry cell (Leclanche cell) and the mercury cell. A secondary cell is one that can be recharged by passing current through it in the opposite direction, so it can be used again and again; e.g. the lead storage battery and the nickel-cadmium cell.

Marking-scheme points

  • Primary cell: cannot be recharged, used once (dry cell)
  • Secondary cell: rechargeable, reusable (lead storage battery)
  • Secondary cells are recharged by passing current in reverse

Chemical Kinetics8 questions

2 markseasyRate of reaction

Define the rate of a chemical reaction. Distinguish between average and instantaneous rate.

Reveal model answer + marking points

The rate of a chemical reaction is the change in the concentration of a reactant or product per unit time. The average rate is the change in concentration over a measurable time interval (delta concentration/delta time). The instantaneous rate is the rate of the reaction at a particular instant of time, obtained by making the time interval very small (the derivative d[concentration]/dt). Its units are usually mol L^-1 s^-1.

rate = -d[R]/dt = +d[P]/dt

Marking-scheme points

  • Rate = change in concentration per unit time
  • Average rate = delta concentration/delta time over an interval
  • Instantaneous rate = rate at a particular instant (d[c]/dt)
2 markseasyFactors affecting rate

State the factors that affect the rate of a chemical reaction.

Reveal model answer + marking points

The rate of a chemical reaction is affected by: (1) the nature and concentration of the reactants (rate usually increases with concentration); (2) temperature (rate generally increases with a rise in temperature); (3) the presence of a catalyst (which increases the rate by providing an alternative path of lower activation energy); (4) the surface area of solid reactants (greater surface area gives a faster rate); and (5) for photochemical reactions, the intensity of light.

Marking-scheme points

  • Concentration of reactants and their nature
  • Temperature (rate increases with temperature)
  • Catalyst, surface area and (for photochemical reactions) light
3 marksmediumOrder and molecularity

Distinguish between order and molecularity of a reaction.

Reveal model answer + marking points

The order of a reaction is the sum of the powers of the concentration terms in the experimentally determined rate law; it can be zero, fractional or a whole number and is an experimental quantity. Molecularity is the number of reacting species (atoms, ions or molecules) that collide simultaneously in an elementary reaction; it is always a whole number (1, 2 or 3) and is a theoretical concept. Order is defined for overall reactions, whereas molecularity is defined only for elementary reactions.

Marking-scheme points

  • Order: sum of powers in the rate law (experimental, can be fractional/zero)
  • Molecularity: number of species in an elementary step (whole number)
  • Order applies to overall reactions; molecularity only to elementary steps
3 marksmediumFirst order reaction

Derive the integrated rate equation for a first order reaction.

Reveal model answer + marking points

For a first order reaction R -> P, the rate = -d[R]/dt = k[R]. Rearranging, d[R]/[R] = -k dt. Integrating between limits [R0] at t = 0 and [R] at time t: ln([R]/[R0]) = -k t, so [R] = [R0] e^(-k t). Converting to base 10 logarithms, k = (2.303/t) log([R0]/[R]). This shows that for a first order reaction, a plot of log[R] against t is a straight line.

k = (2.303/t) log([R0]/[R])

Marking-scheme points

  • Rate = k[R]; integrate d[R]/[R] = -k dt
  • ln([R]/[R0]) = -k t
  • k = (2.303/t) log([R0]/[R])
2 marksmediumHalf-life of first order reaction

The rate constant of a first order reaction is 6.93 x 10^-3 s^-1. Calculate its half-life.

Reveal model answer + marking points

For a first order reaction, the half-life is independent of the initial concentration and is given by t(1/2) = 0.693/k. Substituting k = 6.93 x 10^-3 s^-1: t(1/2) = 0.693/(6.93 x 10^-3) = 100 s. Thus the half-life of the reaction is 100 seconds.

t(1/2) = 0.693/k

Marking-scheme points

  • First order half-life t(1/2) = 0.693/k (independent of concentration)
  • = 0.693/(6.93 x 10^-3)
  • t(1/2) = 100 s
2 marksmediumArrhenius equation

Write the Arrhenius equation and define activation energy.

Reveal model answer + marking points

The Arrhenius equation relates the rate constant to temperature: k = A e^(-Ea/RT), where k is the rate constant, A is the frequency (pre-exponential) factor, Ea is the activation energy, R is the gas constant and T is the absolute temperature. Activation energy (Ea) is the minimum extra energy that the reactant molecules must possess (above their average energy) for a collision to be effective and lead to a reaction. A higher Ea means a slower reaction.

k = A e^(-Ea/RT)

Marking-scheme points

  • k = A e^(-Ea/RT)
  • Ea = minimum extra energy needed for an effective collision
  • Higher Ea -> slower reaction
2 marksmediumEffect of catalyst

How does a catalyst increase the rate of a reaction?

Reveal model answer + marking points

A catalyst increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy. As a result, a larger fraction of the reactant molecules have enough energy to cross the (lowered) energy barrier, so more effective collisions occur and the reaction proceeds faster. The catalyst does not change the enthalpy or the equilibrium position of the reaction, and it is regenerated at the end of the reaction.

Marking-scheme points

  • Provides an alternative path of lower activation energy
  • More molecules can cross the lower energy barrier
  • Does not change the equilibrium; is regenerated
2 marksmediumRate constant units

Write the units of the rate constant for a zero order and a first order reaction.

Reveal model answer + marking points

The units of the rate constant depend on the order of the reaction. For a zero order reaction, the rate = k, so the units of k are the same as the rate: mol L^-1 s^-1 (or mol L^-1 time^-1). For a first order reaction, rate = k[R], so k has units of s^-1 (or time^-1), which are independent of concentration. In general, the units of k are (mol L^-1)^(1-n) time^-1 for an nth order reaction.

units of k = (mol L^-1)^(1-n) time^-1

Marking-scheme points

  • Zero order: units of k are mol L^-1 s^-1
  • First order: units of k are s^-1 (time^-1)
  • General: (mol L^-1)^(1-n) time^-1 for order n

The p-Block Elements9 questions

2 marksmediumGroup 15 elements

Name the elements of Group 15 and give their general valence shell electronic configuration and common oxidation states.

Reveal model answer + marking points

The Group 15 elements (the nitrogen family) are nitrogen (N), phosphorus (P), arsenic (As), antimony (Sb) and bismuth (Bi). Their general valence shell electronic configuration is ns2 np3 (a half-filled p subshell, which gives extra stability). Their common oxidation states are -3, +3 and +5; the stability of the +5 state decreases and that of the +3 state increases down the group due to the inert pair effect.

ns2 np3

Marking-scheme points

  • Group 15: N, P, As, Sb, Bi
  • General configuration ns2 np3 (half-filled p)
  • Oxidation states -3, +3, +5; +3 more stable down the group
2 marksmediumAmmonia

Why does ammonia act as a Lewis base and have a higher boiling point than phosphine (PH3)?

Reveal model answer + marking points

Ammonia (NH3) has a lone pair of electrons on the nitrogen atom which it can donate to an electron-deficient species, so it acts as a Lewis base. It has a higher boiling point than phosphine because nitrogen is small and highly electronegative, so NH3 molecules form strong intermolecular hydrogen bonds, whereas PH3 molecules are held only by weak van der Waals forces; more energy is needed to separate the hydrogen-bonded NH3 molecules.

Marking-scheme points

  • NH3 has a lone pair on N -> donates it -> Lewis base
  • N is small and electronegative -> NH3 forms hydrogen bonds
  • PH3 has only weak van der Waals forces -> lower boiling point
3 marksmediumManufacture of ammonia

Describe the Haber process for the manufacture of ammonia.

Reveal model answer + marking points

Ammonia is manufactured industrially by the Haber process, in which nitrogen and hydrogen combine directly: N2(g) + 3H2(g) <=> 2NH3(g), and the forward reaction is exothermic and proceeds with a decrease in the number of moles. According to Le Chatelier's principle, the optimum conditions are a high pressure (about 200 atmospheres), a moderately low temperature (about 700 K), and a catalyst of finely divided iron with molybdenum (or K2O and Al2O3) as a promoter. The ammonia formed is removed by liquefaction to shift the equilibrium forward.

N2 + 3H2 <=> 2NH3

Marking-scheme points

  • N2 + 3H2 <=> 2NH3 (exothermic, fewer moles on product side)
  • Optimum: about 200 atm pressure, about 700 K temperature
  • Catalyst: finely divided iron with a promoter (e.g. molybdenum)
2 marksmediumManufacture of nitric acid

Describe the Ostwald process for the manufacture of nitric acid.

Reveal model answer + marking points

Nitric acid is manufactured by the Ostwald process, which uses ammonia as the starting material. The steps are: (1) catalytic oxidation of ammonia, 4NH3 + 5O2 -> 4NO + 6H2O, using a platinum-rhodium catalyst at about 500 K; (2) oxidation of nitric oxide, 2NO + O2 -> 2NO2; and (3) absorption of nitrogen dioxide in water, 3NO2 + H2O -> 2HNO3 + NO, where the NO produced is recycled. The dilute acid is then concentrated by distillation.

3NO2 + H2O -> 2HNO3 + NO

Marking-scheme points

  • 4NH3 + 5O2 -> 4NO + 6H2O (Pt-Rh catalyst)
  • 2NO + O2 -> 2NO2
  • 3NO2 + H2O -> 2HNO3 + NO
2 marksmediumAllotropes of phosphorus

Distinguish between white phosphorus and red phosphorus.

Reveal model answer + marking points

White phosphorus consists of discrete tetrahedral P4 molecules; it is soft, poisonous, very reactive, glows in the dark (chemiluminescence), catches fire in air spontaneously and is stored under water. Red phosphorus has a polymeric chain structure of linked P4 units; it is comparatively hard, non-poisonous, much less reactive, does not glow in the dark and does not catch fire spontaneously. Red phosphorus is more stable than white phosphorus.

Marking-scheme points

  • White P: discrete P4 molecules, poisonous, very reactive, glows, stored under water
  • Red P: polymeric, non-poisonous, less reactive, stable
  • White P is converted to red P on heating in the absence of air
2 marksmediumOzone

How is ozone prepared? Why does it act as a powerful oxidising agent?

Reveal model answer + marking points

Ozone (O3) is prepared by passing a silent electric discharge through pure, dry oxygen: 3O2 -> 2O3 (the reaction is endothermic). Ozone acts as a powerful oxidising agent because it is unstable and readily decomposes to give nascent oxygen: O3 -> O2 + [O]. This nascent oxygen is very reactive and readily oxidises other substances (for example, it turns moist starch-iodide paper blue by liberating iodine).

3O2 -> 2O3; O3 -> O2 + [O]

Marking-scheme points

  • Silent electric discharge through dry O2: 3O2 -> 2O3
  • Ozone is unstable and gives nascent oxygen: O3 -> O2 + [O]
  • Nascent oxygen makes it a strong oxidising agent
3 marksmediumManufacture of sulphuric acid

Describe the Contact process for the manufacture of sulphuric acid.

Reveal model answer + marking points

Sulphuric acid is manufactured by the Contact process. The steps are: (1) sulphur or sulphide ore is burnt to form sulphur dioxide, S + O2 -> SO2; (2) sulphur dioxide is catalytically oxidised to sulphur trioxide, 2SO2 + O2 <=> 2SO3, using vanadium pentoxide (V2O5) as catalyst at about 720 K and about 2 atm; (3) sulphur trioxide is absorbed in concentrated sulphuric acid to form oleum, SO3 + H2SO4 -> H2S2O7; and (4) the oleum is diluted with water to give sulphuric acid, H2S2O7 + H2O -> 2H2SO4.

2SO2 + O2 -> 2SO3 (V2O5)

Marking-scheme points

  • S + O2 -> SO2
  • 2SO2 + O2 <=> 2SO3 (V2O5 catalyst)
  • SO3 + H2SO4 -> oleum (H2S2O7); then diluted to H2SO4
2 marksmediumHalogens

Why is fluorine the strongest oxidising agent among the halogens, and why does chlorine act as a bleaching agent?

Reveal model answer + marking points

Fluorine is the strongest oxidising agent among the halogens because of its low bond dissociation energy (weak F-F bond), small atomic size and high hydration energy of the fluoride ion, which together make it accept electrons most readily. Chlorine acts as a bleaching agent because in the presence of moisture it produces nascent oxygen, Cl2 + H2O -> 2HCl + [O], and this nascent oxygen oxidises the coloured substance to a colourless one. The bleaching action of chlorine is permanent.

Cl2 + H2O -> 2HCl + [O]

Marking-scheme points

  • Fluorine: low F-F bond energy, small size, high hydration energy -> strongest oxidiser
  • Cl2 + H2O -> 2HCl + [O] (nascent oxygen)
  • Nascent oxygen bleaches (oxidises) coloured matter permanently
2 marksmediumNoble gases

Why are noble gases chemically inert? Name two compounds of xenon.

Reveal model answer + marking points

Noble gases (Group 18) are chemically inert because they have completely filled valence shells (ns2 np6, except helium which is 1s2), which is a very stable electronic configuration. As a result they have very high ionisation enthalpies and almost zero electron gain enthalpy, so they have little tendency to gain, lose or share electrons. Xenon, being the largest and most easily ionised, does form some compounds, for example xenon difluoride (XeF2) and xenon tetrafluoride (XeF4).

Marking-scheme points

  • Completely filled valence shell (ns2 np6) -> very stable
  • Very high ionisation enthalpy, no tendency to react
  • Xenon compounds: XeF2 and XeF4

The d- and f-Block Elements8 questions

2 markseasyTransition elements

What are transition elements? Why are they called so?

Reveal model answer + marking points

Transition elements are the d-block elements whose atoms or stable ions have partially filled d orbitals (their general configuration is (n-1)d(1-10) ns(1-2)). They are called transition elements because they are placed between the s-block metals and the p-block non-metals in the periodic table and show a gradual transition of properties from the highly electropositive s-block metals to the less electropositive p-block elements.

(n-1)d(1-10) ns(1-2)

Marking-scheme points

  • d-block elements with partially filled d orbitals in atoms/ions
  • General configuration (n-1)d(1-10) ns(1-2)
  • Lie between s-block and p-block, showing transitional properties
2 marksmediumVariable oxidation states

Why do transition elements show variable oxidation states?

Reveal model answer + marking points

Transition elements show variable oxidation states because the energies of the (n-1)d and the ns orbitals are very close to each other. As a result, after the ns electrons are removed, a variable number of the (n-1)d electrons can also take part in bonding. This allows the elements to exhibit several oxidation states that usually differ by one unit (for example, iron shows +2 and +3, and manganese shows +2 to +7).

Marking-scheme points

  • Energies of (n-1)d and ns orbitals are very close
  • A variable number of d electrons can participate in bonding
  • Example: Fe shows +2 and +3, Mn shows +2 to +7
2 marksmediumColoured compounds

Why are most transition metal compounds coloured?

Reveal model answer + marking points

Most transition metal ions are coloured because they have partially filled d orbitals. In the presence of ligands or other ions, the five d orbitals split into two sets of slightly different energy. An electron can absorb a particular wavelength (colour) of visible light and jump from the lower to the higher set of d orbitals (a d-d transition). The colour we see is the complementary colour of the light absorbed. Ions with completely empty or completely filled d orbitals (such as Sc3+ or Zn2+) are colourless.

Marking-scheme points

  • Partially filled d orbitals split into two energy levels
  • Electron absorbs visible light and jumps (d-d transition)
  • Observed colour is complementary to the absorbed light; d0 and d10 ions are colourless
2 marksmediumCatalytic property

Why do transition metals and their compounds act as good catalysts?

Reveal model answer + marking points

Transition metals and their compounds act as good catalysts mainly because: (1) they show variable oxidation states, so they can readily form intermediate compounds with the reactants and provide an alternative path of lower activation energy; and (2) they have the ability to adsorb reactant molecules on their surfaces (due to incomplete d orbitals), which brings the reactants close together and weakens their bonds. Examples: iron in the Haber process and vanadium pentoxide in the Contact process.

Marking-scheme points

  • Variable oxidation states allow formation of intermediates
  • Provide a lower activation energy path
  • Adsorb reactants on their surface (incomplete d orbitals)
3 marksmediumPotassium permanganate

How is potassium permanganate prepared from pyrolusite? State its oxidising action in acidic medium.

Reveal model answer + marking points

Potassium permanganate (KMnO4) is prepared from the mineral pyrolusite (MnO2). MnO2 is fused with potassium hydroxide (KOH) in the presence of air or an oxidising agent like KNO3 to give green potassium manganate (K2MnO4): 2MnO2 + 4KOH + O2 -> 2K2MnO4 + 2H2O. The manganate is then oxidised (electrolytically or by chlorine/ozone) to purple permanganate: 2K2MnO4 + Cl2 -> 2KMnO4 + 2KCl. In acidic medium KMnO4 is a strong oxidising agent: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O (Mn goes from +7 to +2).

MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O

Marking-scheme points

  • 2MnO2 + 4KOH + O2 -> 2K2MnO4 (green manganate)
  • Manganate oxidised to KMnO4 (purple permanganate)
  • Acidic oxidation: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O
2 marksmediumPotassium dichromate

Write the oxidising action of potassium dichromate in acidic medium. What happens to the colour of the solution?

Reveal model answer + marking points

Potassium dichromate (K2Cr2O7) is a strong oxidising agent in acidic medium. Its oxidising half-reaction is Cr2O7(2-) + 14H+ + 6e- -> 2Cr3+ + 7H2O, in which chromium is reduced from the +6 to the +3 oxidation state. During this the colour of the solution changes from orange (dichromate) to green (Cr3+). It is used, for example, to oxidise ferrous ions to ferric ions and iodide to iodine.

Cr2O7(2-) + 14H+ + 6e- -> 2Cr3+ + 7H2O

Marking-scheme points

  • Cr2O7(2-) + 14H+ + 6e- -> 2Cr3+ + 7H2O
  • Chromium reduced from +6 to +3
  • Colour changes from orange to green
2 marksmediumLanthanoid contraction

What is lanthanoid contraction? State one of its consequences.

Reveal model answer + marking points

Lanthanoid contraction is the steady, regular decrease in the atomic and ionic radii of the lanthanoid elements with increasing atomic number, from lanthanum to lutetium. It occurs because as the atomic number increases, electrons are added to the inner 4f subshell, which shields the nuclear charge poorly; so the effective nuclear charge on the outer electrons increases and the radius decreases. A consequence is that the elements of the second and third transition series (such as zirconium and hafnium) have almost the same size and very similar properties, making them difficult to separate.

Marking-scheme points

  • Steady decrease in size of lanthanoids with increasing atomic number
  • Cause: poor shielding by 4f electrons -> higher effective nuclear charge
  • Consequence: Zr and Hf have similar sizes and properties
2 marksmediumMelting points and magnetic properties

Why do transition metals have high melting points, and why are many of their compounds paramagnetic?

Reveal model answer + marking points

Transition metals have high melting and boiling points because their atoms have a large number of unpaired d electrons that form strong metallic bonds (in addition to bonds from the s electrons), so a lot of energy is needed to break them. Many of their compounds are paramagnetic because their ions contain unpaired electrons in the d orbitals; the magnetic moment increases with the number of unpaired electrons and can be estimated by the spin-only formula, magnetic moment = sqrt(n(n+2)) Bohr magnetons, where n is the number of unpaired electrons.

magnetic moment = sqrt(n(n+2)) BM

Marking-scheme points

  • High melting points: strong metallic bonding from unpaired d electrons
  • Paramagnetism due to unpaired d electrons in ions
  • Spin-only magnetic moment = sqrt(n(n+2)) Bohr magnetons

Coordination Compounds9 questions

2 marksmediumWerner's theory

Distinguish between the primary and secondary valency of a metal in Werner's theory.

Reveal model answer + marking points

According to Werner's theory, a metal in a coordination compound has two types of valency. The primary valency is ionisable and corresponds to the oxidation state of the metal; it is satisfied by negative ions and is non-directional. The secondary valency is non-ionisable and corresponds to the coordination number of the metal; it is satisfied by ligands, is directional, and determines the geometry of the complex.

Marking-scheme points

  • Primary valency: ionisable, equals oxidation state, satisfied by anions
  • Secondary valency: non-ionisable, equals coordination number, satisfied by ligands
  • Secondary valency is directional and fixes the geometry
2 markseasyBasic terms

Define ligand and coordination number.

Reveal model answer + marking points

A ligand is an ion or a molecule that has at least one lone pair of electrons which it can donate to the central metal atom or ion, forming a coordinate bond (for example Cl-, NH3, H2O, CN-). The coordination number of the central metal ion is the total number of coordinate bonds it forms with the ligands, that is, the number of ligand donor atoms directly attached to it (for example, in [Cu(NH3)4]2+ the coordination number of copper is 4).

Marking-scheme points

  • Ligand: donates a lone pair to the metal (e.g. NH3, Cl-, CN-)
  • Coordination number: number of donor atoms bonded to the metal
  • Example: [Cu(NH3)4]2+ has coordination number 4
3 marksmediumIUPAC nomenclature

State the main rules for the IUPAC nomenclature of coordination compounds and name [Cu(NH3)4]SO4.

Reveal model answer + marking points

Main rules: (1) the cation is named before the anion; (2) within the complex, the ligands are named in alphabetical order before the central metal; (3) the number of ligands is shown by prefixes di, tri, tetra, etc.; (4) the oxidation state of the metal is written in Roman numerals in brackets; and (5) if the complex ion is an anion, the metal name ends in -ate. For example, [Cu(NH3)4]SO4 is named tetraamminecopper(II) sulphate (copper is in the +2 state).

Marking-scheme points

  • Cation named first; ligands named alphabetically before the metal
  • Number of ligands by di, tri, tetra; oxidation state in Roman numerals
  • [Cu(NH3)4]SO4 = tetraamminecopper(II) sulphate
3 marksmediumIsomerism in complexes

Name the main types of isomerism shown by coordination compounds.

Reveal model answer + marking points

Coordination compounds show two broad types of isomerism. Structural (constitutional) isomerism includes: (1) ionisation isomerism (different ions in solution, e.g. [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br); (2) linkage isomerism (ambidentate ligand attached through different atoms, e.g. -NO2 vs -ONO); (3) coordination isomerism; and (4) hydrate (solvate) isomerism. Stereoisomerism includes: (1) geometrical isomerism (cis and trans forms) and (2) optical isomerism (non-superimposable mirror images).

Marking-scheme points

  • Structural: ionisation, linkage, coordination, hydrate isomerism
  • Stereoisomerism: geometrical (cis-trans) and optical
  • Linkage isomerism needs an ambidentate ligand (e.g. NO2)
2 marksmediumTypes of ligands

What is a chelate ligand and an ambidentate ligand? Give one example of each.

Reveal model answer + marking points

A chelate ligand (chelating ligand) is a polydentate ligand that binds to the same central metal ion through two or more donor atoms, forming a ring structure; this gives extra stability (the chelate effect). An example is ethylenediamine (en), which is bidentate. An ambidentate ligand is a monodentate ligand that has two different donor atoms and can attach to the metal through either one (but only one at a time); an example is the nitrite ion, which can bind through nitrogen (-NO2) or through oxygen (-ONO).

Marking-scheme points

  • Chelate ligand: polydentate, forms a ring (e.g. ethylenediamine)
  • Chelation gives extra stability (chelate effect)
  • Ambidentate ligand: two possible donor atoms, e.g. NO2 (via N or O)
3 markshardValence bond theory

Explain the geometry of a complex using valence bond theory. Distinguish inner and outer orbital complexes.

Reveal model answer + marking points

According to valence bond theory, the central metal ion provides a number of empty hybrid orbitals equal to its coordination number, and each ligand donates a lone pair into these orbitals to form coordinate bonds; the type of hybridisation decides the geometry (e.g. sp3 tetrahedral, dsp2 square planar, sp3d2 or d2sp3 octahedral). An inner orbital complex uses inner (n-1)d orbitals for hybridisation (d2sp3), usually formed with strong field ligands and often low-spin. An outer orbital complex uses the outer nd orbitals (sp3d2), usually formed with weak field ligands and is high-spin.

Marking-scheme points

  • Metal provides empty hybrid orbitals; ligands donate lone pairs
  • Hybridisation decides geometry (sp3, dsp2, d2sp3, sp3d2)
  • Inner orbital (d2sp3, low-spin) vs outer orbital (sp3d2, high-spin)
3 markshardCrystal field theory

Explain the splitting of d orbitals in an octahedral crystal field according to crystal field theory.

Reveal model answer + marking points

According to crystal field theory, the bonding between the metal and the ligands is purely electrostatic. In an isolated metal ion the five d orbitals have the same energy (degenerate). When six ligands approach along the axes in an octahedral field, they repel the d orbitals unequally: the two orbitals pointing along the axes (dx2-y2 and dz2, called eg) are repelled more and rise in energy, while the three orbitals pointing between the axes (dxy, dyz, dzx, called t2g) are repelled less and are lowered in energy. This energy gap between t2g and eg is the crystal field splitting energy (delta o).

Marking-scheme points

  • Metal-ligand bonding treated as electrostatic
  • Six ligands approach along the axes (octahedral)
  • d orbitals split into lower t2g and higher eg; gap = delta o
2 marksmediumSpectrochemical series

What is the spectrochemical series? Distinguish between strong field and weak field ligands.

Reveal model answer + marking points

The spectrochemical series is an arrangement of ligands in order of their increasing crystal field splitting power (the magnitude of delta they produce). A part of it is: I- < Br- < Cl- < F- < H2O < NH3 < en < CN- < CO. Strong field ligands (such as CN- and CO) produce a large splitting (large delta), favouring low-spin complexes; weak field ligands (such as halides and water) produce a small splitting (small delta), favouring high-spin complexes.

Marking-scheme points

  • Ligands arranged by increasing crystal field splitting power
  • Weak field (I-, Br-, H2O): small delta, high-spin
  • Strong field (CN-, CO): large delta, low-spin
2 markseasyImportance of coordination compounds

State any four applications of coordination compounds.

Reveal model answer + marking points

(1) In biological systems: haemoglobin (a complex of iron) carries oxygen, and chlorophyll (a complex of magnesium) is essential for photosynthesis. (2) In metallurgy: metals like silver and gold are extracted and purified using complex formation (cyanide process). (3) In medicine: cisplatin is used in cancer treatment and EDTA complexes are used to treat lead poisoning. (4) In analytical chemistry and electroplating: complexes are used in the estimation of metal ions and in electroplating of metals.

Marking-scheme points

  • Biological: haemoglobin (Fe), chlorophyll (Mg)
  • Metallurgy: extraction of silver and gold (cyanide process)
  • Medicine (cisplatin), analysis and electroplating

Haloalkanes and Haloarenes8 questions

2 markseasyClassification

Distinguish between haloalkanes and haloarenes with one example each.

Reveal model answer + marking points

Haloalkanes (alkyl halides) are compounds in which one or more hydrogen atoms of an aliphatic hydrocarbon (alkane) are replaced by halogen atoms; the halogen is attached to an sp3 carbon (e.g. CH3Cl, chloromethane). Haloarenes (aryl halides) are compounds in which the halogen atom is directly attached to an sp2 carbon of an aromatic ring (e.g. C6H5Cl, chlorobenzene). Haloarenes are much less reactive than haloalkanes towards nucleophilic substitution.

Marking-scheme points

  • Haloalkane: halogen on sp3 (aliphatic) carbon (CH3Cl)
  • Haloarene: halogen on sp2 carbon of an aromatic ring (C6H5Cl)
  • Haloarenes are less reactive to nucleophilic substitution
3 marksmediumPreparation of haloalkanes

How are haloalkanes prepared from alcohols? Write the reactions.

Reveal model answer + marking points

Haloalkanes are commonly prepared from alcohols by replacing the -OH group with a halogen. (1) With hydrogen halides: R-OH + HX -> R-X + H2O (reactivity of alcohols is 3 degrees > 2 degrees > 1 degree). (2) With phosphorus halides: 3R-OH + PCl3 -> 3R-Cl + H3PO3, and R-OH + PCl5 -> R-Cl + POCl3 + HCl. (3) With thionyl chloride (the best method, as the by-products are gases): R-OH + SOCl2 -> R-Cl + SO2 + HCl.

R-OH + SOCl2 -> R-Cl + SO2 + HCl

Marking-scheme points

  • R-OH + HX -> R-X + H2O
  • 3R-OH + PCl3 -> 3R-Cl + H3PO3
  • R-OH + SOCl2 -> R-Cl + SO2 + HCl (best method)
3 markshardSN1 and SN2 mechanisms

Distinguish between the SN1 and SN2 mechanisms of nucleophilic substitution.

Reveal model answer + marking points

In the SN2 (substitution nucleophilic bimolecular) mechanism, the reaction occurs in a single step; the nucleophile attacks the carbon from the side opposite the leaving group, and bond breaking and bond making occur simultaneously. Its rate depends on both the substrate and the nucleophile, and it gives inversion of configuration (Walden inversion); it is favoured by primary halides. In the SN1 (substitution nucleophilic unimolecular) mechanism, the reaction occurs in two steps through a carbocation intermediate; its rate depends only on the substrate, and it usually gives a racemic mixture. It is favoured by tertiary halides and polar protic solvents.

Marking-scheme points

  • SN2: one step, backside attack, inversion; rate depends on both reactants; favoured by primary halides
  • SN1: two steps via carbocation; rate depends only on substrate; racemisation
  • SN1 favoured by tertiary halides and polar protic solvents
2 marksmediumReactivity of haloarenes

Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution?

Reveal model answer + marking points

Haloarenes are less reactive than haloalkanes towards nucleophilic substitution mainly because of: (1) resonance - the lone pair of the halogen delocalises into the ring, giving the carbon-halogen bond a partial double-bond character, so it is shorter and stronger and harder to break; (2) the halogen is attached to an sp2 hybridised carbon which is more electronegative and holds the shared electrons more tightly than the sp3 carbon in haloalkanes; and (3) repulsion between the electron-rich aromatic ring and the approaching nucleophile.

Marking-scheme points

  • Resonance gives the C-X bond partial double-bond character (stronger)
  • Halogen on sp2 carbon (more electronegative, holds electrons tightly)
  • Electron-rich ring repels the incoming nucleophile
2 marksmediumReactions of haloalkanes

What is a dehydrohalogenation (elimination) reaction of a haloalkane? State Saytzeff's rule.

Reveal model answer + marking points

Dehydrohalogenation is a beta-elimination reaction in which a haloalkane loses a hydrogen halide (HX) when heated with an alcoholic solution of potassium hydroxide, forming an alkene. The hydrogen is removed from the beta-carbon (the carbon next to the one bearing the halogen). Saytzeff's rule states that in such an elimination the preferred (major) product is the more highly substituted (more stable) alkene, that is, the alkene formed by removal of the hydrogen from the beta-carbon having the fewer hydrogen atoms.

Marking-scheme points

  • Beta-elimination of HX with alcoholic KOH gives an alkene
  • Hydrogen removed from the beta-carbon
  • Saytzeff's rule: the more substituted (more stable) alkene is the major product
2 marksmediumOptical isomerism

What is a chiral molecule? What is meant by optical activity?

Reveal model answer + marking points

A chiral molecule is one that is non-superimposable on its mirror image, just as the left and right hands are not superimposable; it usually contains at least one carbon atom bonded to four different groups (an asymmetric or chiral carbon). Optical activity is the property of such a chiral substance to rotate the plane of plane-polarised light; the two non-superimposable mirror-image forms are called enantiomers, one rotating the light to the right (dextrorotatory) and the other to the left (laevorotatory).

Marking-scheme points

  • Chiral molecule: non-superimposable on its mirror image (chiral carbon)
  • Optical activity: rotates the plane of plane-polarised light
  • Mirror-image forms = enantiomers (dextro- and laevorotatory)
2 marksmediumGrignard reagent

What is a Grignard reagent? How is it prepared?

Reveal model answer + marking points

A Grignard reagent is an alkyl or aryl magnesium halide, R-Mg-X, which is a very important and reactive organometallic compound used in organic synthesis. It is prepared by the reaction of a haloalkane (or haloarene) with magnesium metal in the presence of dry ether: R-X + Mg -> R-Mg-X (in dry ether). Grignard reagents are highly reactive and must be prepared under anhydrous conditions, because even traces of water or moisture decompose them to alkanes.

R-X + Mg -> R-Mg-X (dry ether)

Marking-scheme points

  • Grignard reagent = alkyl/aryl magnesium halide (R-Mg-X)
  • Prepared: R-X + Mg -> R-Mg-X in dry ether
  • Very reactive; must be kept anhydrous (water decomposes it)
2 markseasyUses of halogen compounds

Why is chloroform stored in dark coloured bottles filled up to the brim?

Reveal model answer + marking points

Chloroform (CHCl3) is stored in dark coloured bottles filled completely up to the brim because in the presence of air (oxygen) and sunlight it undergoes slow oxidation to form a highly poisonous gas, phosgene (carbonyl chloride, COCl2): 2CHCl3 + O2 -> 2COCl2 + 2HCl. Filling the bottle to the brim leaves no air space, and the dark bottle keeps out light; together these prevent the oxidation of chloroform to phosgene.

2CHCl3 + O2 -> 2COCl2 + 2HCl

Marking-scheme points

  • Chloroform is oxidised in air and light to poisonous phosgene (COCl2)
  • 2CHCl3 + O2 -> 2COCl2 + 2HCl
  • Dark bottle filled to the brim excludes light and air

Alcohols, Phenols and Ethers8 questions

2 markseasyClassification of alcohols

Classify alcohols as primary, secondary and tertiary with one example each.

Reveal model answer + marking points

Alcohols are classified according to the type of carbon atom to which the -OH group is attached. In a primary (1 degree) alcohol the -OH is on a carbon attached to only one other carbon, e.g. ethanol (CH3CH2OH). In a secondary (2 degree) alcohol the -OH is on a carbon attached to two other carbons, e.g. propan-2-ol ((CH3)2CHOH). In a tertiary (3 degree) alcohol the -OH is on a carbon attached to three other carbons, e.g. 2-methylpropan-2-ol ((CH3)3COH).

Marking-scheme points

  • Primary: -OH carbon attached to one carbon (ethanol)
  • Secondary: -OH carbon attached to two carbons (propan-2-ol)
  • Tertiary: -OH carbon attached to three carbons (2-methylpropan-2-ol)
3 marksmediumPreparation of alcohols

How are alcohols prepared by the hydration of alkenes and by the reduction of aldehydes and ketones?

Reveal model answer + marking points

(1) Acid-catalysed hydration of alkenes: an alkene adds water in the presence of dilute sulphuric acid following Markovnikov's rule to give an alcohol, e.g. CH2=CH2 + H2O -> CH3CH2OH. (2) Reduction of carbonyl compounds: aldehydes on reduction (with H2/Ni or NaBH4 or LiAlH4) give primary alcohols (R-CHO -> R-CH2OH), and ketones on reduction give secondary alcohols (R-CO-R' -> R-CH(OH)-R'). Alcohols can also be prepared from Grignard reagents reacting with carbonyl compounds.

R-CHO + 2[H] -> R-CH2OH

Marking-scheme points

  • Hydration of alkenes (Markovnikov): CH2=CH2 + H2O -> CH3CH2OH
  • Reduction of aldehyde -> primary alcohol
  • Reduction of ketone -> secondary alcohol
2 marksmediumAcidity of phenol

Why is phenol more acidic than ethanol?

Reveal model answer + marking points

Phenol is more acidic than ethanol because the phenoxide ion formed after the loss of the proton is stabilised by resonance (the negative charge is delocalised into the benzene ring), which makes phenol lose its proton more easily. In contrast, the ethoxide ion from ethanol is not resonance-stabilised, and the alkyl group has an electron-releasing (+I) effect that further destabilises the ethoxide ion. Hence phenol ionises more readily and is a stronger acid than ethanol.

Marking-scheme points

  • Phenoxide ion is stabilised by resonance (charge delocalised into ring)
  • Ethoxide ion is not resonance-stabilised
  • Alkyl (+I) effect destabilises ethoxide, so ethanol is weaker acid
2 marksmediumReactions of alcohols

Write the reaction of ethanol with sodium metal and its dehydration to ethene.

Reveal model answer + marking points

(1) Reaction with sodium: alcohols react with active metals like sodium to liberate hydrogen gas and form sodium alkoxide, 2C2H5OH + 2Na -> 2C2H5ONa + H2. This shows the acidic nature of the -OH group. (2) Dehydration: when ethanol is heated with concentrated sulphuric acid at about 443 K (170 degrees C), it loses a molecule of water to form ethene, C2H5OH -> CH2=CH2 + H2O. Concentrated H2SO4 acts as a dehydrating agent.

C2H5OH -> CH2=CH2 + H2O

Marking-scheme points

  • With sodium: 2C2H5OH + 2Na -> 2C2H5ONa + H2 (shows acidic -OH)
  • Dehydration with conc. H2SO4 at 443 K gives ethene
  • C2H5OH -> CH2=CH2 + H2O
2 marksmediumLucas test

How can you distinguish between primary, secondary and tertiary alcohols using the Lucas test?

Reveal model answer + marking points

The Lucas test uses the Lucas reagent (a mixture of concentrated hydrochloric acid and anhydrous zinc chloride), which converts alcohols into alkyl chlorides that appear as an insoluble oily layer (turbidity). A tertiary alcohol reacts immediately and gives turbidity at once (because it forms the most stable carbocation). A secondary alcohol gives turbidity within about five minutes. A primary alcohol does not react at room temperature and gives no turbidity in the cold. Thus the rate of turbidity distinguishes the three classes.

Marking-scheme points

  • Lucas reagent = conc. HCl + anhydrous ZnCl2
  • Tertiary alcohol: immediate turbidity
  • Secondary: turbidity in about 5 min; primary: no turbidity in the cold
2 marksmediumWilliamson synthesis

What is Williamson's ether synthesis? Write the general reaction.

Reveal model answer + marking points

Williamson's synthesis is an important laboratory method for preparing symmetrical and unsymmetrical ethers. In it, a sodium alkoxide (or sodium phenoxide) reacts with a primary alkyl halide by nucleophilic substitution (SN2) to give an ether: R-O-Na + R'-X -> R-O-R' + NaX. A primary alkyl halide should be used (secondary and tertiary halides tend to undergo elimination). This method is especially useful for preparing mixed (unsymmetrical) ethers.

R-O-Na + R'-X -> R-O-R' + NaX

Marking-scheme points

  • Sodium alkoxide + alkyl halide -> ether (SN2)
  • R-O-Na + R'-X -> R-O-R' + NaX
  • Use a primary alkyl halide; good for mixed ethers
2 marksmediumReimer-Tiemann reaction

What is the Reimer-Tiemann reaction?

Reveal model answer + marking points

The Reimer-Tiemann reaction is used to introduce an aldehyde group onto the benzene ring of phenol. When phenol is treated with chloroform (CHCl3) and aqueous sodium hydroxide (NaOH) and the product is then hydrolysed with acid, an -CHO group is introduced mainly at the ortho position, giving 2-hydroxybenzaldehyde (salicylaldehyde). The reactive intermediate is dichlorocarbene (:CCl2).

phenol + CHCl3 + NaOH -> salicylaldehyde

Marking-scheme points

  • Phenol + CHCl3 + NaOH, then acid hydrolysis
  • Introduces a -CHO group at the ortho position
  • Gives salicylaldehyde (2-hydroxybenzaldehyde); intermediate is dichlorocarbene
2 markshardHydrogen bonding in nitrophenols

Why is ortho-nitrophenol steam volatile while para-nitrophenol is not?

Reveal model answer + marking points

In ortho-nitrophenol, the -OH group and the -NO2 group are close together on adjacent carbons, so they form an intramolecular hydrogen bond (within the same molecule), a process called chelation. As a result the molecules do not associate with one another and it is volatile (steam volatile). In para-nitrophenol, the two groups are far apart, so they form intermolecular hydrogen bonds between different molecules, which associate them into larger units, making it less volatile and not steam volatile.

Marking-scheme points

  • ortho-nitrophenol forms intramolecular hydrogen bonding (chelation)
  • So its molecules do not associate -> volatile/steam volatile
  • para-nitrophenol forms intermolecular hydrogen bonds -> less volatile

Aldehydes, Ketones and Carboxylic Acids9 questions

2 marksmediumPreparation of carbonyl compounds

How are aldehydes and ketones prepared by the oxidation of alcohols?

Reveal model answer + marking points

Aldehydes and ketones can be prepared by the controlled oxidation of alcohols. Oxidation of a primary alcohol gives an aldehyde (which can be further oxidised to a carboxylic acid), R-CH2OH -> R-CHO. To stop at the aldehyde stage, a mild oxidising agent such as pyridinium chlorochromate (PCC) is used. Oxidation of a secondary alcohol gives a ketone, R-CH(OH)-R' -> R-CO-R'. Tertiary alcohols are not easily oxidised as they have no hydrogen on the carbon bearing the -OH group.

R-CH2OH -> R-CHO; R2CHOH -> R2CO

Marking-scheme points

  • Primary alcohol -> aldehyde (use mild oxidant like PCC to stop there)
  • Secondary alcohol -> ketone
  • Tertiary alcohols resist oxidation (no H on the -OH carbon)
3 marksmediumNucleophilic addition

Why do aldehydes and ketones undergo nucleophilic addition reactions? Give one example.

Reveal model answer + marking points

The carbonyl group (C=O) is polar because oxygen is more electronegative than carbon, so the carbon carries a partial positive charge and the oxygen a partial negative charge. This makes the carbonyl carbon electrophilic, so it is readily attacked by nucleophiles, leading to nucleophilic addition. For example, with hydrogen cyanide (HCN), the addition gives a cyanohydrin: R-CHO + HCN -> R-CH(OH)-CN. Aldehydes are more reactive than ketones towards nucleophilic addition due to less steric hindrance and a greater positive charge on the carbonyl carbon.

R-CHO + HCN -> R-CH(OH)-CN

Marking-scheme points

  • C=O is polar; carbonyl carbon is electrophilic (partial positive)
  • Nucleophile attacks the carbonyl carbon (nucleophilic addition)
  • Example: R-CHO + HCN -> R-CH(OH)-CN (cyanohydrin)
2 marksmediumAldol condensation

What is the aldol condensation reaction?

Reveal model answer + marking points

The aldol condensation is a reaction of aldehydes or ketones that have at least one alpha-hydrogen atom, in the presence of a dilute base (such as dilute NaOH). Two molecules combine: the alpha-carbon of one adds to the carbonyl carbon of the other to give a beta-hydroxy aldehyde or ketone (an aldol). For example, two molecules of acetaldehyde give 3-hydroxybutanal (CH3CHO + CH3CHO -> CH3CH(OH)CH2CHO). On heating, the aldol loses water to give an alpha, beta-unsaturated carbonyl compound.

2CH3CHO -> CH3CH(OH)CH2CHO

Marking-scheme points

  • Needs an alpha-hydrogen and a dilute base
  • Product is a beta-hydroxy aldehyde/ketone (aldol)
  • Two acetaldehyde molecules -> 3-hydroxybutanal; loses water on heating
2 marksmediumCannizzaro reaction

What is the Cannizzaro reaction?

Reveal model answer + marking points

The Cannizzaro reaction is a disproportionation (self-oxidation-reduction) reaction shown by aldehydes that do not have an alpha-hydrogen atom, when they are treated with a concentrated alkali (such as concentrated NaOH). One molecule of the aldehyde is oxidised to a carboxylate salt and another is reduced to an alcohol. For example, two molecules of formaldehyde give methanol and sodium formate: 2HCHO + NaOH -> CH3OH + HCOONa. Benzaldehyde behaves similarly.

2HCHO + NaOH -> CH3OH + HCOONa

Marking-scheme points

  • Shown by aldehydes without an alpha-hydrogen, with conc. alkali
  • Disproportionation: one molecule oxidised, another reduced
  • 2HCHO + NaOH -> CH3OH + HCOONa
2 marksmediumTests for aldehydes

How can you distinguish between an aldehyde and a ketone using chemical tests?

Reveal model answer + marking points

Aldehydes are easily oxidised and give positive results with mild oxidising agents, whereas ketones do not. (1) Tollens' test: on warming with Tollens' reagent (ammoniacal silver nitrate), an aldehyde gives a bright silver mirror on the walls of the test tube, while a ketone gives no reaction. (2) Fehling's test: on warming with Fehling's solution, an aliphatic aldehyde gives a brick-red precipitate of cuprous oxide (Cu2O), while a ketone does not react. Thus these tests distinguish aldehydes from ketones.

Marking-scheme points

  • Aldehydes are oxidised by mild oxidants; ketones are not
  • Tollens' test: aldehyde gives a silver mirror
  • Fehling's test: aliphatic aldehyde gives a brick-red precipitate (Cu2O)
2 marksmediumPreparation of carboxylic acids

How are carboxylic acids prepared from primary alcohols and from Grignard reagents?

Reveal model answer + marking points

(1) From primary alcohols (or aldehydes) by oxidation: a primary alcohol is oxidised by a strong oxidising agent such as acidified potassium permanganate or potassium dichromate to give a carboxylic acid, R-CH2OH -> R-CHO -> R-COOH. (2) From Grignard reagents: a Grignard reagent reacts with carbon dioxide (dry ice) and the product is hydrolysed with acid to give a carboxylic acid with one more carbon atom, R-MgX + CO2 -> R-COOMgX, then hydrolysis gives R-COOH.

R-MgX + CO2 -> R-COOMgX -> R-COOH

Marking-scheme points

  • Oxidation of primary alcohol/aldehyde: R-CH2OH -> R-COOH
  • Grignard reagent + CO2 then hydrolysis -> carboxylic acid
  • Grignard method adds one carbon atom
3 marksmediumAcidity of carboxylic acids

Why are carboxylic acids more acidic than phenols? How do electron-withdrawing groups affect their acidity?

Reveal model answer + marking points

Carboxylic acids are more acidic than phenols because the carboxylate ion formed after the loss of the proton is stabilised by resonance in which the negative charge is spread equally over two electronegative oxygen atoms, making it very stable. In the phenoxide ion the negative charge is mainly on one oxygen and partly on less electronegative ring carbons, so it is less stabilised. Electron-withdrawing groups (like -Cl or -NO2) increase acidity because they further disperse and stabilise the negative charge of the carboxylate ion (for example, chloroacetic acid is stronger than acetic acid), while electron-releasing groups decrease acidity.

Marking-scheme points

  • Carboxylate ion: charge spread equally over two oxygen atoms (very stable)
  • Phenoxide ion is less stabilised
  • Electron-withdrawing groups increase acidity; electron-releasing groups decrease it
2 marksmediumReactions of carboxylic acids

Write the esterification reaction of a carboxylic acid and the decarboxylation reaction.

Reveal model answer + marking points

(1) Esterification: a carboxylic acid reacts with an alcohol in the presence of a little concentrated sulphuric acid (as catalyst) to form an ester and water, R-COOH + R'-OH -> R-COO-R' + H2O; this reaction is reversible and gives esters that have fruity smells. (2) Decarboxylation: the sodium salt of a carboxylic acid, when heated with soda lime (NaOH + CaO), loses carbon dioxide to give a hydrocarbon (alkane) with one carbon less, R-COONa + NaOH -> R-H + Na2CO3.

R-COOH + R'-OH -> R-COO-R' + H2O

Marking-scheme points

  • Esterification: R-COOH + R'-OH -> ester + water (conc. H2SO4 catalyst)
  • Decarboxylation: R-COONa + NaOH (soda lime) -> R-H + Na2CO3
  • Decarboxylation removes CO2 and gives an alkane with one less carbon
2 marksmediumIodoform test

What is the iodoform test? Which compounds give a positive result?

Reveal model answer + marking points

The iodoform test is used to detect the presence of a methyl ketone group (CH3-CO-) or a group that can be oxidised to it, such as CH3-CH(OH)-. When such a compound is warmed with iodine and sodium hydroxide (or sodium hypoiodite), it gives a yellow precipitate of iodoform (CHI3) with a characteristic smell. Compounds like acetaldehyde, acetone, ethanol and isopropanol give a positive iodoform test.

Marking-scheme points

  • Detects a methyl ketone (CH3-CO-) or CH3-CH(OH)- group
  • Warm with I2 and NaOH -> yellow precipitate of iodoform (CHI3)
  • Positive for ethanol, acetaldehyde, acetone, isopropanol

Amines8 questions

2 markseasyClassification of amines

How are amines classified? Give one example of each.

Reveal model answer + marking points

Amines are derivatives of ammonia in which one, two or three hydrogen atoms are replaced by alkyl or aryl groups. They are classified as: primary (1 degree) amine, in which one hydrogen of ammonia is replaced (e.g. CH3NH2, methylamine); secondary (2 degree) amine, in which two hydrogens are replaced (e.g. (CH3)2NH, dimethylamine); and tertiary (3 degree) amine, in which all three hydrogens are replaced (e.g. (CH3)3N, trimethylamine).

Marking-scheme points

  • Amines are derivatives of ammonia (H replaced by alkyl/aryl)
  • Primary: one H replaced (CH3NH2)
  • Secondary: two replaced ((CH3)2NH); Tertiary: three replaced ((CH3)3N)
3 marksmediumPreparation of amines

How are primary amines prepared by the reduction of nitro compounds and by Hofmann's bromamide reaction?

Reveal model answer + marking points

(1) Reduction of nitro compounds: a nitro compound is reduced (with H2/Ni, or Sn/HCl, or Fe/HCl) to a primary amine; for example, nitrobenzene is reduced to aniline, C6H5NO2 + 6[H] -> C6H5NH2 + 2H2O. (2) Hofmann's bromamide degradation: an amide is treated with bromine and a strong alkali (Br2 + NaOH/KOH) to give a primary amine with one carbon atom less than the amide, R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O.

R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O

Marking-scheme points

  • Reduction of nitro compound -> primary amine (nitrobenzene -> aniline)
  • Hofmann bromamide: amide + Br2 + NaOH -> amine with one carbon less
  • R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O
3 marksmediumBasicity of amines

Why are amines basic in nature? Compare the basic strength of amines in the gaseous phase.

Reveal model answer + marking points

Amines are basic because the nitrogen atom has a lone pair of electrons which it can donate to a proton (or to a Lewis acid), forming a bond; thus amines can accept a proton. In the gaseous phase (or a non-aqueous medium), only the inductive effect of the alkyl groups operates: more alkyl groups increase the electron density on nitrogen, so the basic strength follows the order tertiary > secondary > primary > ammonia. In aqueous solution the order changes because of the combined effect of inductive effect, solvation (hydrogen bonding of the cation) and steric hindrance.

Marking-scheme points

  • Amines are basic: nitrogen lone pair accepts a proton
  • Gas phase (inductive effect only): tertiary > secondary > primary > ammonia
  • Aqueous order differs due to solvation and steric effects
2 marksmediumHinsberg test

How does the Hinsberg test distinguish between primary, secondary and tertiary amines?

Reveal model answer + marking points

The Hinsberg test uses Hinsberg's reagent (benzenesulphonyl chloride, C6H5SO2Cl). A primary amine reacts to form a sulphonamide that is soluble in alkali (because the N-H hydrogen is acidic). A secondary amine forms a sulphonamide that is insoluble in alkali (it has no N-H hydrogen left to ionise). A tertiary amine does not react at all with the reagent (it has no replaceable hydrogen on nitrogen). Thus the three classes of amine are distinguished by their behaviour with Hinsberg's reagent.

Marking-scheme points

  • Reagent: benzenesulphonyl chloride (Hinsberg's reagent)
  • Primary amine: product soluble in alkali; secondary: product insoluble
  • Tertiary amine: does not react
2 marksmediumCarbylamine reaction

What is the carbylamine reaction? What is it used for?

Reveal model answer + marking points

The carbylamine reaction (isocyanide test) is a reaction in which a primary amine (aliphatic or aromatic) is heated with chloroform and alcoholic potassium hydroxide to form an isocyanide (carbylamine), which has an extremely unpleasant (foul) smell: R-NH2 + CHCl3 + 3KOH -> R-NC + 3KCl + 3H2O. Secondary and tertiary amines do not give this reaction. Because only primary amines respond, the carbylamine reaction is used as a test to detect primary amines.

R-NH2 + CHCl3 + 3KOH -> R-NC + 3KCl + 3H2O

Marking-scheme points

  • Primary amine + CHCl3 + alcoholic KOH -> isocyanide (foul smell)
  • R-NH2 + CHCl3 + 3KOH -> R-NC + 3KCl + 3H2O
  • Given only by primary amines -> test for primary amines
3 marksmediumDiazonium salts

How is benzenediazonium chloride prepared? Write one of its reactions (Sandmeyer reaction).

Reveal model answer + marking points

Benzenediazonium chloride is prepared by diazotisation: aniline is treated with nitrous acid (formed from sodium nitrite and hydrochloric acid) at a low temperature of 0 to 5 degrees C, C6H5NH2 + NaNO2 + 2HCl -> C6H5N2Cl + NaCl + 2H2O. In the Sandmeyer reaction, the diazonium group is replaced by a halogen or cyanide using the corresponding copper(I) salt; for example, C6H5N2Cl + CuCl -> C6H5Cl + N2. Diazonium salts are very useful for introducing many groups into the benzene ring.

C6H5NH2 + NaNO2 + 2HCl -> C6H5N2Cl + NaCl + 2H2O

Marking-scheme points

  • Diazotisation: aniline + NaNO2 + HCl at 0-5 degrees C -> C6H5N2Cl
  • Sandmeyer reaction: replace N2+ with Cl/Br/CN using Cu(I) salts
  • C6H5N2Cl + CuCl -> C6H5Cl + N2
2 marksmediumBasicity of aniline

Why is aniline less basic than ethylamine (or ammonia)?

Reveal model answer + marking points

Aniline is less basic than ethylamine (and than ammonia) because in aniline the lone pair of electrons on the nitrogen atom is delocalised (drawn) into the benzene ring by resonance. As a result the lone pair is less available for donation to a proton, so aniline accepts a proton less readily and is a weaker base. In ethylamine, the alkyl group has an electron-releasing (+I) effect that increases the electron density on nitrogen and makes the lone pair more available, so it is more basic.

Marking-scheme points

  • In aniline the N lone pair is delocalised into the ring (resonance)
  • Lone pair is less available for protonation -> weaker base
  • In ethylamine, +I effect of the alkyl group makes N more basic
2 marksmediumCoupling reaction

What is the coupling reaction of diazonium salts? Give one example.

Reveal model answer + marking points

The coupling reaction is a reaction in which a diazonium salt reacts with an electron-rich aromatic compound such as phenol or an aromatic amine to form a brightly coloured azo compound (containing the -N=N- linkage). For example, benzenediazonium chloride reacts with phenol in a mildly alkaline medium to give p-hydroxyazobenzene (an orange dye). These coupling reactions are used to prepare a large number of azo dyes.

Marking-scheme points

  • Diazonium salt + phenol/aromatic amine -> coloured azo compound (-N=N-)
  • Example: benzenediazonium chloride + phenol -> p-hydroxyazobenzene
  • Used to make azo dyes

Biomolecules9 questions

2 markseasyClassification of carbohydrates

Classify carbohydrates into three types based on hydrolysis, with one example of each.

Reveal model answer + marking points

Based on their behaviour on hydrolysis, carbohydrates are classified as: (1) monosaccharides - simple sugars that cannot be hydrolysed further into smaller units (e.g. glucose, fructose); (2) oligosaccharides - which give 2 to 10 monosaccharide units on hydrolysis, the most common being disaccharides (e.g. sucrose, which gives glucose and fructose, and maltose); and (3) polysaccharides - which give a large number of monosaccharide units on hydrolysis (e.g. starch, cellulose and glycogen).

Marking-scheme points

  • Monosaccharides: cannot be hydrolysed further (glucose, fructose)
  • Oligosaccharides (disaccharides): give 2-10 units (sucrose, maltose)
  • Polysaccharides: give many units (starch, cellulose)
2 marksmediumReducing and non-reducing sugars

Distinguish between reducing and non-reducing sugars with examples.

Reveal model answer + marking points

A reducing sugar is a carbohydrate that has a free aldehyde or ketone group (a free hemiacetal), so it can reduce Tollens' reagent and Fehling's solution; all monosaccharides (glucose, fructose) and many disaccharides (maltose, lactose) are reducing sugars. A non-reducing sugar has no free aldehyde or ketone group (the reducing groups of both units are involved in the glycosidic bond), so it does not reduce Tollens' or Fehling's reagents; sucrose is the common example of a non-reducing sugar.

Marking-scheme points

  • Reducing sugar: has a free aldehyde/ketone group; reduces Tollens'/Fehling's
  • Examples: glucose, fructose, maltose, lactose
  • Non-reducing sugar: no free reducing group (e.g. sucrose)
2 marksmediumGlucose

State two evidences that show the presence of an aldehyde group and an alcohol group in glucose.

Reveal model answer + marking points

Evidence for an aldehyde (-CHO) group in glucose: (1) it reduces Tollens' reagent to give a silver mirror and Fehling's solution to give a red precipitate; (2) it reacts with hydroxylamine to form an oxime and adds one molecule of HCN to form a cyanohydrin, confirming a carbonyl group. Evidence for alcohol (-OH) groups: glucose reacts with acetic anhydride to form a penta-acetate, showing the presence of five -OH groups. Thus glucose is a polyhydroxy aldehyde.

Marking-scheme points

  • Aldehyde group: reduces Tollens' and Fehling's; forms oxime and cyanohydrin
  • Alcohol groups: forms penta-acetate with acetic anhydride (five -OH groups)
  • Glucose is a polyhydroxy aldehyde
2 marksmediumStarch and cellulose

State two differences between starch and cellulose.

Reveal model answer + marking points

(1) Starch is a polymer of alpha-glucose units and consists of two components, amylose (a linear chain) and amylopectin (a branched chain), joined by alpha-glycosidic linkages; cellulose is a straight-chain polymer of beta-glucose units joined by beta-glycosidic linkages. (2) Starch is the main storage carbohydrate (food reserve) in plants and can be digested by humans, whereas cellulose is a structural material (the main component of plant cell walls) and cannot be digested by humans (we lack the enzyme cellulase).

Marking-scheme points

  • Starch: alpha-glucose units (amylose + amylopectin), alpha-linkages
  • Cellulose: beta-glucose units, straight chain, beta-linkages
  • Starch is a food reserve (digestible); cellulose is structural (indigestible by humans)
3 marksmediumProteins and denaturation

What are proteins? What is meant by denaturation of a protein?

Reveal model answer + marking points

Proteins are naturally occurring biomolecules that are polymers of alpha-amino acids joined by peptide bonds (-CO-NH-); they are essential for growth and maintenance of the body and act as enzymes, hormones and structural materials. Denaturation of a protein is the process in which a protein loses its natural three-dimensional shape (its secondary and tertiary structure) due to heat, change in pH, or addition of chemicals, while the primary structure (sequence of amino acids) remains intact. On denaturation the protein loses its biological activity; for example, the coagulation of egg white on boiling.

Marking-scheme points

  • Proteins are polymers of alpha-amino acids linked by peptide bonds
  • Denaturation: loss of secondary and tertiary structure (shape)
  • Caused by heat/pH/chemicals; loses biological activity (e.g. boiling egg white)
2 marksmediumStructure of proteins

Name and briefly describe the four levels of protein structure.

Reveal model answer + marking points

(1) Primary structure - the specific sequence in which the amino acids are linked in the polypeptide chain. (2) Secondary structure - the local folding of the chain into shapes such as the alpha-helix or beta-pleated sheet, held together by hydrogen bonds. (3) Tertiary structure - the overall three-dimensional folding of the whole polypeptide chain, giving a globular or fibrous shape. (4) Quaternary structure - the arrangement and association of two or more polypeptide chains (subunits), as in haemoglobin.

Marking-scheme points

  • Primary: sequence of amino acids
  • Secondary: alpha-helix or beta-pleated sheet (hydrogen bonds)
  • Tertiary: overall 3D fold; Quaternary: association of subunits (haemoglobin)
2 markseasyEnzymes

What are enzymes? State two of their characteristics.

Reveal model answer + marking points

Enzymes are biological catalysts that speed up the biochemical reactions occurring in living organisms; chemically almost all enzymes are proteins (globular proteins). Characteristics: (1) they are highly specific, each enzyme usually catalysing only one particular reaction or type of reaction (lock-and-key specificity); and (2) they are highly efficient and work best under mild conditions of an optimum temperature (around body temperature) and an optimum pH; they are denatured and lose activity outside these conditions.

Marking-scheme points

  • Enzymes = biological catalysts (mostly proteins)
  • Highly specific (one enzyme, one reaction)
  • Work best at an optimum temperature and pH
2 markseasyVitamins

How are vitamins classified? Name one deficiency disease caused by a lack of vitamin C and vitamin D.

Reveal model answer + marking points

Vitamins are organic compounds required in small amounts in the diet for normal growth and health. They are classified into two groups: (1) fat-soluble vitamins (A, D, E and K), which are soluble in fats and oils and are stored in the liver and fatty tissues; and (2) water-soluble vitamins (the B group and C), which are soluble in water and must be supplied regularly in the diet as they are not stored. Deficiency of vitamin C causes scurvy, and deficiency of vitamin D causes rickets in children.

Marking-scheme points

  • Fat-soluble vitamins: A, D, E, K (stored in the body)
  • Water-soluble vitamins: B group and C (not stored)
  • Vitamin C deficiency: scurvy; vitamin D deficiency: rickets
2 marksmediumNucleic acids

State three differences between DNA and RNA.

Reveal model answer + marking points

(1) DNA (deoxyribonucleic acid) contains the sugar deoxyribose, whereas RNA (ribonucleic acid) contains the sugar ribose. (2) DNA contains the nitrogenous bases adenine, guanine, cytosine and thymine, whereas RNA contains adenine, guanine, cytosine and uracil (uracil replaces thymine). (3) DNA is usually a double-stranded helix and stores the genetic (hereditary) information, whereas RNA is usually single-stranded and mainly takes part in protein synthesis.

Marking-scheme points

  • Sugar: deoxyribose in DNA, ribose in RNA
  • Bases: DNA has thymine, RNA has uracil (instead of thymine)
  • DNA double-stranded (stores genes); RNA single-stranded (protein synthesis)

One chapter at a time. You’ve got this.

Star this page, do a few questions each day, and by exam week Class 12 Chemistry will feel like an old friend. Know someone else sitting the same paper? Send it — you both walk in calmer.