1 Ionic compounds consist of cations whose total positive charge is balanced by the negative charge of anions. However, a new class of compounds called electrides have been synthesised that have ‘trapped’ or localised electrons taking on the role of anions. Electrides are being studied as they have unusual properties not normally found in ordinary salts.

Electrides have the general formula M+(L)ne-, in which e- represents the trapped electron, M+ is an alkali metal cation and L is an organic molecule that binds the cation in a cage.

(a) Crown ethers are a group of synthetic large-ring polymers which are cyclic polymers of ethane-1,2-diol and are named in the form x-crown-y, where x is the total number of atoms in the ring and y is the total number of oxygens.

(i) Write an equation to show the formation in solution of an electride from a caesium atom and two molecules of the crown ether, 15-crown-5, represented by 15C5. Use the state symbol (solv) to indicate salvation by a non-aqueous solvent.

Cs (s) + 2 15C5 (solv) Cs+(15C5)2e- (solv).................................................................................................................................................

(ii) Calculate the mass of caesium metal needed to react with 5.0 cm3 of the crown ether, 15C5.

[Mr 15C5 = 220, Cs = 132.9, density of 15C5 = 1.1 g cm-3]

Mass of 15C5 = 5.0 x 1.1 = 5.5 g

No of moles of 15C5 = 5.5 / 220 = 0.025 mol

Mass of Cs metal = 0.025 / 2 x 132.9 = 1.66 g

(iii) Calculate the theoretical yield in cm3 of the electride product (density 1.30 g cm-3) formed from the above reaction.

No of moles of electride = 0.025 / 2 = 0.0125 mol

Mass of electride = 0.0125 x (220 x 2 + 132.9) = 7.16 g

Theoretical yield = 7.16125 / 1.30 = 5.51 cm3

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Class Reg NumberAnswer Scheme

Candidate Name .......................................................................

Chemistry H2 9746Tutor TuteeRevision Exercise 20: Paper 2 Revision

(b) A generalised Born-Haber cycle for electrides is shown below, where M is the metal, L is a ligand and e- is an electron.

Cryptands are similar to crown ethers but have a third organic chain attached to two trivalent nitrogen atoms. The prefix numbers of a cryptand name indicate the number of oxygen atoms in each of the three chains.

(i) Explain what is meant by the term standard enthalpy change of formation.

It refers to the enthalpy change when 1 mole of compound is formed from its constituent elements in their standard states under standard conditions of 298 K and 1 atm.................................................................................................................................................

(ii) Using the data tabulated below, construct a Born-Haber cycle and calculate a value for the enthalpy of formation of Cs+(C222)e-(s) from Cs(s) and C222(s), where C222 is a cryptand.

Equations H / kJ mol-1

Cs(s) Cs(g) +76.6

Cs(g) Cs+(g) + e- +378.1

C222(s) C222(g) +54.3

Cs+(g) + C222(g) Cs+(C222)(g) -276.5

Cs+(C222)(g) + e- Cs+(C222)e-(g) -292.3

Show correct Born-Haber cycle with annotated arrows and respective values, label 0 at Cs (s) + C222(s), Hf accordingly

Using Hess’ Law,

Hf = 76.6 + 54.3 + 378.1 – 276.5 – 292.3 = -59.8 kJ mol-1

A Level Chemistry Notes and Questions Chapter 22 Sulphur hexafluoride, SF6, was first synthesised by Henri Moissan in 1900 by burning

2

sulphur in fluorine gas. The synthesis involves breaking the weak S–S and F–F covalent bonds followed by the formation of strong S–F bonds, making the process extremely exothermic.

(a)(i) Write an equation to show the standard enthalpy change of formation of sulphur hexafluoride from fluorine and sulphur molecules (S8).

3F2 (g) + 1/8 S8 (g) SF6 (g).................................................................................................................................................

(ii) Calculate an approximate value for the standard enthalpy change of formation of sulphur hexafluoride using the following bond dissociation energies.

S–S 264 kJ mol-1 F–F 158 kJ mol-1 S–F 156 kJ mol-1

Hr = HBEreactants - HBEproducts

= 3(158) + 264 – 156(6) = -198 kJ mol-1

(b) The industrial synthesis of SF6 involves the combustion of molten sulphur in fluorine gas. The product is contaminated with traces of disulfur decafluoride (S2F10). To remove the impurity, the product is heated to 500oC at which temperature it disproportionates.

S2F10 (g) SF6 (g) + SF4 (g)

(i) Draw a labelled sketch of the bonding present in SF4 and SF6, showing bond angles and the molecular shapes where appropriate.

SF4: see-saw (4 bond pairs, 1 lone-pair)

SF6: octahedral (6 bond pairs, 0 lone-pair)

(ii) Predict whether PF5 would be expected to be more or less chemically reactive than SF6. Explain your answer with reference to the relative sizes of the central atoms of the two

3

molecules and the strength of the bonds formed with fluorine.

PF5 is expected to be chemically more reactive than SF6. The phosphorus atom of PF5 is larger than the corresponding sulphur atom of SF6, hence it is more ‘exposed’ to attacking chemical species. The P-F bonds are expected to be weaker due to the greater size of the phosphorus atom compared to the sulphur atom. As a consequence the bond is longer and hence weaker since the 2 atoms cannot approach each other closely. .................................................................................................................................................

(iii) Define the term disproportionation.

It refers to the simultaneous oxidation and reduction of a single chemical species..................................................................................................................................................

(iv) Deduce the oxidation numbers for sulphur in all three sulphur containing species present during disproportionation of disulphur decafluroide.

S2F10: +5, SF6: +6, SF4: +4.................................................................................................................................................

(c) Industrial production has resulted in atmospheric concentrations increasing rapidly over the past 45 years. In 1978, the level of SF6 in the atmosphere was 0.6 ppt (parts per trillion, where 1 trillion is 1012) and by 1998, the level had reached 4 ppt.

(i) Calculate the number of SF6 molecules present in 1 dm3 of the atmosphere measured at 273 K, 1 atmospheric pressure and at s.t.p. in the year 1978.

No of SF6 molecules = (6.02 x 1023 / 22.4) x (0.6 / 1012) = 1.16125 x 1010 mol

(ii) Calculate the total percentage increase in the SF6 concentration in the atmosphere from 1978 to 1998.

% increase = (4 – 0.6) / 0.6 x 100% = 5.66%

(iii) Assuming the current rate of increase remains constant, calculate the SF6 concentration in the atmosphere by 2010.

4 x (0.25/100) x 12 = 0.12

0.12 + 4.0 = 4.12 ppt

A Level Notes and Questions Chapter 26 3 Coffee is a highly popular drink as it provides a pleasant-tasting and warming drink. The

delicious aroma of roast coffee beans has been found to contain over 800 different

4

compounds. They include melanoidins, citric acid and a class of polyhydroxy compounds called chlorogenic acids, as shown below.

(a) Draw the structures of the organic product(s) when chlorogenic acid is reacted with the following compounds, stating also the type of reaction undergone.

(i) Aqueous bromine,

Electrophilic addition (across C=C via Markonikov’s rule) and electrophilic substitution (2 phenol groups)

(ii) Acidified 2,4-dinitrophenylhydrazine,

Hydrolysis of ester bond

No orange precipitate observed due to absence of C=O functional group

(iii) Phosphorus(V) chloride,

Steamy white fumes of HCl observed, nucleophilic substitution

(iv) Cold, dilute potassium manganate(VII)

5

Oxidation (formation of diol)

(v) Excess aqueous sodium hydroxide in reflux.

Hydrolysis of ester bond, acid-base reaction with phenol and carboxylic acid group, NO reaction with –OH groups on cyclohexane ring

(b) Chlorogenic acid, C16H18O9, behaves effectively as a weak monobasic acid according to the equation:

C16H18O9 (aq) + H2O (l) ⇌ H3O+ (aq) + C16H17O9- (aq)

[Ka = 3.00 x 10-4 mol dm-3]

For 250 cm3 of solution containing 3.54 g chlorogenic acid, calculate

(i) The concentration of chlorogenic acid in mol dm-3,

No of moles of chlorogenic acid = 3.54 / (12.0 x 16 + 1.0 x 18 + 9 x 16.0) = 0.01 mol

[chlorogenic acid] = 0.01 / (250/1000) = 0.04 mol dm-3

(ii) The pH of the chlorogenic acid solution.

Ka = [H+]2 / [HA] [H+] = 3.46 x 10-3 mol dm-3

pH = -lg(3.46 x 10-3) = 2.46

(iii) A 15 cm3 sample of the chlorogenic acid solution is titrated with 0.02 mol dm-3 aqueous sodium hydroxide. Calculate the volume of sodium hydroxide solution needed to react with

6

the chlorogenic acid present.

No of moles of chlorogenic acid = 15/1000 x 0.04 = 6 x 10-4 mol

Volume of NaOH = (6 x 10-4 / 0.02) x 1000 = 30.0 cm3

(iv) Predict the pH at the equivalence point, explaining your answer.

pH will be greater than 7 (about 8) due to the strong conjugate base of C16H17O9- of

the weak acid C16H18O9 having moderate tendency to undergo hydrolysis in water to produce an excess of OH- over H+..................................................................................................................................................

(v) Hence, sketch a graph showing the change in pH when 0.1 mol dm -3 aqueous chlorogenic acid is gradually added to 10 cm3 of 0.1 mol dm-3 aqueous barium hydroxide solution.

Volume of chlorogenic acid at end point = 20.0 cm3

Maximum buffering capacity = pKa = 3.52

(c) Tea leaves contain about 4% by mass of caffeine while coffee beans contain between 1 to 2%. Caffeine has the molecular structure as shown below. Another similar compound is adenosine found in DNA.

Describe a chemical test to distinguish between both compounds, stating any observations.

Add HCl (aq) and reflux to break the amide linkage in caffeine. Add Na2CO3 (aq) at room temperature to the remaining compound. If effervescence which forms a white precipitate in limewater is observed (CO2), caffeine is present. If no effervescence is observed, adenosine is present. .................................................................................................................................................

(d) When coffee beans or tea leaves are immersed in hot water, the concentration of caffeine increases with time. The concentration c rises rapidly initially, but then increases at a slower rate before reaching a constant value c at equilibrium.

Research on coffee has shown that that the rate constant, k, does not depend on the degree of stirring but rather k varies inversely with r2, where r is the radius of the ground coffee particles, infusion is therefore faster the more finely the beans are ground; k increases by roughly 3% for every 1oC rise in temperature.

7

(i) Sketch a graph of concentration of caffeine, c, against time t when coffee beans or tea leaves are immersed in hot water.

(ii) If the radius r, of coffee beans in separate samples are reduced by a factor of half and a quarter respectively, what will be the effect on the rates of infusion of caffeine, k?

Rate of infusion of caffeine will increase by 4 (1/(0.52)) and 16 (1/(0.252) times respectively since k (1/r2), resulting in a faster rate of infusion.................................................................................................................................................

A Level Notes and Questions Chapter 304(a) In the past decade, scientists have learnt that ancient DNA extracted from mummified

tissue samples of extinct animals and even fossils can be analysed. Tris (hydroxymethyl) aminomethane, or tris, is a biochemical buffer whose buffering range is pH 7 to 9 and Kb is 1.19 x 10-6 for the following reaction.

(HOCH2)3CNH2 (aq) + H2O (l) ⇌ (HOCH2)3CNH3+ (aq) + OH- (aq)

Tris TrisH+

(i) Explain what is meant by the term buffer.

A buffer is a solution which can maintain a fairly constant pH when a small amount of acid or base is added to it. .................................................................................................................................................

(ii) With reference to the buffering range of tris, account for its basicity.

The electron-donating inductive effect of the alkyl group increases the availability of the lone pair of electrons on N for donation, leading to basicity of tris .................................................................................................................................................

(ii) The optimum pH of blood is 7.40. Calculate the ratio of tris / trisH+ at this pH.

Ka = Kw / Kb = (10-14) / (1.19 x 10-6) = 8.40 x 10-9 mol dm-3

pH = pKa + log10 ([tris]/[trisH+])

7.4 = 8.08 + log10 ([tris]/[trisH+])

Ratio = 10-0.68 = 0.20

8

c

c

t

(iii) Calculate the pH after 1 cm3 of 6 mol dm-3 HCl(aq) is added to a 200 cm3 sample of the pH 7.4 tris buffer.

No of moles of HCl = 1/1000 x 6 = 6 x 10-3 mol

The H+ from HCl will convert tris to trisH+

mol/dm-

3Tris H+ trisH+

I 0.1 (6 x 10-3) / 0.2 = 0.03

0.1

C -0.03 -0.03 +0.03E 0.07 0 0.13

pH = 7.4 + log10 (0.07/0.13) = 7.13

(b) The bleach solution used in ancient DNA laboratories contains sodium hypochlorite, NaClO.

(i) With the aid of an equation, outline the procedure for making a solution of sodium hypochlorite in the laboratory.

2NaOH + Cl2 NaCl + NaClO + H2ONaClO can be prepared by bubbling chlorine gas through cold dilute aqueous NaOH..................................................................................................................................................

(ii) Calculate the number of moles of sodium hypochlorite in 100 cm3 of 20% industrial strength bleach solution, whereby 20% means 20 g of NaClO is present in 100 cm3 of bleach solution. Hence, calculate the concentration of the bleach solution of NaClO.

No of moles of NaClO = 20 / (23+35.5+16.0) = 0.268 mol

[NaClO] = 0.268 / (100/1000) = 2.68 mol dm-3

(iii) Suggest why ancient DNA researchers take care not to mix bleach solution with acid.

Poisonous chlorine gas which is not very soluble in water will be evolved..................................................................................................................................................

Industrial strength bleach is unstable and on standing, particularly in sunlight, evolves oxygen.

(iv) Write an equation for the above reaction.

2ClO- O2 + 2Cl-

.................................................................................................................................................

(v) Suggest how bleach should be stored in ancient DNA labs to minimise its decomposition.

Bleach should be stored in light proof containers and stored in cold conditions to minimise its decomposition.................................................................................................................................................

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(c) Another buffer system in the blood involves carbonic acid and hydrogencarbonate. When carbon dioxide is dissolved in water, the following equilibria are established:

The first and slowest step in the reaction is the dissolution of the gas:

CO2 (g) ⇌ CO2 (aq)

The next reaction produces hydrogencarbonate ions

CO2 (aq) + H2O (l) ⇌ H+ (aq) + HCO3- (aq)

which then dissociate to give carbonate ions,

HCO3- (aq) ⇌ H+ (aq) + CO3

2- (aq)

The table below shows some physical properties of pure carbon dioxide under commercial carbonating conditions.

Molar mass 44.01 g mol-1

Vapour pressure (15oC) 5.085 x 106 Pa

Density of gas (15oC) 1.85 g dm-3

Temperature 31.1 oC

Pressure 7.383 x 106 Pa

Solubility of gas in water (15oC, 105 Pa) 1.98 g dm-3

The relationship between gas pressure and concentration of mass of dissolved carbon dioxide is given by Henry’s Law:

PCO2 = Kh[CO2(aq)]

Where PCO2 represents the partial pressure of CO2 above the drink, [CO2(aq)] the concentration of dissolved CO2 gas and Kh, the Henry’s law constant for CO2.

Henry’s Law states that the amount of gas dissolved in a solution is directly proportional to the pressure of the gas in the head space above the solution.

(i) It was found that Henry’s Law is not exactly obeyed by carbon dioxide, especially under commercial carbonating conditions. With reference to the data above, explain why Henry’s Law may not be obeyed under them.

The high pressures and low temperatures used in the carbonating process cause the CO2 to deviate from ideal behaviour and the interaction between CO2 molecules in solution affects its solubility in water. CO2 reacts chemically with the water and a proportion forms carbonic acid, increasing the solubility of the gas..................................................................................................................................................

(ii) Suggest why the solubility of carbon dioxide and nearly all other gases decreases with increasing temperature.

The dissolution of carbon dioxide is an exothermic process. Increasing the temperature causes the equilibrium position to shift to the left by LCP to favour the undissolved CO2 gas and water, since the backward reaction is endothermic where the excess heat is removed. Hence, solubility decreases with increasing

10

temperature. .................................................................................................................................................The Henry’s Law constant, Kh, for carbon dioxide in water at 25oC is 3.8 x 10-2 mol dm-3

atm-1.

(iii) Air contains 0.0035% of carbon dioxide by volume. Calculate the solubility in mol dm -3 of carbon dioxide at 25oC.

Kh = [CO2(aq)] / PCO2 = 3.8 x 10-2 mol dm-3 atm-1

PCO2 = 0.0035 / 100 x 1 = 3.5 x 10-4 atm-1

[CO2(aq)] = Kh x PCO2 = (3.8 x 10-2) (3.5 x 10-4) = 1.33 x 10-5 mol dm-3

(iv) A fizzy drink is produced in a factory using a carbonating process which supplies carbon dioxide at a pressure of 12 atmospheres. Calculate the volume of carbon dioxide measured at 25oC and 1 atmosphere that would be dissolved in a 500 cm3 bottle of the carbonated drink, assuming that Henry’s Law is obeyed.

PCO2 = 10 atm

[CO2(aq)] = Kh x PCO2 = (3.8 x 10-2) x 12 = 0.456 mol dm-3

In 500 cm3, the amount of undissolved CO2 = 0.456 x (500/1000) = 0.228 mol

Vol of CO2 dissolved = 0.228 x 24 = 5.47 dm3 (to 3 sf)

(v) Assuming Henry’s Law is obeyed by carbon dioxide, deduce the pH change of the resulting carbonated drink if the pressure is doubled from 12 to 24 atmospheres.

Assuming Henry’s Law is obeyed by CO2, a doubling of the pressure employed in the carbonating process will increase [CO2(aq)], and hence H3O+ ions by a factor of 2. A doubling of H+ (aq) concentration will increase the pH by about 0.3 (log10(2)) units.

A Level Notes and Questions Chapter 32, 335 In recent years, there has been a growing demand for reduced alcohol-beers but

unfortunately ethanol (alcohol) is a major contributor to beer flavour and drinkability and so beers that contain very little alcohol will taste very different from one containing the usual 4 to 5% alcohol, but beers contain many other substances that contribute to their flavour which must be taken into account when producing low-alcohol beers.

(a) A typical Australian beer contains 5% ethanol by volume. Given that the densities of ethanol and water are 0.789 and 0.998 g cm-3 respectively, calculate

(i) The concentration of ethanol in mol dm-3,

Composition by volume of beer = 5 cm3 ethanol : 95 cm3 H2O

Composition by mass of beer = (5 x 0.789) : (95 x 0.998) = 3.95 g ethanol : 94.81 g H2O

% by mass of ethanol = 3.95 / 94.81 x 100 = 4.16 %

Amount of ethanol in 100 cm3 = 3.95 / 46 = 0.086 mol

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[ethanol] = 0.086 x 10 = 0.860 mol dm-3

(ii) The number of ethanol molecules present in a pint of low alcohol beer that contains 1% by volume of ethanol.

[1 dm3 = 1.8 pints, Avogadro’s constant = 6.02 x 1023 mol-1]

[ethanol] = 0.086 / 5 = 0.0172 mol dm-3

No of molecules of ethanol in 1 litre = 0.0172 x 6.02 x 1023 = 1.04 x 1022

No of molecules of ethanol in 1 pint = (1.04 x 1022) / 1.8 = 5.7 x 1021

(b) Bacteria like yeast used in the fermentation of glucose to ethanol contain a ‘proton pump’ to absorb sugars and other nutrients. The simplest method of restricting the amount of alcohol produced during fermentation is to use strains of yeast that are unable to convert the sugars that brewing yeasts would use to convert to alcohol.

(i) Account for the higher than expected pH of a low-alcohol beer (ethanol concentration less than 1.2%) compared to that of a typical beer where the ethanol concentration is between 4 to 5.

Due to lower concentrations of ethanoic acid formed by the bacterial and chemical oxidation of ethanol. Alternatively, beers with limited fermentation may be high as yeast uses ‘proton pumps’ to assimilate sugars. With less assimilation, there will be less protons and hence higher pH..................................................................................................................................................

(ii) Although some beers are carbonated with dissolved carbon dioxide, this is not a significant contributor to the low pH of beers. Comment on this statement.

A small proportion of the dissolved CO2 reacts with water to form carbonic acid which is a weak acid and only weakly dissociated to give a relatively low concentration of H+ ions..................................................................................................................................................

(c) Hops were first added to beer in the 15th century and not only give beer its bitter taste but also to add special flavours and aromas. Hops contain compounds that react on heating to form other, even more bitter compounds. One such reaction, in which humulone is converted to the extremely bitter compound, iso-humulone, is shown below.

12

(i) Examine the structure of iso-humulone and note the presence of two hydroxyl groups within the molecule. Explain why the iso-humulone molecules do not undergo simple oxidation when beer is left to stand in air.

Iso-humulone does not undergo simple oxidation due to the absence of primary or secondary alcohol (-OH) groups. Only tertiary alcohol and –OH bonded to C atom of C=C are present which are unable to undergo oxidation under mild conditions..................................................................................................................................................

(ii) Account for the greater solubility of iso-humulone compared to humulone, despite the presence of only two hydroxyl groups in the former.

The greater solubility of iso-humulone compared to humulone is due to the loss of the aromatic ring system..................................................................................................................................................

(iii) Bitterness is a flavour, not an aroma and cannot be smelt. Suggest why it is unlikely that the R group (CH2CH(CH3)2) present in humulone and iso-humulone is the specific functional group responsible for the bitter taste of beer.

It is unlikely as it undergoes no change during the conversion of tasteless humulone to the bitter iso-humulone..................................................................................................................................................

(iv) State the structural relationship between humulone and iso-humulone.

Structural isomers..................................................................................................................................................

(v) Predict if iso-humulone makes a significant contribution to the smell of beer, explaining your answer.

Iso-humulone makes little or no contribution to the smell of beer due to its relativel low volatility which reflects it relatively high molecular mass. Moreover, bitterness is not an aroma, since there are no bitterness receptors in the nose. .................................................................................................................................................

A Level Notes and Questions Chapter 256(a) Both esters and amides undergo acid-catalysed hydrolysis. The mechanism for the

13

hydrolysis of either functional group involves the following four steps:

protonation; nucleophilic addition of water; proton transfer; elimination of an alcohol or an amine molecule.

For esters, the protonation that occurs during the first step produces the more stable cation A rather than cation B.

(i) Suggest why cation A is more stable than cation B.

Cation A is more stable than cation B since the lone pair of electrons on the O in A can be delocalised into the C=O+, hence the positive charge is spread over 3 atoms:

On the other hand, for cation B, delocalisation involving the C=O is not possible..................................................................................................................................................

The length of the C-N bond in amides (0.132 nm) is significantly shorter than an ordinary C-N bond (0.147 nm). Furthermore, the amide group is planar.

(ii) Suggest an explanation for these observations.

This is due to the partial donation of the nitrogen atom’s lone pair of electrons to the δ+ carbon of the carbonyl group, though the overlap of a filled p-orbital on nitrogen with the π bond of the carbonyl group:

Due to the partial double bond character of the C–N bond, it is shorter and stronger than normal C–N bond. In order for the delocalisation to be effective, the p-orbital carrying the lone pair on N must be parallel to the p-orbitals of C and O in the C=O. Hence, the N must be sp2 hybridised, like the carbonyl carbon, leading to a planar

14

structure..................................................................................................................................................

(iii) Use this information, together with your mechanism in part (i), to suggest why amides undergo hydrolysis at a much slower rate than esters.

The extent of overlap with esters is much smaller than in amides because the oxygen atom in esters is much more electronegative than is the nitrogen atom in amides, so it is less likely to donate its lone electron pair. Hence the carbonyl carbon in amides are less electron deficient than that in esters, leading to slower reaction with the water nucleophiles, and hence slower rate of hydrolysis..................................................................................................................................................

(c) Gel electrophoresis is often employed to analyze the mixtures of amino acids and small peptides obtained by the hydrolysis of proteins. In a particular experiment, glycine, H2NCH2COOH, is an amino acid obtained from the hydrolysis of a polypeptide protein molecule.

(i) Using the amino acid glycine (H2NCH2COOH, isoelectronic point = 6.1) as an example, explain how the pH of the solution used for electrophoresis can influence the results.

When pH < 6.1, glycine molecules exist predominantly as positively charged species. Hence it is attracted and will migrate towards the negative electrode.When pH > 6.1, glycine molecules exist predominantly as negatively charged species. Hence it is attracted and will migrate towards the positive electrode.When pH = 6.1, glycine exists predominantly as a zwitterions and will be electrically neutral. It will migrate to neither of the electrodes..................................................................................................................................................

The diagram below shows the result of carrying out electrophoresis on a sample of amino acids obtained from hydrolyzing a protein.

Assuming all the amino acids shown above are overall singly charged species at a particular operating pH,

(ii) which amino acid has the lowest relative molecular mass?

S.................................................................................................................................................

(iii) which amino acid has a positive charge?

P.................................................................................................................................................

A Level 2007 H3 Q4a VJC, RJC H3 Chemistry 2008 Prelim Q4, 5(b)

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7(a) The substance GHB was originally designed for use in sleeping pills. However, other drug-abuse issues were identified with the substance and its sale was restricted in 2003. GHB stands for hydroxybutyric acid, an old name for the structure shown below.

(i) Name the two functional groups in GHB.

Hydroxyl (alcohol) and carboxylic acid functional groups.................................................................................................................................................

(ii) Give the systematic name for GHB.

4-hydroxy butanoic acid.................................................................................................................................................

(b) A substance known as GBL is converted into GHB in the body. The structure of GBL is shown below.

GBL

(i) Name the functional group in GBL.

Ester functional group.................................................................................................................................................

(ii) Name the type of reaction by which GBL forms GHB in the body.

Hydrolysis.................................................................................................................................................

(iii) Describe a chemical test to distinguish between GHB and GBL, stating any observations.

Add aqueous Na2CO3 at room temperature to both GHB and GBL. If effervescence which forms a white ppt in limewater (CO2) is observed, GHB is present. If no effervescence is observed, GBL is present..................................................................................................................................................

(c) GHB is a weak acid. Weak acids can be represented as HA.

(i) Write an equation to show how a weak acid HA behaves when dissolved in water.

16

HA + H2O ⇌ H3O+ + A- or HA ⇌ H+ + A-

.................................................................................................................................................

(ii) Use ions and molecules from this equation to explain the meaning of the term conjugate base.

A– is the conjugate base of HA; related by loss of proton.................................................................................................................................................

(iii) Write an expression for the acidity constant Ka of an acid HA.

Ka = [H+] x [A–]/ [HA]

(iv) A 0.10 mol dm–3 solution of GHB has a pH of 2.9.Calculate the value of Ka for GHB and give its units.

Ka = (10-2.9)2 / 0.10 = 1.58 x 10-5 mol dm-3

(v) State one simplifying assumption that you made when carrying out your calculation in (iii).

[HA] initial = [HA] at equilibrium.................................................................................................................................................

(d) A mixture of GHB and its sodium salt acts as a buffer solution.

(i) Explain the meaning of the term buffer solution and explain why buffer solutions are found in our bodies.

A buffer solution is one which can maintain a fairly constant pH when a small amount of acid or base is added to it. They are found in our bodies so as to maintain the optimum pH of the body / blood / enzymes. .................................................................................................................................................

(ii) Calculate the pH of a buffer solution containing equal amounts of GHB and its sodium salt.

pH = pKa = -lg (1.58 x 10-3) = 4.80

OCR AS Level Chemistry by Design SAM Unit F335 Q48(a) Explain what is meant by ligand exchange.

substitution of one ligand for another.................................................................................................................................................

(b) Describe all the colour changes and observations that take place when an aqueous solution of ammonia is gradually added to a solution of Cu2+(aq), until the ammonia is in excess. Write equations for these transformations.

When NH3 (aq) is added, a blue ppt of Cu(OH)2 is first formed as follows when its IP > Ksp

NH3 + H2O NH4+ + OH–

17

[Cu(H2O)6]2+ + 2OH– Cu(OH)2 + 6H2O (1)

When excess NH3 is added, the ppt dissolves to give a more stable deep blue solution due to the formation of the complex ion [Cu(NH3)4(H2O)2]2+

Cu(OH)2 + 4NH3 + 2H2O [Cu(NH3)4(H2O)2]2+ + 2OH– (2)

Overall reaction is obtained by adding (1) and (2)

[Cu(H2O)6]2+ + 4NH3 [Cu(NH3)4(H2O)2]2+ + 4H2O (3)

Reaction (3) is a ligand exchange reaction since NH3 displaces H2O from [Cu(H2O)6]2+ to form [Cu(NH3)4(H2O)2]2+

Hence, the decrease in concentration of [Cu(H2O)6]2+ in (3) causes eqm position (1) to shift backwards such that IP < Ksp Cu(OH)2, causing Cu(OH)2 to dissolve

.................................................................................................................................................

(c) Blood gets its colour from oxygen-carrying molecules with organic groups surrounding a transition metal ion. In humans this transition metal is iron, and the blood is red. In horseshoe crabs, the metal is copper and the blood is blue, and in sea squirts the metal is vanadium and the blood is green.

The sketch below shows the major absorption peak for human blood.

On this sketch, show and label the corresponding absorption peaks for the blood of horseshoe crabs and sea squirts. Explain your answer.

Horseshoe crab blood is blue due to absorption at orange/red region and hence the complementary colour (blue) is shown. Sea squirts blood is green and hence the complementary region will be orange/red and blue/violet will be absorbed..................................................................................................................................................

(d) A 0.0100 mol sample of an oxochloride of vanadium, VOCIx required 20.0 cm3 of 0.100 mol dm-3 acidified potassium manganate(VII) for oxidation of the vanadium to its +5 oxidation

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state.

Deduce the value of x in the formula VOCIx.

No of moles of KMnO4 = (20/1000) x 0.100 = 0.002 mol

0.01 mol VOClx 0.002 mol KMnO4

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

5 mol VOClx 1 mol MnO4- 5 mol e-

1 mol VOClx 1 mol e-

Initial OS = +5 – 1 = +4

x = 2 (VOCl2)L_A Level Chemistry SAM

End of Paper

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