1997理論及實作解答

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8 theoretical problems

2 practical problems

THE TWENTY-NINTH

INTERNATIONAL CHEMISTRY OLYMPIAD

13–22 JULY 1997, MONTREAL, CANADA

THEORETICAL PROBLEMS

PROBLEM 1

Compound X is a trisaccharide which occurs principally in cottonseed meal. Compound X does not react with Benedict’s or Fehling’s solutions nor does it mutarotate. Acid-catalyzed hydrolysis gives three different D-hexoses, A, B, and C. Compounds A and B, as well as compound 1 (see below), all give the same osazone upon reaction with excess acidic phenylhydrazine. Compound C reacts with nitric acid to give an optically inactive compound D. The Kiliani-Fischer approach is used to establish the configurational relationship between D-glyceraldehyde and C. The intermediate aldotetrose which leads to C does not give a meso compound when oxidized by nitric acid. When A is treated with nitric acid, the dicarboxylic acid (aldaric acid) produced is optically active. Both A and B react with 5 moles of HIO4; one mole of A gives 5 moles of methanoic (formic) acid and

one mole of methanal (formaldehyde) while one mole of B gives 3 moles of methanoic (formic) acid and 2 moles of methanal (formaldehyde) and one mole of carbon dioxide. Both A and B are related to the same aldotetrose which is the diastereoisomer of the one to which C is related. Methylation of X followed by hydrolysis gives a 2,3,4-tri-O-methyl-D-hexose (E) (derived from A), a 1,3,4,6-tetra-O-methyl-D-2,3,4-tri-O-methyl-D-hexose (F) (derived from B), and a 2,3,4,6-tetra-O-methyl-D-hexose (G) (derived from C).

1.1 On the answer sheet, draw Fischer projection formulas of A, B, C, and D.

1.2 On the answer sheet, complete the appropriate Haworth projection formulas to clearly show the ring size and absolute stereochemistry of E, F, and G. Either of the anomeric forms are acceptable as an answer.

C H HO H OH H OH H CH ₂OH HO O H C CH ₂OH OH H O H D-Glyceraldehyde Compound 1 _______________

S O L U T I O N

Compound X is a trisaccharide which does not react with Benedict’s solution nor does it mutarotate. This shows that X is a non-reducing sugar and therefore only acetal or ketal linkages exist at all of the anomeric carbons. Of the three monosaccharides, A and B give the same osazone and therefore have identical stereochemistry at C-3, C-4, and C-5 (and C-6). A and B are also different from compound 1 (i.e. D-mannose) yet give the same osazone, and thus one of them must be the C-2 epimer of D-mannose (i.e. D-glucose) and the other must be the corresponding keto sugar at C-2 (i.e. D-fructose). (This deduction is confirmed later in the oxidative cleavage reactions.) Compound C, after reaction with nitric acid, gives an optically inactive aldaric acid D. The two possible aldaric acids which could be D are thus:

CO _{2 H} H H OH OH CO _{2 H} H HO HO H CO _{2 H} OH OH OH OH CO _{2 H} H H H H AA1 (=D) AA2

The aldotetrose which is the precursor of C (and thus also of D) does not give a meso compound after reaction with nitric acid and therefore must be the D-threose:

CHO OH CH ₂OH H OH CH _{2 OH} H H CHO HO OH CH _{2 OH} H OH CHO H OH CO ₂H H OH CO 2H H OH CO ₂H H H CO 2H HO D-Glyceraldehyde meso; thus not part of chiral; thus part of (D-threose) C C

It follows from this that the aldaric acid D produced from C above is AA1 and thus that C must be D-galactose. Compound A reacts with 5 moles of HI to give 5 moles of methanoic (formic) acid and one mole of methanal (formaldehyde) suggesting that it is an aldohexose while B reacts with 5 moles of HI to give 4 moles of methanoic (formic) acid, one mole of methanal (formaldehyde) and one mole of CO2 suggesting that it is a ketohexose.

Compounds A and B are related to the same tetrose which is not the same as that of C (i.e. are related to D-erythrose). The tetrose which is related to A and B must therefore have the following structure and accordingly A is D-glucose and B is D-fructose.

CHO CH 2 OH OH H H OH D-erythrose CHO OH H OH OH CH ₂OH H H HO H CH ₂OH H O OH H HO OH CH ₂OH H A (D-glucose) B (D-fructose) CHO H H OH OH CH ₂OH H HO HO H C (D-galactose)

1.2 CHO OCH ₃ H OCH 3 OH CH 2 OH H H H 3 CO H E derived from A H H 3 CO OCH 3 H HOH _{2 C} OCH 3 H H H OH O H H OCH 3 H _{3 CO} H O H 3 COH 2 C CH 2 OCH 3 OH CH 2 OCH 3 H O OCH ₃ H H _{3 CO} OH CH 2 OCH 3 H F derived from B H OCH ₃ H H _{3 COH 2 C} OCH 3 H H _{3 CO} H OH O H CHO H H OCH 3 OH CH _{2 OCH 3} H H 3 CO H _{3 CO} H G derived from C 1.3

During methylation, only hydroxyl groups not involved in acetal/ketal formation (either intra- or intermolecular) will be etherified. From the methylation data, only E has two free hydroxyl groups with which to link to the other carbohydrates. Thus A must be the central carbohydrate.

These results indicate that the sequence of monosaccharides in X is C-A-B (or B-A-C).

If: A5 represents the furanose (5-membered ring) form of carbohydrate A. A6 represents the pyranose (6-membered ring) form of carbohydrate A.

B5 represents the furanose (5-membered ring) form of carbohydrate B, etc. then the trisaccharide X would be represented as: C6-A6-B5

One of the 4 possible variations in the structure of X is given below.

C A H H OH HO H O HOH _{2 C} CH _{2 OH} O O H H H H OH OH H OH CH ₂ O H OH H CH _{2 OH} OH H HO H H O B Trisaccharide X

Note: The nature of the anomeric linkages was not specified in the problem. The linkage arrangement of A to B and C may also be reversed (i.e. a 1,1' linkage between C and A and a 1,6 linkage between A and B.

PROBLEM 2

Professor Molina of the Massachusetts Institute of Technology won the 1995 Nobel Prize in Chemistry for his work on atmospheric chemistry. One reaction that he has studied in detail is the acid rain reaction which produces H2SO4 in the atmosphere. He has

proposed two possible stoichiometric reactions:

Proposal A: H2O (g) + SO3 (g) → H2SO4 (g)

Proposal B: 2 H2O (g) + SO3 (g) → H2SO4 (g) + H2O (g)

2.1 Using simple collision theory, what reaction orders would be expected for Proposal A and for Proposal B?

Proposal B is thought to proceed by the following two-step process:

SO3 + 2 H2O SO3 • 2 H2O (fast)

SO3 • 2 H2O

2

k

→ _{H2SO4 + H2O (slow)}

(SO3 • 2 H2O is a complex which is stabilized by hydrogen bonds and k2 << k1 or k-1)

2.2 By applying the principle of stationary (steady) states, derive the appropriate rate law and hence the reaction order of the two-step mechanism for Proposal B.

2.3 Recent quantum chemical calculations have shown that the overall reaction activation energies for the two proposals are:

EA = + 80 kJ mol -1

for Proposal A EB = – 20 kJ mol -1

for Proposal B

State the relationship between the rate constant and the temperature (Arrhenius relationship) for each proposal and predict the temperature dependence of the rate constants for each proposal.

2.4 The formation of H2SO4 is faster in the upper atmosphere (T = 175 K) than at the

earth’s surface (T = 300 K). Which proposed pathway must dominate in the upper atmosphere given the activation energies in part (iii) and your understanding of the Arrhenius equation?

_______________

k₁ k _-1

S O L U T I O N

Summing the exponents gives a second order process

Proposal B: 2 4 2 3 2 [H SO ] [SO ][H O] d k d t =

Summing the exponents gives a third order process

2.2 The steady state approximation gives:

The rate law here will be similar to that found for an enzymatic process which proceeds through a rapid reversible binding step followed by a slower reaction process.

2

3 2

1 3 2 -1 3 2 2 3 2

[SO ·2 H O]

[SO ][H O] [SO ·2 H O] [SO ·2 H O] 0

d

k k k

d t = − − =

Thus: 2

-1[SO ·2 H O] +3 2 2[SO ·2 H O]3 2 1[SO ] [H O]3 2

k k =k and therefore: 2 1 3 2 3 2 -1 2 [SO ][H O] [SO ·2 H O] + k k k = Also: 2 4 2 3 2 [H SO ] [SO ·2 H O] d k

d t = and substituting from above yields 2 1 2 3 2 2 4 2 3 2 -1 2 [SO ] [H O] [H SO ] [SO ·2 H O] + k k d k d t = = k k

However, since k2 << k-1 the above reduces to:

2 1 2 3 2 2 2 2 4 eq 2 3 2 3 2 -1 [SO ][H O] [H SO ] [SO ][H O] [SO ][H O] k k d K k k d t = k = =

which is also third order

2.3 Knowing the Arrhenius relationship: _k =_Ae- /E RT

For proposal A: _k =_Ae-EA/RT =_Ae-83.6/RT _{which increases with increasing T.}

For proposal B only the slow step is critical in determining the dependence of the rate constant on the temperature. The complexation step is very stable which

2.4 Assuming that the pre-exponential factors are comparable in magnitude as is usually the case, the reaction will be faster at the lower temperatures found in the upper atmosphere due to the temperature dependence deduced above, and thus Proposal B must be operating. The rational for Proposal B which involves the relative sizes of the rate constants is nonsense and was included to balance the choices. The rational for Proposal A involving collision probabilities appears plausible but is not a factor — it is the massive negative activation energy which controls the situation.

PROBLEM 3

Chemists at Merck Frosst Canada in Montréal have developed a promising drug which is useful against asthma. The structure of MK-0476 is shown below.

MK-0476 COOH H₃C H₃C OH S Cl N

During their investigation, they devised a simple and efficient synthesis, depicted below, for the thiolated portion of MK-0476 starting from diethyl ester A.

3.1 Give the structures of the intermediate products B - F prepared during this synthesis.

COOC2H5 COOC2H5 1) LiAlH4 2) H3O+ A B CO Cl N

,

1) KOH (aq), then H+ 2) CH2N2 E F (C9H14O3S) G CH2COOH CH2SH 1) CH3-SO2Cl (C2H5)3N 2) CH3 S- Cs+ O 1) NaOH (aq) 2) H3O+

In one of the last steps of the synthesis on MK-0476, the dilithium salt of the above thiol acid (G) was coupled with the side chain of the rest of the molecule as shown below.

H MK-0476 COOH H₃C H₃C OH S Cl N H₃C H₃C OH Cl N O O H₃C S O 1) 2) H + CO₂- Li+ Li+ -S

3.2 Based on the observed stereochemistry of the above reaction, what is the mechanistic designation of this coupling process?

3.3 If the process proceeds by your proposed mechanism, what change would occur to the overall rate of the reaction if the concentration of both the thiolate salt and the substrate H were simultaneously tripled?

3.4 For the nucleophilic substitution reaction, model studies were carried out using bromoethane as the substrate to perfect the above coupling. Draw only the structure of the major product of the reaction of one molar equivalent of bromoethane with:

a) G plus two molar equivalents of base b) G plus one molar equivalent of base

3.5 A side reaction of G is its oxidative dimerization.

Draw the structure of the dimeric product, showing all non-bonded electrons. _______________

S O L U T I O N

3.2 SN2 Bimolecular Nucleophilic Substitution

3.3 Rate = k[substrate] [nucleophile]

The overall rate is directly dependent on the concentration of both the substrate and the nucleophile. Thus tripling the concentration of both of the reactants will result in a 9-fold increase in the overall reaction rate.

3.4

CH _{2 SCH 2}CH ₃ CH _{2 CO 2 H} From 2 equiv base

CH 2 SH

CH 2 CO 2 CH 2 CH 3

From 1 equiv base 3.5 CH _{2 S} CH 2 CH 2 SCH ₂ C HO C OH O O

oxidative coupled product

PROBLEM 4

Graph paper is provided for your optional use in this question.

If you choose to use it, print your name and identification code in the upper right corner of the graph paper.

****************************

HIn is a weakly acidic indicator. HIn + Na+OH- Na+In- + H2O

also written as HIn In- + H+

At normal temperatures, the acid dissociation constant for this indicator is Ka = 2.93 × 10-5.

The absorbance data (1.00 cm cells) for 5.00 × 10-4 M (mol dm-3) solutions of this indicator in strongly acidic and strongly alkaline solutions are given in the following table.

Absorbance Data (A)

λ, nm pH = 1.00 pH = 13.00 400 0.401 0.067 470 0.447 0.050 485 0.453 0.052 490 0.452 0.054 505 0.443 0.073 535 0.390 0.170 555 0.342 0.342 570 0.303 0.515 585 0.263 0.648 615 0.195 0.816 625 0.176 0.823 635 0.170 0.816 650 0.137 0.763 680 0.097 0.588 Yellow Green Blue Indigo Violet Red Orange

4.1 Predict the observed colour of the a) acidic and b) basic forms of the indicator.

Using a “50 nm wide bar”, shade the appropriate area of the wavelength scale on the answer sheet which would correspond to the colour of the indicator at the pH values

given in the table.

For example, if observed colour is green, your answer would appear as:

violet blue green yellow red

400 500 600 700 nm

wavelength (nm)

4.2 A filter is located between the light source and the sample. What colour filter would be most suitable for the photometric analysis of the indicator in a strongly acidic medium?

4.3 What wavelength range would be most suitable for the photometric analysis of the indicator in a strongly basic medium?

4.4 What would be the absorbance of a 1.00 × 10-4 M (mol dm-3) solution of the indicator in alkaline form if measured at 545 nm in a 2.50 cm cell?

4.5 Solutions of the indicator were prepared in a strongly acidic solution (HCl, pH = 1)

and in a strongly basic solution (NaOH, pH = 13). Perfectly linear relationships

between absorbance and concentration were observed in both media at 490 nm and 625 nm, respectively.

The molar absorptivities at the two wavelengths are:

λ = 490 nm λ = 625 nm HIn (HCl) 9.04 x 102 M-1 cm-1 3.52 x 102 M-1 cm-1 In- (NaOH) 1.08 x 102 M-1 cm-1 1.65 x 103 M-1 cm-1

(M = mol dm-3)

Calculate the absorbance (1.00 cm cell) at the two wavelengths for an aqueous 1.80 × 10-3 M (mol dm-3) solution of the indicator HIn.

S O L U T I O N

4.1 The observed colour will be the complementary colour to that of the absorption maximum.

a) Acidic conditions (pH 1):

The sample absorbs at 490 ± 25 (blue-green) and thus will transmit the complementary colour and will appear to be yellow-orange (625 ± 25 nm). b) Basic conditions (pH 13):

The sample absorbs at 625 ± 25 (yellow-orange) and thus will transmit the complementary colour and will appear to be blue-green (490 ± 25 nm).

4.2 The filter should transmit the colour that the sample will absorb most efficiently. The acidic sample absorbs most strongly in the blue range (490 ± 25 nm) and thus a similar colour filter would be most suitable for the photometric analysis of the sample.

4.3 The wavelength range to be used for maximum sensitivity should correspond to that at which the sample absorbs most strongly. The maximum absorbance for the basic form of the indicator in solution occurs at 625 ± 25 nm and this is the most suitable wavelength for the analysis.

4.4 From a graph of A versus wavelength, the absorbance of a 5.00×10-4 M basic solution at 545 nm is 0.256. From the plot, it is clear that this region of the graph is linear and thus the above value can also be interpolated from the data table.

A = ε l c (Beer’s Law)

where l = length of cell, c = concentration of analyte, ε = molar absorptivity. Therefore ε = A = 0.256 = 5.12×102 M-1 cm-1

l c = 1.0 × 5.00×10-4

Absorbance of a 1.00×10-4 M basic solution of the indicator using a 2.50 cm cell is:

A = 5.12×102 × 2.50 × 1.0×10-4 = 0.128

4.5 The dissociation reaction of the indicator is: [HIn] = [H+] + [In-]

[H+] = [In-] (1) and [HIn] + [In-] = 1.80×10-3 M (2) a [H ][In ] [HIn] K + − = (3)

Substitute (1) and (2) into (3)

2 -5 a -3 -[In ] 2.93 10 1.8 10 [In ] K − = = × × −

Rearrangement yields the quadratic expression [In-]2 + 2.93×10-5 [In-] – 5.27×10-8 = 0

which results in [In-] = 2.15×10-4 M

[HIn] = 1.80×10-3 M – 2.15×10-4 M = 1.58×10-3 M The absorbance at the two wavelengths are then:

A490 = (9.04×102× 1 × 1.58×10-3) + (1.08×102× 1 × 2.15×10-4 ) = 1.45

A625 = (3.52×10 2_×

PROBLEM 5

Iron metal melts at 1811 K. Between room temperature and its melting point, iron metal can exist in different allotropic or crystalline forms. From room temperature to 1185 K, the crystal structure of iron metal exists as a body-centred cubic (bcc) lattice known as

α-iron. From 1185 K to 1667 K, the structure becomes face-centred cubic (fcc) and is called γ-iron. Above 1667 K, and up to its melting point, iron reverts to a bcc structure similar to that of α-iron. The latter phase is called δ-iron.

Given that the density of pure iron metal is 7.874 g cm-3 at 293 K: 5.1 Calculate the atomic radius of iron (expressed in cm).

5.2 Calculate its density (expressed in g cm-3) at 1250 K.

Notes: Ignore the small effects due to the thermal expansion of the metal. Clearly define any symbols which you use, e.g. r = atomic radius of Fe.

Steel is an alloy of iron and carbon in which some of the interstitial spaces (“holes”) of the crystal lattice (iron) are occupied by small atoms (carbon). Its carbon content typically ranges from 0.1 % to 4.0 %. In a blast-furnace, the melting of iron is facilitated when it contains 4.3 % of carbon by mass. If this mixture is cooled too rapidly the carbon atoms remain dispersed within the α-iron phase. This new solid, called martensite, is extremely hard and brittle. Although is it slightly distorted, the size of the unit cell of this solid (martensite) is the same as that of α-iron (bcc).

Assuming that the carbon atoms are evenly distributed in the iron structure:

5.3 Calculate the average number of carbon atoms per unit cell of α-iron in martensite containing 4.3 % C by mass.

5.4 Calculate the density (expressed in g cm-3) of this material.

Molar masses and constants:

M(Fe) = 55.847 g mol-1

M(C) = 12.011 g mol-1

NA = 6.02214×1023 mol-1

S O L U T I O N

5.1 Calculation of the atomic radius of iron (expressed in cm). 5.2 Calculation of its density (expressed in g cm-3) at 1250 K.

Expected steps of the calculation:

1. Define the length a, b, c, d1, d2, and r and volume V1 and V2 parameters for both bcc

and fcc structures of iron (cf. Details below).

2. Calculate the volume V1 of the unit cell of α-iron from its density ρbcc at 293 K, the

molar weight M(Fe) of iron, and Avogadro’s number NA.

3. Calculate the length d1 of the edge of the bcc unit cell from its volume V1.

4. Calculate the atomic radius r of iron from the length “d1”.

5. Calculate the length d2 of the edge of the fcc unit cell at 1250 K from the atomic

radius r of iron.

6. Calculate the volume V2 of the fcc unit cell of γ-iron from the length d2 of its edge.

7. Calculate the mass m of the number of iron atoms in a unit cell of γ-iron from the molar weight M(Fe) of iron and Avogadro’s number NA.

8. Calculate the density ρfcc of γ-iron from the values of “m” and “V2”.

An alternative route to ρfcc of γ-iron, involving the percent space filling ratios of both

the bcc and fcc unit cells, can replace the aforementioned steps 5 through 8. In this route, these steps are labeled from 5' through 8' as listed below:

5'. Calculate the percent space filling ratio of the bcc unit cell. 6'. Calculate the percent space filling ratio of the fcc unit cell.

7'. Calculate the ρfcc/ρbcc density ratio from the fcc/bcc space filling ratios.

Face-centered cubic structure (fcc) Calculations in details

At 293 K, α-iron has a bcc crystal structure.

Each unit cell possesses 2 atoms and one of them is in the center of the cell. At 1250 K, γ-iron has a fcc crystal structure.

Each unit cell possesses 4 atoms and each side has one-half an atom at its center. r = atomic radius of iron.

a = length of the diagonal on one side of the bcc unit cell.

b = length of the diagonal passing through the center of the bcc unit cell. c = length of the diagonal on one side of the fcc unit cell.

d1 = length of the edge of the bcc unit cell of α-iron.

d2 = length of the edge of the fcc unit cell of γ-iron.

V1 = Volume of the bcc unit cell of α-iron.

V2 = Volume of the fcc unit cell of γ-iron.

Va = Volume of one atom.

Va1 = Volume occupied by 2 atoms in one bcc unit cell.

Va2 = Volume occupied by 4 atoms in one fcc unit cell.

R1 = Percent space filling ratio in a bcc unit cell.

R2 = Percent space filling ratio in a fcc unit cell.

Va = (4/3) π r 3 Va1 = 2 Va Va2 = 4 Va b = 4 r; a2 = 2 d12; b2 = d12 + a2 = 3 d12 d1 = (b 2 /3)1/2 = (16 r2/3) ½; V1 = d1 3 = [(16 r2/3)1/2]3 c = 4 r; c2 = 2 d22 d2 = (c2/2)1/2 = (16 r2/2) ½; V2 = d23 = [(16 r2/2) ½]3 2. 1.000 cm3 of iron weights 7.874 g at 293 K (ρbcc).

1 mole of iron weights 55.847 g (MFe).

Thus, 0.1410 mol (7.874 g / 55.847 g mol-1) of iron occupy a volume of 1.000 cm3 or 1 mole of iron will occupy a volume of 7.093 cm3

1 mole corresponds to 6.02214×1023 atoms

V1 = (7.093 cm3 mol-1) × (2 atoms/unit cell) / (6.02214×1023 atoms mol-1)

V1 = 2.356×10-23 cm3 per unit cell

3. d1 = (V1)1/3 = (2.356×10-23 cm3)1/3

d1 = 2.867×10 -8

cm

4. For a bcc structure, the value of d1 can be expressed as: d1 = [(16 r2)/3]1/2

so the value of “r” will be: r = (3 d1 2

/16)1/2

r = [3 (2.867×10-8 cm)2/16]1/2

r = 1.241×10-8 cm

5. At 1250 K, in the fcc structure, the value of “d2” is given by: d2 = (16 r2/2)1/2

d2 = [16 (1.241×10-8 cm)2/2]1/2

d2 = 3.511×10-8 cm

6. V2 = d23 = (3.511×10-8 cm)3

V2 = 4.327×10-23 cm3

7. The mass “m” of the 4 iron atoms in the fcc unit cell will be:

m = (55.847 g mol-1) × (4 atoms/unit cell) / (6.02214×1023 atoms mol-1)

m = 3.709×10-22 g per unit cell

8. ρfcc = m / V2 = (3.709×10-22 g) / (4.327×10-23 cm3)

ρfcc = 8.572 g/cm3

Alternative route to ρfcc of γ-iron:

5'. R1 = [(Va1) / (V1)] × 100% = [(2 Va) / (V1)] × 100% R1 = ([2 × (4/3) πr3] / [(16 r2/3)1/2]3) × 100% R1 = ([(8/3) πr3] / [(16/3)3/2r3]) × 100% R1 = ([(8/3) π] / [(16/3)3/2]) × 100% R1 = [(8.378) / (12.32)] × 100% R1 = 68.02 % 6'. R2 = [(Va2) / (V2)] × 100% = [(4 Va) / (V2)] × 100%

R2 = ([(16/3) π] / [8 3/2 ]) × 100% R2 = [(16.76) / (22.63)] × 100% R2 = 74.05% 7'. ρfcc / ρbcc = (74.05%) / (68.02%) ρfcc / ρbcc = 1.089 8'. ρfcc = 1.089 ×ρbcc ρfcc = 1.089 × 7.874 g cm-3 ρfcc = 8.572 g cm -3

5.3 Calculation of the average number of carbon atoms per unit cell of α-iron in martensite containing 4.3 % C by mass.

5.4 Calculation of the density (expressed in g cm-3) of this material.

Expected Steps of the Calculation:

1. From the percent composition of martensite (by mass), calculate the relative amounts of moles of carbon and iron.

2. Bring the C/Fe molar ratio to one (1) unit cell (Note: 2 Fe atoms per unit cell).

3. Find the smallest whole number of C atoms for the smallest whole number of unit cell (facultative).

4. Calculate the mass of iron per unit cell. 5. Calculate the mass of carbon per unit cell.

6. Calculate the total mass of carbon and iron in one unit cell.

7. Calculate the density of martensite [ρ(martensite at 4.3 %C)] from the total mass of C and Fe and volume (V1) of α-iron bcc unit cell.

Details:

1. In 100.0 g of martensite at 4.3 % C: (4.3 g C) / (12.011 g mol-1) = 0.36 mol C (95.7 g Fe) / (55.847 g mol-1) = 1.71 mol Fe

So we have 1 carbon atom for 4.8 iron atoms or 0.21 carbon atoms per iron atom.

2. Martensite has a "bcc" crystal structure (2 iron atoms per unit cell). [(1 C atom) / (4.8 Fe atoms)] × (2 Fe atoms / unit cell)

3. 5 carbon atoms [(0.42 C atom / 0.42) × 5] in 12 unit cells [(1 unit cell/0.42) × 5]. 5 carbon atoms dispersed in 12 unit cells

4. [(55.847 g/mol) / (6.02214×1023 atoms/mol)] × (2 atoms/unit cell of α-iron) 1.8547×10-22 g Fe per unit cell of α-iron

5. (12.011 g/mol) / (6.02214×1023 atoms/mol) 1.9945×10-23 g C per atom

6. [1.8547×10-22 g Fe + (0.42 C at. × 1.9945×10-23 g / C at.)] per unit cell 1.938×10-22 g C and Fe per unit cell

7. Each unit cell of α-iron occupies a volume, V1 = 2.356×10 -23

cm3 (cf. Question i)

ρ(martensite at 4.3 % C) = (1.938×10-22 g C and Fe) / (2.356×10-23 cm3)

PROBLEM 6

6.1 Much of the world’s supply of platinum group metals is derived from the residues recovered from the electrolytic refining of copper and nickel. A flow chart for the recovery of platinum and palladium is shown on the following page.

i) Clearly draw the shape (geometry) of both the PtCl62- and the PdCl42- anions.

ii) Clearly draw all possible stereoisomeric structures of monomeric Pd(NH3)2Cl2.

Label the structures that you have drawn with their correct stereochemical descriptors.

iii) What is the role of the FeSO4 in the second step of the flow chart? Write a

balanced equation for the reaction of FeSO4 in this step.

iv) Write a complete balanced equation for the ignition of Pd(NH3)2Cl2 in air to give

Pd metal. In this reaction, what is being oxidized and what is being reduced?

6.2 Reaction of a main group chloride (24.71 g) with ammonia (10.90 g) gave a mixture of products consisting of NH4Cl (25.68 g), a solid element A (2.57 g) and a yellow

crystalline nitride of this element (7.37 g) according to the equation below. n AwClx + m NH3 → p NH4Cl + q A + r AyNz

(where n, m, p, q, r, w, x, y and z are coefficients to be determined)

A sample of the nitride exploded violently when struck with a hammer, but it underwent controlled polymerization on heating to give a bronze-coloured, fibrous solid which exhibits metallic conductivity. Element A also undergoes polymerization to a high molecular weight linear polymer upon heating.

Molar masses:

M(Cl) = 35.453 g mol-1 M(N) = 14.007 g mol-1 M(H) = 1.008 g mol-1 i) Identify element A.

ii) Write a complete balanced equation for the reaction of the chloride with ammonia.

iii) Assuming conventional oxidation states, write the balanced redox half-reaction equations involved in this reaction.

Method of Purification of Platinum and Palladium

platinum metals concentrate

ignition

filtrate precipitate

precipitate filtrate

filtrate precipitate

filtrate precipitate aqua regia

precipitate filtrate impure impure precipitate pure pure impure sponge Pt + NaCl Na₂PtCl ₆

( Na2PtCl 6) (Ir, Rh & Pd hydroxides)

NH 4 Cl

(NH4) 2 PtCl6

(HAuCl _{4 , H2}PtCl ₆, H₂PdCl ₄) _{(Rh, Ru, Ir, AgCl)} FeSO4 (H _{2 PtCl 6}, H _{2 PdCl 4 ) (Au)} NH _{4 Cl} (H 2 PdCl 4) ( (NH4 ) 2 PtCl 6 ) NH _{4 OH, HCl} (traces of

Rh, Ru, Ir & Ag) ( Pd(NH 3) 2Cl 2)

Zn

(Rh, Ru, Ir & Ag)

NH ₄OH, HCl

Pd(NH 3) 2Cl 2

Pd sponge

S O L U T I O N

6.1

i) The PtCl62- anion consists of a Pt(IV) centred in a regular octahedron of Cl- ions. The

PdCl4

anion consists of a Pt(II) centred in a square of Cl- ions.

Pt Cl Cl Cl Cl Cl Cl 2 - PtCl 62- Pd Cl Cl Cl Cl 2 - PdCl ₄2-

ii) Like PdCl42-, Pd(NH3)2Cl2 is also square planar. However, in this case there are two

distinct ways in which the two different substituent groups (ligands) can be arranged. One places the two Cl- on adjacent corners of the square (and axiomatically, the two NH3 on the other two adjacent corners). This arrangement is called the cis isomer.

The second arrangement has the pairs of the same ligand placed on diagonally opposite corners of the square. This arrangement is called the trans isomer. There

are only these two stereoisomers possible for a monomeric form of the complex.

Pd NH 3 NH ₃ Cl Cl Pd Cl NH ₃ Cl H ₃ N cis _Pd(NH trans 3 ) 2 Cl 2

iii) The FeSO4 [i.e. Fe(II)] acts as a reducing agent. Under the conditions used in the

process, the Fe(II) is a strong enough reducing agent to reduce Au(III) to Au(0), but not to reduce Pd(II) or Pt(IV).

HAuCl4 + 3 FeSO4 → Au0 + HCl + FeCl3 + Fe2(SO4)3

iv) Pd(NH3)2Cl2 + O2 → Pd0 + N2 + 2 H2O + 2 HCl

Pd2+ + 2 e– → Pd0 reduction 2 O0 + 4 e– → 2 O2- reduction 2 N3- – 6 e– → 2 N0 oxidation

or Pd(NH3)2Cl2 + 2 O2 → Pd0 + 2 NO + 2 H2O + 2 HCl Pd2+ + 2 e– → Pd0 reduction 4 O0 + 8 e– → 4 O2– reduction 2 N3– – 10 e– → 2 N2+ oxidation or Pd(NH3)2Cl2 + 3 O2 → Pd 0 + 2 NO2 + 2 H2O + 2 HCl Pd2+ + 2 e– → Pd0 reduction 6 O0 + 12 e– → 6 O2– reduction 2 N3– – 14 e– → 2 N4+ oxidation

In this reaction the ammonia is oxidized to water and dinitrogen (or nitrogen oxides) and the Pd(II) and dioxygen are being reduced.

Part of the oxidation is due to the Pd(II) acquiring two electrons to go to Pd(0), and part by dioxygen which is reduced to water. In the presence of noble metal catalysts, NH3 can also be oxidized to (NO)x. Thus other nitrogen species are also in principle

possible in the above ignition.

6.2

i) A = sulphur

All Cl is located in the NH4Cl, and thus the weight of Cl is found by:

53.492 g NH4Cl → 35.453 g Cl

25.68 g NH4Cl → ? g Cl

? = 25.68 × 35.453 / 53.492 = 17.02 g Cl

Total amount of A in the reaction is 24.71 g Cl – 17.02 g Cl = 7.69 g A There is 2.57 g

free A and (7.69 – 2.57) = 5.12 g A in the nitride. The amount of N bound in nitride is therefore

7.37 g nitride – 5.12 g A bound in nitride = 2.25 g N bound in nitride The amount of nitrogen bound in NH4Cl is 25.68 – 17.02 = 8.66 g

The rule of multiple proportions is applied:

Proportions of 1 : 1

? g A binds → 35.453 g Cl

? = 35.453 × 7.69/17.02 = 16.02 g

A could be oxygen, but it is a main group element and it is a gas and thus it can be excluded.

Nitride:

5.12 g A binds → 2.25 g N ? g A binds → 14.007 g N

? = 14.007 × 5.12/2.25 = 31.87* g

A could be is sulphur, fits the physical description and Mr(S) = 32.064

(*Phosphorus, Mr(P) = 30.97, is also possible, but the highest degree of

“polymerization” known is 4, in the P4 molecule and thus P must also be excluded.)

Proportions of 1 : 2

Chloride:

7.69 g A binds → 17.02 g Cl ? g A binds → 2 × 35.453 g Cl

? = 2 × 35.453 × 7.69 / 17.02 = 32.03 g. Again A could be sulphur. Nitride:

5.12 g A binds → 2.25 g N ? g A binds → 2 × 14.007 g N

? = 2 × 14.007 × 5.12 / 2.25 = 63.75 g

A could be Cu but it is a Group B element and thus can be excluded. Therefore element A must be sulphur.

ii) 3 SCl2 + 8 NH3 → 6 NH4Cl + S + S2N2

or

6 SCl2 + 16 NH3 → 12 NH4Cl + 2 S + S4N4

In fact the elemental S is in the form of S8 and so the equation should be multiplied

by a factor of 8. Although S2N2 roughly fits the description of colour and explosive

instability, it is actually S4N4 that is produced in this reaction. Both of these ring

compounds under carefully controlled heating polymerize to give the high molecular weight linear polymer (SN)n, which is one of the rare examples of a metal containing

is due to the relatively weak S-N bonds and the large amount of strain energy in the ring. The S-S bond is also quite weak and can be broken by heating. For this reason the S8 ring also undergoes polymerization at high temperature. However, there is

hardly any energy stored in the form of ring strain in this ring and so the polymerization is not highly exothermic or explosive. Other sulphur chlorides (S2Cl2

and SCl4) do not fit the stoichiometry of the reaction.)

iii) A disproportionation reaction involving sulphur occurs: 2 S2+ – 2 e–→ 2 S3+

PROBLEM 7

7.1 One mole of Cl2 (g), which may be assumed to obey the ideal gas law, initially

at 300 K and 1.01325×107 Pa, is expanded against a constant external pressure of 1.01325×105 Pa to a final pressure of 1.01325×105 Pa. As a result of the expansion, the gas cooled to a temperature of 239 K (which is the normal boiling point of Cl2),

and 0.100 mol of Cl2 condensed.

The enthalpy of vaporization of Cl2 (l) is 20.42 kJ mol -1

at the normal boiling point, the molar heat capacity of Cl2 (g) at constant volume is Cv = 28.66 J K-1 mol-1

and the density of Cl2 (l) is 1.56 g cm-3 (at 239 K). Assume that the molar heat

capacity at constant pressure for Cl2 (g) is Cp = Cv + R.

(1 atm = 1.01325×105 Pa, R = 8.314510 J K-1 mol-1 = 0.0820584 L atm K-1 mol-1) i) Either draw a complete molecular orbital energy diagram or write the complete

electronic configuration of Cl2. Predict the bond order of Cl2 and thus whether

this molecule will be diamagnetic, ferromagnetic, or paramagnetic.

ii) For the changes described above, calculate the change in the internal energy (∆E) and the change in the entropy (∆Ssys) of the system.

7.2 For the following reactions occurring in dilute aqueous solution at 298 K: [Ni(H2O)6] 2+ + 2 NH3 [Ni(NH3)2(H2O)4] 2+ + 2 H2O (1) lnKc = 11.60 and ∆H 0 = –33.5 kJ mol-1

[Ni(H2O)6]2+ + en [Ni(en)(H2O)4]2+ + 2 H2O (2)

lnKc = 17.78 and ∆H 0 = –37.2 kJ mol-1

Note: en is ethylenediamine (a neutral bidentate ligand)

(R = 8.314510 J K-1 mol-1 = 0.0820584 L atm K-1 mol-1)

Calculate ∆G0, ∆S0 , and Kc at 298 K for reaction [3] occurring in a dilute aqueous

solution:

[Ni(NH3)2(H2O)4]2+ + en [Ni(en)(H2O)4]2+ + 2 NH3 (3)

4σ* 4σ 5σ 5σ* 3σ 2π* 3σ* 2σ 2σ* 1σ* 1σ 1π* 2π 1π 1s 1s 2s 2s 3s 2p0, 2p± 1 2p0, 2p± 1 3p0, 3p± 1 3p0, 3p± 1 3s

S O L U T I O N

i) Electronic configuration of a Cl atom: 1s22s22px 2 2py 2 2pz 2 3s23px 2 3py 2 3pz 1

Significant atomic orbitals (AO) = 1(K) + 4(L) + 4(M) = 9 AO Number of electrons in these AOs: 17

Number of molecular orbitals (MO) equals number of AOs:

Thus 2 x [1(K) + 4(L) + 4(M)] = 18 MOs are present in a Cl2 molecule

In the formation of Cl2: 2 x 17 = 34 electrons to go into the 18 MOs.

1σ21σ*22σ22σ*23σ21π41π*43σ*24σ24σ*25σ22π42π*4 or (KK)(LL)(σ3s)2(σ*3s)2(σ3p)2(π3p)4 (π*3p)4 or (σ1s)2(σ*1s)2(σ2s)2(σ*2s)2(σ2pz)2(π2px)2(π2py)2(π*2px)2(π*2py)2(σ*2pz)2 (σ3s)2(σ*3s)2(σ3pz)2(π3px)2(π3py)2(π*3px)2(π*3py)2(σ*2pz)0 or (KK)(LL) (σ3s)2(σ*3s)2(σ3pz)2(π3px)2(π3py)2(π*3px)2(π*3py)2(σ*2pz)0

*assumption: - bond formation is along the z-axis (equivalent formulae for x or y axes are accepted)

Bond order is given by (n-n*)/2: n = 18; n* = 16

(18 – 16) / 2 = 1 (1 σ bond, no π bond)

The Cl2 molecule has a bond order of 1.

The Cl2 molecule is diamagnetic since there are no unpaired electrons.

ii) Summary of the changes involved:

Cl ₂(g) 1 mol 300 K ∆E ₁ cooled 1.013×10 7 (100 atm) Cl ₂ (g) 1 mol 239 K 1.013×105 (1 atm) ∆ E ₂ Cl ₂ (l) 0.1 mol 239 K Pa Pa

The total process is an expansion plus an isobaric change of phase (gas to liquid) and since the internal energy (E) is a function of state, the total change in the internal energy is ∆E = ∆E1 + ∆E2.

Process 1:

1 v = 1 28.66 239 300 =

E n C dT

∆ =

_∫

Note: a) ∆E for a perfect gas is a function only of T

b) Cv is constant

c) “–” sign means a loss of energy due to the work needed for expansion of 1 mole of gas

Process 2: For convenience, the data were manipulated in atm; equivalent procedure in Pa will require the appropriate conversion factor

From an energetic point of view, the liquid formation Process 2 can be split into two separate steps :

• the vaporization heat loss (decreased internal energy, -) from the system into surroundings (since the process takes place at constant pressure, the heat is equal to the change in the enthalpy)

• the work done by the surroundings in compressing the system to a smaller volume (increased internal energy, +).

Volume of gas which condensed is

V = n R T / P = (0.1 × 0.0820584 × 239) / 1 = 1.96 dm3 Volume of liquid Cl2: (0.1 × 2 × 35.454) / 1.56 = 4.54 cm3

2 2 ext (phase change) = 2 ext( l g)

E H P V H P V V

∆ = ∆ −

_∫

but Vl is approximately 0 and can be neglected

(ca. 4.5 cm3 liquid volume vs. ca. 17.6 dm3; ca. 0.03 % error)

∆E2 = (0.1) ( – ∆Hvap) + PextVg

= 0.1 × (– 20420) + (1 × 1.96 L) × 101.325 J dm-3 atm-1 = –2042.0 + 198.5 = –1843.5

∆E = ∆E1 + ∆E2 = –1748.3 + (–1843.5) = –3591.8

Entropy S is a function of two variables of state. Since in Process 1 the known variables are T and P, expression of S is chosen as S(T,P).

∆Ssys = ∆S 1 + ∆S 2 and -1 -1 28.66 + 8.314 = 36.97 J K mol p v C = C +R = 2 2 1 1 1 239 1 ln ln 1.0 36.97 ln 8.314 ln – 8.40 38.29 300 100 p T P S n C n R T P ∆ = − = × − = = + = = 29.89 J K-1

For the phase transition (constant temperature), by definition ∆S2 = Q / T

Since the pressure is constant in this case, Q / T = Qp / T = ∆H / T 1 2 2 0.1 ( 20420) 8.54 J K 239 H S T − ∆ × − ∆ = = = − ∆Ssys = 29.89 – 8.54 = 21.35 J K-1

7.2 Beware of round-off errors in the variations to the solution to this problem: One can get small differences due to conversion into and out of the ln relationships. It is the approach which matters.

One reverses the signs of lnKc and ∆H0 for Reaction 1 when it is reversed.

Equilibrium constants are multiplied when equations are added, thus lnK’s will add.

Reaction 3 = Reaction 2 - Reaction 1 Thus ∆S3 = ∆S2 – ∆S1 and ∆G3 = ∆G2 – ∆G1 ∆G01 = – R T lnKc1 = – 8.314 × 298 × 11.60 = – 28740 J mol-1 = –28.74 kJ mol-1 ∆H01 = – 33.5 kJ mol-1 ∆S01 = (∆H01 – ∆G01) / T = (– 33.5) – (– 28.74)) / 298 = – 0.0161 kJ K-1 mol-1 = –16.1 J K-1 mol-1 Similarly: ∆G02 = – 44.05 kJ mol -1 ∆H02 = – 37.2 kJ mol-1 ∆S02 = – 22.98 J K-1 mol-1

Reaction 3 = Reaction 2 – Reaction 1, thus

∆H03 = ∆H02 – ∆H01 = – 3.7 kJ ∆S03 = ∆S02 – ∆S01 = 39.08 J K-1 ∆G03 = ∆H 0 3 – T ∆S 0 3 = –15.35 kJ mol -1 15.35 3 Thus e RT c K = = 4.90×102 Alternatively: ∆G03 = ∆G 0 2 – ∆G 0 1 = – 44.05 – (–28.74) = –15.31 kJ mol -1 thus K = 4.82×102) ∆S0 = (∆H0 – ∆G0) / T= (– 3700 – (–15311)) / 298 = 38.96 J K-1

PROBLEM 8

An electrolyte is prepared from H2SO4, CuSO4 and distilled water and its volume is

100.0 cm3. The concentrations of H+ and Cu2+ in the electrolyte are c(H+) = 1.000 M (mol dm-3) and c(Cu2+) = 1.000×10-2 M (mol dm-3), respectively. Two cubic platinum electrodes are immersed in the electrolyte. Both of the electrodes are single crystals with only one face (100) exposed to the electrolyte (the other five faces are blocked physically by an insulator which is stable in the electrolyte). The exposed surface area of each electrode is equal to 1.000 cm2. During an electrolysis a total charge of 2.0000 C is passed between the cathode and the anode. At the cathode, two simultaneous processes are occurring: deposition of an epitaxial (layer-by-layer) Cu layer and H2 gas generation.

At the anode, O2 gas is generated. The H2 gas is collected in a flask under the following

conditions (assume ideal gas behaviour):

T = 273.15 K and P(H2) = 1.01325×10 4

Pa; the volume of H2 is equal to 2.0000 cm 3

8.1 Write equations of the processes taking place at the electrodes.

8.2 Calculate the number of moles of H2 gas generated at the cathode and the number

of moles of Cu deposited on the electrode.

8.3 Calculate the number of Cu monolayers formed on the Pt (100) cathode.

Note that the lattice constant of Pt is a(Pt) = 3.9236×10-8 cm.

Both Pt and Cu have the fcc (face centred cubic) crystallographic structure. Molar masses and constants:

M(H) = 1.00795 g mol-1 M(Cu) = 63.546 g mol-1 e = 1.60218×10-19 C

F = 96485.3 C mol-1

R = 8.314510 J K-1 mol-1 = 0.0820584 L atm K-1 mol-1

Vm = 22.4141 dm3

1 atm = 1.01325×105 Pa

NA = 6.02214×1023 mol-1

S O L U T I O N

Approach

• Determine the number of H2 moles generated by the electrolysis.

• Calculate the charge required for the H2 production and the charge of the formation of

the Cu deposit and thus the number of moles of Cu in the deposit.

• Calculate the surface concentration of atoms in the Pt (100) face thus the number of Pt atoms per 1 cm2; during an epitaxial growth the number of Cu atoms per 1 cm2 equals the number of Pt atoms

• Determine the charge necessary to form one monolayer of Cu and subsequently the number of Cu monolayers on Pt (100)

Calculations

8.1 Balanced electrode equations (a) Anode: 2 H2O → H + + O2 + 4 e – (b) Cathode:

Two reactions occur simultaneously at the cathode:

+

-2

2 H + 2 e → H

2+

-Cu + 2 e → Cu

8.2 Determination of the charge necessary to generate 2.0000 cm3 of H2 gas

(T = 273.15 K, p = 10.1325 kPa)

Two approaches to determination of the number of H2 gas moles

(a) Determination of n(H2): p V = n(H2) R T 6 6 2 10132.5 2.0000 10 (H ) 8.9230 10 mol 8.314510 273.15 n − − × × = = × × 2 2 (H ) (H ) m V n V =

Vm = 22.4141 dm3 mol-1 (Vm refers to the pressure of 1 atm or at p(H2) = 101.325 kPa

and because the pressure of H2 is ten times smaller, one knows right away that the

volume occupied under 1 atm would be 0.2000 cm3)

3 6 2 0.20000 10 (H ) 8.9230 10 mol 22.4141 n − − × = = ×

Determination of the charge necessary to generate 8.9230×10-6 moles of H2 gas

Two approaches to determination of the charge:

(a) the number of H atoms, NH, equals twice the number of H2 molecules; if one

multiplies NH by e, then one gets the sought charge, QH

6 19

H 2 8.9230 10 A 1.60218 10

Q = × × − ×N × × −

H 1.7219 C

Q =

(b) one may use the Faraday law

H H H

m =k Q

where kH is the electrochemical equivalent of H thus the mass of H generated

by 1 C; to use this formula one has to calculate kH; knowing that the charge of

1 F = 96485.3 C results in formation of 1 mole of H (1/2 mole of H2), one may

easily determine kH H 96 485.3 1.0000 1.00795 = k 5 -1 H 1.04467 10 g C k = × − Subsequently 6 H H 5 H 2 8.9230 10 1.00795 1.04467 10 m Q k − − × × × = = × H 1.7219 C Q =

Determination of the Cu deposition charge

Cu 2.0000 H

Q = −Q

Cu 2.0000 1.7219 0.2781 C

Q = − =

The moles of Cu is thus 0.2781 / 2 F = 1.4412×10-6

8.3 Determination of the charge of formation of 1 monolayer (ML) of the Cu deposit and the number of Cu monolayers on the Pt (100) substrate

Calculate the number of surface Pt atoms in the (100) face Surface area of the fundamental unit:

Number of atoms per fundamental (100) unit: nu = 2

Surface atom concentration:

15 2 Pt(100) 15 2 2 1.2991 10 cm 1.5395 10 cm u u n s A − − = = = × ×

The number of Cu atoms per 1 cm2 equals the number of Pt atoms - epitaxial growth

σ_Cu(100) = σ_Pt(100) =1.2991 10 cm× 15 -2

The charge of formation of one monolayer (ML) of Cu equals:

15 ML 2 1.2991 10 q = × ×e × 4 ML 4.1628 10 C q = × −

Determination of the number of Cu monolayers on the Pt (100) substrate

ML ₄ 0.2780C 4.1628 10 C n = ₋ × ML 668 ML n =

One can also calculate the number of Cu atoms (8.6802×1017) formed from the number of moles produced and divide this by the number of atoms (1.2991×1015) on the exposed Pt surface to also arrive at 668 monolayers.

PRACTICAL PROBLEMS

PROBLEM 1

(Practical)

Determination of Mg2+ and Ca2+ in Bottled Water

• The Ksp for calcium oxalate is 2.3×10-9 and the Ksp for magnesium oxalate is

8.6×10-5.

• In a solution buffered to maintain pH 10, Calmagite indicator is pink when bound to Mg2+ and blue in the absence of available magnesium ions. Calcium ions are not bound by Calmagite.

• EDTA binds to Mg2+ and Ca2+ even in the presence of Calmagite. The stoichiometry of the EDTA-metal complex formed with both Mg2+ and Ca2+ is 1:1.

• Molar masses: M(Ca) = 40.08 g mol-1 M(Mg) = 24.31 g mol-1

Chemicals Available

500 cm3 sample of “Bottled Water” aqueous buffer (pH 10)

Calmagite indicator

aqueous saturated ammonium oxalate aqueous ethylenediaminetetraacetic acid aqueous standardized* Mg2+ distilled water

*0.928 mg Mg2+/cm3 solution, 0.0382 moles Mg2+/dm3

Procedure

A. Precipitation of calcium ions

Precipitate the calcium ions in a 25.00 cm3 aliquot of the “Bottled Water” sample by accurately adding approximately 0.50 cm3 of saturated ammonium oxalate solution (from the common burettes in each lab room). Carefully swirl the solution to ensure uniform mixing. Allow at least 45 minutes for complete precipitation to occur.

B. Standardization of the EDTA solution

Using distilled water, dilute 5.00 cm3 of the standardized magnesium solution to a final volume of 100.0 cm3. Add 40 cm3 of distilled water, 5 cm3 of pH 10 buffer solution, and some Calmagite indicator to 5.00 cm3 of diluted magnesium solution. Titrate this sample with EDTA solution to a clear blue end point.

Repeat as necessary.

C. Titration of Mg2+ and Ca2+

Add 40 cm3 of distilled water, 5 cm3 of pH 10 buffer solution, and some Calmagite indicator to 5.00 cm3 of the “Bottled Water” sample. Titrate this sample with EDTA solution to a clear blue end point.

Repeat as necessary.

D. Titration of Mg2+

Add 40 cm3 of distilled water, 5 cm3 of pH 10 buffer solution, and some Calmagite indicator to 5.00 cm3 of the calcium-free “Bottled Water” sample prepared in part A. The presence of a small amount of calcium oxalate will not interfere with your titration. Titrate this sample with EDTA solution to a clear blue end point.

Repeat as necessary.

Calculations

Calculate the concentration of Mg2+ (in mg dm-3) in the “Bottled Water” sample. Calculate the concentration of Ca2+ (in mg dm-3) in the “Bottled Water” sample.

PROBLEM 2

(Practical)

Organic qualitative analysis

You have six bottles containing six different organic compounds. From the list of eight compounds given below, identify the contents of each bottle using the reagents available.

Many of these compounds have strong odours. To prevent the laboratory from becoming too odorous, you must keep each bottle tightly capped when it is not in use. Dispose of any waste produced in the bottle labelled “ORGANIC WASTE” at your station. Also place used litmus paper in this bottle. Keep the waste bottle capped when not in use.

Chemicals Available litmus paper, red and blue

aqueous ceric ammonium nitrate aqueous chromic-sulfuric acid aqueous 2,4-dinitrophenylhydrazine aqueous 0.2% KMnO4 acetone (2-propanone) Possible Unknowns* 2-butanone 1-decene 2,3-diamino-2,3-dimethylbutane hexane 3-methyl-1-butanol 2-methyl-2-butanol nonanal propanoic acid

*Several of the unknowns are present as dilute aqueous solutions. This will not interfere with the test results.

PROBLEM 3

(Practical)

Synthesis of the Substituted Dihydro-1,3-benzoxazine (C)

N H 3 C

O

C

Benzoxazines have long been recognized as useful biologically-active compounds. One such compound (C) will be prepared using the three-step synthesis described below. All of the product obtained in Step I should be used in Step II and similarly all of the product from Step II should be used in Step III. You will be evaluated on both the yield and purity of the final product.

Chemicals available

5 cm3 reaction vial containing 2.5 cm3 of ethanolic 1-amino-4-methylbenzene (0.22 g), vial containing 0.25 g of 2-hydroxybenzaldehyde,

vial containing 0.1 g of sodium borohydride, vial containing 0.042 g of paraformaldehyde, test tube containing dilute ethanolic KOH, (50 mg of KOH dissolved in 10 cm3 of ethanol), wash bottle containing dry ethanol,

Ice is available in each laboratory room.

Molar masses:

M(H) = 1.008 g mol-1 M(C) = 12.011 g mol-1 M(N) = 14.007 g mol-1 M(O) = 15.999 g mol-1 M(Na) = 22.990 g mol-1 M(B) = 10.811 g mol-1

Procedure STEP I CH 3 NH ₂ + OH C H O OH C N CH 3 H (i) (ii) A

1. Place the small magnetic stirbar in the 5 cm3 reaction vial containing solution (i) and stir.

2. Add the 2-hydroxybenzaldehyde from vial (ii) dropwise to the stirred solution in vial (i). After a short period of time a yellow solid will crystallize out. This is intermediate product A.

3. Isolate the yellow solid (A) by vacuum (suction) filtration and wash it with ice-cold ethanol. STEP II OH C H N CH 3 OH NH CH ₃ NaBH 4 (iii) A B

1. Add the impure Product A from Step I to a 5 cm3 reaction vial containing approximately 1.5 cm3 of ethanol.

2. Surround the vial with ice/water, and stir the reaction vigorously with the spatula while carefully adding small amounts of sodium borohydride (iii) over a period of about

5 minutes until the bright yellow colour disappears. The reaction will bubble.

Note that you have been given more sodium borohydride than is required for this reaction.

STEP III OH NH CH 3 N O CH 3 (CH 2 O) n (iv) B C

1. Dissolve all of the paraformaldehyde (iv) in approximately 2.5 cm3 of ethanolic potassium hydroxide (v) in a 5 cm3 vial. Stir to dissolve all of the solid.

2. Add all of product B from Step II to the vial. Stir and gently reflux the mixture for 15 minutes. A clear solution should be obtained.

3. Concentrate the solution by carefully boiling off some of the ethanol leaving approximately 1 cm3 in the vial and allow the vial to cool. The crystals which form are the required product C.

4. Isolate the crude product C by vacuum (suction) filtration and air dry the crystals. 5. Recrystallize the crude product from ethanol. Air dry the crystals for 15 minutes. 6. Determine the melting point* and then mass of the final product.

7. Place all of your remaining product in the numbered vial labelled “PRODUCT C” and hand it in for evaluation.

* Note: A melting point is always recorded as a range -- from when the crystals first begin to melt until the last crystal has melted. The melting point apparatus should be allowed to cool to approximately 50 degrees before you use it. The supervisors will be rechecking both your reported melting point and mass for product C.