1

NH

NHNH

NH

cyclam

5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (31th Mar 2012)

Theoretical tasks and solutions

TASK 1

Complexes with cyclam

14811-tetraazacyclotetradecane (cyclam) is a macrocyclic ligand with four

coordination sites at nitrogen atoms

This ligand can form complexes with heavy metal cations (with 11 stoichiometry)

These complexes are applicable as catalysts they can be also applied in medicine as

they affect important processes of biological significance

These complexes are stable in pH range limited by disadvantageous effect of

nitrogen atoms protonation (only neutral molecule exhibits complexing properties) and possibility

of heavy metal hydroxides deposition

This task concerns general conditions of stability of complexes with cyclam in solutions of

constant pH = 7 Basing on obtained data it would be possible to predict stability of complexes of

specific metal cations By solving this task please assume predomination of some protonated forms

of the ligand

Problems

a Calculate concentration of the neutral ligand in solution with total concentration of cyclam

(protonated and non-protonated forms) equal to 001 M (pH = 7)

b Determine the ratio of stability constant and so-called bdquoconditionalrdquo stability constant rsquo

where rsquo = [ML]([M]cL) [ML] is concentration of the complex [M] is concentration of free

metal ions (in symbols M ML the charge is omitted) cL is total concentration of cyclam

(protonated and non-protonated forms) not bound with metal ions

c Derive equation and calculate the minimal value of stability constant for which in solution of pH =

7 at least 999 of metal ions is complexed for cL = 001 M

d Heavy metal ions of charge 2+ can form hydroxide deposits What is the minimal value of

stability constant to avoid hydroxide deposition in solution containing cyclam and metal ions

(at pH = 7) Express the stability constant as a function of solubility product of the hydroxide

Ks0 and other selected equilibrium constants given in the task and concentrations ML and cL

e Are the conditions from sections (c) and (d) fulfilled for cations Cu2+

i Ni2+

for [ML] = 001 M

and cL = 001 M For the case of the complex with Ni(II) ions calculate the concentration of non-

complexed nickel ions How does this concentration change after 2-fold dilution of the solution

Dissociation constants of acids Ka1 (for H4L4+

) = 410

3 Ka2 (for H3L

3+) = 3

10

2

Ka3 (for H2L2+

) = 310

11 Ka4 (dla HL

+) = 3

10

12 (L is the form presented in the figure)

Stability constants of complexes with cyclam for Cu2+

210

27 for Ni

2+ 2

10

22

Solubility products Ks0 for Cu(OH)2 310

19 for Ni(OH)2 2

10

15

wwwShimiPediair

2

TASK 2

Phosphorus oxide ndash acidic or alkaline

Compound A is formed in the reaction of white phosphorus with oxygen carried out at lowered

pressure and at a temperature below 50 degC Reactive compound A can be isolated via vacuum

distillation from the products mixture in the form of white crystalline substance The compound

melts at ca 24 degC and can be easily dissolved in a series of organic solvents eg in benzene Only

one signal was found in the 31

P NMR spectrum of its benzene solution It was also observed that the

solution obtained by dissolution of 0185 g of compound A in 1000 g of benzene freezes at a

temperature that is by 043deg lower than melting point of pure benzene The cryoscopic constant Et

for benzene equals 512 K middot kgmol

Compound A reacts very easily with water to form a solution that contains acid B which forms a

precipitate of an anhydrous barium salt upon addition of the excess of Ba(OH)2 solution

A sample of compound A with a mass m1 = 126 g was reacted with a substantial excess of nickel

carbonyl Ni(CO)4 The reaction was proceeding at room temperature and a colourless gas was being

evolved It was found that after compound A had been consumed the mass of the reaction mixture

decreased by 0642 g 45 g of pure compound C was obtained after removal of the unreacted

carbonyl The 31

P NMR spectrum of this compound comprised only one signal as well

Compound A in the solid state (308 g) was also reacted with gaseous diborane (B2H6) at room

temperature After 24 hours products of the reaction were dissolved in n-pentane and left for

crystallisation 267 g of pure crystalline compound D were collected

It was found that this compound reacts violently with water and in the course of reaction colourless

gas is liberated and the resulting solution contains acids B and E In a carefully controlled reaction

0147 g of compound D were used and ca 80 mL (equivalent for 0 degC and atmospheric pressure) of

gas were given off Compound D reacts easily with nickel carbonyl too and using an excess of

carbonyl leads to the previously described compound C

Problems

a Determine compound A formula and confirm it with appropriate calculations Write the

equation of white phosphorus reaction with oxygen

b Draw Lewis electron structure of compound A molecule taking into account all of the lone

valence electron pairs Justify your answer

c Write the chemical equation in molecular form (reagent form) of compound A reaction with

water

d Write the chemical equation in molecular form of acid B reaction with barium hydroxide Draw

the molecular structure of anion present in the structure of barium salt Justify your answer

e Determine the formula of compound C and write the equation of its formation reaction Justify

your answer and confirm it with calculations

f Draw and describe the structure of nickel coordination sphere in compound C Justify your answer

g Determine the formulae of compounds D and E Write the equation of compound D reaction

with water Justify your answer and confirm it with calculations

h Basing on the appropriate acids and bases definition determine the chemical character of

reagents in the formation reaction of compound D and calculate the reaction yield

i Draw and describe the boron coordination sphere in compound D Justify your answer

Use the following values of molar masses in your calculations (gmol)

B ndash 1081 C ndash 1201 H ndash 1008 Ni ndash 5869 O ndash 1600 P ndash 3097

and the molar volume of gases at 0 degC and under atmospheric pressure Vm = 2241middot10minus3

m3mol wwwShimiPediair

3

Fig 1 Molecular

formula of a 12C4

crown ether

TASK 3

Determination of stoichiometry and stability of complexes with an NMR method

Application of approximations in limiting conditions

One of the problems in the 2nd

stage of the 58 Chemistry Olympiad was

related to the crown ethers ability to complex metal ions Now we

present the results of 1H NMR study on formation reaction of a

12-crown-4 ether (labelled as 12C4) with Na+ cation introduced into the

system in a form of sodium thiocyanate NaSCN

Three independent experiments have been carried out the first and the

second at temperature 23degC and the third one at temperature ndash50degC In

the two first experiments the spectra were recorded at varying NaSCN

concentration and a single averaged signal coming from both the

complexed molecules and the free ether was monitored The signal chemical shift δobs was

dependent on the ratio of sodium ions [Na+]0 and crown ether [E]0 concentrations The appearance

of the averaged signal proves that the complexation reaction is a so called fast exchange reaction

Let us remind that for such reactions the observed chemical shift δobs is an average of characteristic

chemical shifts of the ligand δE and the complexes δEnM multiplied by corresponding

concentration ratios of ligands unbound and bound in complexes to the total concentration of

ligands in the system By writing the general complex formula as EnM (where E means ether and

M means metal) the relationship can be presented as follows

E3M

0

3E2M

0

2EM

0

E

0

obs]E[

]ME[3

]E[

]ME[2

]E[

]EM[

]E[

]E[

(1)

The total concentration of ligands in the system [E]0 = [E] + [EM] + 2[E2M] + 3[E3M]+

Experiment 1 The characteristic chemical shift δE = 2761 ppm has been determined from the

NMR spectrum of the pure crown ether Then small amounts of solid sodium thiocyanate were

added to a 12C4 solution (in deuterated methanol CD3OD) with concentration [E]0 =

0219 moldm3 and the chemical shift δobs was recorded as a function of total concentrations

[NaSCN][12C4] The measurement results are shown in Fig 2 and additionally the data for points

within the linearity range ie at [NaSCN][12C4] le 03 are given in Table 1

The experimental points form a

curve showing two characteristic

bends which prove that two

complexes k1 and k2 with different

stoichiometry and different overall

complex formation constants for

metal ion and for ether βn are

present in the system Based on the

plot given in Fig 2 one can

determine the stoichiometry of the

two complexes being formed and

remaining in equilibrium

Fig 2

complex 1 complex 2

wwwShimiPediair

4

Table 1 Results of experiment 1 [E]0 = 0219 moldm3

[NaSCN][12C4] δobs ppm

0

0045

0096

0150

0240

0300

2761

2846

2937

3025

3190

3282

Experiment 2 In all measurements the initial concentration of the crown ether was kept constant at

[E]0 = 0050 moldm3 while the thiocyanate concentration varied but was much higher than that of

the emerging complex The sodium ion concentrations c and the corresponding resultant chemical

shifts δobs are given in Table 2

Table 2 Results of experiment 2 [E]0 = 0050 moldm3

Total Na+ ion concentration

c (moldm3)

Observed chemical shift

δobs ppm

0204

0271

0359

0495

0646

0768

4437

4558

4660

4757

4821

4856

Experiment 3 The spectrum of the pure 12C4 ether was recorded again at temperature ndash50degC then

a small amount of NaSCN was added and the spectrum was recorded again The spectrum showed

distinctly resolved ether and complex peaks (the fast exchange reaction does not occur here) It

turned out that the chemical shift of the pure ether virtually did not change with decreasing

temperature One can therefore assume that the chemical shift of the k2 complex that is stable in the

presence of a large excess of ether is the same at both 23degC and at ndash50degC and is δk2 = 4575 ppm

Using the results of the three experiments described above determine the stoichiometry of complexes

that are formed in the system calculate their stability constants and characteristic chemical shift of the

k1 complex by following the directions given below

In the solution one should assume the following symbols for the major quantities occurring in the problem

[E]0 ndash total concentration of the crown ether constant in each experiment

[Na+] = c ndash total concentration of sodium ions varying during titration

δE ndash chemical shift of the pure 12C4 crown ether

δk1 δk2 ndash chemical shifts of two complexes

ck1 ck2 ndash concentrations of EnM complexes respectively in equilibrium

To simplify the notation of equations the following symbols can be also introduced

Eobsobs Ek1k1 and Ek2k2

wwwShimiPediair

5

Problems

a Based on the plot in Fig 2 determine the stoichiometric coefficients of the resulting complexes Justify

the answer with a short comment

b Give chemical equations for reactions taking place in the system and the formulae for stability

constants of the complexes by writing them down in a form containing the initial ether concentration

[E]0 and the concentration of sodium ions c

c Using appropriate equations determine complex concentrations ck1 and ck2 by introducing the

simplifications proposed in Note 1

d Give an equation describing the dependence of the averaged chemical shift δobs on corresponding

chemical shifts of the ether δE (data) and the two complexes δk1 (this should be treated as an unknown)

and δk2 (data) Present the equation as a dependence of corresponding differences of chemical shifts that

is obs k1 and k2 Write the two equations in two forms corresponding to the limiting cases by

introducing corresponding complex concentrations determined in Direction c

e Calculate β2 from the simplified equation from Direction d for the limiting case [E]0 gtgt c by

determining the slope of the straight line (δobsndashδE) = f(c) from two selected points from Table 1 and using

known value of δk2 (see Note 2)

f Solve the equation from Direction d in the high Na+ concentration limit (c) by converting the equation

so as to get a linear dependence of 1(δobs ndash δE) on 1c Calculate β1 and δk1 constants (see Note 3)

Notes

1 The equations for equilibrium constants can be simplified if the experiment conditions can

be considered as the limiting ones

In experiment 1 [E]0 gtgt c so the equilibrium is strongly shifted towards the k2 complex and

we can assume that [E]0 gtgt ck2 gtgt ck1 asymp 0

In experiment 2 [E]0 ltlt c so the equilibrium is strongly shifted towards the k1 complex and

we assume that c gtgt ck1 gtgt ck2 asymp 0

Corresponding approximations are best introduced by determining ck1 and ck2 as functions of

complex stability constants β1 and β2

2 Fig 3 shows linear dependence of (δobs ndash δE) on c in the limit [E]0 gtgt c The plot should facilitate the

selection of points of which the coordinates allow to calculate the slope of the straight line

Fig 3

Experiment 1

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6

3 To solve the problem in the high Na+ concentration limit ([E]0 ltlt c) it is necessary to

ldquolineariserdquo the dependence of the corresponding difference of chemical shifts (δobsndashδE) on

varying [Na+] = c concentration The dependence of 1(δobsndashδE) on 1c should be linear

(1(δobsndashδE) = a(1c) + b) in the limit of large values of the ratio c[E]0 ie in the conditions of

experiment 2 Fig 4 shows the relationship plotted using the data from Table 2

Fig 4

TASK 4

Biologically active indole derivatives

Tryptamines are derivatives of indole (a heterocyclic compound consisting of two condensed rings

a benzene ring and a pyrrole ring) abundantly present in living organisms To tryptamines belong

inter alia neurotransmiters as important as melatonin or serotonin Due to that also other

tryptamine derivatives are not neutral to human organism and may have medicinal use or exert

hallucinogenic effects

The phosphorylated tryptamine derivative F initially isolated from plant sources has been obtained

through the reaction sequence depicted in Scheme 1 The starting material for the synthesis was 4-

hydroxyindole It is known that the most reactive position of indole is the C-3 within the pyrrole

ring which easily undergoes aromatic electrophilic substitution but may also be involved in

reactions typical for enamines

First the phenol group of 4-hydroxyindole was protected by reacting it with benzyl chloride in the

presence of sodium methanolate This led to compound A which was then subjected to Friedel-

Crafts acylation with oxalyl chloride leading to compound B Subsequently B without isolation

was reacted with dimethylamine which resulted in compound C In the next step C was reduced to

D (molecular formula C19H22N2O) by means of LiAlH4 and then catalytically hydrogenated to yield

compound E and toluene as a by-product In the last step E was converted into F by applying

following reagents 1 n-buthyllithium 2 tetrabenzyl pyrophosphate (TBPP) 3 catalytic

hydrogenation (PdC H2)

A homologue of compound E known as 4-hydroxygramine and having molecular formula of

C11H14N2O may be obtained in a multi-component reaction (Mannich reaction) by reacting

compound A with formaldehyde and dimethylamine followed by a catalytic hydrogenation

Exploiting a similar multi-component reaction but from different starting materials one can obtain

compound X as a mixture of several stereoisomers (Scheme 2 depicts only one of them)

Experiment 2

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7

O

ClO

Cl NH

A

NH

OH

Cl

OP

OP

O

OO

OO

C19H22N2O

TBPP =Bn

Bn

Bn

Bn

Scheme 1

MeOHMeO-

F

H2 Pd-C

1 n-BuLi2TBPP3 H2 Pd-C

[ B ]C

DE

LiAlH4

CH3

O

CH3

N O

O

NH

OCH

3Chiral

Scheme 2

342

1

X

Problems

a Draw structural formulas for compounds A-F and 4-hydroxygramine

b Draw structural formulas of the starting materials for the synthesis of compound X

c Compound X could be obtained from the same starting materials but in a stereochemically pure

form (as depicted in Scheme 2) Choose which conditions should be applied to achieve this

I the reaction temperature should be increased

II the reagents should be mixed in a certain order

III the reaction should be carried out in the presence of L-proline

IV the reaction mixture should be stirred always in the same direction

d Determine the absolute configuration of all stereogenic centers in compound X

e Explain the reactivity of the C-3 position of indole by means of appropriate mesomeric structures

TASK 5

The Diels-Alder Reaction

Cycloaddition reactions including Diels-Alder reaction in most cases are stereospecific what

means that geometry of substrates (eg dienophile or diene) determine the structure of the formed

product

X

Y Y

X+

General Scheme of the Diels-Alder Reaction

I The carboxylic acid with molecular formula C4H4O4 as well as its dimethyl esters (A B and C)

exist as two geometric isomers and one regioisomer and all of them undergo the Diels-Alder

reaction The reaction of geometric isomers A and B mixture with diene D composed entirely of

carbon and hydrogen atoms provides diastereomeric products E1 E2 and F Ozonolysis of E and F

mixture with zinc dust form product G (diastereomeric mixture) and subsequent reduction with

LiAlH4 excess leads to the mixture of diastereomers of polihydroxylic alcohol H

wwwShimiPediair

8

OHOH

OHOHLiAlH

4A i B

D E1 + E2 i F(diastereomers)

1) O3

2) ZnG

Hmixture of three diastereomers

1) O3

2) Zn

ID

J

mixture of three diastereomers

Compounds A and B after ozonolysis with zinc dust form the same product I which also undergo

cycloaddition reaction with diene D The reaction leads to the cyclic ether J The molecular mass of

J is equal to the sum of mass of reactants D and I

II Stereochemistry in Diels-Alder reaction Isomer A in the reaction with diene D forms the

mixture of two achiral diastereomers E1 and E2 The reduction of diastereomeric mixture of E1 and

E2 with hydrogen on palladium catalyst leads to only one product K

The Diels-Alder reaction of isomer B with diene D provides exclusively racemic mixture of

cycloadduct F The product F after ozonolysis with zinc dust and subsequent reduction with LiAlH4

excess leads to one diastereomer of polihydroxylic alcohol H (as a racemic mixture)

Problems

I

a Draw the structure of geometric isomer A or B and structure of regioisomer C

b Draw the structure of diene D

c Determine general structure of Diels-Alder product EF (no stereochemistry required)

d Draw the general structure of ozonolysis product G (no stereochemistry required)

e Draw the structure of compound I and general structure of J (no stereochemistry required)

II

a Determine unambiguously geometry of isomers A and B

b Draw stereochemical structures of compounds E (E1 and E2) and their reduction product K

c Draw the stereochemical structure of compound F and stereochemical structure of

corresponding diastereomer of polihydroxylic alcohol H

d Draw all possible stereoisomers of the product J Determine enantiomers and diastereomers

SOLUTIONS

SOLUTION OF TASK 1

a The total concentration of cyclam cL = [L] + [HL+] + [H2L

2+] + [H3L

3+] + [H4L

4+] Using acidic

dissociation constants after rearrangement one can write

cL = [L]1 + [H+]Ka4 + [H

+]2(Ka4 Ka3) + [H

+]3(Ka4 Ka3 Ka2) + [H

+]

4(Ka4 Ka3 Ka2 Ka1)

Taking into account the values of dissociation constants it can be assumed that at pH = 7 the

form H2L2+

predominates in the solution ie the above equation can be simplified to the form

cL = [L][H+]

2(Ka3 Ka4) thus [L] = cL Ka3 Ka4 [H

+]

2 After introducing numerical results for

cL = 001 M and [H+] = 10

7 M the obtained [L] = 9

10

11 M

wwwShimiPediair

9

OP

O

PO

P

O O

O

P

b Because stability constant = [ML]([M][L]) the ratio rsquo = cL[L] Basing

on previous data cL[L] = 001 910

11 = 11

10

8

c Because [ML][M] = 999 01 rsquo = [ML]([M]cL) = 999 (01 001) = 1

10

5

Thus = rsquo11

10

8 = 11

10

13

d = [ML]([M][L]) = [ML][H+]2([M]cL Ka3 Ka4) Assuming that a deposit of

hydroxide does not precipitate Ks0 = [M][OH]

2 the above equation can be written in the form

= [ML][H+]

2[OH

]

2(cL Ks0 Ka3 Ka4) = KW

2 ([ML]cL) ( Ks0 Ka3 Ka4)

e For [ML] = cL = 001 M using appropriate constants the minimal values of fulfilling the

conditions from (d) are 410

12 for the Cu

2+ complex and 6

10

8 for the Ni

2+ complex Comparison

of these results with experimental data shows that Cu2+

and Ni2+

complexes obey the conditions

from (c) and (d)

Because for Ni2+

- cyclam complex rsquo = [ML]([M]cL) for [ML] = cL [M] = 1rsquo is obtained

For the Ni2+

complex rsquo = 210

22 11

10

8 = 2

10

14 Thus [M] = [Ni

2+] = 1 (2

10

14) ie

[Ni2+

] = 5

10

15 M

2-fold dilution results in the same value of the [ML]cL ratio therefore Ni2+

ions concentration does

not change as well

SOLUTION OF TASK 2

a White phosphorus is built of P4 molecules Depending on the reaction conditions it may form

P4On (n = 6 7 8 9 or 10) oxides in the reaction with oxygen whose structures stem from P4

molecules structure Freezing point depression of benzene allows one to calculate the molality of

oxide A solution (cmA) and consequently its molar mass

0840125

430A

t

t

m

E

Tc molkg czyli 220

08400100

1850A

mAb

A

cm

mM gmol

This value corresponds to P4O6 molar mass (21988 gmol) Thus compound A is

phosphorus(III) oxide P4O6 Phosphorus reaction with oxygen proceeds according to the

following equation

P4 + 3O2 P4O6

b The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule There is

one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs

This structure agrees with 31

P NMR spectrum indicating all phosphorus nuclei are chemically

equivalent

c P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation

P4O6 + 6H2O 4H3PO3

d Phosphoric(III) acid contains PndashH bond and is a diprotic acid of an average strength It forms

therefore two series of salts and sparsely soluble barium salt containing HPO32minus

anion will be

formed with the excess of barium hydroxide

H3PO3 + Ba(OH)2 3 + 2H2O

HPO32minus

anion exhibits tetrahedral structure (sp3 hybridisation of phosphorus) There are three

oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron wwwShimiPediair

10

Nickel coordination sphere

O

H

O

PO

O

PO

H

O

or

2 2

e Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base

in chemical reactions Therefore Lewis adduct is formed in the reaction with nickel carbonyl in

which phosphorus forms a chemical bond with nickel replacing one of the CO ligands Assuming

that the only gaseous product of the reaction is CO the molar ratio of the evolved CO to the used

P4O6 equals

0642 g 126 g

002292 000573 4 12801 gmol 21988 gmol

which indicates that four nickel atoms

were bound by a P4O6 molecule The reaction proceeds according to the following equation

4Ni(CO)4 + P4O6 P4O6middot4Ni(CO)3 + 4CO

The formula of compound C is P4O6middot4Ni(CO)3 (or P4O6[Ni(CO)3]4) The molar mass of

compound C equals 79076 gmol Such a reaction course is confirmed by the approximate molar

mass of compound C calculated from the reaction product mass

4 6

C

P O

45785 gmol

000573

CmM

n

f It follows from the fact that all of phosphorus nuclei are chemically

equivalent in the molecule of compound C that every phosphorus atom is

bound to one nickel atom and three oxygen atoms The coordination

centres in the form of nickel atoms satisfy the 18 electron rule (10 valence

electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons

coming from phosphorus) Therefore the structure of nickel coordination

sphere is tetrahedral and ligands bound to nickel atom are localised in

vertices of a slightly deformed tetrahedron

g Similarly as in the reaction with nickel carbonyl P4O6 phosphorus oxide serves as a Lewis base

in the reaction with B2H6 forming an adduct with acid (BH3) As a result of the reaction with

water the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and

H3BO3 boric acid (acid E) The liberated gas is molecular hydrogen and its quantity depends on

the amount of reacted diborane

2

2 6

H 3

B H 3

800595 10

6 2241 10 6m

Vn

V

mole which correspond to B2H6 mass equal to

2 6 2 6 2 6

3

B H B H B H 0595 10 2767 00165m n M g The molar ratio of P4O6 to B2H6 in the Lewis

adduct equals 3(0147 00165) g 0595 10 0594 0595 11

21988 gmol

Only two phosphorus atoms are involved in bond formation with BH3 and the formula of

compound D is P4O6 middot 2BH3 (or P4O6 middot

B2H6) The hydrolysis reaction proceeds according to the

following equation

P4O6 middot 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2

P Ni

C

C

CO

OO

O

O

O

wwwShimiPediair

11

h The reaction between P4O6 and B2H6 leading to the formation of compound D

(P4O6 + B2H6 P4O6middot2BH3) is an acid-base reaction according to the Lewis definition

Phosphorus oxide being electron pair donor is a Lewis base and B2H6 (acceptor) is a Lewis acid

The reaction product compound D is a Lewis adduct The yield of compound D formation

reaction equals

4 6 4 6 4 6

D D

P O P O P O

267 24755100 100 100 770

308 21988

Dn m Mw

n m M

i There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D

Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed

tetrahedron

B P

O

O

OH

HH

or

B P

O

O

OH

HH

SOLUTION OF TASK 3

a The experimental curve δobs as a function of the concentration ratio [NaSCN][12C4] shows

clear changes of the slope at [NaSCN][12C4] = c[E]0 asymp 05 and at c[E]0 asymp 10 This means

that complexes with EM = 2 and EM = 1 ie E2M and EM are formed in the system In the

range c[E]0 gtgt 10 (thiocyanate excess) the equilibrium is shifted towards the EM complex

and in the range c[E]0 ltlt 05 towards the E2M complex

b The complex formation reactions occurring at the equilibrium are as follows

(k1) E + M EM and (k2) 2E + M E2M

The complex formation equilibrium constants k1 and k2 also referred to as the complex

stability constants can be written as

)()2]E([]T[]E[

]ET[β

k21kk21k0

k11

ccccc

c

(1)

)()2E]([]T[]E[

]TE[β

k2k1

2

k21k0

k2

2

22

ccccc

c

(2)

c From the equations for the equilibrium constants β1 and β2 ck1 and ck2 respectively are to be

determined by introducing the simplifications suggested in Note 1 In equation (1) we assume

that c gtgt ck1 gtgt ck2 and consequently cminusck1minusck2 asymp c We can further neglect subtraction of ck2

in the first term of the denominator and get

cc

c

)]E([β

k10

1k1

(3)

In equation (2) we assume that [E]0 gtgt ck2 gtgt ck1 and we can assume that [E]0 minus ck2 minus 2ck1 asymp [E]0

and in the second term of the denominator we can neglect subtraction of ck1 and we get

)(]E[

βk2

2

0

k22

cc

c

(4)

Then from equations (3) and (4) we determine

1

101k

β1

β]E[

c

cc (5)

wwwShimiPediair

12

and 2

2

0

2

2

0k2

β]E[1

β]E[

cc (6)

d The resultant chemical shift resulting from chemical shifts of the crown ether δE and the

complexes δk1 and δk2 can be written as follows

k2

0

k1k1

0

k1

0

k2k10

[E]2

[E][E]

2[E]δ

cδ

cδ

ccδ Eobs

(7)

remembering that [E]0 = [E] + ck1 + 2 ck2

By performing corresponding multiplications reducing terms in equation (7) and introducing

substitutions to simplify the notation Eobsobs Ek1k1 and Ek2k2

we obtain

k2

0

k2k1

0

k1obs

]E[2

]E[

cc (8)

Equation (8) should be written as two equations for limiting conditions that is by substituting

equations (5) and (6) to corresponding terms of equation (8) we obtain

k1

1

11obs_k

β1

β

c

c (9)

and k2

2

2

0

20obs_k2

β]E[1

β]E[2

c (10)

e In the case when [E]0 gtgt c and the equilibrium is strongly shifted towards the k2 complex

(E2M) (experiment 1) we solve equation (10) written as a function of c

c

2

2

0

k220obs_k2

β]E[1

β]E[2

(11)

One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = mc The factor

m can be determined from the point that is farthest away from the 00 point but still

belonging to the line that is point 5 in Table 1 One only has to recalculate the coordinates

of the point to c Δδobs_k2 For point 024 3190 we obtain the values c = 00526 moldm3

and Δδobs = 0429 ppm as well as m = 8155 ppmmoldm3 From equation (11) we determine

β2 and having substituted the data we get β2 = 201 (moldm3)-2

f Solution in the case c gtgt [E]0 (experiment 2) boils down to the case solved in Folder B of the

58 Chemistry Olympiad ie the solution of equation (9) To convert equation (9) into a linear

function one must write the equation as an equality of reciprocals as suggested by Note 4 (by

rising both sides of the equation to the power minus1)

Ek11Ek11

1

Eobs_k1

11

β

11

β

β11

cc

c (12)

And convert to the form

Ek1Ek11

1

Eobs_k1

11

β

11

c (13)

The plot in Fig 4 shows ideal linearity of data Eobs_k1

1

= a(1c)+b so comparing

corresponding parameters of the straight line with equation (13) we get the following

relationships

wwwShimiPediair

13

)(β

1

Ek11 a and

Ek1

1

b

For two extreme points from Table 2 we calculate corresponding reciprocals as given in the

Table below

c (moldm3) (1c) (moldm

3)-1

δobs ppm (1(δobs

ndash δE)

ppm

-1

0204 4902 4437 0597

0768 1302 4856 0477

Then we calculate parameters of the straight line a = 00332 ppm-1(moldm

3) and b = 0434 ppm

-1

and substitute them to formulae δk1 =1b+ δE and β1= ba obtaining δk1 = 5065 ppm and

β1= 131 (moldm3)-1

SOLUTION OF TASK 4

a A

B

C

D

E

F

4-hydroksygramina

b Reactants for the synthesis of X

c Conditions for

stereoselective

reaction III

d Absolute

configuration of X

3S 4S

e Reactivity of position C-3 of indole comes from high electron density in this position In

contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is

not connected with dearomatization of the benzene ring which would be energetically unfavorable

struktura I struktura II struktura III

wwwShimiPediair

14

SOLUTION OF TASK 5

Part I

I a

CO2Me

MeO2C

CO2Me

CO2Me

or

(A or B)

I a CO

2MeMeO

2C

C I b D

I c

CO2Me

CO2Me

CO2Me

CO2Me

or

E or F

I d

CO2Me

CO2Me

O

OG

I e

OO

OMeI

I e

O

CO2Me

O

CO2Me

or

J

Part II

II a CO

2Me

MeO2C

CO2Me

CO2Me

A B

(Z)-isomer (E)-isomer

II b

CO2Me

CO2Me

CO2Me

CO2Me

H

CO2Me

CO2Me

H

H

H

CO2Me

CO2Me

CO2Me

CO2Me

CO2Me

CO2Me

+

+

+

cycloadducts cis E (E1 i E2)

two diastereomers (achiral)

(from ester A)

or

or

II c

OHOH

OHOH

OHOH

OHOHor

diastereomer of compound H

obtained from cykloadduct F

II b

CO2Me

CO2Me

CO2Me

CO2Me

K IIc

CO2Me

CO2Me

H

CO2Me

H

CO2Me

CO2Me

CO2Me

CO2Me

MeO2C

MeO2C

MeO2C

H

CO2Me

H

MeO2C

enantiomers

cycloadduct trans F

(obtained from ester B)

or

or

wwwShimiPediair

5 8 C h e m i s t r y O l y m p i a d F i n a l c o m p e t i t i o n (30th March 2012)

Practical tasks and solutions

TASK 1

Aluminium polychloride analysis

Suspensions are removed during the purification of water and sewage using the so-called

coagulants They form sols having high surface which detain suspension particles and after addition

of flocculants they form precipitates Aluminium polychloride (PAC) is one of the coagulants and it

is obtained via a reaction between aluminium hydroxide and hydrochloric acid but only some of its

hydroxyl groups are substituted with chloride ions and as a result its composition is variable

A PAC sample whose mass (m1) is given on a 100 mL volumetric flask labelled A was dissolved

in water with a small amount of nitric acid and a clear solution was obtained 5000 mL of EDTA

solution whose concentration is given on a bottle was added The resulting solutionrsquos pH was

adjusted to ca 45 using methyl orange as an indicator Afterwards the solution was heated and

boiled for 10 minutes while the pH was corrected using diluted ammonia solution After cooling

the solution was transferred to flask A and water was added to the graduation mark

A second PAC sample whose mass (m2) is given on a 100 mL volumetric flask labelled B was

dissolved in 10 mL of 2 molL nitric acid solution The solution was transferred to flask B and

water was added to the graduation mark

Glassware and reagents at your disposal

burette two Erlenmeyer flasks with ST

beaker 25 mL volumetric pipette

graduated cylinder 10 mL volumetric pipette

small funnel wash bottle with distilled water

ca 0050 molL KSCN solution

ca 0025 molL MgCl2 solution

Reagents at disposal of all participants

EDTA solution (concentration given on a bottle) ammonium buffer with pH = 10

ca 005 molL AgNO3 solution 10 NH4Fe(SO4)2 solution

eriochrome black T mixed with NaCl and a spatula chloroform with a pipette

Attention The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in

your answer sweet

Additional information The EDTA complex of aluminium exhibits lower conditional stability

constant that the MgY2minus

complex but the aluminium complex is inert The eriochrome black T

magnesium complex is much less stable than the MgY2minus

complex The AgNO3 solution is acidified

with nitric acid (its concentration is 05 molL) The AgCl solubility product is higher than that of

AgSCN which is higher than AgBr Kso wwwShimiPediair

2

Problems

a Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the

procedures given below and the information contained in the problem

b Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using

the information contained in the problem the procedures given below and the available reagents

c Write the equations (at least 6) of chemical reactions taking place during sample analysis as well

as of the reactions carried out before and described in the problem

d Derive the formulae to calculate the concentrations of titrants necessary for the completion of

tasks a and b

e Give the determined concentrations of solutions form task d

f Determine the percentage of aluminium in PAC

g Determine the percentage of chloride ions in PAC

h Determine the stoichiometric formula of PAC Al(OH)xCly

i Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem

description)

j Why is chloroform introduced into the flask during chloride ions determination Is it necessary

to do it when determining bromide ions Justify your answer

Procedures

Complexometric determination of aluminium

Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer

flask Introduce a precisely known amount of EDTA solution to the flask excessive with respect to

the expected amount of aluminium Add one drop of methyl orange and diluted ammonia solution

until the colour changes to yellow-orange Heat the resulting solution and boil it for 10 minutes

correcting its pH with diluted ammonia solution After cooling add 5 mL of ammonium buffer

solution with pH = 10 a pinch of eriochrome black T and titrate the navy solution quickly with

MgCl2 solution until the colour changes to violet-blue

Repeat the titration

Attention The titration has to be carried out quickly and as soon as a noticeable change of colour is

observed the titrantrsquos volume has to be read The titration must not be continued even though the

analyte solution may recover the blue colour

Argentometric determination of chloride ions

Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer

flask with ST Add a precisely known volume of AgNO3 solution excessive with respect to the

expected amount of chloride ions If necessary add nitric acid so that its concentration lies in the

range 02 ndash 05 molL Add ca 2 mL of chloroform ca 1 mL of NH4Fe(SO4)2 solution close the

flask with a stopper and shake it Open the flask wash the stopper with distilled water and titrate the

obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate

settles down)

Repeat the titration

wwwShimiPediair

3

TASK 2

Distinguishing surface active agents (surfactants)

Owing to their characteristic structures and specific behaviour in aqueous solutions surface active

agents (surfactants) are used diversely in analytics The interactions between cationic surfactants

and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the

chelates are not very stable (eg complexes with metal titration indicators) they may be

decomposed Metal chelates which are insoluble in water may be dissolved in aqueous solutions of

non-ionic surfactants due to the fact that they form micelles at appropriate concentrations Cationic

and anionic surfactants may form precipitates that are soluble in the excess of surfactant with

suitable dyes (acidic or alkaline) The mentioned phenomena make it possible to differentiate

between cationic anionic and non-ionic surfactants

The solutions of substances given in the table below are placed in test tubes labelled A-J The

solutions concentrations are also given

Tes

t tu

bes

A-J

Substance Concentration

Iron(III) chloride 210-5

molL

Mercury(II) nitrate 210-5

molL

Dithizone HDz 210-4

molL

Eriochrome cyanine R ECR 210-4

molL

Safranin T SFT 210-4

molL

Rose Bengal RB 210-4

molL

Potassium palmitate PK 110-2

molL

Sodium dodecyl sulphate SDS 110-2

molL

Triton X-100 TX 110-2

molL

Cetyltrimethylammonium chloride CTA 110-2

molL

Dithizone is present in a slightly alkaline aqueous solution and is used for extraction-

spectrophotometric determination of mercury or silver (it forms orange and yellow chelates

respectively in strongly acidic solutions) In slightly acidic solutions eriochrome cyanine R forms

violet complexes with aluminium and iron(III) ions which turn blue under the influence of cationic

surfactants Safranin T is an alkaline whereas rose Bengal acidic non-chelating dye Triton X-100

poly(ethelene oxide) substituted octylphenol is a non-ionic surfactant

Glassware and reagents at your disposal

8 empty test tubes

6 polyethylene Pasteur pipettes

wash bottle with distilled water

sulphuric acid 1molL

indicator paper

You may use the solutions for problem 1

wwwShimiPediair

4

Problems

a Give the probable arrangement of substances in test tubes A-J taking into account solutionsrsquo

colour and their pH and carrying out simple tests for the presence of surfactants

b Derive a plan that will allow you to identify substances in the solutions

c Identify the substances present in the solutions in test tubes A-J using the available reagents and

the given procedure

d Give justification for your identification confirming it with two observations

e Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator

if the analysed solution contains cationic surfactants Justify your answer with appropriate

observations

Investigating the influence of surfactants on coloured systems

Transfer ca 1 mL of the analysed dye solution into a test tube and introduce the same amount of

metal solution Add surfactant solution drop by drop shake the test tube and carefully watch what is

taking place in the solution Carry out a blind test for comparison

USE YOUR SOLUTIONS ECONOMICALLY DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS

Eriochrome cyanine R ECR Rose Bengal RB

Safranin T SFT Dithizone H2Dz

wwwShimiPediair

5

SOLUTION OF PROBLEM 1

a Analysis plan for the determination of aluminium percentage in PAC sample

Upon heating of the m1 sample with EDTA solution according to the procedure aluminium ions

reacted with some of the EDTA forming AlY The remaining EDTA has

to be titrated with MgCl2 solution In order to determine the number of aluminium moles in the

sample a portion of 2500 mL from flask A has to be titrated with MgCl2 solution using V1 mL of

the titrant Taking into account the fact that metal ions react with EDTA in a 11 molar ratio and

the relative volumes of the flask and pipette one may write

22 14504 MgClEDTAMgClEDTAAl cVcnnn

b Analysis plan for the determination of chloride ions percentage in PAC sample

Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the

resulting mixture in flask B is acidic In order to determine the amount of chloride

ions one has to take 2500 mL of the solution from flask B add 2500 mL of acidified AgNO3

solution with a known concentration (one gets the appropriate concentration of acid according to

the given procedure) and determine chloride ions according to the procedure using V2 mL of

KSCN solution Taking into account the relative volumes of the flask and pipette one gets the

following relationship

)cVc(258)n(n8n KSCN2AgNOKSCNAgNOCl 33

c Equations of the reactions taking place during the chemical analysis

PAC reaction with nitric acid OxHyClAlxHClAl(OH) 2

3

yx

Aluminium ions reaction with EDTA H2AlYYHAl 2

2

3 OHOHH 2

Titration of EDTA excess with magnesium chloride solution

2HMgYMgYH 222

2 OHNHOHNHH 244 HMgInHInMg2

Reactions taking place during chloride ions determination

AgClAgCl AgSCNAgSCN 23 FeSCNFeSCN

m1 = 02718 g

m2 = 05895 g S = 0800

Titration Titrant volumes mL

Conc of MgCl2

Determination of Al

Conc of AgNO3 and

KSCN

Determination of Cl

V0 2000 1990 average 1995

V1 1000 990 average 995

V3 1870 1880 average 1875

V2 1605 1615 average 1610

EDTA concentration

004990 molL

wwwShimiPediair

6

d Derivation of the formulae for solutions concentrations

Titration of 1000 mL of EDTA solution ammonium buffer solution eriochrome black T using V0 mL of

MgCl2 solution 0

EDTA

0

MgCl

V

c01

V

n2

2

MgClc

Titration of 1000 mL of MgCl2 solution + 2500 mL of AgNO3 solution + chloroform + iron(III)

solution using V3 mL of KSCN solution

KSCNClAgNO nnn3

KSCN3MgClKSCNMgClAgNO cVc102nn2c25223

From the information in the problem KSCNAgNO cS

25c25

3

S

KSCN

AgNO

cc

3

KSCN3MgCl

KSCN cVc102S

c25

2

3

MgCl

3

MgCl

KSCNVS-25

cS102

V-S

25

c102c 22

e Concentrations of MgCl2 KSCN and AgNO3 solutions

2MgClc 002501 molL cKSCN = 004002 molL

3AgNOc 005003 molL

f Determination of aluminium percentage in PAC

PAC mass in flask A equals m1 = 02718 g

mAl = (50004990 - 4995002501)2698 = 4047 mg Al percentage = 1489

g Determination of chloride ions percentage in PAC

PAC mass in flask B equals m2 = 05895 g

mCl = 8(25005001 ndash 1610004003)3545 = 1720 mg Cl percentage = 2918

h Stoichiometric formula of PAC

PAC formula ndash Al(OH)xCly x + y = 3

molar ratio f Al 14892698 = 05517 molar ratio of Cl 29183545 = 08232

y = 0823205517 =1492 x = 3 ndash 149 = 1508 Al (OH)151Cl149

i Explanation of the colour change during heating of PAC with EDTA

In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium

cations are formed acidifying the solution 32

2 AlYH 2HAlY

H+ ions have to be neutralised by adding ammonia solution so that the reaction equilibrium is

shifted towards AlY complex formation

j Justification of chloroform usage

Chloroform is added to separate AgCl precipitate from the solution Owing to the fact that AgCl is

more soluble than AgSCN the unwanted reaction could take place ClAgSCNSCNAgCl When determining bromide ions it is not necessary to

introduce chloroform as AgBr is less soluble than AgSCN

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7

SOLUTION OF PROBLEM 2

An exemplary arrangement of solutions

Substance Substance

A Potassium palmitate PK F Iron(III) chloride

B Sodium dodecyl sulphate SDS G Mercury(II) nitrate

C Triton X-100 TX H Eriochrome cyanine R ECR

D Cetyltrimethylammonium chloride CTA I Safranin T SFT

E Dithizone HDz J Rose Bengal RB

a Probable arrangement of substance in test tubes A-J

Colourless solutions having nearly neutral pH may contain metal ions or surfactant solutions

They are in test tubes A B C D F and G

Soap solution ie potassium palmitate solution may be slightly opalescent and alkalinendash test tube

A Dithizone eriochrome cyanine R Rose Bengal and safranin T are orange-red (test tubes E H I

and J) and the HDz solution is alkaline ndash test tube E Surfactant solutions froth upon shaking

which is visible for test tubes B C and D

b Identification plan

Reaction with sulphuric acid with frothing solutionsndashPK identification (palmitic acid insoluble in

water ndash the only one from surfactants) with coloured solution -

dithizonate anion reacts with acid to form dark precipitate of H2Dz Rose Bengal changes its

colour to light yellow

Mercury(II) ions may be identified with dithizone (one of the colourless not frothing solutions) ndash

formation of the red-orange precipitate soluble in any surfactant

The other colourless not frothing solution containing Fe(III) ions (may be identified with

potassium thiocyanate solution from problem 1) allows one to identify ECR ndash formation of violet

solution This allows one to identify CTA ndash colour change to violet CTA gives precipitate with

Rose Bengal as a result of ion-pair adduct formation The remaining coloured solution (SFT) may

be used for SDS identification basing on precipitate formation with a small amount of surfactant

Remaining TX does not form precipitates with any of the dyes

wwwShimiPediair

8

c and d identification of substances in test tubes A-J and justification

Identification Justification

A PK

Opalescent and slightly alkaline solution

Froths when shook with distilled water

+ K rarr precipitate the only one of surfactants

+ MgCl2 or running water rarr white precipitate

B SDS

Colourless and neutral solution

Upon shaking with distilled or running water froths abundantly

+ K rarr no changes

SFT (tt I) + SDS rarr precipitate + SDS rarr precipitate dissolution

RB (tt J) + SDS rarr no changes Fe-ECR + SDS rarr no changes

C TX

Colourless and neutral solution

Upon shaking with distilled or running water froths abundantly

+ K rarr no changes

Dissolves dithizone mercury and silver dithizonates precipitates

SFT (tt I) or RB (tt J) + TX rarr no changes

D CTA

Colourless and neutral solution

Upon shaking with distilled or running water froths abundantly

+ K rarr no changes

Fe-ECR (violet) + CTA rarr blue the only one of surfactants

RB (tt J) + CTArarr precipitate + CTA rarr precipitate dissolution

E HDz

Orange and slightly alkaline solution

+ K rarr brown precipitate soluble in TX (SDS CTA)

Hg(II) + HDz +K rarr orangedarr soluble in TX (SDS CTA)

Ag(I) + HDz +K rarr orangedarr soluble in TX (SDS CTA)

F FeCl3

Colourless and slightly acidic solution does not froth

+ KSCN (problem 1) rarr orange solution

+ ECR (tt H) rarr violet solution + CTA rarr blue solution

+ HDz +K rarr brown precipitate soluble in TX (SDS CTA)

G Hg(NO3)2

Colourless and slightly acidic solution does not froth

+ KSCN (problem 1) rarr no changes

+ ECR (tt H) rarr no changes

+ HDz +K rarr orange precipitate soluble in TX (SDS CTA)

wwwShimiPediair

9

Identification Justification

H ECR

Orange and neutral solution

+ K rarr no changes

+ Fe(III) rarr vilet solution + CTA rarr blue solution

+ Al(III) (diluted flask B) rarr violet solution + CTA rarr blue solution

I SFT

Red and neutral solution

+ K rarr no changes

+ SDS (1-2 drops)rarr precipitate + SDS rarr precipitate dissolution

+ CTA rarr no changes

+ TX rarr no changes

J RB

Red and neutral solution

+ K rarr turns colourless

+ SDS rarr no changes

+ CTA (1-2 drops)rarr precipitate + CTArarr precipitate dissolution

+ TX rarr no changes

e Complexometric determination of Mg(II) in the presence of CTA

The presence of cationic surfactant renders magnesium determination via EDTA titration with

eriochrome black T as an indicator impossible In the ammonium buffer solution the violet

magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution

colour changes to blue just as at the end of magnesium titration with EDTA solution

Used abbreviations + K ndash addition of sulphuric acid

+ CTA ndash addition of cetyltrimethyl ammonium solution

+ SDS ndash addition of sodium dodecyl sulphate solution

+ Fe(III) ndash addition of iron(III) chloride solution

+ Hg(II) ndash addition of mercury(II) nitrate solution

+ TX ndash addition of Triton X-100 solution

tt ndash test tube

wwwShimiPediair

10

Comments to the solution of task 2

b Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely

soluble in water) this is the only surfactant solution that becomes cloudy Dithizonate anion one of

the orange solutions forms dark precipitate of H2Dz upon reaction with acid The second of the

orange dyes Rose Bengal turns practically colourless upon reaction with acid The remaining dyes

do not change their colour upon acidification One may identify mercury(II) ions (one of the not

frothing colourless solutions) using dithizone which form complexes with them in acidic solutions

as opposed to iron(III) ions Dithizone and mercury dithizonate precipitates are dissolved upon

addition of surfactants The remaining clear not frothing solution contains iron(III) ions which can

be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour)

Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns

violet-blue upon addition of CTA) One may find the acidic dye using CTA thanks to the forming

precipitate ndash Rose Bengal The other red-orange alkaline dye safranin T can be identified as it

forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant ndash

Triton X-100

c d The presence of potassium palmitate in test tube A is confirmed by the precipitation of

magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and

calcium palmitate precipitates from tap water Other surfactants do not give such precipitates The

presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO3 solution ndash

formation of yellow chelate precipitate soluble in surfactants (or chloroform) Silver nitrate

solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate Triton

X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the

dyes SDS does not form precipitate with Rose Bengal neither does it affect iron(III) or aluminium

complexes (from flask B after substantial dilution with water) Mercury(II) dithizonate is not

decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver

dithizonate

wwwShimiPediair