Key to Preparatory Problem of IChO 2020 - Istanbul, Turkey

(version 1, published on January 31, 2020)
keys are provided by Young, Alec, and Dr. Chen, thanks for their contributions.

https://icho2020.tubitak.gov.tr/icho-2020-hazirlik-sorular%C4%B1

Problem 1. Salvia Species Growing in Turkey: Isolation and Total Synthesis of Abietane Diterpenoids

The genus Salvia, named after a Latin word, salvare (“healer”), has a variety of species with important medicinal activities. They have been used for the treatment of colds, flu, and menstrual disorders in most regions of the world since ancient times. In Turkish folk medicine, Salvia L. species have also been used as a carminative, diuretic, hemostatic, spasmolitic, and stomachic, and in the treatment of mouth and throat irritations due to their antibacterial and wound healing properties. The genus Salvia includes over 900 species across the world, 58 of which are endemic in Turkey.

Female Turkish scientists Ulubelen & Topçu with co-workers have studied Anatolian _Salvia _plants growing in Turkey, and isolated and characterized more than 320 natural products, most of which are terpenoids, while one third are new diterpenoids.

In one of their studies on Salvia multicaulis Vahl., Ulubelen & Topçu isolated four new aromatic abietane norditerpenoids (1–4), which showed strong antituberculous activity. In addition to the antibacterial and antifungal activities of the isolated diterpenoids, the plant extracts also showed antioxidant, antiinflammatory, and cholinesterase inhibitory activities. _S. multicaulis _has folkloric use in Anatolia, such as an appetizer, for wound healing, against scorpion stings, and in the treatment of respiratory and urinary infections and diabetes.

Later, a research group in Turkey developed a synthetic route to obtain derivatives of natural products 1–4. This problem covers the synthesis of related compounds. The following reaction schemes illustrate the total synthesis of diterpenoids 1 and 5.

1.1. Draw the structure of the products A–M, without any stereochemical detail. Hint: In second step

, combination of lithium bromide and cerium(IV) ammonium nitrate (CAN) is used as a brominating reagent. Compound C is a benzaldehyde derivative and used in the synthesis step of compound M.

1.2. During the cyclization of H to I-1, another isomeric compound, I-2, with the formula CHO, is also formed. Draw the structure of I-2.

1.3. The following reaction scheme is related to the synthesis of 6, a desmethyl derivative of the diterpenoids 1 and 2. Draw the structures of products N–Y, without any stereochemical detail. Hint: Compounds R, S and T exhibit acidic character. The transformation of compound V to W includes Robinson annulation and a possible deformylation reaction steps.

1.4. During the transformation of compound V to W (Robinson annulation step), the use of a precursor of the α,β-unsaturated ketone, such as a β-chloroketone or N,N,N,-trialkyl-3-oxobutan-1-aminium halide (as used in the reaction scheme), can be more favorable. Explain.

1.5. Draw possible tautomeric forms of compound V.

1.6. Compound Y can be also obtained via ring-closing (electrocyclization) of the compound Z. Draw structure of Z.
1.7. For the transformation of X to Y, which of the following reagents can also be used? (Ignore S2’ type reactions).

Key to Problem 1 - by Young

Problem 2. Istanbulins and RelatedSesquiterpene Natural Products

Some elements received their names from different places around the world. In this respect, the record belongs to the Swedish village of Ytterby, after which four elements were named: ytterbium (Yb), yttrium (Y), erbium (Er), and terbium (Tb). However, elements are not the only chemical entities that owe their names to such places. Interestingly, a class of natural products, istanbulins A–E, received their names from the city of Istanbul. The first two members of this family, istanbulins A and B, were first isolated by Prof. Dr. Ayhan Ulubelen and co-workers from the plant Smyrnium olusatrum in 1971. The isolation of the remaining members, istanbulins C–E, was reported by Ulubelen and co-workers between 1979 and 1982.

Istanbulins constitute a subclass of a much larger family of natural products called sesquiterpenes. Two important sesquiterpene natural products with a similar 6-6-5 fused ring system are vernolepin (1) and vernomenin (2). Danishefsky and co-workers reported an elegant total synthesis of these two natural products in 1976 via the utilization of the Diels–Alder (DA) chemistry of the so-called Danishefsky’s diene.

Please note that all formulae depicting chiral molecules in this question refer to racemic mixtures.



In this context, Danishefsky’s diene (3) and the Rawal–Kozmin diene (4) are two electron-rich dienes that found widespread use in organic synthesis, and their structures are shown below.

TMS: trimethylsilyl; TBS: tert-butyldimethylsilyl

2.1. Draw the major resonance structures of dienes 3 and 4. Indicate the carbon atoms with higher electron density on each diene.

2.2. Compounds 3 and 4 have been extensively used as diene components in Diels–Alder reactions. Draw the conformations of 3 and 4 required to be able to enter a DA reaction. Predict which compound is a more reactive diene in a DA reaction with maleic anhydride (5).

2.3. When a mixture of Danishefsky’s diene (3) and compound 6 was heated followed by treatment with acid (TsOH, p-toluenesulfonic acid), compound A was obtained as the major product.

Draw the structures of all possible Diels–Alder products with the molecular formula of CHO that can be obtained from the reaction of 3 and 6. Drawing only one enantiomer of an enantiomeric pair is sufficient.

2.4. Determine the structure of the major product A.

2.5. Diels–Alder adduct A was converted to compound 7 via a sequence of 4 steps as shown below. Compound B is known to be acidic. Draw the structures of B–D.

2.6. When compound 7 is reacted with 1 equiv of m-CPBA, product E was obtained as a major product. Circle the functional group that reacts selectively with m-CPBA, and draw the structure of E.

2.7. The syntheses of vernolepin (1) and vernomenin (2) were completed as shown in the scheme below. Draw the structures of compounds F–J. In the final step, compound I is the precursor of 1.

Key to Problem 2

Problem 3. Çay, Cha, Chai, Te, Tea,Tee, Thé, Thee, and Earl Grey Tea Flavor: Bergamot

| Cha | Chinese, Japanese, Korean, Portuguese….. |

| | —- | —- | :—-: | | Chai | Russian, Persian… | | | Çay | Turkish, Azerbaijani… | | | čaj | Bosnian, Croatian, Czech, Serbian, Slovak.. | | | Shay | Arabic… | | | Te | Italian, Spanish… | | | Tea | English… | | | Tee | German… | | | Thé | French… | | | Thee | Dutch… | | | Chaay | Hindi… | | | ….. | … | |

Tea (in Turkish: çay) is popular throughout Turkey and the Turkish diaspora. Turkish tea culture also extends from Azerbaijan to some countries in the Balkan Peninsula. Turkey has the highest per capita tea consumption in the world, i.e. 2.5 kg/person per year, followed by the United Kingdom (2.1 kg/person per year).

Bergamotene and derivatives (1-4), sesquiterpenes, are analogues of pinnae monoterpenes. Found in bergamot oil, the bergamotenes contribute to the aroma and flavor of Earl Grey tea.

3.1. The following reaction scheme illustrates the synthesis of α-trans-bergamotene (1). Draw the structures of products A~G.

3.2. What is the function of MeNO reagent in the transformation of A to B?

Key to Problem 3

Problem 4. Early Russian Organic Chemistsand Markovnikov’s Rule

The last year was devoted to the 150 anniversary of the discovery of Markovnikov’s rule, formulated by Vladimir V. Markovnikov in 1869. Markovnikov was a PhD student of the famous early Russian scientist Alexander Butlerov. In his PhD thesis in 1869, Markovnikov discovered the famous rule that exists in almost every textbook on organic chemistry. According to Markovnikov’s rule, when an unsymmetrical alkene or alkyne reacts with a hydrogen halide (hydrogen chloride, hydrogen bromide, or hydrogen iodide), the hydrogen atom of HX adds to the carbon atom having the highest number of hydrogen atoms. However, depending on the reagent or substrate, in some cases, opposite results could also be possible, and these kinds of reactions are called anti-_Markovnikov addition. Although Markovnikov’s rule was developed for and is specifically applied to the addition of hydrogen halides to alkenes or alkynes, many other additions are also described as Markovnikov or _anti-_Markovnikov depending on the regioselectivity of the addition reaction.

Actually, the rule should be revised as follows: “_addition to this kind of double or triple bond proceeds through more stable intermediates”. In some cases, besides electronic effects, steric effects can also affect the formation of Markovnikov or anti-Markovnikov addition products.

The following problems are mainly related to discoveries described by the student of the more distinguished organic chemist Alexander Butlerov or his colleagues at Kazan University, Tatarstan, Russia.
4.1. Draw the structures of major products A-E, including the appropriate stereochemistry (ignore optical isomerism).



4.2. Draw the structures of major products F and G for the following reactions.

Wagner–Meerwein Rearrangement (WMR)
Wagner is another famous scientist who worked at Kazan University contemporaneously with Butlerov and Markovnikov. Wagner proposed that bornyl chloride undergoes an internal rearrangement to form pinene. Meerwein then generalized this type of rearrangement. Thus, this kind of reaction was named Wagner–Meerwein rearrangement. These reactions take place when a carbocation is formed. Generally, a carbocation is rearranged to a more stable carbocation, if possible, by neighboring group migration. In addition, if the reaction does not proceed through a carbocation or borderline carbocation intermediates, rearrangements do not take place.
4.3. Considering the formation of intermediates for every reaction, draw the structures of reagents H and I and major products J–M.

Acid-catalyzed Wagner–Meerwein Rearrangement

The acid-catalyzed reaction of 4,4-dimethylcyclohexa-2,5-dien-1-one resulted in the formation of a compound, the NMR data of which are given below.

For N; H NMR (300 MHz, CDCl): δ = 6.95 (d, J = 8.0 Hz, 1H), 6.61 (d, J = 2.8 Hz, 1H), 6.57 (dd, J = 8.0, 2.8 Hz, 1H), 5.39 (bs, 1H), 2.16 (s, 3H), 2.14 (s, 3H). C NMR (100 MHz, CDCl): δ 153.4, 137.9, 130.4, 128.6, 116.6, 112.3, 19.8, 18.7.

4.4. Find the structure of product N and propose a plausible mechanism.

4.5. What kind of difference do you expect in the H NMR spectrum after a drop of DO is added to the solution in the NMR tube?

Zaitsev’s Rule

Zaitsev, who described a rule named after him (Zaitsev’s or Saytzeff’s or Saytzev’s rule), was another PhD student of Butlerov’s. Zaitsev’s rule is an empirical rule for estimating preferred alkene product(s) in elimination reactions. At Kazan University, the chemist Alexander Zaitsev studied various elimination reactions and observed a general trend in the resulting alkenes. More generally, Zaitsev’s rule stipulates that in an elimination reaction the most substituted product will be formed. The following problem is mainly related to Zaitsev’s rule.
4.6. Draw the structures of elimination products O–Q and compound R. What is the major product formed by the thermal reaction of R described in the following scheme?

4.7._ _Which base(s) can be used to increase the ratio of Q relative to EtONa?

Problem 5. Arndt–Eistert Homologation

Fritz Georg Arndt (6 July 1885–8 December 1969) was a German chemist who had a great influence on the development of chemistry in Turkey. He was employed for two decades of his professional life at Istanbul University in two distinct periods. He discovered the Arndt–Eistert synthesis with Bernd Eistert. The Arndt–Eistert synthesis is the chemical reaction for one-carbon homologation (i.e. the conversion of RCOH to RCHCOH) of carboxylic acids and is called the homologation process. In the Arndt–Eistert homologation, the key step is the Wolff rearrangement of diazoketones to ketenes, which can be achieved thermally, photochemically, or by silver (I) catalysis. The reaction is conducted in the presence of nucleophiles such as water, alcohols, or amines to capture the ketene intermediate to yield carboxylic acids, esters, or amides, respectively. In this problem, synthesis of indolizidine alkaloids is studied.



5.1. As depicted in the scheme below, the synthesis of indolizidines 167B and coniceine could be easily and concisely achieved from β,γ-unsaturated ester B. The key step (A → B) is the Wolff rearrangement. Compound C has a lactam core, which is a bicyclic heterocycle containing a six-membered ring fused to a saturated five-membered ring, one of the bridging atoms being nitrogen.

Draw the structures of A–D without any stereochemical detail.

5.2. In the Arndt–Eistert homologation reaction, an α-diazo ketone can undergo photochemical Wolff rearrangement to form α-ketocarbene via nitrogen extrusion. This intermediate undergoes a 1,2-alkyl shift to give the ketene product.



Draw the structures of the α-ketocarbene and ketene intermediates in the second step (A → B).

5.3. Addition of propylmagnesium bromide to compound C, followed by AcOH/NaBH, is the last step in the total synthesis of indolizidine 167B.

Draw the structure of an intermediate (CHN) in the fourth step (C → D).

5.4. An alternative synthesis of coniceine is depicted below. Draw the structures of E–J.

Key to Problem 5 - by Alec

Problem 6. Atovaquone

Atovaquone, an approved drug, is used to treat pneumocystosis and malaria. Ketoester 1 and aldehyde 2 are key compounds in the synthetic process of atovaquone.

6.1. The synthesis of key compound ketoester 1 is shown below. A mixture of phthalic anhydride and EtN is treated with diacid. Gas evolution is observed during this period. Treatment of the reaction mixture with aq. HCl solution provides formation of acid 3 through intermediate A with two carboxylic acid groups. Acid 3 is converted to the isomeric intermediate B, containing both hemiacetal and ester functionalities, followed by dehydration to the alkene C, which is then brominated to give D under acidic condition. Dibromide D undergoes solvolysis in a hot mixture of HO/AcOH to give tertiary carbocation intermediate E, which is then trapped with water to give intermediate hemiacetal F. Finally, rearrangement of intermediate hemiacetal F provides key compound 1.

Note: The square brackets denote that the product was not isolated but reacted further without purification. The conversion of 3 to 1 is a one-pot reaction that involves a series of reactions occurring one after another in the same vessel without isolation and purification of intermediates.
_

Spectroscopic data for intermediates B and C: B: H NMR  = 7.86–7.52 (4H), 4.13 (bs, 1H, exchangeable with DO), 1.97 (s, 3H). C: H NMR  = 7.92–7.58 (4H), 5.24 (m, 2H); C NMR  = 166.8, 151.8, 139.0, 134.4, 130.4, 125.3, 125.1, 120.6, 91.3; MS m/z = 146.0

Draw the structure of intermediates A–F in the synthesis of 1.

6.2. The synthesis of aldehyde 2 starts from cyclohexene by key steps including Friedel–Crafts acylations, haloform, reduction, and oxidation. Friedel–Crafts acylation of cyclohexene with acetyl chloride yields chlorocyclohexyl methyl ketone J. Reaction of cyclohexene with acetyl chloride produces an initial carbocation G that undergoes two successive Wagner–Meerwein hydride migrations to form isomeric carbocations H and I, respectively. Trapping of carbocation I with chloride ion produces J, the Friedel–Crafts reaction of which with chlorobenzene provides K. Haloform reaction of methyl ketone K using sodium hypochlorite (NaOCl) gives the corresponding acid L. Acid L is converted into the aldehyde 2 in a several-step reaction sequence.

Draw structure of isomeric carbocations G**I formed in this reaction.



6.3.** Are these carbocations chiral?

6.4. Draw the structure of JL.
6.5. Choose all correct statements for L.

6.6. Which of the following compound(s) result(s) in the haloform reaction of K?

6.7. Which of the following reagents are appropriate to form aldehyde 2 from L?

Choose all correct reactions.

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Key to Problem 6 - by Alec

Problem 7. Which is (±)-Trikentrin A?

Although the indole skeleton is ubiquitous in nature, annulated indoles at any of the benzenoid positions are uncommon. The trikentrins and the structurally similar herbindoles represent fascinating such examples of 6,7-annulated indole or polyalkylated cyclopent[g]indole natural products. The trikentrins were isolated from the marine sponge Trikentrion flabelliforme and possess antibacterial activity. Possible structures for trikentrin A are shown in the Figure below. In this problem, we will find out which of these structures is trikentrin A.

There are several ways to synthesize trikentrin A. Two routes below involve aryne-based and hydrovinylation strategies and both finally lead to the formation of trikentrin A. The first step for problems 8.1 and 8.2 is the Bartoli reaction or Bartoli indole synthesis, which is the organic reaction of ortho-substituted nitroarenes with vinyl Grignard reagents to yield substituted indoles. In particular, it is the most efficient route to 7-substituted indoles.



(±)-Trikentrin A: H NMR (CDCl): δ 8.08 (bs, NH, 1H), 7.156.59 (3H), 3.44 (dt, J = 8.8, 7.5 Hz, 1H), 3.22 (dt, J = 8.8, 7.5 Hz, 1H), 2.94 (dq, J = 15.0, 7.5 Hz, 1H), 2.93 (dq, J = 15.0, 7.5 Hz, 1H), 2.60 (dt, J = 12.3, 7.5 Hz, 1H), 1.50 (d, J = 6.8 Hz, 3H), 1.37 (d, J = 7.0 Hz, 3H), 1.36 (t, J = 7.5 Hz, 3H), 1.32 (dt, J = 12.3, 8.8 Hz, 1H); C NMR (CDCl): δ 143.4-101.6 (8 signals), 44.8-15.1 (7 signals).

Aryne-based strategy
**

7.1. Draw the structures of A–I.

7.2. Draw the structure of the aryne involved as a reaction intermediate in step D → E.

Hydrovinylation strategy

7.3. Chemical transformation of bromo-nitrobenzene to corresponding 7-vinylindole J includes in Bartoli reaction followed by the vinylation step with vinylstannane. Draw the structure of J.

7.4. The second step is the Ni(II)-catalyzed asymmetric hydrovinylation of J. The ligands (K1–K4) used for hydrovinylation are given above.

Note: ee = enantiomeric excess; % ee = % major enantiomer - % minor enantiomer

Choose the correct statement(s):

| ☐ | Ligand 3 gave the best enantioselectivity. | | —- | —- | | ☐ | Ligand 4 gave a racemic mixture. | | ☐ | Each of the ligands K1–K4 is chiral. | | ☐ | Each of the ligands K1–K4 gave excellent yield (>95%) of the product. |

7.5. For the hydrovinylation step, choose the correct statement(s):

| ☐ | (allyl)NiBr or [(allyl)NiBr] is a source of vinyl. | | —- | —- | | ☐ | In this Ni-allyl complex, each nickel has oxidation number +2. | | ☐ | In this Ni-allyl complex, the electron count of Ni is 18. | | ☐ | This complex has a square planar geometry. |

7.6. Draw the structures of L–P. The absolute configuration of the asymmetric center in the hydrovinylation product is S. Hint: In the C NMR spectrum of compound M, one carbonyl carbon signal was observed at δ = 178.3 ppm.

Key to Problem 7 -by Alec

Correction: The last product need one more Me group, similar to its precursor.

Problem 8. Stereoisomers of 1,2,3-Triphenylpropane-1,3-diol

8.1. Draw all possible stereoisomers of 1,2,3-triphenylpropane-1,3-diol.

8.2. List all the achiral compounds.

8.3. List all the chiral compounds.

8.4. Which of the following properties or methods can be used to distinguish between the chiral compounds from question 8.3? Choose all correct statements.

Problem 9. NMR, Symmetry, and Structural Analysis

Naphthalene halides: Key compounds for many applications

Besides benzene, naphthalene is one of the best-known aromatic hydrocarbons. Therefore, the chemistry of naphthalene (1) has been extensively studied and many naphthalene derivatives have been synthesized. Halogen derivatives of this kind of compound are key for many transformations. For this reason, nearly all halogenated derivatives of naphthalene are known in the literature. Both H and C NMR spectra of symmetric compounds are characteristic, and allow researchers to exclude possible non-symmetrical structures to analyze the correct structures. Let us consider naphthalene tetrabromide isomers 2.

9.1._ _Draw the structures of all naphthalene tetrabromide(s) with 3 signals in C NMR and one signal (singlet) in H NMR spectra.

9.2._ _Draw the structures of all naphthalene tetrabromide(s) with 5 signals in C NMR spectra.

9.3._ _Draw the structures of all naphthalene tetrabromide(s) with 6 signals in C NMR and a doublet (J = 8–9 Hz) in H NMR spectra.

9.4._ _Draw the structures of all naphthalene tetrabromide(s) with 6 signals in C NMRand a doublet (J = 1.5–2.0 Hz) in H NMR spectra.

Dynamic NMR: fast transformation between tautomeric forms and identical nuclei in NMR

Bullvalene (3) is very suitable for degenerate Cope rearrangements. Without counting enantiomers, the number of possible valence tautomers of a bullvalene with ten distinguishable positions is 10!/3 = 1,209,600. This arrangement enables all carbon and hydrogen atoms to appear equivalent on the NMR timescale. At sufficiently high temperature, both H NMR and C NMR spectra of bullvalene show only one signal, average to a rounded peak. However, at −60 °C, as Cope rearrangements do not take place, olefinic and aliphatic protons are observed separately.



9.5. At low temperature, ignoring any Cope rearrangement, how many carbon signals do you expect from the C NMR spectrum of bullvalene?

Label identical carbon atoms with letters a, b, c… on the molecular structure.

9.6 Owing to fast tautomerism, some molecules give clearer spectra due to apparent symmetry. In light of this information, how many signals do you expect from the C NMR spectra of the following compounds?



9.7. In the literature, it has been shown that the tropolone diacetate derivative 4 has fewer signals than expected in C NMR spectroscopy.

Draw reasonable resonance structure(s) and/or transformation(s) responsible for this symmetry. How many signals do you expect for this molecule in the C NMR spectrum?




Stereochemistry of the epoxidation reaction of bicyclic alkenes.

9.8. Considering the following pieces of information, draw the structures of all possible stereoisomers formed under the given reaction conditions.

Hint: A and B are isomers with 3 signals and C is an isomer with 4 signals in C NMR spectroscopy.



9.9. Draw the structures of the stereoisomer(s) formed under the given reaction conditions. How many signals do you expect for the epoxide product(s) in C NMR spectra?

Problem 10. Woodward–Hoffmann Rules and Pericyclic Reactions

The Woodward–Hoffmann rules (or the pericyclic selection rules), developed by Robert B. Woodward and Roald Hoffmann, are used to rationalize or predict some stereochemical aspects and the activation energy of pericyclic reactions. They are for all classes of pericyclic reactions (and their reverse ‘retro’ processes), such as cycloadditions, sigmatropic shift, electrocyclization, ene, and cheletropic reactions.

(∆) | conrotatory (con) | | | photochemical (hν) | disrotatory (dis) | | 4n+2 | thermal | disrotatory | | | photochemical | conrotatory |

10.1. Thermal reaction of compound 1 results in the formation of endiandric acid 2 by a series of pericyclic reactions. Show all steps and classify their pericyclic processes.



How many π electrons are involved in the following reactions? Are these reactions thermally or photochemically allowed according to the Woodward–Hoffmann rules?

10.2.


10.3.

10.4. Domino Diels–Alder reaction of A with succinimide results in the formation of adduct 3. Draw the structures of A–C.

10.5. The following reaction scheme illustrates the synthesis of endo-isomer of benzenoid tetracyclic hydrocarbon I starting from o-xylene. Br-elimination of tetrabromo-o-xylene D with sodium iodide leads to a reactive intermediate which undergoes a 4 electrocyclization to yield compound F. Draw the structures of intermediates and products D–I.




retro-Diels–Alder Reaction

The retro-Diels–Alder (r_DA) reaction is the reverse of the Diels–Alder reaction, i.e., the formation of diene and dienophile from cyclohexene. Generally, an _r_DA reaction is initiated by heating. In some cases, low temperature is sufficient for this transformation, depending on the nature of the substrate.

10.6. Cyclopentadienes are very useful synthetic intermediates in the fields of organic and coordination chemistry. Parent (unsubstituted) cyclopentadiene is obtained by the thermal decomposition of dicyclopentadiene. However, substituted cyclopentadienes are generally unstable due to the facile migration of the endocyclic double bonds. Consequently, practical and general methods for the synthesis of substituted cyclopentadienes are limited. In the following reaction scheme, the synthesis of a substituted cyclopentadiene derivative is given. Besides _r_DA, some steps also involve the _inverse-Diels–Alder reaction, which is a cycloaddition between an electron-rich dienophile and an electron-poor diene (such as tetrazine 4), through the interaction of the HOMO orbital of dienophile and the LUMO orbital of diene.

Draw the structures of the intermediates and products J–N.

10.7. Nucleophilic aromatic substitution reactions constitute an important class of reactions in synthetic organic chemistry. In the following scheme, the reactions of aryl halide 5 proceed via two different kinds of intermediates in presence of a cyclic 1,3-diene depending on the reaction conditions and the nature of the substituent on the aromatic ring. Draw the structures of products (O and P), and discuss possible intermediates responsible for the formation of these products.

Problem 11. Benzoporphyrin

The name “porphyrin” derives from the Greek word porphyra, meaning purple. Porphyrins are a group of macrocycle organic compounds, composed of four modified pyrrole subunits. They have a total of 26 π-electrons, 18 of which form a planar porphyrin ring structure. They are often described as aromatic. Metal complexes derived from porphyrins occur naturally. One of the best-known families of porphyrin complexes is heme, the pigment in red blood cells. A benzoporphyrin is a porphyrin with a benzene ring fused to pyrrole unit(s).

11.1. Benzoporphyrins can be prepared starting from a masked pyrrole derivative E. The synthesis of E starts with a reaction of cis-1,2-dichloroethene and thiophenol to give A. Oxidation of A yields B having phenylsulfonyl units. The cis-product B is then converted to its trans isomer C when treated with a catalytic amount of Br under UV light. The Diels–Alder reaction between C and 1,3-cyclohexadiene under thermal conditions gives the product D, which is converted to a pyrrole carboxylic acid ester when reacted with ethyl isocyanoacetate. Ester then is treated with TFA to give the pyrrole derivative E.

Draw the structures of compounds A–E including stereochemistry when necessary.

11.2. Porphyrins can easily be prepared via a cyclization reaction of pyrrole derivatives with aldehydes. Draw the structure of aldehyde F and determine the oxidation state of zinc in compound H.

11.3. When H is heated under vacuum, it can give a more conjugated product through a retro-Diels–Alder reaction.

To complete the structure of I, draw the structures of the dashed circle part of I (all the circles are identical) and J.



Ammonia is a major metabolic compound and the importance of its sensitive detection has been emphasized recently because of its correlation with specific diseases. In normal physiological conditions, ammonia can be expelled from slightly alkaline blood and emanated through the skin or exhaled with the breath. Dysfunction in the kidney or liver that converts ammonia to urea can result in an increase in the ammonia concentration in breath or urine. Consequently, the detection of the ammonia present in breath or urine can be used for the early diagnostics of liver or stomach diseases. The development of sensor devices for measuring ammonia with a sensitivity of 50 ppb–2 ppm and with a fast response time is highly desired.

For that purpose, I was used to prepare a fiber-optic ammonia gas sensor. Exposure of this sensor to ammonia changes the transmittance of the fiber-optic. By using an appropriate spectrometer, ammonia gas in different concentrations was passed through the sensor and the change in transmittance was measured. The results of these measurements are listed in the Table below.

		[NH] (ppm)	Sensor response (%)
0.500 1.00 2.00 4.00 7.00 9.00 11.0 20.0 25.0 30.0	–0.2540 –0.7590 –1.354 –1.838 –2.255 –2.500 –2.600 –2.947 –3.152 –3.256

| | :—-: | —- |

11.4. Using the linear region of sensor response data prepare a calibration curve and find the calibration equation as

.

11.5. This sensor is then used for the detection of ammonia in human breath. When a kidney patient’s breath was fed into the sensor, a –3.812% change in the response is observed. Calculate the ammonia concentration in the patient’s breath.

Problem 12. Blue to Green, Turquoise

The beauty of the turquoise color of Lake Salda, where blue meets white sands, fascinates those who see it. Lake Salda, in the southern province of Burdur’s Yeşilova district, has been referred to as “Turkey’s Maldives” in recent years for its white sandy beaches and turquoise water. In fact, turquoise is an opaque, blue to green mineral that is a hydrated phosphate of copper and aluminum with the chemical formula of CuAl(PO)(OH)·4HO, and is known as a gemstone. The word turquoise dates back to the 17 century and is derived from the French turquois, meaning “Turkish” because the mineral was first brought to Europe through Turkey, from mines in the historical Khorasan Province of Persia. Phosphorus, which is also in the structure of turquoise, is an essential part of life. Without the phosphates in biological molecules such as ATP, ADP, and DNA, we would not survive. Phosphorus compounds can be found in the minerals in our bones and teeth. With few exceptions, minerals containing phosphorus are in the maximally oxidized state as inorganic phosphate rocks, which are partially made of apatite, and they are today the chief commercial source of this element. Phosphate products are used as fertilizers in agriculture. They are also used in animal feeds, as a leavening agent in baking powder and flour, as an additive to beverages, and in pharmaceuticals. Industrial uses include water softening, rust proofing, fire proofing, in insecticides and detergents, and for the manufacture of elemental phosphorus.


Lake Salda

There are three important allotropes of phosphorus: X, Y, and Z. However, another form of phosphorus, W, also exists (given below). X is a soft, waxy solid. It is exceptionally harmful and to a great degree reactive and also displays chemiluminescence. Crystals of X are composed of P molecules. Y is obtained by heating X to 250 °C within the sight of daylight. It is nonpoisonous and odorless. Y does not show chemiluminescence. It exists as a polymeric solid. Z is produced from X under inert atmosphere. Z is the most stable allotrope of phosphorus and has a layered structure. W is a form of phosphorus that can be produced by day-long annealing of Y above 550 °C.

The interconvertible forms of all allotropes of phosphorus

12.1. Identify allotropes of phosphorus indicated by X, Y, Z, and W.

12.2. Draw the structure of X, Y, Z allotropes of phosphorus and sketch the geometry of X.

12.3. P ignites suddenly in air at around 35 °C to form a phosphorus oxide derivative. Thus, it is kept under water. When P reacts with different amounts of dry halogens, phosphorus trihalides (PX) or phosphorus pentahalides (PX) are obtained. PX can also be obtained by the reaction of the halogens with PX. The phosphorus pentahalides undergo hydrolysis in two steps to form acid. The phosphoryl halides can be prepared by the hydrolysis of the appropriate pentahalides in a limited amount of water or by the reaction of the trihalides with oxygen. Dropping of the oxide derivative of phosphorus into water produces a hissing sound, heat, and acid product. The reaction of P with sodium or potassium hydroxide produces phosphine gas as the major product and potassium or sodium hypophosphite as a by-product. Phosphine burns in chlorine spontaneously, forming a phosphorus trihalide (PX) or phosphorus pentahalide (PX).

Write the formulas of products A–F.



When phosphorus reacts with excess of halogens, it can form five-coordinated compounds such as PCl. Phosphorus mixed pentahalides like PFCl are prepared by the addition of one halogen to the phosphorus trihalide of a second halogen.

12.4. Draw the Lewis structures of PCl and PFCl molecules.

12.5. By using VSEPR theory, predict the molecular geometries of PCl and PFCl.

12.6. Estimate the polarity of PCl and PFCl molecules.

12.7. Compare the axial P–Cl bond length to the equatorial P–Cl bond length in PCl.

12.8. Draw the hybridization scheme of the PFCl molecule and estimate which hybrid orbitals are used to form the axial and equatorial bonds.

12.9. The synthesis of PH from hydrogen with white phosphorus is given below. Calculate ΔH for the following reaction, using bond energies (single bond energies (BE) (in kJ.mol) for P–P: 213, H–H: 435, P–H: 326).

Organophosphorus compounds are organic compounds containing phosphorus. Phosphorus can adopt a variety of oxidation states, and organophosphorus compounds are generally classified based on their derivatives of phosphorus(V) or phosphorus(III), which are the predominant classes of compounds. Organophosphorus compounds are widely used as nucleophiles and ligands. Two major applications are as reagents in the Wittig reaction and as supporting phosphine ligands in homogeneous catalysis. Their nucleophilicity is evidenced by their reactions with alkyl halides to give phosphonium salts. Phosphines are nucleophilic catalysts in organic synthesis, e.g., the Rauhut–Currier reaction and Baylis–Hillman reaction.

Triphenylphosphine (PPh) is a common organophosphorus compound and it is widely used in the synthesis of organic and organometallic compounds. When a toluene solution of compound 1 and excess of PPh are heated to reflux, first compound 2 is formed and then compound 3.

Spectral data of compounds 1–3 are given below (for H NMR and C NMR data [δ values (relative area)]:


12.10. Identify the structures of 2 and 3.

Hint: The C NMR signal of 1 at 224.3 ppm is similar to the chemical shift observed for carbene carbons; the peaks between 184 and 202 ppm correspond to carbonyls; and the peak at δ 73.3 is typical for CHCH bridges in dioxycarbene complexes.
12.11. Determine if 2 is more likely to be the facial (fac) or meridional (mer) isomer.

Hint: The three ν(CO) bands with equal intensities are observed in the IR spectrum of compound 2. Protons of the carbene ligand occur as a multiplet in the H NMR spectrum.

12.12. Determine if 3 is more likely to be cis or trans isomer.

Hint: The two ν(CO) bands are of approximately equal intensity at 1944 and 1860 cm in the IR spectrum of compound 3. The P NMR spectrum shows a single resonance signal.

Some organophosphorus compounds such as sarin, soman, and VX are often referred as “nerve gases” despite the fact that they are liquids at room temperature. Each country signing the 1997 Chemical Weapons Convention agreed to ban the development of chemical weapons and to destroy chemical weapons and associated production facilities by 2012. Sarin can be destroyed by room temperature hydrolysis using aqueous NaCO to give NaF and the sodium salt of an organophosphate. The hydrolysis of nerve agent VX is more difficult. It reacts slowly with aqueous NaOH at room temperature, and the reaction has to be carried out at 360 K over several hours.

12.13. Determine the organophosphorus salt formed in the following hydrolysis reaction.



Two chromium complexes containing the ligands CO, PF, and PCl in octahedral geometry are given below. In an octahedral complex, the molecular orbitals created by coordination can be seen as resulting from the donation of two electrons by each of six σ-donor ligands to the d-orbitals on the metal, called σ-bonding. π-bonding (Pi bonding) in octahedral complexes is also possible when the ligand has p, d _or π_ molecular orbitals available. Ligands such as CO, CN and phosphines (of formula PR) are π acceptor, with empty orbitals that can interact with metal d orbitals in a π fashion. In most cases, the net back π bonding predominates, and electron density is transferred from the metal to the ligand. π-bonding can affect metal-ligand bond energy and bond length in carbonyl and phosphine complexes.

Answer the following questions considering the π interaction.

12.14. In which complex is the C–O bond shorter, Cr(CO)(PF) or Cr(CO)(PCl)?

12.15. In the infrared spectrum of which complex do the C–O stretching bands have *higher energy, Cr(CO)(PF) or Cr(CO)(PCl)?

Key to Problem 12 by Dr. Chen

12.1 & 12.2

X	Y	Z	W
white phosphous	red phosphous	black phosphous	purple phosphous
soft, waxy, toxic, reactive, chemiluminescent, P4	nonpoisonous, nonchemiluminescent, polymer	most stable, layer structure, similar to graphite	monoclinic phosphorus, is also known as Hittorf’s metallic phosphorus

reference:

https://en.wikipedia.org/wiki/Allotropes_of_phosphorus

https://www.sciencedirect.com/science/article/pii/S245226271730020X

12.3

Problem 13.

The simple d-block oxides such as FeO and CoO and many related mixed metal compounds have important properties. They have structures related to the mineral spinel, MgAlO, and may be given a general formula of ABO.

Stoichiometric amounts of two aqua complexes of transition metal (A and B) nitrate salts are thermally reacted to form a spinel ABO crystalline solid that has a face-centered cubic (fcc) structure with a unit cell composition of 8 ABO. Depending on the location of these two cations (A and B), the spinel structures are divided into two categories as normal and inverse spinels. In a normal spinel, the A ions occupy the tetrahedral holes and the B ions occupy the octahedral holes, but in the inverse spinel structures, the 2+ ions are replaced by half of the 3+ ions in the structure.
Crystalline solid has an ordered structure in which the unit cell repeats along all 3 principal axes of a three-dimensional matter. The smallest group of atoms in the material that constitute this repeating pattern is the unit cell of the structure. The unit cell completely reflects the geometry and structure of the entire crystal, which is built up by repetitive translation of the unit cell along the principal axes. Face centered cubic (fcc) is one of a common structure type of crystalline solid. Anions (X) are in the corners and faces of a cube (1/8 from each corners and ½ from each faces, because the corners and faces are shared by 8 and 2 unit cells, respectively) in the simplest fcc structure. The cations (M) occupy the holes among the anions. There are 8 tetrahedral (corners) and 4 octahedral holes (1 at the middle and 3 on the edges, each edge has ¼ octahedral hole) in a fcc _structure. Therefore, the unit cell composition is MX with an empirical formula of MX. However, the unit cell of a spinel structure is constructed by 8 of these _fcc units.

**

29.746 g salt of A was mixed with 58.202 g salt of B in a thermal process to produce 24.724 g pure product, ABO. In the spinel formation process, the metal ion of salt A keeps its oxidation state but the metal ion B undergoes oxidation. Both salts contain the same number of the water molecule(s), metal ion, and nitrate ion(s). Elemental analysis of the spinel provided the following data: 6.538 g metal A and 11.786 g metal B. Assume the end product is a diamagnetic solid matter. Considering the information provided above. Answer the following questions.

13.1. Suggest formulas for the salts of A and B.

13.2. Draw the structure of one of the complex ions i) without and ii) with one of the nitrates being in the coordination sphere as a bidentate ligand and identify if the inversion center is present in the complexes. Inversion is a symmetry operation that translates every atom through the center to the opposite side.

13.3. Place the metal ions in an appropriate location in the crystal structure and suggest if it is a normal or inverse spinel.

The x-ray diffraction data of ABO provides a unit cell parameter of 8.085 Å, which is constructed from 8 _fcc _units and corresponds to a length of the edges of the cube.

13.4. Sketch one of the ffc units of ABO and place the atoms in the unit.

13.5 What is the density of ABO? (hint: 1 Å is 1.0 x 10 m)

Reacting this spinel with other transition metals (M) produces M doped ABO, where M has a choice of occupying the place of either A or B-sides. The side product is AO (mono-oxide of A).

13.6. M is Mn in compound C and Ni in compound D, suggest the location of Mn and Ni ions in the structure of C and D, respectively. Assume splitting energy in Ni and B are 11500 cm and 20800 cm in the octahedral field, respectively, and the pairing energy is 19500 cm.

If the doping is in a small quantity or in some cases the doped metal ion behaves like a free ion in the lattice (it means, electrons of M only feel the surrounding atoms and localized to M and its 1 shell of atoms in the structure). Assume Mn is behaving like a free ion in the lattice and creating its own localized electronic energy levels.

13.7. Draw** the d-orbital splitting and identify if the Mn species are paramagnetic or diamagnetic.

Magnetic susceptibility could be calculated from the spin only formula:



where n is a number of unpaired electrons. However, some other electronic couplings affect the magnetic moment such that a correction term is needed. The correction term α is related to ground state (where α = 4 for a non-degenerate and 2 for a degenerate ground state, degeneracy of a ground state can be determined from the electron configurations, such as completely filled and half-filled set of orbitals creates a non-degenerate and a partially filled set of orbitals create degenerate states) and λ = 88 cm for Mn and -315 cm for Ni), and splitting energy (Δ is 5000 cm for Mn and 11500 cm for Ni) and the magnetic moment is:

,

The magnetic susceptibility can be experimentally determined and it is interrelated with the magnetic moment (if we ignore the diamagnetic contributions) with the following formula:

where T is temperature in Kelvin and X is the molar magnetic susceptibility.

13.8. What is the magnetic susceptibility of the products at 25 C, if the samples C and D weigh 25.433 and 25.471 g, respectively (each obtained from 24.724 g ABO)?

13.9. Place all the metal ions (A, B, Mn, and Ni) into their appropriate locations in the lattice and fill up the following table. Use t for d, d, and d and e for d, d orbitals in octahedral (O) and t and e orbitals in tetrahedral (T) cases. If there is distortion, predict the type of distortion(s) and show the d-orbital splitting.

Problem 14.

Medicinal inorganic chemistry based on metal‐based drugs is broadly defined as the area of research related to metal ions and metal complexes and their clinical applications. It is a new research area that developed from the discovery of the anticancer agent cisplatin. Cisplatin, cis-diamminedichloroplatinum(II), is a yellow powder and an anticancer drug widely used in the treatment of a variety of tumors, especially those of the testes, ovaries, head, and neck.

The synthesis of cisplatin starts with K[PtCl], but has undergone several improvements since it was published more than 100 years ago. The main problem is the occurrence of impurities and the formation of the by-product trans-platin. Nowadays, the synthetic routes are mostly based on a method published in the 1970s by Dhara. In the initial step, K[PtCl] is reacted with excess KI, and the platinum complex is converted into the iodo analogue (A). Subsequently, NH is added to the compound A and compound B is formed by ligand exchange in which two NH ligands are exchanged with two iodo ligands. B is a yellow solid that is filtered, dried, and mixed with the aqueous solution of AgNO. The insoluble AgI can be filtered off and cis-diamminediaquaplatinum(II) nitrate (C) is formed; then excess KCl is added to the solution of C to yield cisplatin (D).

The success of the synthesis relies on the strong trans effect of the iodo ligands. The spectator ligands T that are trans _to the leaving group in square-planar complexes influence the rate of substitution. This phenomenon is called the _trans _effect. Key point is that a strong σ-donor ligand or π-acceptor ligand greatly accelerates substitution of a ligand situating in the trans position. _Trans _effects follow the order given below.

For a T σ -donor: OH- < NH < Cl- < Br- < CN-, CH-< I- < SCN- < PR, H-
For a T π-acceptor: Br- < I- < NCS- < NO- < CN- < CO, CH



14.1. Write the formulas of A–D.

14.2. Draw molecular structures of A–D.

14.3. Is the complex D polar?

14.4. Sketch the d-orbital splitting of cisplatin complex D in view of Crystal Field Theory and show the electron distribution diagram.

14.5. Determine magnetic nature of complex A.

The platinum complex binds to DNA and causes cross-linking, which triggers the programmed cell death (apoptosis). However, the other geometrical isomer of the square planar structure transplatin, _trans-diamminedichloroplatinum(II) (F), is not effective for the treatment of cancer. Transplatin is synthesized starting from [Pt(NH)] to which the first and second Cl ligands are added to form transplatin (F) as represented in the scheme below.



14.6. Draw the molecular structures of E and F.

The most important classes of antitumor agents, cisplatin, carboplatin, and oxaliplatin as platinum(II) diamines are widely used in chemotherapy to treat a wide variety of cancers.



However, the therapeutic index of these agents is relatively narrow; their use is often plagued with severe toxicity and the development of resistance, which leads to disease progression. Recently, oxoplatin, iproplatin, ormaplatin and satraplatin are Pt complexes that have been used clinically (oxoplatin) or in clinical trials.

14.7. All complexes have the same geometry and oxidation number for the Pt central atom.

Write the oxidation state of Pt and geometry of the complexes.

14.8. Which Pt complex, cisplatin or satraplatin, is kinetically more inert for substitution reactions?

14.9 Oxaplatin is an isomer of [Pt(NH)Cl(OH)] complex. Draw all stereoisomers and indicate the chiral one(s).

Platinum complexes (oxoplatin, iproplatin, ormaplatin, and satraplatin) can be considered prodrugs that are primarily intracellularly activated by biological reducing agents such as thiols, ascorbic acid, and glutathione (GSH) to kill cancer cells.

In a study, for example, the reduction of cis,trans,cis-[PtCl(OCOCH)(NH)] (G, prodrug), which has a similar structure to satraplatin, by aqueous extract of cancer cells (A2780, A2780cisR, and HT-29) yields cisplatin (D, drug) and free acetate ion as given below.

14.10. Draw the molecular structure of G.

14.11. Sketch the d-orbital splitting of the metal ion in G and write the electronic configuration.

14.12. Decide whether G is paramagnetic or diamagnetic.

14.13. The complex G crystallizes into a monoclinic crystal system of parameters: the lengths of the unit cell: a = 14.9973, b = 8.57220, c = 11.1352 Å, the β angle in the unit cell = 126.7690°, the number of the molecules in the unit cell (Z) = 4, M = 436.16 g/mol (the complex has one water molecule in the crystal structure).

Calculate the density (ρ) of the complex.

Hint: the volume of a monoclinic crystal unit cell is

Problem 15.

The Salt Lake basin in Turkey is of great importance for the conservation of biological diversity and is classed as a wetland according to international criteria. It is also one of Turkey’s richest lakes for the presence of birds. There are 85 bird species, 129 insect species (4 of which are endemic), 15 mammal species, and 38 endemic plant species. Some 40% of Turkey’s salt needs (as table salt) are supplied from this lake. Salt in the Salt Lake is formed by meteorological waters draining underground and melting the previously formed salt domes and carrying them along the tectonic lines. Salt production in the Salt Lake is done by evaporation of lake water under the sun. A pooling system is used in the salt production with solar energy.

Salt Lake
Table salt is one of the most common household chemicals. It is 97% to 99% sodium chloride, which is an ionic compound with the chemical formula NaCl, representing a 1:1 ratio of sodium and chloride ions. NaCl is the compound most responsible for the salinity of seawater and of the extracellular fluid of many multicellular organisms. In its edible form of table salt, it is commonly used as a condiment and food preservative. A second major application of sodium chloride is de-icing of roadways in subfreezing weather. Large quantities of sodium chloride are also used in many industrial processes such as the chloro-alkaline industry and soda-ash industry as well as in miscellaneous industrial uses: water softening, medicine, agriculture, firefighting, and cleanser. NaCl is used, directly or indirectly, in the production of many sodium compounds, which consume most of the world’s production. The scheme below shows the preparation of some sodium compounds starting from NaCl.

Preparation of some sodium compounds starting from NaCl.

15.1. Write the formulas of products A–G.

Sodium carbonate (NaCOsoda ash) is used primarily in the manufacture of glass, which is produced mostly from natural sources, such as the mineral trona, NaCONaHCOnHO. It can be also manufactured mostly from NaCl, CaCO, and NH using a process introduced by the Belgian chemist Ernest Solvay in 1863. The key step involves the reaction of NH(g) and CO(g) in saturated NaCl(aq). Of the possible ionic compounds that could precipitate from such a mixture (NaCl, NHCl, NaHCO, and NHHCO), the least soluble is sodium hydrogen carbonate (sodium bicarbonate, NaHCO). It is isolated from solution by filtration and then converted to sodium carbonate (NaCO) by heating. According to this explanation,

15.2. Balance the reactions given below.

15.3. Using CaCO (limestone), how can you produce the CO gas you need to prepare NaHCO?

15.4. Write the Lewis structure of CaCO with all resonances and show formal charges for each atom in the structure.

15.5. Describe the molecular geometry and propose a plausible hybridization scheme for the central atom in the ion CO.

15.6. Compare the bond lengths of CO, CO, and CO in increasing order.

NaCl crystallizes in a face-centered cubic (fcc) structure. The density of NaCl is 2180 kg/m and the ionic radius of Na is 99 pm.

15.7. How many atoms are there in the unit cell? Which atoms occupy octahedral holes?

15.8. Calculate the length of the unit cell of NaCl and the ionic radius of the Cl ion (as pm).

15.9. The alkali metals react rapidly with oxygen to produce several different ionic oxides. Under appropriate conditions, generally by carefully controlling the supply of oxygen, the oxide MO can be prepared for each of the alkali metals. Lithium reacts with excess oxygen to give ……(A)…. and a small amount of ……(B)……. Sodium reacts with excess oxygen to give mostly ….(C)…… and a small amount of ……(D)……..Potassium, rubidium, and cesium react with excess oxygen to form …….(E)….., ……(F)….., and ……(G)…..

Fill in the blanks above (for A–G) with convenient formulas of metal oxides.

15.10. Draw the Lewis structures of oxide, peroxide, and superoxide ions.

15.11. Draw the molecule orbital energy level diagram of peroxide and superoxide ions and compare their bond lengths and energies.

When LiClO, NaClO, and KClO crystallize from an aqueous solution that may or may not contain water molecules called water of crystallization as part of the solid structures, although no simple rule exists for predicting with certainty whether the ions will retain all or part of their hydration spheres in the solid state, cations with high charge densities tend to retain all or part of their hydration spheres in the solid state. When the cations have low charge densities, the cations tend to lose their hydration spheres; thus, they tend to form anhydrous salts. The ionic radius of Li, Na, and K is 76 pm, 102 pm, and 138 pm, respectively.
~~ ~~
15.12. Calculate the charge densities of the ions in C mm.

15.13. Which perchlorate salt is most susceptible to form an anhydrous compound?

Problem 16.

Turkey is one of the 7 countries in the world in terms of thermal source richness with almost 1300 thermal springs throughout Anatolia. There are thermal hotels in many cities such as Ankara, Bursa, Balıkesir, Yalova, Erzurum, Sivas and Afyonkarahisar. Afyonkarahisar, located in the Aegean region, is the most famous city in Turkey for its thermal springs. The thermal waters of Afyonkarahisar contain over 42 different minerals and many trace elements. The most concentrated ones are sulfur, calcium, chloride, sodium, and carbonates. Among these minerals, sulfur is important as “nature’s beauty mineral” because the human body needs it to manufacture collagen, which keeps human skin elastic, beautiful, and young looking. Moreover, sulfur is used to minimize the symptoms of many skin diseases including dermatitis, eczema, dandruff, and warts. People with arthritis may obtain pain relief from taking a soothing bath in thermal sulfur springs. Mineral water containing sulfur compounds is also shown to decrease cholesterol and blood pressure. Therefore, sulfur chemistry is an important topic. In this question, you will explore sulfur chemistry by studying its different reactions and compounds.

Hot spring

Sulfur is extracted as the element from underground deposits. It has many allotropes and its allotropy is complicated, but the most common sulfur allotrope is the puckered rings of S (orthorhombic sulfur, -form).

16.1. Sketch the molecular structure of S and indicate whether the molecule has a horizontal mirror plane or not.
Upon the burning of S with oxygen, compound A is produced. Further catalytic oxidation of compound A yields compound B. The reaction of A and B with water (hydrolysis) yields C and D. Compound D is an oxoacid and a central substance of the chemical industry worldwide.

16.2. Write the formulas of compounds A–D.

16.3. Draw molecular shape of the compounds by giving the name of geometries.

16.4. Write the oxidation state of the sulfur atoms in C and D.

16.5. Give balanced chemical equations for the synthesis of A–D.

Compound A can also be obtained by heating alkaline or alkaline earth sulfide minerals like CaS in an excess of air.

16.6. Write the balanced chemical equation for the synthesis of A from CaS.

Upon the reaction of D and B, compound E which is a dense and corrosive liquid that is used as a basic chemical for sulfonation processes is produced.

16.7. Give a balanced chemical equation for the synthesis of E from D.

16.8. Write molecular formula and draw the molecular shape of E.

16.9. Determine the oxidation state of the sulfur atoms in E.

The reaction of S with a stoichiometric amount of chlorine gas yields compound F and the further reaction of F with excess chlorine gas results in the formation of G, which is used as a precursor for the synthesis of sulfur dyes and synthetic rubber. The reaction of G with B yields the compounds H and A. H is a toxic compound used as the chlorinating agent in organic synthesis.

16.10. Write molecular formulas and draw the molecular shapes of F, G, and H.

16.11. Give balanced chemical equations for the synthesis of compounds F, G, and H.

One of the most common naturally occurring sulfur minerals is pyrite (FeS: iron(II) disulfide), called fool’s gold because it is a brass-yellow mineral and thus most people suppose that it is gold ore. The treatment of pyrite with hydrochloric acid results in the formation of a colorless, flammable, water-soluble gas with a “rotten egg” odor, compound I. Compound I is dissolved in thermal waters for spa applications since it is reported that the therapeutic effects of thermal water are directly correlated to its sulfur concentration. Compound I is slightly heavier than air and can be detected by lead(II) acetate paper strip test in which a reaction occurs between lead(II) acetate and I, producing compound J. Moreover, upon the oxidation of I, compound A can be yielded.

16.12. Write the molecular formulas of I and J.

16.13. Draw the molecular shape of I and write the name of the shape.

16.14. Give balanced chemical equations for the synthesis of I and J.

The sulfur oxoacids are chemical compounds that contain sulfur, oxygen, and hydrogen atoms. Sulfur has several oxoacids; one of them is thiosulfuric acid, with the molecular formula HSO, which can be synthesized by the reaction of sulfite with I. On the other hand, the controlled oxidation of sulfite by MnO in acidic solution yields another sulfur oxoacid, called dithionic acid, HSO.

16.15. Give balanced chemical equations for the synthesis of HSO and HSO.

16.16. Draw the molecular shapes of HSO and HSO.

On the other hand, the thiosulfate ion (SO) is a very good complexing agent for Ag and thus it is used in photography for removing unchanged AgBr from exposed photographic film. Upon the reaction of sodium thiosulfate ion with AgBr, sodium salt of a coordination compound with coordination number 2 is yielded.

16.17. Give a balanced chemical equation for the reaction of AgBr with NaSO.

16.18. Draw the molecular structure of the yielded coordination complex considering its geometry.

16.19. Write the electron configuration of the silver ion in the coordination compound.

The determination of HS content in thermal waters is important for spa applications. An iodometric titration method can be utilized for this purpose. In a typical experiment, 500 mL of sample is collected from a thermal water source and purged with N(g) to ensure the transfer of all HS gas into 50 mL of 0.010 M NaOH solution in a closed system. After adjusting pH of the solution to 6.0, 1.25 mL of 0.0030 M I solution and 1.0 g of KI are added to this solution and the resultant solution is stored in the dark for 15 minutes after sealing it with parafilm. After adding 1.0 mL of 20 mg/mL starch solution, the resultant solution is titrated with 0.0500 M NaSO until the end-point and consumed NaSO volume is recorded as 95.62 mL.

16.20. Write all balanced equations of this experiment.

16.21. Calculate HS concentration in the thermal water source in ppm by assuming that there is no interfering species in the water source and all HS content of thermal water is swept into the NaOH solution.

Problem 17.

Flavonoids are a group of natural products with some phenolic groups that are present in many fruits and vegetables. Flavonoids are commonly used in our daily life due to their antioxidant and anticarcinogenic properties. Rutin is a flavonoid class substance composed of the flavonol quercetin and disaccharide rutinose.


Chemical structure of rutin.

It is of very low toxicity for human health and it is known that rutin can supply electrons to reactive free radicals to produce more stable and healthy structures. Rutin is also known as vitamin P, in which P is due to its permeability. Rutin is an electrochemically active material and many researchers have extensively studied its electrochemical behavior using different electrochemical techniques.

Cyclic voltammetry is a useful technique for electrochemical measurement of an analyte, which is dissolved in a useful electrolyte solution. There are three electrodes in an electrochemical cell solution; working, counter, and reference electrodes. The potential of working electrode is scanned versus reference electrode because reference electrode has a constant potential value. Reverse electrochemical reactions of the working electrode occur at the counter electrode. Therefore, current flows between working and counter electrodes. Reference electrode is used to adjust the potential of the working electrode at a known value. This technique is applied based on potentiodynamic application. Potential of the working electrode is scanned versus reference electrode between two potential values depending on time. Cyclic voltammetry application results in a graphic (voltammogram) of current versus scanned potential. There are two important parameters to evaluate a voltammogram; peak potential and peak current. The peak potential and peak current are calculated using x-axis and y-axis of a voltammogram at the peak maximum, respectively.

The cyclic voltammetry (CV) behavior of rutin at 25 °C has been tested using a glassy carbon electrode, a saturated calomel electrode (SCE), and a Pt wire as working, reference, and counter electrodes, respectively. In this study, CV data for 1.0 × 10 mol/dm rutin solutions at different pH values have been obtained by scanning the potential between 0.00 and 0.80 V at a scan rate of 100 mV/s. Anodic peak potential (Ep), cathodic peak potential (Ep), anodic peak current (Ip), and cathodic peak current (Ip) values supplied from related CVs depending on the pH are presented in the following Table.

Table. Some CV parameters depending on the pH of a solution containing 1.0×10 mol/dm rutin.

17.1. In a three-electrode system, electrochemical oxidation or reduction of an analyte in the electrochemical cell occurs on the __ because its potential is adjusted against the _.

Which of the following words fit into the blanks in the above sentence?

a)
working electrode / reference electrode
b)
counter electrode / working electrode
c)
reference electrode / working electrode
d)
working electrode / counter electrode

17.2. Both anodic and cathodic peak potentials shift to negative potential values by increasing the pH because the electrochemical reaction of rutin includes __.

Which of the following words fits into the blank in the above sentence?

a)
Na
b)
K
c)
H
d)
I

17.3. Electrochemical oxidation of rutin is _ because of the fact that Ip/Ip is about 1 and ΔE_p is almost 0.0592/_n V.

Which of the following words fits into the blank in the above sentence?
a)
irreversible
b)
reversible
c)
quasi-reversible
d)
catalyzed

17.4. How long does it take to obtain each CV value?

17.5. Calculate the number of transferred electrons for the electrochemical reaction of rutin including 2 H.

17.6. Propose an electrochemical redox mechanism for rutin.

17.7. The SCE reaction is and the SCE contains saturated KCl solution prepared by dissolving 342 g of KCl in 1.0 L of aqueous solution. How does the potential of the SCE change (decrease or increase) in the case of 1.0 M KCl?

In order to determine the amount of rutin in a vitamin P tablet, the following procedures have been used:

i) A 500 mg vitamin P tablet is dissolved in deionized water, pH is adjusted to 2.0, and total volume is completed to 500 mL in a volumetric flask. A 10 mL part of this solution is placed in a three-electrode cell. CV is obtained with an anodic peak current (Ip) of 2.26 A.

ii) A solution in the absence of rutin has been prepared at pH 2.0. After placing all electrodes into this solution, CV has been recorded three times by cleaning the electrode with deionized water for each measurement. Then Ip values have been read as 0.16, 0.11, and 0.18 μA, respectively.

iii) The standard rutin solutions of 1.0, 5.0, 10.0, 20.0, 30.0, and 50.0 mM have been prepared and Ipvalues of these solutions have been obtained from related CVs as demonstrated in the following Table.

Table. Ip values for various rutin standard solutions.

Note that all of the CVs have been obtained by using same working electrode beyond this experiment.

17.8. Draw a calibration curve for the rutin determination method.

17.9. Write a mathematical equation for the calibration curve.

17.10. Calculate the rutin amount in the vitamin P tablet as wt %.

17.11. Calculate the calibration sensitivity and limit of detection (LOD) of this method for a signal to noise ratio (S/N) of 3.0.

Note: Limit of detection:

Problem 18.

The particle in an one-dimensional box model is a crude approximation for conjugated molecules. In this model, π electrons are assumed to move freely over the carbon framework of conjugated bonds. Therefore, this model is also called the free electron model (FEM). The length of the box may be approximated via

, where L is the box length and

is the number of carbons. Furthermore, the Pauli principle is applied when electrons are filled to the energy levels. The energy of a particle in an one-dimensional box can be written as follows:

where m is the mass of the particle, h is the Planck constant, and n is a positive integer.

For the 1,3,5,7-octatetraene molecule assuming FME:

18.1. Draw an energy diagram, fill the electrons, and calculate orbital energies.

18.2. Calculate the total π energy of the molecule.

18.3. Determine the wave length of the light (in nm) that require to excite an electron from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO).

For two-dimensional conjugated systems, we may use the particle in a two-dimensional box model. In this case the total energy can be written as follows:

where _L_1 and _L_2 are the lengths and

and

are the quantum numbers of the first and second dimensions, respectively.

Graphene is a sheet of carbon atoms in the form of a two-dimensional hexagonal lattice in which one atom forms each vertex.

For a square shaped graphene sheet with L_1 = _L_2 = 11 Å:

18.4. The distance between two adjacent carbons in the hexagonal 6-carbon unit is approximately 1.4 Å. Calculate the number of electrons in a (×) sheet of graphene. For this problem you may ignore edge electrons. (Area of a regular hexagon with a side of _L is

).

18.5. Calculate the energy of the HOMO.

18.6. Calculate the energy of the LUMO.

18.7. The difference between energies of the LUMO and HOMO is called the band gap (E). Calculate the band gap.

The models for a particle in a one- and two-dimensional box can be extended to a three-dimensional rectangular box of dimensions

,

, and

, yielding the following expression for the allowed energy levels:

where _n_1, _n_2, and _n_3 are the quantum numbers of the first, second, and third dimensions, respectively. For a particle in a cubic box of length

:

18.8. Give the expressions for the five different lowest energies.

18.9. Draw a diagram showing all the five energy levels. Indicate degeneracy of each level.

Problem 19.

Vibration of a diatomic molecule is reminiscent of two masses on a spring with a potential energy that is a function of the displacement from equilibrium. Hence, the harmonic oscillator model is utilized to compute vibrational frequencies. These frequencies are called harmonic vibrational frequencies. The energy of a harmonic oscillator can be written as follows:

where ν is the harmonic vibrational frequency, h is the Planck constant, and n is a nonnegative integer. The harmonic vibrational frequency can be calculated as follows:

where k is the force constant and μ is the reduced mass:

where m_1 and _m_2 are the masses of the first and the second atoms, respectively.

For the CO molecule the value of the force constant is 1902.4 N/m. For this problem, the atomic masses of isotopes can be approximated by their mass numbers.

19.1. Calculate the harmonic vibrational frequency of the CO molecule in Hz.

19.2. Express the harmonic vibrational frequency of the CO molecule in cm.

19.3. Calculate the zero-point vibrational energy (ZPVE) of the CO molecule in kcal/mol.

19.4. Calculate the harmonic vibrational frequency of the CO molecule in cm.

19.5. Calculate the harmonic vibrational frequency of the CO molecule in cm.

The harmonic oscillator model can readily be extended to polyatomic molecules. In this case, the total vibrational energy of a molecule with _n vibrational frequencies can be written as follows:

where νi are the harmonic vibrational frequencies, h is the Planck constant, and ni are nonnegative integers.

For the water molecule the harmonic vibrational frequencies are 1649, 3832, and 3943 cm. Using the harmonic oscillator model, for the water molecule:

19.6. Calculate the ZPVE value (in J and cm units).

19.7. Calculate the first 5 energy levels (in cm).

To describe the rotational motion of a diatomic molecule, the rigid rotor model is used. In this model the bond length (R) of the diatomic molecule is kept constant during the rotational motion. Using the rigid rotor model, the rotational energy of a diatomic molecule can be written as follows:

where I is the moment of inertia and l is a nonnegative integer. The moment of inertia can be written as follows:

where μ is the reduced mass and R is the bond length of the diatomic molecule.

In the microwave spectrum of the CO molecule the value of frequency for the lowest energy transition is 115.270 GHz.

19.8. Calculate the bond length of the CO molecule in Å.

19.9. For the CO molecule predict the frequency of the next two absorptions (selection rule is

.

19.10. For the CO molecule, calculate the frequency of the lowest energy absorption.

Problem 20.

In the future, humankind will most likely consume all resources that are necessary for life on earth and will have to relocate to an earth-like planet. Assume that you have started to live on a new planet where standard pressure condition is 2 bar, standard concentration is 1 mol dm, and all types of gases behave as an ideal gas. On this planet, you are asked to determine equilibrium conditions for the reaction below:

20.1. Calculate the change in standard enthalpy of the reaction at 298 K by using the following information:

| X4Y8(s) = 4X(g) + 4Y2(g) |

| | —- | —- | |

|

| |

|

| |

|

|

20.2. Calculate ΔG° of the reaction at 298 K.
20.3. Calculate K° of the reaction at 298 K.
20.4. Assume that ΔH° of the reaction does not depend on temperature. Find K of the reaction at 50 °C.
20.5. Calculate the percent degree of dissociation for XY at 298 K where total pressure is 0.2 bar.
20.6. In order to increase the amounts of products, which one do you choose to increase (if you choose both, put a cross next to both of them):

Moreover, in this future, the Earth will have a very unstable climate. The surface temperature could increase or decrease all of a sudden. Suppose that you travelled through time to the era in which the Earth’s climate is extremely unstable. Your task in this era is to observe the thermodynamics of phase transitions of water, the most precious substance where all life has originated. Suppose that the temperature suddenly decreased to −20 °C.

One mole of water becomes supercooled liquid water at −20 °C and 1 bar pressure and then turns into ice at the same temperature (note that the temperature of the surroundings is constant at −20 °C).

By using the following data for water:

The heat of fusion (ΔH°) of ice at 0 °C and 1 bar is 6020 J mol

During the conversion of supercooled liquid water to ice at −20 °C:

20.7. Calculate the total entropy change in the system.
20.8. Calculate the total entropy change in the surroundings.
20.9. Calculate the total entropy change in the universe.

and