IChO 2001 · Theory problem 6

Organic Chemistry of Indian Spices

Download PDFIChO 2001
Figure 1

The rhizomes of ginger (Zingiber officinale) are well known for their medicinal and flavouring properties. In Ayurveda (the traditional system of medicine in India) different formulations of ginger are used for the treatment of gastrointestinal problems, common cold and other ailments. Several compounds are responsible for the pungency of ginger. Many are simple substituted aromatic compounds with different side chains. Three of them, Zingerone, (+)[6] Gingerol (to be referred hereafter as Gingerol only), and Shogaol are particularly important.

Zingerone: \(\ce{C11H14O3}\)
Gingerol: \(\ce{C17H14O4}\)
Shogaol: \(\ce{C17H24O3}\)

6.1 Zingerone gives positive \(\ce{FeCl3}\) and 2,4-DNP (2,4-dinitrophenylhydrazine) tests. It does not react with Tollen's reagent. Therefore, Zingerone contains the following functional groups: (Mark X in the correct boxes.)

(a) alcoholic hydroxyl \(\square\) (e) ester \(\square\)
(b) aldehydic carbonyl \(\square\) (f) alkoxyl \(\square\)
(c) ketonic carbonyl \(\square\) (g) unsaturation \(\square\)
(d) phenolic hydroxyl \(\square\)

The data obtained from the \(\ce{^1H}\) NMR spectrum of Zingerone are shown in Table 1. Some other relevant information is given in Table 2.

Table 1: \(\ce{^1H}\) NMR spectral data on Zingerone

Chemical shifts \((\delta)\) Multiplicity Relative intensity
2.04 singlet 3
2.69, 2.71 two (closely spaced) triplets of equal intensity 4
3.81 singlet 3
5.90 broad singlet (\(\ce{D2O}\) exchangeable) 1
6.4 – 6.8 two doublets with similar chemical shifts and one singlet 3

*) For clarity, some of the data have been altered slightly.)

Table 2: Approximate \(\ce{^1H}\) chemical shifts \((\delta)\) and spin-spin coupling constants \((J)\) of some protons

\(\ce{^1H}\) Chemical shifts

alkyl - H 0.9 - 1.5 [S1] 3.4 - 3.6
[S2] 2.0 - 2.7 [S3] 4.0 - 7.0
[S4] 2.3 - 2.9 [S5] 6.0 - 8.0
[S6] 3.3 - 3.9 [S7] 4.0 - 11.0

[S1]

Figure 2

[S2]

Figure 3

[S3]

Figure 4

[S4]

Figure 5

[S5]

Figure 6

[S6]

Figure 7

[S7]

Figure 8

Spin–spin coupling constants \((J)\)

Alkenes cis 5 – 14 Hz (commonly around 6 – 8 Hz)
trans 11 – 19 Hz (commonly around 14 – 16 Hz)

Zingerone on bromination with bromine water gives only one nuclear mono brominated product. The IR spectrum of Zingerone indicates the presence of a week intramolecular hydrogen bond. The same is present even after Clemmensen reduction (\(\ce{Zn}\) – \(\ce{Hg}/\ce{HCl}\)) of Zingerone.

6.2 From the information above deduce the following:

i) side chain in Zingerone

ii) substituent on the aromatic ring

iii) relative positions of the substituents on the ring

6.3 Draw a possible structure of Zingerone based on the above inferences.

6.4 Complete the following reaction sequence for the synthesis of Zingerone.

Figure 9

6.5 Zingerone can be easily converted into Gingerol by the following reaction sequence:

Figure 10

Notes: (1) \(\ce{Me3SiCl}/\ce{(Me3Si)2NH}\) is used to convert OH into \(\ce{-OSiMe3}\); the group \(\ce{-SiMe3}\) can be removed by acid hydrolysis.

(2) LDA is lithium disopropylamine, a strong, very hindered, non.nucleophilic base.

i) Draw the structure of D.

ii) Draw the structure of Gingerol.

iii) Complete the Fischer projection of the \(R\)-enantiomer of Gingerol.

iv) In the above reaction sequence (6.5), about 2 – 3 % another constitutional isomer (E) of Gingerol is obtained. Draw the likely structure of E.

v) Will the compound E be formed as

(a) a pair of enatiomers? \(\square\)

(b) a mixture of diastereomers? \(\square\)

(c) a mixture of an enantiomeric pair and a meso isomer? \(\square\)

(Mark X in the correct box.)

vi) Gingerol (\(\ce{C17H26O4}\)) when heated with a mild acid (such as \(\ce{KHSO4}\)) gives Shogaol (\(\ce{C17H24O3}\)). Draw the structure of Shogaol.

6.6 Turmeric (Curcuma longa) is a commonly used spice in Indian food. It is also used in Ayurvedic medicinal formulations. Curmunin (\(\ce{C21H20O6}\)), an active ingredient of turmeric, is structurally related to Gingerol. It exhibits keto-enol tautomerism. Curcumin is responsible for the yellow colour of turmeric and probably also for the pungent taste.

The \(\ce{^1H}\) NMR spectrum of the keto form of Curcumin shows aromatic signal similar to that of Gingerol. It also shows a singlet around \(\delta\) 3.5 (2H) and two doublets (2H each) in the region \(\delta\) 6 – 7 with \(J=16\ \mathrm{Hz}\). It can be synthesized by condensing TWO moles of A (refer to 6.4) with one mole of pentan-2,4-dione.

i) Draw the stereochemical structure of Curcumin.

ii) Draw the structure of the enol form of Curcumin.

iii) Curcumin is yellow in colour because it has

(a) a phenyl ring \(\square\)

(b) a carbonyl group \(\square\)

(c) an extended conjugation \(\square\)

(d) a hydroxyl group \(\square\)

(Mark X in the correct box.)

Official English text of the problem.

Answers

Hereafter, \(\ce{Ar}\) denotes the 4-hydroxy-3-methoxyphenyl group, fragment SMILES [*]c1cc(OC)c(O)cc1. The wildcard marks the bond to the side chain.

  • 6.1: Mark (c) ketonic carbonyl and (d) phenolic hydroxyl.

  • 6.2(i): A 3-oxobutyl side chain, \(\ce{-CH2CH2COCH3}\); fragment SMILES [*]CCC(=O)C.

  • 6.2(ii): A methoxy group, fragment SMILES [*]OC, in addition to the established phenolic hydroxyl, fragment SMILES [*]O.

  • 6.2(iii): Taking the side-chain attachment as ring C1, put methoxy at C3 and hydroxyl at C4. Thus hydroxyl and methoxy are ortho, hydroxyl and side chain para, and methoxy and side chain meta.

  • 6.3: Zingerone, 4-(4-hydroxy-3-methoxyphenyl)butan-2-one: \[ \ce{ArCH2CH2COCH3}. \] SMILES: CC(=O)CCc1cc(OC)c(O)cc1.

  • 6.4 — A: Vanillin, 4-hydroxy-3-methoxybenzaldehyde.
    SMILES: O=Cc1cc(OC)c(O)cc1.

  • 6.4 — B: Acetone, propan-2-one.
    SMILES: CC(=O)C.

  • 6.4 — C: Dehydrozingerone, shown as the usual \(E\) condensation product, (3E)-4-(4-hydroxy-3-methoxyphenyl)but-3-en-2-one.
    SMILES: CC(=O)/C=C/c1cc(OC)c(O)cc1.

  • 6.5(i): D is the methyl-side lithium enolate of O-trimethylsilylzingerone. The phenolic oxygen is protected, and the terminal methyl group has become the terminal enolate methylene.
    SMILES: C=C([O-])CCc1cc(OC)c(O[Si](C)(C)C)cc1.[Li+].

  • 6.5(ii): [6]-Gingerol, 5-hydroxy-1-(4-hydroxy-3-methoxyphenyl)decan-3-one: \[ \ce{ArCH2CH2COCH2CH(OH)(CH2)4CH3}. \] SMILES, configuration unspecified: CCCCCC(O)CC(=O)CCc1cc(OC)c(O)cc1.
    Its correct formula is \(\ce{C17H26O4}\). The stated achiral synthesis gives a racemate.

  • 6.5(iii): (5R)-[6]-Gingerol.
    SMILES: CCCCC[C@@H](O)CC(=O)CCc1cc(OC)c(O)cc1.
    In a Fischer projection with \(\ce{CH2COCH2CH2Ar}\) above and \(\ce{(CH2)4CH3}\) below the stereocentre, place H on the left and OH on the right.

  • 6.5(iv): E, 3-(4-hydroxy-3-methoxybenzyl)-4-hydroxynonan-2-one: \[ \ce{ArCH2CH(COCH3)CH(OH)(CH2)4CH3}. \] SMILES, configurations unspecified: CC(=O)C(Cc1cc(OC)c(O)cc1)C(O)CCCCC.

  • 6.5(v): Mark (b), a mixture of diastereomers. There are two enantiomeric pairs and no meso form.

  • 6.5(vi): [6]-Shogaol, 1-(4-hydroxy-3-methoxyphenyl)dec-4-en-3-one: \[ \ce{ArCH2CH2COCH=CH(CH2)4CH3}. \] The expected major \(4E\) form has SMILES CCCCC/C=C/C(=O)CCc1cc(OC)c(O)cc1.

  • 6.6(i): Diketo curcumin, (1E,6E)-1,7-bis(4-hydroxy-3-methoxyphenyl)hepta-1,6-diene-3,5-dione: \[ \ce{ArCH=CHCOCH2COCH=CHAr}, \] with both outer double bonds \(E\).
    SMILES: COc1cc(/C=C/C(=O)CC(=O)/C=C/c2cc(OC)c(O)cc2)ccc1O.

  • 6.6(ii): The hydrogen-bonded enol tautomer of curcumin, (1E,4Z,6E)-5-hydroxy-1,7-bis(4-hydroxy-3-methoxyphenyl)hepta-1,4,6-trien-3-one: \[ \ce{ArCH=CHCOCH=C(OH)CH=CHAr}. \] SMILES: COc1cc(/C=C/C(=O)/C=C(O)/C=C/c2cc(OC)c(O)cc2)ccc1O.
    Orient the enolic OH toward the remaining carbonyl oxygen to form an intramolecular hydrogen bond.

  • 6.6(iii): Mark (c), an extended conjugation.

Solution

6.1 — Functional-group tests

Iron(III) chloride, [Fe+3].[Cl-].[Cl-].[Cl-], gives the characteristic positive test for a phenolic hydroxyl group.

2,4-Dinitrophenylhydrazine, NNc1ccc([N+](=O)[O-])cc1[N+](=O)[O-], detects an aldehyde or ketone carbonyl. The negative Tollens test excludes the aldehyde alternative in the intended interpretation. The carbonyl is therefore ketonic.

The active Tollens ion is diamminesilver(I), [NH3]->[Ag+]<-[NH3]; its coordination geometry is linear.

Consequently, the boxes justified by these tests are (c) and (d). The methoxy group is established by the subsequent NMR evidence.

6.2 — Side chain, substituents, and their relative positions

For zingerone, \(\ce{C11H14O3}\), the degree of unsaturation is

\[ \mathrm{DBE}=\frac{2(11)+2-14}{2}=5. \]

A benzene ring accounts for four degrees of unsaturation, and the ketone accounts for the fifth. Thus the side chain requires no additional carbon–carbon double bond or ring.

The NMR assignments are:

\(\delta\), ppm Integral Assignment and deduction
2.04 3 A methyl directly attached to a carbonyl, \(\ce{-COCH3}\). It is a singlet because there is no proton on the adjacent carbonyl carbon.
2.69, 2.71 4 Two adjacent methylene groups, \(\ce{-CH2CH2-}\), each appearing approximately as a triplet. One is benzylic and the other is adjacent to the ketone.
3.81 3 A methoxy group, \(\ce{-OCH3}\).
5.90 1 The exchangeable phenolic OH proton.
6.4–6.8 3 Three aromatic protons: two adjacent protons and one effectively isolated proton in the simplified spectrum.

The integrated proton count is

\[ 3+4+3+1+3=14, \]

matching the molecular formula.

6.2(i). The acetyl methyl and the two coupled methylene groups require

\[ \boxed{\ce{Ar-CH2-CH2-C(=O)-CH3}}. \]

The two closely spaced methylene resonances cannot individually be assigned to the benzylic and carbonyl-adjacent positions from the supplied shifts alone.

6.2(ii). The \(3\mathrm H\) singlet at \(\delta 3.81\) identifies methoxy. Together with the phenolic OH and the side chain, this gives three substituents on the aromatic ring.

6.2(iii). The intramolecular hydrogen bond survives removal of the ketone oxygen by Clemmensen reduction. Its acceptor must therefore be the methoxy oxygen. The hydroxyl and methoxy groups must be adjacent, allowing the phenolic OH to point toward the ether oxygen.

For completeness, the reduced compound is 4-butyl-2-methoxyphenol, CCCCc1cc(OC)c(O)cc1. A balanced representation of the Clemmensen reduction is

\[ \ce{C11H14O3 + 2Zn + 4HCl ->[Hg] C11H16O2 + 2ZnCl2 + H2O}. \]

The aromatic NMR pattern and the bromination result complete the assignment. The phenolic OH strongly directs bromination to its ortho and para positions. To leave only one such position available:

  • one ortho position is occupied by methoxy;
  • the para position is occupied by the side chain;
  • the other ortho position remains available for bromination.

With the side chain numbered C1, methoxy is therefore at C3, hydroxyl at C4, and bromination occurs at C5.

The product, 4-(3-bromo-4-hydroxy-5-methoxyphenyl)butan-2-one, has SMILES CC(=O)CCc1cc(OC)c(O)c(Br)c1. Its formation is balanced as

\[ \ce{C11H14O3 + Br2 -> C11H13BrO3 + HBr}. \]

In the original ring numbering, H5 and H6 give the two principal aromatic doublets, while H2 gives the approximate singlet. Small longer-range couplings are omitted in the simplified data.

6.3 — Zingerone

All the evidence is satisfied by

\[ \boxed{\ce{ArCH2CH2COCH3}}, \]

where \(\ce{Ar}\) is 4-hydroxy-3-methoxyphenyl.

To draw it, attach a four-carbon chain to ring C1 in the order methylene–methylene–carbonyl carbon–methyl, place methoxy at ring C3, and place hydroxyl at ring C4. This is zingerone, with the name and SMILES given in the Answers.

6.4 — Synthesis of zingerone

The final hydrogenation suggests that C is the corresponding conjugated enone:

\[ \ce{ArCH=CHCOCH3}. \]

Disconnecting its newly formed carbon–carbon double bond gives the aldehyde vanillin, A, and the ketone acetone, B.

The balanced aldol-condensation equation is

\[ \ce{ArCHO + CH3COCH3 ->[NaOH(aq)] ArCH=CHCOCH3 + H2O}. \]

In molecular formulas,

\[ \ce{C8H8O3 + C3H6O -> C11H12O3 + H2O}. \]

Hydroxide generates a small equilibrium concentration of acetone enolate. Its carbon attacks the aldehyde carbonyl carbon, while the aldehyde \(\pi\) electrons move onto oxygen. Proton transfer gives a β-hydroxy ketone, and elimination of water produces the conjugated enone C, dehydrozingerone.

Selective hydrogenation of the carbon–carbon double bond then gives zingerone:

\[ \ce{ArCH=CHCOCH3 + H2 ->[catalyst] ArCH2CH2COCH3}, \]

or

\[ \ce{C11H12O3 + H2 -> C11H14O3}. \]

The \(E\) enone is the usual favored condensation product; either alkene geometry would give the same saturated product on hydrogenation.

6.5(i) — Formation and structure of D

The phenolic OH must first be protected so that it does not consume the strong base.

The relevant reagents and protected intermediate are:

Name SMILES
Chlorotrimethylsilane C[Si](C)(C)Cl
Hexamethyldisilazane C[Si](C)(C)N[Si](C)(C)C
O-Trimethylsilylzingerone CC(=O)CCc1cc(OC)c(O[Si](C)(C)C)cc1
Lithium diisopropylamide, LDA CC(C)[N-]C(C)C.[Li+]
Diisopropylamine CC(C)NC(C)C

A balanced net representation of protection by hexamethyldisilazane is

\[ \ce{2C11H14O3 + (Me3Si)2NH ->[Me3SiCl] 2C14H22O3Si + NH3}. \]

The bulky base preferentially removes a proton from the less hindered terminal methyl group under the intended kinetic-enolate conditions. The deprotonation is

\[ \ce{C14H22O3Si + LiN[CH(CH3)2]2 -> C14H21LiO3Si + HN[CH(CH3)2]2}. \]

Thus D is represented by the resonance forms

\[ \ce{R-C(O^-)=CH2 <-> R-C(=O)-CH2^-}, \]

with a lithium counterion, where \(\ce{R}\) is the protected aromatic–ethyl residue.

To draw D, retain the aromatic methoxy group, replace phenolic OH by \(\ce{-OSiMe3}\), and replace the terminal \(\ce{-COCH3}\) unit by \(\ce{-C(O^-)=CH2}\), accompanied by \(\ce{Li+}\).

Its SMILES in the Answers specifies the formal ion pair. Lithium coordination and aggregation are not specified by the problem.

The usual temperature for this kinetic enolization is \(-78\,^\circ\mathrm C\); the supplied scheme prints \(78\,^\circ\mathrm C\).

6.5(ii) — Formation and structure of gingerol

Hexanal, CCCCCC=O, provides an aldehyde carbon bearing a pentyl substituent.

The terminal carbon of D attacks the carbonyl carbon of hexanal. The aldehyde carbonyl becomes an alkoxide; acid workup protonates it and removes the phenolic trimethylsilyl protecting group.

The resulting carbon skeleton is

\[ \ce{ArCH2CH2COCH2CH(OH)(CH2)4CH3}. \]

The net carbon-skeleton reaction is an aldol addition:

\[ \ce{ArCH2CH2COCH3 + CH3(CH2)4CHO -> ArCH2CH2COCH2CH(OH)(CH2)4CH3}. \]

Therefore,

\[ \boxed{\ce{C11H14O3 + C6H12O -> C17H26O4}}. \]

This establishes the correct gingerol formula independently of the conflicting printed formulas.

Including the lithium counterion and hydrolysis of the silyl group, a balanced net workup equation is

\[ \ce{C14H21LiO3Si + C6H12O + H2O + H+ -> C17H26O4 + (CH3)3SiOH + Li+}. \]

The silicon-containing product is trimethylsilanol, C[Si](C)(C)O.

Only one stereocentre is formed in this major product. Because the starting materials and reagents are achiral, the synthesis gives equal amounts of the two enantiomers unless a resolution or asymmetric step is added. The sign \((+)\) in the introduction alone does not establish an \(R\) or \(S\) configuration.

6.5(iii) — Fischer projection of the R enantiomer

At gingerol C5, the CIP priorities are

\[ \ce{OH} \;>\; \ce{CH2COCH2CH2Ar} \;>\; \ce{(CH2)4CH3} \;>\; \ce{H}. \]

The two carbon chains initially tie: each begins with a methylene carbon. At the next comparison, the carbonyl carbon has the effective attached-atom list \((\ce{O},\ce{O},\ce{C})\), which outranks the alkyl-chain list \((\ce{C},\ce{H},\ce{H})\).

A suitable Fischer projection is

\[ \begin{array}{ccc} & \ce{CH2COCH2CH2Ar} &\\ & | &\\ \ce{H} & -\ce{C}- & \ce{OH}\\ & | &\\ & \ce{(CH2)4CH3} & \end{array} \]

The horizontal bonds point toward the observer; the vertical bonds point away. Here the lowest-priority group, H, points toward the observer, so the apparent sense of the \(1\to2\to3\) sequence is reversed, giving \(R\).

The corresponding stereochemical SMILES is CCCCC[C@@H](O)CC(=O)CCc1cc(OC)c(O)cc1.

6.5(iv) — Constitution of the minor product E

Zingerone also has an enolizable methylene on the other side of its carbonyl. Minor deprotonation there, followed by addition to hexanal, gives

\[ \boxed{\ce{ArCH2CH(COCH3)CH(OH)(CH2)4CH3}}. \]

The balanced net addition is

\[ \ce{ArCH2CH2COCH3 + CH3(CH2)4CHO -> ArCH2CH(COCH3)CH(OH)(CH2)4CH3}. \]

This is 3-(4-hydroxy-3-methoxybenzyl)-4-hydroxynonan-2-one, E. It has the same molecular formula, \(\ce{C17H26O4}\), as gingerol, but different connectivity.

6.5(v) — Stereoisomers of E

E has two stereogenic centres, C3 and C4 in its nonan-2-one parent chain. There is no internal symmetry that could produce a meso form.

The four configurations and their explicit SMILES are:

Stereoisomer of E SMILES
\((3R,4R)\)-E CC(=O)[C@H](Cc1cc(OC)c(O)cc1)[C@H](O)CCCCC
\((3S,4S)\)-E CC(=O)[C@@H](Cc1cc(OC)c(O)cc1)[C@@H](O)CCCCC
\((3R,4S)\)-E CC(=O)[C@H](Cc1cc(OC)c(O)cc1)[C@@H](O)CCCCC
\((3S,4R)\)-E CC(=O)[C@@H](Cc1cc(OC)c(O)cc1)[C@H](O)CCCCC

The first two form one enantiomeric pair, and the last two form another. Members belonging to different pairs are diastereomers. The relative amounts of the two pairs need not be equal.

Thus the correct answer is (b), a mixture of diastereomers.

6.5(vi) — Dehydration to shogaol

Gingerol is a β-hydroxy ketone. Acid-catalysed dehydration removes the hydroxyl at C5 and a hydrogen from C4, producing a carbon–carbon double bond conjugated with the ketone:

\[ \ce{ArCH2CH2COCH2CH(OH)(CH2)4CH3 -> ArCH2CH2COCH=CH(CH2)4CH3 + H2O}. \]

The molecular equation is

\[ \boxed{\ce{C17H26O4 -> C17H24O3 + H2O}}. \]

The product is [6]-shogaol. Draw the double bond between side-chain C4 and C5, immediately next to the C3 carbonyl.

The less crowded \(E\) enone is the expected favored product. The supplied data establish the conjugated alkene position but do not independently measure its geometry.

6.6(i) — Structure and stereochemistry of diketo curcumin

Pentan-2,4-dione, also called acetylacetone, has SMILES CC(=O)CC(C)=O.

Condensation of its two terminal methyl groups with two molecules of vanillin gives

\[ \ce{2ArCHO + CH3COCH2COCH3 -> ArCH=CHCOCH2COCH=CHAr + 2H2O}. \]

The molecular balance is

\[ \boxed{\ce{2C8H8O3 + C5H8O2 -> C21H20O6 + 2H2O}}. \]

The structure explains the NMR data:

  • The two identical substituted aromatic rings give equivalent sets of aromatic signals resembling those of gingerol.
  • The central \(\ce{-COCH2CO-}\) group gives a \(2\mathrm H\) singlet around \(\delta 3.5\), consistent with the supplied \(3.4\)–\(3.6\) range for a carbon between two carbonyl groups.
  • Molecular symmetry makes the four vinylic protons occur as two sets of two, giving two \(2\mathrm H\) doublets.
  • \(J=16\ \mathrm{Hz}\) lies in the supplied trans range and exceeds the supplied cis range. Both outer double bonds are therefore \(E\).

Its degree of unsaturation also agrees:

\[ \mathrm{DBE}=\frac{2(21)+2-20}{2}=12. \]

These comprise eight from the two benzene rings, two from the carbonyls, and two from the additional carbon–carbon double bonds.

6.6(ii) — Enol form

Enolization of one carbonyl using a hydrogen from the central methylene gives

\[ \ce{ArCH=CHCOCH2COCH=CHAr <=> ArCH=CHCOCH=C(OH)CH=CHAr}. \]

The molecular formula remains \(\ce{C21H20O6}\). Acid- or base-catalysed proton transfer changes one carbonyl into an enolic OH while forming the central carbon–carbon double bond.

Draw the enolic OH and the remaining carbonyl oxygen on the same side of the central framework. The favorable central \(Z\) arrangement permits an \(\ce{O-H...O}\) hydrogen bond closing a six-membered pseudoring. The two outer double bonds retain their \(E\) configurations.

The SMILES given in the Answers encodes \(1E,4Z,6E\). It does not itself encode the hydrogen bond or the approximately planar conformation that supports conjugation.

6.6(iii) — Origin of the yellow colour

The correct choice is (c), extended conjugation.

The aromatic rings, alkene bonds, and carbonyl/enol groups form extended conjugated systems. In the enol tautomer, conjugation extends continuously through the central chain. This lowers the energy of electronic excitation enough for absorption to reach the visible region; absorption toward the violet–blue part of the spectrum gives the observed yellow colour.

For completeness, the small species used in the equations above have the following names and SMILES:

Species SMILES
Water O
Dihydrogen [H][H]
Sodium hydroxide [Na+].[OH-]
Bromine BrBr
Hydrogen bromide Br
Zinc [Zn]
Mercury [Hg]
Hydrogen chloride Cl
Zinc chloride [Zn+2].[Cl-].[Cl-]
Ammonia N
Hydrogen ion [H+]
Lithium ion [Li+]
Potassium hydrogen sulfate [K+].[O-]S(=O)(=O)O

The solution was checked against the official solution.