1 Compound A has the molecular formula C8H10. When A is treated with chlorine in the presence of ultraviolet light, compound B having the molecular formula C8H9Cl is formed.

A also reacts with chlorine under a different condition to produce compound E which is isomeric with B. When B is heated with aqueous NaOH, compound C with molecular formula C8H10O is formed. However, when E is subjected to heating with aqueous NaOH, no new compound is formed.

When C is heated with acidified potassium manganate (VII), compound D is formed, which gives an orange precipitate when heated with 2,4-dinitrophenylhydrazine. When E is heated with acidified potassium manganate (VII), effervescence is observed and compound F with molecular formula C7H5O2Cl is formed.

Deduce the structural formulae of compounds A to F, explaining your reasoning.

Since A has C:H ratio close to 1, there is likely to be a benzene ring. A undergoes free radical substitution with Cl2 in uv light to form B (monosubstitution). A reacts with Cl2 via electrophilic substitution to form E.

B undergoes nucleophilic substitution with NaOH to form C. However, since Cl atom is directly bonded to C atom on benzene ring, the lone pair on Cl is delocalised into the benzene ring forming partial double bond character which strengthens the C-Cl bond, making it not susceptible to nucleophilic attack by OH -, hence no new compound is formed. (Note: Cl atom can be at C2 or C4 with respect to –CH 2CH3

group since ethyl group is 2,4 directing)

When C undergoes oxidation by acidified KMnO4 to form D, a carbonyl compound as indicated with positive condensation reaction with 2,4-DNPH, it indicates that C is a secondary alcohol which is oxidised to a ketone. NOT primary alcohol since an aldehyde will be oxidised to carboxylic acid directly under strong oxidation without

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Chemistry H2 9746Tutor TuteeRevision Exercise 21: Structural Elucidation

distillation.

When E is heated with acidified KMnO4, effervescence observed indicates a loss of CO2 (oxidative cleavage) to form chlorobenzoic acid.

2 A certain organic solvent was distilled to produce a single compound A. When A was reacted with 2,4-dinitrophenylhydrazine, an orange precipitate was produced. With aqueous alkaline iodine, A gave a pale yellow precipitate. A did not react with either hot acidified potassium dichromate (VI) nor with aqueous bromine. Reduction of A with LiAlH4

in dry ether yielded an equimolar mixture of two isomers, B and C, with the molecular formula C8H16O.

Suggest the structural formulae of A, B and C, explaining the reactions involved.

A is a carbonyl compound (either aldehyde or ketone) due to positive condensation reaction with 2,4 DNPH. Indeed, A is a methyl ketone as further affirmed by positive tri-iodoform test (oxidative cleavage) to form CHI3 as the pale yellow precipitate.

Since A did not react with hot acidified K2Cr2O7, it does not contain primary nor secondary alcohol groups and hence cannot be oxidised. Since A did not react with Br2(aq), it does not contain phenol (no electrophilic substitution), phenylamine (no electrophilic substitution) nor C=C (no electrophilic addition).

Reduction of A with LiAlH4 in dry ether reduced the ketone functional group to a corresponding secondary alcohol B and C. Since an equimolar mixture of isomers are obtained, it is likely that B and C are enantiomers of each other and hence contain a chiral carbon.

However, due to C:H ratio of 1:2 but A is not an alkene, a cyclic structure must be present. Hence, to fulfil structures of A, B and C, a cyclohexane ring must be present.

3 Alcohol A has the molecular formula C6H14O. Reaction of A with acidified potassium manganate (VII) produces compound B. B forms a yellow precipitate when warmed with aqueous alkaline iodine. Heating A with concentrated H2SO4 at 170oC produces C, C6H12. Vigorous oxidation of C with hot concentrated acidified potassium manganate (VII) forms 3-methylbutanoic acid as one of the products.

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Deduce the structural formulae of compounds A to C, explaining your reasoning.

A undergoes oxidation with acidified KMnO4 to give B, a methyl ketone as indicated by positive tri-iodoform test with aqueous alkaline iodine (oxidative cleavage) (NOT –CH(CH3)(OH) group due to oxidation to methyl ketone already). A undergoes elimination with concentrated H2SO4 at 170oC to give C, an alkene.

Vigorous oxidation of C (oxidative cleavage) by acidified KMnO4 forms 3-methylbutanoic acid, and the other product has to be CO2 and hence terminal alkene due to molecular formula of C6H12.

4 Compound A, C6H7ON, is sparingly soluble in water but dissolves in hydrochloric acid. It also dissolves in aqueous NaOH but not in aqueous Na2CO3.

On reaction with 1 mol ethanoyl chloride, A forms compound B, C8H9O2N. B is no longer soluble in hydrochloric acid but is still soluble in aqueous NaOH. On reacting B with aqueous bromine, compound C with molecular formula C8H6O2NBr3 is formed.

When A is treated with 2 mol ethanoyl chloride, compound D with molecular formula C10H11O3N is formed. D is neither soluble in hydrochloric acid nor in aqueous NaOH.

Deduce the structural formulae of compounds A to D, explaining your reasoning.

Since A has a C:H ratio of close to 1, a benzene ring is likely to be present. As A is sparingly soluble in water but dissolves in HCl (acid-base neutralisation), an amine (more specifically phenylamine) group is present. Since it also dissolves in NaOH(aq) (acid-base neutralisation), a phenol group is present but not a –COOH group.

Reacting A with 1 mol CH3COCl forms B via nucleophilic substitution to form an amide with the phenylamine group since it is still soluble in NaOH(aq) but not with HCl. On reacting B with Br2(aq), electrophilic substitution occurs and tri-bromination (as seen from 3 H replaced by 3 Br in formula) results in C formed where Br replaces H atoms at C2, C4 and C6, implying that –NH2 and –OH groups are on C1 and C3.

When A is treated with 2 mol CH3COCl, 1 ester group and 1 amide group are now formed via nucleophilic substitution since the resultant compound D is neither soluble in HCl nor NaOH(aq).

5 A neutral compound A has the molecular formula, C14H18O3NBr. When A is refluxed with

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aqueous NaOH, three compounds B, C and D are obtained.

Upon analysis, B is found to be a straight-chain molecule with molecular formula C3H9N. When B is treated with excess chloroethane, E is formed. E has a relative molecular mass of 179.5 and gives a white precipitate immediately when treated with aqueous silver nitrate.

C has the molecular formula C3H5O3Na. Upon acidification, it yields F. F is also obtained when A is refluxed with dilute H2SO4. It was found that 2 mol of aqueous NaOH is required for complete neutralisation of 1 mol of F.

D (C8H7O3Na), upon addition of acidified potassium dichromate(VI) and heating with immediate distillation, yields G. Treating G with 2,4-dinitrophenylhydrazine gives an orange precipitate with the following structure:

Deduce the structural formulae of compounds A to G, explaining your reasoning.

Since A is neutral and can be hydrolysed to 3 compounds by refluxing with NaOH(aq), it contains ester bonds.

B is an amine (CH3CH2CH2NH2) which undergoes nucleophilic substitution with excess chloroethane to give the quaternary ammonium salt E. Since E gives an immediate white precipitate with silver nitrate, it further confirms the presence of a chloride salt. Subtracting [35.5 (Ar of chlorine) + (12x3+7) (Mr of propyl group -C3H7, (7 not 9 H since 2 H on –NH2 group are replaced by ethyl groups)) + 14 (Ar of nitrogen)] from 179.5, subtracting and dividing it by 3 indicates substitution of 3 ethyl groups at the N atom on B.

C is a carboxylate salt and F is the corresponding carboxylic acid, and hence obviously will be obtained upon acid hydrolysis of A with H2SO4. Since 2 mol of NaOH(aq) is needed for complete neutralisation of F, F contains 2 –COOH groups.

D is also a carboxylate salt and is oxidised to the aldehyde G (due to heating with immediate distillation with K2Cr2O7). G undergoes condensation reaction with 2,4-DNPH to give the formula shown above.

6 P, C10H16O, is an insect sex attractant. When P is gently oxidised with warm acidified potassium dichromate (VI) it gives Q, C10H14O, which gives an orange precipitate with 2,4-dinitrophenylhydrazine reagent, but does not give an orange precipitate with Fehling’s reagent.

Treatment of Q with cold dilute acidified potassium manganate (VII) gives R, C10H20O7, but Q with hot concentrated acidified potassium manganate (VII) gives S, T and carbon dioxide in the ratio 1:1:2.

S, C3H6O, gives a yellow precipitate with alkaline aqueous iodine.

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T can be synthesised from the following reaction sequence.

HCN H+/K2Cr2O7 boilingCH2(CHO)2 U V T, C5H4O6

NaCN (aq) warm H2SO4 (aq)

Identify compounds S, T, U and V. Suggest a possible structure for Q and thus for R and P. Explain your reasoning.

P is oxidised to Q, a ketone / non-aliphatic aldehyde due to positive test with 2,4-DNPH (condensation) but negative test with Fehling’s reagent (oxidation; test for aliphatic aldehyde).

Q reacts with cold dilute KMnO4 and undergoes oxidation and diol formation indicated by 6 –OH groups increase and hence 3 C=C bonds are present. Strong oxidation (oxidative cleavage) of Q with hot concentrated acidified KMnO4 gives S, T and CO2 in 1:1:2 ratio, indicating presence of 2 terminal alkene groups.

S is a methyl ketone or contains –CH(CH3)(OH) group but is the former due to the molecular formula; S is propanone.

CH2(CHO)2 undergoes nucleophilic addition with HCN to form U (CH2(CH(OH)(CN))2), a di-cyanohydrin due to 2 aldehyde functional groups. U is then oxidised to V (CH2(COCN)2) , a diketone since U is a secondary alcohol and then acidified to form the di-carboxylic acid T (CH2(COCOOH)2).

7 An optically active compound F, C9H12O, reacts with acidified K2Cr2O7 to give G, C9H10O. F reacts with iodine in aqueous sodium hydroxide to give a yellow precipitate and the sodium salt of the acid H, C8H8O2. H is oxidised by acidified KMnO4 to form I, C8H6O4.

Suggest the structures of F, G, H and I, giving relevant equations to explain your reasoning.

F is optically active, indicating presence of chiral centres. F undergoes oxidation with acidified K2Cr2O7 to form G, a ketone, since there is a loss of 2 H and no change in O atoms, thereby indicating that F is a secondary alcohol. F gives a positive tri-iodoform test (oxidation – oxidative cleavage) and hence contains –CH(CH3)(OH) group, further supporting that F is a secondary alcohol. H is the carboxylic acid formed after the methyl group is oxidatively cleaved off from F.

8 J, C9H11I, on boiling with aqueous potassium hydroxide gave a compound K, C9H12O. With acidified sodium dichromate(VI), compound K yielded L, C9H10O. With hot acidified potassium manganate(VII), compound K yielded M, C7H6O2. K when warmed with potassium hydroxide and iodine is able to produce yellow precipitate.

Suggest the structures of J, K, L and M, giving relevant equations to explain your reasoning.

J undergoes nucleophilic substitution upon boiling with KOH(aq) as I is replaced by –OH group to form K, which is oxidised by acidified Na2Cr2O7 to give L. However, upon strong oxidation (oxidative cleavage), K yields M and there is a decrease in 2 C and 6 H but increase in 1 O, indicating the presence of a carboxylic acid group. M is benzoic acid (note the standard formula C7H6O2 – close C:H ratio indicating presence of benzene ring). K when warmed with alkaline I2(aq), undergoes oxidative cleavage

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to give yellow precipitate, indicating presence of either methyl ketone or –CH(CH3)(OH) group, but it is the latter since K is an alcohol which is formed via nucleophilic substitution using KOH and J.

9 Compound C, C6H10O, is an insect pheromone. C reacts with Fehling’s solution, decolourises Br2(aq) and gives an orange precipitate with 2,4-dinitrophenylhydrazine reagent. Treatment of C with H2 over a nickel catalyst produces D, C6H14O, which reacts with hot acidified K2Cr2O7 to give E, C6H12O2. Neither D nor E reacts with 2,4-dinitrophenylhydrazine, but both react with sodium metal.

Compound E can be obtained from 3-bromopentane by the following route.

Identify the lettered compounds D to F, explaining your reasoning and suggest a possible structure for C.

Since C reacts with Fehling’s solution, it is an aliphatic aldehyde (oxidation), Br2(aq) C contains C=C, 2,4-DNPH carbonyl compound (but it is aldehyde due to positive test with Fehling already). C undergoes reduction with H2/Ni to form D, a primary alcohol. Since there is an increment of 4 H, C contains 1 C=C bond and 1 CHO group in order to satisfy both abovementioned conditions. D undergoes oxidation with hot K2Cr2O7/H+ to give E, a carboxylic acid due to increase in 1 O atom. Since neither D nor E reacts with 2,4 DNPH, there are no carbonyl carbon functional group but contain –OH or –COOH group due to positive test (redox) with Na metal. 3-bromopentane undergoes nucleophilic substitution with ethanolic KCN to give F, a nitrile, which undergoes acidification to give E, a carboxylic acid.C contains –CHO and C=C groups which can occur randomly about the structure

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and hence above are some possible structures of C. 10 The molecules of compound K, C7H15Br, are chiral. On treatment with NaOH(aq), K

produces alcohol L, C7H15OH, which does not react with hot acidified Na2Cr2O7(aq). Treating compound K with ethanolic NaOH produces a mixture of four different isomeric alkenes with the formula C7H14, only two of which are cis-trans isomers of each other.

Suggest the structural formulae of compound K and the four alkenes.

Refer to N08/3 answer scheme.

11 A, C20H32O2, is a naturally occurring compound that could affect blood pressure.

A is reduced by LiAlH4 in dry ether to give a neutral compound B, C20H34O.

When A is heated under reflux with ethanol and concentrated sulphuric acid it gives C, C22H36O2. A solution of C shaken with finely divided palladium under an atmosphere of hydrogen gives D, C22H44O2.

Treatment of C with hot concentrated acidified potassium manganate (VII) breaks it down into three products, CH3(CH2)4CO2H, HO2C(CH2)3CO2C2H5 and E, C3H4O4, in the ratio of 1:1:3 respectively.

Identify compounds B to E and suggest a structure for A, explaining your reasoning.

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Since there is a decrease in 1 oxygen atom after reduction of A to form B, A is a carboxylic acid which is reduced to a primary alcohol by LiAlH4 in dry ether (further supported by an increase in 2 H atoms; -COOH to –CH2OH)

When A is heated under reflux with ethanol and concentrated H2SO4, it undergoes esterification to give C, further supporting that it is a monocarboxylic acid since there is an increment of 2 C atoms in C. As C undergoes reduction by H2/Pd to give D which has 8 H atoms more, there are 4 C=C bonds in C.

C is then strongly oxidised (oxidative cleavage) to give the three products, of which E is a dicarboxylic acid and hence has to appear within the long chain alkene since it is cleaved at both ends to give rise to a dicarboxylic acid. Hence, the sequence is one of the other two products followed by 3 E then the last product.

Note: ester group in the second product is not hydrolysed probably due to complications involved should there be subsequent hydrolysis giving rise to more products and hence confusion, but there should be hydrolysis owing to acid and heat being present.

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And should there be an extension, let E be HCOOH / H2C2O4 (and may well be oxidised directly to CO2, tricking one into thinking that there is a terminal alkene) – note that 2 acids can be further oxidised – methanoic acid and ethanedioic acid to CO2 and H2O, and CO2 respectively

End of Paper

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