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Example an Organic Structure Elucidation Question from a School Preliminary Examination Paper

(The question stem is highlighted in green.) 2011 Hwa Chong Institution / Paper 3 / Question 2(d) The distinctive scent of rose oil comes mainly from a family of closely related chemicals called the rose ketones. Compounds C, D and E belong to this family. C and D have the same molecular formula, C13H20O, while E has molecular formula, C13H18O. C can exist as a pair of enantiomers, but D and E do not show any optical activity. 1 mol of C and D each reacts with 2 mol of liquid bromine at room temperature. 1 mol of E, however, reacts with 3 mol of liquid bromine.

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C, D and E react with hydrogen in the presence of nickel to give the same compound, C13H26O:

C and D each undergo oxidative cleavage with acidified KMnO4 to give ethanoic acid as one of the two organic products. In addition, C also gives compound C10H14O4, while D gives compound C11H16O5 respectively. E, on the other hand, gives three organic products on reaction with KMnO4: ethanoic acid, CH3COCO2H and compound F, C8H10O6. 1 mol of F reacts with 1 mol of Na2CO3 to produce CO2 gas. 1 mol of F also reacts with 2 mol of HCN under cold conditions in the presence of trace base. Deduce the structures of compounds C, D, E, and F, giving explanations of the reactions that occurred. [8]

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Proposed Solution Observations Deductions C can exist as a pair of enantiomers but D and E does not show any optical activity.

C contains a chiral carbon. D and E do not contain chiral carbon (or are symmetrical).

1 mol of C and D reacts with 2 mol of liquid bromine. 1 mol of E reacts with 3 mol of liquid bromine.

Electrophilic addition of Br2 to alkene functional group. C and D have 2 alkene groups, while E has 3 alkene groups.

C, D and E react with hydrogen in the presence of nickel to give the same compound of molecular formula C13H26O.

(Catalytic) reduction / hydrogenation occurs.

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Observations Deductions C undergoes oxidative cleavage with acidified KMnO4 to give: CH3CO2H compound with

molecular formula C10H14O4

Oxidation results in loss of 3 carbon atoms: 2 as CH3COOH, 1 as CO2 (⇒ C

contains a terminal alkene) addition of 3 oxygen atoms:

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Observations Deductions D undergoes oxidative cleavage with acidified KMnO4 to give: CH3CO2H compound with

molecular formula C11H16O5

Oxidation results in loss of 2 carbon atoms: 2 as CH3COOH (⇒ D does not

contain a terminal alkene) addition of 4 oxygen atoms:

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Observations Deductions E undergoes oxidative cleavage with acidified KMnO4 to give: CH3CO2H CH3COCO2H Compound F with

molecular formula C8H10O6

Oxidation results in loss of 5 carbon atoms: 2 as CH3COOH, 3 as CH3COCO2H

to form F.

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Observations Deductions F reacts with 1 mole of Na2CO3 to produce CO2.

F contains 2 carboxylic acid groups which undergo acid-base reaction with Na2CO3 to produce CO2.

4 of the oxygen atoms are from the 2 carboxylic acid groups.

F reacts with 2 moles of HCN under cold conditions with base as the catalyst.

Nucleophilic addition of HCN across C=O occurs. F contains 2 ketone groups. (Aldehydes are oxidised further

to carboxylic acid groups.) Remaining 2 oxygen atoms are from these 2 ketone groups.