[ Related Blog post: http://www.wewwchemistry.com/2013/09/the-mind-boggling-h2-chemistry ]

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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.