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RAFFLES INSTITUTION
YEAR 6 H2 CHEMISTRY 2026
Tutorial 18 Carbonyl Compounds
Reaction Mindmaps
❑ Aldehydes
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❑ Ketones
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Answers to self-check questions
Q1
Q2(a) LiAlH4 OR H2(g), Ni catalyst
Q2(b)
Heat propan-1-ol with acidified K2Cr2O7 carefully with distillation of the product formed to collect mainly propanal
as distillate.
Q2(c)
Observations: Orange ppt is formed.
Answers to MCQ: 3) B 4) C 5) B 6) D 7) B
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Suggested Solutions to Practice Questions:
1 (a) (i) Mechanism: Nucleophilic addition
KCN(aq) ⎯→ K+(aq) + CN−(aq)
Step 1
Step 2
(ii) KCN provides the initial CN– ions for the nucleophilic attack on the carbonyl carbon.
HCN is used in the second step as a (Bronsted) acid to protonate the anionic
intermediate.
Note: Do NOT merely say that KCN helps to increase the rate of reaction and HCN is
the reagent, as the question is about the roles of KCN and HCN in the mechanism.
(iii) The carbonyl carbon in CH 3CHO is bonded to one alkyl group while that in (CH 3)2CO
is bonded to two alkyl groups. Hence, the carbonyl carbon in (CH3)2CO is less electron
deficient (less +) due to the additional electron donating alkyl group . There is also
greater steric hindrance about the carbonyl carbon in (CH 3)2CO which hinder the
approach of the attacking nucleophile. Therefore, propanone , (CH3)2CO, reacts at a
slower rate than ethanal, CH3CHO.
(b) Hydrolysis reaction
OH
O
OH
(c) (i) Lactic acid has a chiral carbon and no plane of symmetry through the molecule. Lactic
acid in milk rotates the plane of polarised light because only 1 enantiomer is present
(i.e. all the molecules present are of the same chirality).
(ii) In the slow step, the CN− nucleophile can attack the planar carbonyl carbon of ethanal
from either side with equal probability, resulting in the formation of a racemic mixture
of two enantiomers, thus optical activity cancels out.
Checklist for nucleophilic addition mechanism:
Name of mechanism
δ+ on C, δ– on O in Step 1 and δ+ on H, δ– on C of HCN in Step 2
Lone pair of electrons on Nu
Curly arrows to show flow of electrons
- Lone pair on Nu to δ+ C
- C=O bond to δ– O
- Lone pair on O− to H (of HCN)
- H−C bond to C
Label slow/fast steps
NOTE: HCN is a weak acid and hence a
poor source of CN− ion. The CN− ion comes
initially from the catalytic amount of
KCN(aq) added.
δ– δ+
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2 (a)
(b)
Note that the second step cannot be carried out using H2 and Ni, as the alkene group will also
be reduced.
Alternatively, you may carry out reduction in the first step followed by elimination in the second
step.
(c)
(d)
Note: The reagents and conditions for step 1 and step 2 cannot be interchanged as shown below.
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3 (a) Test
Add I2(aq) and NaOH(aq) to each compound in a test-tube and heat in a hot water bath.
Observation
gives yellow ppt of CHI3 while gives no yellow ppt of CHI3
IMPORTANT NOTE:
Use of hot acidified KMnO 4 is not accepted because both compounds will undergo side-chain
oxidation.
will give benzoic acid while will give 1,2-benzenedicarboxylic acid.
(b) Test
Add Tollens’ reagent to each compound in a test-tube and heat in a hot water bath.
Observation
gives a silver mirror.
gives no silver mirror.
OR
Test
Add K2Cr2O7 (aq), H2SO4 (aq), heat (DO NOT heat under reflux)
Observation
will turn orange K2Cr2O7 green while
will not produce a colour
change.
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4 (a)
Evidence/Information Deductions
A has molecular formula C8H8O C to H ratio = 1:1
A likely contains a benzene ring
A orange ppt
Condensation (or addition-elimination)
reaction
A is an aldehyde or ketone.
A Ag
A (no reaction)
A undergoes oxidation with Tollens’ but not
with Fehling’s.
A is an aromatic aldehyde.
A
Oxidation reaction. A has 2 substituents on
the ring in the 1 and 4 position.
A is
(b)
Evidence/Information Deductions
C orange ppt
Condensation (or addition-elimination)
reaction
C is an aldehyde or ketone.
B (C3H8O) C (C3H6O)
Oxidation reaction. B is a secondary alcohol
and C is a ketone. (since there is no increase
in number of oxygen atoms)
B is
, C is
.
Note:
• If B was a primary alcohol, C would be a
carboxylic acid with two oxygen atoms.
• Also, C cannot be an aldehyde as any
aldehyde group formed (from oxidation of
a primary alcohol using KMnO4) would be
further oxidised to a carboxylic acid.
Tollens’
2,4 DNPH
acidified
KMnO4
heat
Fehling’s
2,4 DNPH
acidified
KMnO4
heat
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5 (a) Evidence/Information Deductions
A B (orange ppt)
Condensation (or addition-elimination) reaction
A is an aldehyde or ketone.
A (C5H10O) C (C5H10O2)
Compound A undergoes oxidation to give compound C.
A is an aldehyde and C is a carboxylic acid.
(Note: 1 oxygen atom in A to 2 oxygen atoms in C)
A D
Compound A undergoes reduction to give compound
D, which is a primary alcohol.
D E
Elimination of water from the primary alcohol to form a
terminal alkene, E
E CH3CH(CH3)COOH
Oxidation of the C=C in alkene E. Loss of one C atom
as CO2.
Since CO2 and 2-methylpropanoic acid are products of
the oxidative cleavage of an alkene, E is
H
C
H
C
H
CH(CH3)2
Hence, alcohol D is
OH
(a primary alcohol)
A is
H
O
B is
C is
OH
O
acidified
K2Cr2O7
heat
LiAlH4
2,4 DNPH
conc
H3PO4
heat
acidified
KMnO4
heat
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6 (a) Molecular mass of K is 86.
86 – 16 = 70 = 12n + 2n ⇒ n = 5
(b) Evidence/Information Deductions
J K
Since there is no change in number of O atom
J is a secondary alcohol and K a ketone.
K yellow ppt
K undergoes oxidation and contains CH3CO-
group
J 2 alkenes
(no cis-trans)
Elimination of H2O
J is
K is
For explanation (not required in answer):
From the first evidence, J could be CH3CH(OH)CH2CH2CH3 or
The two alkenes from the elimination of J are:
(no cis-trans isomerism in both compounds)
[O]
alkaline
I2(aq)
conc. H3PO4 catalyst
heat
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7
1,3-butadiene does not undergo nucleophilic addition reaction with HCN due to the absence of
an electron deficient C (+ C) to attract the CN− nucleophile.
(Note: Both C=C and C=O bonds are electron rich. Hence it is insufficient to state that the
C=C electron cloud repel the nucleophile CN − as carbonyl with C=O group can undergo
nucleophilic addition reaction).
In 4-methyl-1-penten-3-one, the nucleophilic attack on the + carbonyl C is sterically hindered
by the –CH(CH3)2 group.
Also, the p orbitals of the sp2 C in C=C and C=O overlap to form a delocalised electron cloud.
Due to the highly electronegative O atom, the delocalised electron cloud is pulled towards the
O atom and the terminal alkene C becomes +.
Hence, the nucleophile will attack the + terminal alkene C instead, resulting in the nucleophilic
addition occurring at the alkene group to form C instead of at the carbonyl group to form B.
Resonance structure and mechanism (not required in answer, only for further understanding)
