www.wewwchemistry.com Page 1 of 12
Example of a Recent H2 Chemistry Paper 3 Free-Response Question
from a School Preliminary Examination
(The question stem is highlighted in green.)
2011 Anglo-Chinese Junior College / Paper 3 / Question 1
(a) Samples of 2-bromo-2-methylpropane were dissolved in dilute aqueous ethanol
(80% ethanol and 20% water by volume) and reacted with sodium hydroxide
solution. Several experiments were carried out at constant temperature. The initial
rate of reaction was determined in each case.
Expt [(CH3)3CBr] / mol dm–3
] / mol dm–3
Rate / mol dm–3
1 0.020 0.010 20.2
2 0.020 0.020 20.2
3 0.040 0.030 40.4
Calculate a value for the rate constant. Hence write the rate equation for the
www.wewwchemistry.com Page 2 of 12
Comparing Expt 1 & 2,
] increases by 2 times as [(CH3)3CBr] is kept constant, rate remains
∴ Order of reaction with respect to [OH–
] is 0.
Comparing Expt 1 & 3,
Since order of reaction with respect to [OH–
] is 0, changes in [OH–
] does not affect rate.
When [(CH3)3CBr] increases by 2 times, rate increases by 2 times.
∴ Order of reaction with respect to [(CH3)3CBr] is 1.
Rate equation: Rate = k[(CH3)3CBr]
Substituting values from Expt 1,
Rate constant, k =
= 1010 s–1
∴ Rate equation for reaction at temperature of experiments:
Rate = (1010)[(CH3)3CBr]
www.wewwchemistry.com Page 3 of 12
(b) The hydrolysis of 2-bromopropane is now investigated. Samples were dissolved in
dilute aqueous ethanol (80% ethanol and 20% water by volume) and reacted with
sodium hydroxide solution. The rate equation is found to be as follows:
Rate = (0.24 × 10–5
)[CH(CH3)2Br] + (4.7 × 10–5
(i) The rate equation obtained indicates that the hydrolysis of 2-bromopropane
exhibits a mixture of both first order and second order kinetics. Suggest why
this may be so.
2-bromopropane is a secondary halogenoalkane.
The two –CH3 groups in 2-bromopropane hinder the backside attack of the OH–
nucleophile on its electrophilic carbon. 2-bromopropane is thus less likely to undergo
SN2 reaction than a primary halogenoalkane.
2-bromopropane is also less likely to undergo SN1 reaction when compared to a
tertiary halogenoalkane, as the secondary carbocation intermediate formed from 2-
bromopropane is less stable than a tertiary carbocation formed from a tertiary
www.wewwchemistry.com Page 4 of 12
(ii) By deriving an expression in terms of [OH–
], show that the % rate due to SN2
Rate equation for SN2 reaction: Rate = (4.7 × 10–5
∴ Rate due to SN2
www.wewwchemistry.com Page 5 of 12
(iii) Using the expression derived from (b)(ii), calculate the % rate due to SN2 for
] of 0.01 mol dm–3
, 0.1 mol dm–3
and 1.0 mol dm–3
] / mol dm–3
www.wewwchemistry.com Page 6 of 12
(iv) In light of your answer to (b)(iii), state how the % rate due to SN2 depends on
the concentration of [OH–
] and explain why it varies in that manner.
For SN2 reaction, Rate = (4.7 × 10–5
For SN1 reaction, Rate = (0.24 × 10–5
SN2 reaction is favoured by high nucleophile concentration, [OH–
], while SN1 reaction
is independent of [OH–
Thus when [OH–
] increases, % rate due to SN2 increases.
www.wewwchemistry.com Page 7 of 12
(v) By comparing the magnitude of the two rate constants in the rate equation
given in (b), comment on the ease of hydrolysis by the two different
The Arrhenius Equation is given as follows:
The Arrhenius Equation may be written as: ln k = ln A –
⇒ If k is large, Ea is small.
k for SN2 reaction (= 4.7 × 10–5
) > k for SN1 reaction (= 0.24 × 10–5
⇒ Ea for SN2 reaction < Ea for SN1 reaction
At the same temperature, reaction proceeds by the SN2 pathway more than the SN1
www.wewwchemistry.com Page 8 of 12
(vi) Draw a labelled Maxwell-Boltzmann Curve for this reaction to illustrate the
distribution of energy. Based on your answer to (b)(v), indicate the activation
energies clearly on the diagram for the two different reaction pathways.
www.wewwchemistry.com Page 9 of 12
(vii) On the same axes, draw the energy profile illustrating both the SN1 kinetics
and the SN2 kinetics components observed for this reaction. Indicate the
activation energies clearly on the diagram, for the two reaction pathways.
BE(C–Br) = 280 kJ mol–1
< BE(C–O) = 360 kJ mol–1
⇒ Energy produced during formation of C–O bond > energy absorbed during
breaking of C–Br bond.
Thus, the hydrolysis of 2-bromopropane is an exothermic reaction.
www.wewwchemistry.com Page 11 of 12
(viii) Explain how the rate constant will change if CH(CH3)2Cl is used instead of
In both the SN1 and SN2 reaction pathway, the slow step involves the breaking of
C–X (where X = halogen) bond.
⇒ Rate of nucleophilic substitution, whether by SN1 and SN2 mechanism, depends
on the strength of the C–X bond.
Since the C–Cl bond is stronger than the C–Br bond, a higher activation energy is
needed for the slow step if CH(CH3)2Cl is used.
Rate constant is thus smaller if CH(CH3)2Cl is used.
www.wewwchemistry.com Page 12 of 12
(ix) Draw the structures of two possible organic by-products that may be formed
in this reaction, other than CH(CH3)2(OH). Briefly account for their formation.
Possible by-products :
A is formed when CH(CH3)2Br undergoes elimination in the presence of ethanolic
Some of the product CH(CH3)2(OH) undergoes an acid-base reaction with OH–
to generate another nucleophile, CH(CH3)2O:–
, which attacks CH(CH3)2Br to form B.