2011 ACJC Preliminary Examination, H2 Chemistry Syllabus 9647, Paper 3, Question 1

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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 [OH–] / mol dm–3 Rate / mol dm–3 s–1

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

reaction. [3]


Proposed Solution Comparing Expt 1 & 2, When [OH–] increases by 2 times as [(CH3)3CBr] is kept constant, rate remains constant. ∴ 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 =

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20.20.020 = 1010 s–1

∴ Rate equation for reaction at temperature of experiments: Rate = (1010)[(CH3)3CBr]


(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)[CH(CH3)2Br][OH–] (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.

Proposed Solution 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 halogenoalkane.


(ii) By deriving an expression in terms of [OH–], show that the % rate due to SN2

is

Proposed Solution Rate equation for SN2 reaction: Rate = (4.7 × 10–5)[CH(CH3)2Br][OH–] ∴ Rate due to SN2


(iii) Using the expression derived from (b)(ii), calculate the % rate due to SN2 for

various [OH–] of 0.01 mol dm–3, 0.1 mol dm–3 and 1.0 mol dm–3. Proposed Solution [OH–] / mol dm–3 % rate

0.01

0.1 66.2 1.0 95.1


(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. Proposed Solution For SN2 reaction, Rate = (4.7 × 10–5)[CH(CH3)2Br][OH–] For SN1 reaction, Rate = (0.24 × 10–5)[CH(CH3)2Br] 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.


(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 pathways. The Arrhenius Equation is given as follows:

Proposed Solution The Arrhenius Equation may be written as: ln k = ln A –

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lnEaRT

⇒ 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 pathway.


(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.

Proposed Solution


(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.

Proposed Solution Note: 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.


Ent

halp

y, H

Reaction coordinate

CH(CH3)2Br + OH–

CH(CH3)2OH + Br–

Ea(SN2)

Ea(SN1)

!Hr


(viii) Explain how the rate constant will change if CH(CH3)2Cl is used instead of

CH(CH3)2Br. Proposed Solution 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.


(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. [17]

[Total: 20]

Proposed Solution

Possible by-products :

A B A is formed when CH(CH3)2Br undergoes elimination in the presence of ethanolic

NaOH. Some of the product CH(CH3)2(OH) undergoes an acid-base reaction with OH– ions

to generate another nucleophile, CH(CH3)2O:–, which attacks CH(CH3)2Br to form B.

Education

2011 ACJC Preliminary Examination, H2 Chemistry Syllabus 9647, Paper 3, Question 1