ASR BPOC Notes

2025 JC2 H3 BASIC PHYSICAL ORGANIC CHEMISTRY 2025/ASRJC/Chemistry 1 ANDERSON SERANGOON JUNIOR COLLEGE H3 CHEMISTRY BASIC PHYSICAL ORGANIC CHEMISTRY CONTENT 1 Introduction 2 Hammond Postulate 3 Bells-Evan-Polanyi Principle 4 The relationship between Hammond Postulate and the Bells-Evan-Polanyi Principle LEARNING OUTCOMES Students should be able to understand and apply the following concepts involving kinetic control and thermodynamic control to the study of reaction mechanisms: (i) the Hammond postulate: relationship between the transition state and the nearest stable species (ii) the Bell–Evans–Polanyi principle ➢ relationship between activation energy and enthalpy change of reaction ➢ quantitative calculations based on EA = A + B∆Hr 1 INTRODUCTION 1.1 Reaction Mechanism For a chemist involved in synthesis, mechanistic knowledge of a reaction allows intelligent variation of reaction conditions, temperatures and proportions of reagents to maximise yield of pure products. Reaction mechanism is a sequence of elementary steps that are involved in the pathway from reactants to products. Elementary step refers to the simplest step which cannot be further broken down into simpler steps. E.g. Overall reaction: (CH3)3CCl + OH– (CH3)3COH + Cl– (Reactants) (Products) The SN1 mechanism involves 2 elementary steps: Step 1 C Cl CH3 CH3 + − C CH3 CH3 CH3 carbocation intermediate slow CH3 + Cl− + Step 2 OH fast CH3 C CH3 CH3 OH C CH3 CH3 CH3 +

2025 JC2 H3 BASIC PHYSICAL ORGANIC CHEMISTRY 2025/ASRJC/Chemistry 2 1.2 Energy Profile Diagram An energy profile (or reaction profile) diagram is a graph showing how the energy of the reacting system changes as a function of the reaction coordinate. A reaction intermediate is a species that is formed in one step of a reaction mechanism and consumed in a subsequent step. This species can be a molecule, ion or a free radical. A transition state (TS) is defined as the state corresponding to the highest energy along the reaction coordinate of an elementary reaction. TS can neither be isolated nor can it be studied by spectroscopic method. However, TS does have a _______________________ _______________________ which allow chemists to investigate how changes in the reagents and solvent influence the difference in energy between reagents and TS. If these influences make the transition state more stable compared with the reagents, the reaction will be faster. Energy ΔH

2025 JC2 H3 BASIC PHYSICAL ORGANIC CHEMISTRY 2025/ASRJC/Chemistry 3 2 HAMMOND POSTULATE 2.1 Definition Hammond postulate is helpful to understand the relationship between the rate of reaction and the stability of the products/intermediates formed. In 1955, George S. Hammond postulated that: If two states, as for example, a transition state and an unstable intermediate, occur consecutively during a reaction process and have nearly the same energy content, their interconversion will involve only a small re -organisation of the molecular structures. This means that the geometrical structure of the transition state resembles the structure of the intermediate species. Hence, the structure of a transition state resembles the structure of the nearest stable species. Endothermic reaction Exothermic reaction __________ transition state • Energy level of the TS is closer to that of the __________. • Structure of the TS resembles the structure of the __________ of that step. __________ transition state • Energy level of the TS is closer to that of the __________. • Structure of the TS resembles the structure of the __________ of that step E.g. For a SN1 reaction for the hydrolysis of C(CH3)3Cl, TS of step 1 is __________TS. TS of step 2 is __________TS. Energy C(CH3)3Cl C(CH3)3OH +C(CH3)3

2025 JC2 H3 BASIC PHYSICAL ORGANIC CHEMISTRY 2025/ASRJC/Chemistry 4 2.2 Application of Hammond Postulate The Hammond Postulate suggests that high-energy intermediates and the transition states that lead to their formation are affected by similar substituent effects. Example In an electrophilic addition of unsymmetrical alkene, consider the reaction of methylpropene and HBr: + HBr 3o carbocation1o carbocation Br Br A B Why is product B formed in a greater proportion as compared to product A? From H2 Chemistry: 3o carbocation is more stable than 1 o carbocation since it has ____________________ alkyl groups attached to the C+ which __________ the positive charge on the carbocation to a greater extent. Hence 3o carbocation is more stable and is formed at a faster rate leading to a greater proportion of product B being formed. Why does a more stable carbocation lead to a faster rate of reaction? What is the link between stability and rate of reaction? TS of step 1 is a __________ TS which means that TS is closer in energy to the ____________________ and its structure resembles that of the ____________________. Hence any factor that stabilises the ____________________ will also stabilise the TS. Since ____________________ is stabilised by greater number of electron-donating R groups, the 3o TS is also stabilised by ____________________. Therefore, the more stable 3o TS has a ____________________ for its formation leading to a __________________ rate of formation.

2025 JC2 H3 BASIC PHYSICAL ORGANIC CHEMISTRY 2025/ASRJC/Chemistry 5 3. BELL-EVANS-POLANYI (BEP) PRINCIPLE 3.1 Definition The BEP Principle describes the experimental observation that a linear relationship exists between the activation energy (E A) and the enthalpy change of reaction (Δ H) for similar reactions. This relationship can be expressed as: EA = A + BΔHr where EA is the activation energy of a reaction, ΔHr is the enthalpy change of reaction, and A and B are constants How is the expression (EA = A + BΔHr) derived? The BEP Principle was derived from the observation that, for certain closely-related reactions, the rate constant (k) and equilibrium constant (Keq) are related by the equation ln(k) = ln() + ln(Keq) where  and  are experimentally determined constants. An example of such reaction is RCOOH + N(CH3)3 RCOO– + N(CH3)H+ From the graph, a linear relationship is observed between ln(k) and ln(Keq). The gradient of the line will give the value for  while the y-intercept will give the value for ln(). ln(k) ln(Keq)

2025 JC2 H3 BASIC PHYSICAL ORGANIC CHEMISTRY 2025/ASRJC/Chemistry 6 Using the Arrhenius equation ( AE RTk Ae − = ) and Go = –RT In(Keq), ln(k) = ln() + ln(Keq) can be manipulated to ln ln AE GA RT RT   − = − ln lnAE GART RT   = − + )ln (A AE T S TRH = −+ lnA TAE RT S H  − += For closely-related reactions, ln ART S T  − is constant, hence the equation is simplified to AEH = +

2025 JC2 H3 BASIC PHYSICAL ORGANIC CHEMISTRY 2025/ASRJC/Chemistry 7 3.2 Application of BEP Principle (a) Predicting the rate of reaction This relationship provides an efficient way to estimate the activation energy of reactions, and hence, reaction rate within the same family. endothermic reaction exothermic reaction An __________ in the positive enthalpy of reaction results in an __________ in the activation energy. This thus __________ the rate of the reaction. An __________ in the negative enthalpy of reaction results in a __________ in the activation energy. This thus __________ the rate of reaction. Example In the following reaction: N NR R 2R + N2 From the d iagram, it can be observed that the ____________________ the reaction, the ______________