Question 1

Why is anti-Markovnikov addition only observed with HBr and not HCl or HI?

Accepted Answer

Anti-Markovnikov addition is thermodynamically favorable only with HBr because both propagation steps yield negative Gibbs free energy values. With HCl, the second propagation step has positive ∆G, making it unfavorable. With HI, both the first and second propagation steps are thermodynamically unfavorable regardless of temperature, since both enthalpy and entropy terms are positive.

Question 2

How do enthalpy and entropy terms affect the spontaneity of radical addition reactions?

Accepted Answer

Spontaneity depends on Gibbs free energy (∆G = ∆H − T∆S). For HBr and HCl in the first propagation step, the enthalpy term dominates at low temperatures, making ∆G negative and the reaction favorable. However, at high temperatures, the entropy term becomes significant. With HI, both terms are positive, so ∆G remains positive regardless of temperature, making the reaction always unfavorable.

Question 3

What determines the sign of Gibbs free energy in the second propagation step?

Accepted Answer

In the second propagation step, the entropy term is nearly zero because reactant and product molecule counts are equal. Therefore, the enthalpy term dictates the ∆G sign. For HBr, the reaction is exothermic with negative ∆G, making it favorable. For HCl, ∆G is positive and endothermic, rendering the reaction thermodynamically unfavorable.

Question 4

Why does temperature affect the thermodynamic favorability of HBr addition differently than HI addition?

Accepted Answer

With HBr and HCl, temperature influences which term dominates: at low temperatures, enthalpy controls ∆G; at high temperatures, entropy becomes significant. With HI, both enthalpy and entropy terms are positive, so ∆G is always positive regardless of temperature, making the reaction permanently unfavorable across all temperature ranges.

Question 5

What is the relationship between the number of molecules and entropy in radical propagation steps?

Accepted Answer

When reactant and product molecule counts are equal, the entropy term approaches zero, as seen in the second propagation step. This means enthalpy alone determines ∆G. In contrast, the first propagation step involves different molecule numbers, allowing entropy to compete with enthalpy in determining reaction spontaneity and thermodynamic favorability.

Question 6

How can Gibbs free energy predict whether radical addition to alkenes will occur?

Accepted Answer

A negative ∆G indicates a spontaneous, thermodynamically favorable reaction. By evaluating ∆G for each propagation step using ∆G = ∆H − T∆S, chemists can predict reaction feasibility. Only HBr yields negative ∆G values for both propagation steps, explaining why anti-Markovnikov addition succeeds with HBr but fails with HCl and HI.

Question 7

Why is the first propagation step thermodynamically favorable for HBr but unfavorable for HI?

Accepted Answer

In the first propagation step, ∆G depends on competition between enthalpy and entropy terms. For HBr and HCl, the enthalpy term dominates at low temperatures, yielding negative ∆G. For HI, both enthalpy and entropy are positive, so ∆G is always positive, making the first step inherently unfavorable and preventing the radical addition mechanism from proceeding.

Radicale anti-Markovnikov-toevoeging aan alkenen: thermodynamica

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