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Question #44505

Difficulty: 8/10 | 🟢 Free

The Question

The anti-Markovnikov addition of \ce{HBr} to an alkene ( $A$ ) in the presence of a peroxide initiator ( $I$ ) proceeds via a radical chain mechanism. Consider the following elementary steps at 298 K:

\begin{align*} \text{Initiation:} \quad & I \xrightarrow{k_d} 2RO^\bullet \ & RO^\bullet + \ce{HBr} \xrightarrow{k_1} \ce{ROH} + Br^\bullet \ \text{Propagation:} \quad & Br^\bullet + A \xrightarrow{k_2} R^\bullet \ & R^\bullet + \ce{HBr} \xrightarrow{k_3} P + Br^\bullet \ \text{Termination:} \quad & 2Br^\bullet \xrightarrow{k_t} \ce{Br2} \end{align*}

Assuming the Steady-State Approximation (SSA) applies to all radical intermediates ( $RO^\bullet$ , $Br^\bullet$ , $R^\bullet$ ) and the kinetic chain length is large, which of the following represents the correct rate law for the overall formation of the alkyl bromide product ( $P$ )?

Options

A.
$k_2 \left( \frac{k_d}{k_t} \right)^{1/2} [I]^{1/2} [A]$
B.
$k_3 \left( \frac{k_d}{k_t} \right)^{1/2} [I]^{1/2} [\ce{HBr}]$
C.
$k_2 \left( \frac{k_d}{k_t} \right) [I] [A]$
D.
$k_3 \left( \frac{k_2}{k_t} \right)^{1/2} [I]^{1/2} [A]^{1/2} [\ce{HBr}]$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Peroxide Effect (Kharasch)|Rate Law Derivation from Mechanism|Chemical Kinetics

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous application of the Steady-State Approximation (SSA) to radical chain mechanisms, specifically the Peroxide Effect (Kharasch addition).

A common trap is assuming that the overall rate depends directly on the concentration of the initiator (first-order) or the reagent (HBr).

By deriving the rate law, students must recognize that bimolecular termination of the chain-carrying radical ( $Br^\bullet$ ) leads to a square-root dependence (0.5 order) on the initiator concentration.

Furthermore, the rate is independent of [HBr] because the propagation step involving the alkene ( $A$ ) dictates the steady-state concentration of the carbon-centered radical.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Hydrocarbons - Class 11 Part 2
Concept	Peroxide Effect (Kharasch)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free