Let's break down the **free-radical bromination of ethylbenzene** step-by-step and provide the two required **propagation steps**. --- ## **Step 1: Understanding Initiation** The initiation step (already provided) is: \[ \text{Br}_2 \xrightarrow{hv} 2\ \text{Br}^\bullet \] Light (\(hv\)) splits Br\(_2\) into two bromine radicals. --- ## **Step 2: Propagation Steps** ### **Propagation 1** The bromine radical abstracts a benzylic hydrogen from ethylbenzene, forming a benzylic radical and HBr: \[ \mathrm{C_6H_5CH_2CH_3} + \mathrm{Br}^\bullet \rightarrow \mathrm{C_6H_5CH^\bullet CH_3} + \mathrm{HBr} \] - **Ethylbenzene** (\(\mathrm{C_6H_5CH_2CH_3}\)) - **Bromine radical** (\(\mathrm{Br}^\bullet\)) - **Benzylic radical** (\(\mathrm{C_6H_5CH^\bullet CH_3}\)) - **Hydrogen bromide** (\(\mathrm{HBr}\)) --- ### **Propagation 2** The benzylic radical reacts with Br\(_2\), forming the brominated product and regenerating a Br radical: \[ \mathrm{C_6H_5CH^\bullet CH_3} + \mathrm{Br_2} \rightarrow \mathrm{C_6H_5CHBrCH_3} + \mathrm{Br}^\bullet \] - **Benzylic radical** (\(\mathrm{C_6H_5CH^\bullet CH_3}\)) - **Bromine** (\(\mathrm{Br_2}\)) - **Brominated product** (\(\mathrm{C_6H_5CHBrCH_3}\)) - **Bromine radical** (\(\mathrm{Br}^\bullet\)) --- ## **Final Answers** ### **Propagation 1:** \[ \boxed{ \mathrm{C_6H_5CH_2CH_3} + \mathrm{Br}^\bullet \rightarrow \mathrm{C_6H_5CH^\bullet CH_3} + \mathrm{HBr} } \] ### **Propagation 2:** \[ \boxed{ \mathrm{C_6H_5CH^\bullet CH_3} + \mathrm{Br_2} \rightarrow \mathrm{C_6H_5CHBrCH_3} + \mathrm{Br}^\bullet } \] These are the two propagation steps for the free-radical bromination of ethylbenzene.

# Free-Radical Bromination of Ethylbenzene This guide outlines the steps involved in the free-radical bromination of ethylbenzene, focusing on the initiation and propagation steps. ## Step 1: Initiation The initiation step involves the generation of bromine radicals from bromine molecules: \[ \text{Br}_2 \xrightarrow{hv} 2\ \text{Br}^\bullet \] - **Br\(_2\)**: Bromine molecule - **Br\(^\bullet\)**: Bromine radical - **\(hv\)**: Energy from light ## Step 2: Propagation Steps ### Propagation Step 1 In the first propagation step, a bromine radical abstracts a hydrogen atom from ethylbenzene, forming a benzylic radical and hydrogen bromide: \[ \mathrm{C_6H_5CH_2CH_3} + \mathrm{Br}^\bullet \rightarrow \mathrm{C_6H_5CH^\bullet CH_3} + \mathrm{HBr} \] - **Reactants**: - Ethylbenzene (\(\mathrm{C_6H_5CH_2CH_3}\)) - Bromine radical (\(\mathrm{Br}^\bullet\)) - **Products**: - Benzylic radical (\(\mathrm{C_6H_5CH^\bullet CH_3}\)) - Hydrogen bromide (\(\mathrm{HBr}\)) ### Propagation Step 2 In the second propagation step, the benzylic radical reacts with another bromine molecule to produce the brominated product and regenerate a bromine radical: \[ \mathrm{C_6H_5CH^\bullet CH_3} + \mathrm{Br_2} \rightarrow \mathrm{C_6H_5CHBrCH_3} + \mathrm{Br}^\bullet \] - **Reactants**: - Benzylic radical (\(\mathrm{C_6H_5CH^\bullet CH_3}\)) - Bromine molecule (\(\mathrm{Br_2}\)) - **Products**: - Brominated product (\(\mathrm{C_6H_5CHBrCH_3}\)) - Bromine radical (\(\mathrm{Br}^\bullet\)) ## Final Answers ### Propagation Step 1: \[ \boxed{ \mathrm{C_6H_5CH_2CH_3} + \mathrm{Br}^\bullet \rightarrow \mathrm{C_6H_5CH^\bullet CH_3} + \mathrm{HBr} } \] ### Propagation Step 2: \[ \boxed{ \mathrm{C_6H_5CH^\bullet CH_3} + \mathrm{Br_2} \rightarrow \mathrm{C_6H_5CHBrCH_3} + \mathrm{Br}^\bullet } \] These propagation steps are essential for understanding the mechanism of free-radical bromination in organic chemistry.