Reactivity 3.3. Electron sharing reactions

Which equation represents a propagation step in the reaction of chloromethane with  bromine? 

A. CH₃Cl → •CH₂Cl + H• 

B. CH₃Cl + Br• → •CH₂Cl + HBr 

C. •CH₂Cl + Br• → CH₂ClBr 

D. CH₃Cl + Br• → CH₂ClBr + H• 

Answer: B. CH₃Cl + Br• → •CH₂Cl + HBr 

Propagation must: 

   • Start with a radical, and 

   • End with a new radical so the chain continues. 

For bromination of chloromethane: 

Br• + CH₃Cl → •CH₂Cl + HBr 

•CH₂Cl + Br₂ → CH₂ClBr + Br• 

Both are propagation steps. 

A. Incorrect: CH₃Cl → •CH₂Cl + H• 

No radical at start → initiation-type 

B. Correct: CH₃Cl + Br• → •CH₂Cl + HBr 

Matches propagation step 1 

C. Incorrect: •CH₂Cl + Br₂ → CH₂ClBr 

Missing the Br• product → incomplete propagation 

D. Incorrect: CH₃Cl + Br• → CH₂ClBr + H• 

Wrong products → doesn’t follow mechanism 

Iron(II) salts catalyse the decomposition of aqueous hydrogen peroxide to water and oxygen. The reaction mechanism involves •OH and HOO• free radicals.

Fe²⁺ + H₂O₂ → Fe³⁺ + HO• + OH⁻ 

Fe³⁺ + H₂O₂ → Fe²⁺ + HOO• + H⁺ 

a. Define the term radical. 

b. Draw the Lewis formula for the HOO• free radical. 

c. State the type of catalysis that takes place between iron(II) sulfate and hydrogen peroxide and state the property of iron(II) sulfate that allows it to carry out its function  as a catalyst in the above decomposition reaction. 

a. Define the term radical. 

A radical (or free radical) is an atom or molecule that has an unpaired electron. Because of this unpaired electron, radicals are highly reactive. 

b. Draw the Lewis formula for the HOO• radical. 

Hydroperoxyl radical: H–O–O• 

Lewis structure:

Explanation: 

   • Each oxygen atom has 6 valence electrons. 

   • The O–O single bond shares 2 electrons. 

   • One oxygen (the one bonded to hydrogen) has two lone pairs; the other oxygen has one lone pair and one unpaired electron (•). 

c. Type of catalysis and property of Fe²⁺ 

   • Type of catalysis: 

→ Homogeneous catalysis, because both the iron(II) ions and hydrogen peroxide are in the same (aqueous) phase. 

   • Property of Fe²⁺ that enables catalysis: 

→ Iron can alternate between Fe²⁺ and Fe³⁺ oxidation states easily, allowing it to repeatedly form and regenerate radicals (•OH and HOO•), thus maintaining the catalytic cycle. 

Bromine, Br₂, reacts with butane, C₄H₁₀, when exposed to ultraviolet light.

a. State the type of reaction that occurs between bromine and butane and give the name of the product formed which does not contain a carbon atom. 

Some of the products produced in the mixture have the molecular formula C₄H₈Br₂. 

b. Give a single equation for the formation of C₄H₈Br₂ from bromine and butane. 

c. State the number of structural isomers with the formula C₄H₈Br₂ that could have been formed in the reaction between bromine and butane. 

One of the structural isomers with the molecular formula C₄H₉Br displays optical isomerism. 

d. Draw the displayed formula of the radical which leads to the formation of the two optical isomers. 

a. Reaction type: free-radical substitution (photochemical, UV). Product with no carbon:  hydrogen bromide, HBr. 

b. Overall equation for a dibromide (C₄H₈Br₂) from butane and bromine: C₄H₁₀ + 2Br₂ → C₄H₈Br₂ + 2HBr 

c. Number of structural (constitutional) isomers of C₄H₈Br₂ from butane: 6 (1,1-; 2,2-; 1,2-; 2,3-; 1,3-; 1,4-dibromobutane; 2 ↔ 3 and 1 ↔ 4 are equivalent by  symmetry.) 

d. The C₄H₉Br isomer that is optically active is 2-bromobutane. The radical leading to the two enantiomers is the sec-butyl radical at C-2: 

CH₃ − ĊH − CH₂ − CH₃

(dot on C-2; attack of Br• on either face gives the pair of enantiomers). 

Methylbenzene, C₆H₅–CH₃, reacts with chlorine to form different products depending on the reaction’s conditions. 

Gaseous methylbenzene and chlorine in the presence of ultraviolet radiation will form chloromethyl benzene, C₆H₅–CH₂Cl. 

a. Give the displayed formula of the product. 

b. Give equations for the steps which lead to the production of chloromethyl benzene and explain the role of the ultraviolet radiation. 

c. A small amount of by-product with the relative formula mass of 182.3 g mol^-1 is found. Suggest the molecular formula of this product. 

a. Product (chloromethyl benzene, benzyl chloride): C₆H₅–CH₂Cl 

b. Radical substitution mechanism (gas phase, UV): 

   • Initiation (role of UV): UV light photolyzes Cl₂ to make radicals

   • Propagation: 

1. Cl• + C₆H₅CH₃ → C₆H₅CH₂• + HCl 

2. C₆H₅CH₂• + Cl₂ → C₆H₅CH₂Cl + Cl• 

(Chain continues via the regenerated Cl•) 

   • Termination (examples): 

     o Cl• + Cl• → Cl₂ 

     o C₆H₅CH₂• + Cl• → C₆H₅CH₂Cl 

     o C₆H₅CH₂• + C₆H₅CH₂• → C₆H₅CH₂CH₂C₆H₅ 

c. By-product with Mᵣ ≈ 182.3: C₁₄H₁₄ (bibenzyl, C₆H₅CH₂CH₂C₆H₅), formed by coupling of two benzyl radicals. 

Which of the following reactions is a radical substitution reaction?

A. CH₄ + 2Cl₂ → CH₂Cl₂ + 2HCl 

B. CH₃CH₂OH → CH₂CH₂ + H₂O 

C. CH₂CH₂ + Br₂ → CH₂BrCH₂Br 

D. CH₃Cl + NaOH → CH₃OH + HCl

Answer: A. CH₄ + 2Cl₂ → CH₂Cl₂ + 2HCl 

A. Correct: CH₄ + 2Cl₂ → CH₂Cl₂ + 2HCl 

This is a substitution reaction between methane and chlorine that occurs via a free radical mechanism under UV light. → Radical substitution reaction.

B. Incorrect: CH₃CH₂OH → CH₂CH₂ + H₂O 

C. Incorrect: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br 

This is an addition reaction (electrophilic addition) to an alkene, not radical substitution.

D. Incorrect: CH₃Cl + NaOH → CH₃OH + HCl 

This is a nucleophilic substitution reaction (SN₁ or SN₂), not radical.