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Reactivity 3. What are the mechanisms of chemical change? - Electron sharing reactions

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Bond Fission: Chemical reactions involve the breaking of covalent bonds. This can occur via homolytic fission, where each atom retains one electron from the shared pair, forming radicals (X⋅X\cdot), or heterolytic fission, where one atom retains both electrons, forming ions (X+X^+ and :Y−:Y^-).

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Lewis Acids and Bases: A Lewis acid is an electron pair acceptor (e.g., BF3BF_3, AlCl3AlCl_3, or H+H^+), while a Lewis base is an electron pair donor (e.g., :NH3:NH_3, H2O:H_2O:, or OH−OH^-).

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Nucleophiles and Electrophiles: Nucleophiles (Nu−Nu^-) are electron-rich species with a lone pair that seek electron-deficient centers. Electrophiles (E+E^+) are electron-deficient species that seek electron-rich centers like double bonds.

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Curly Arrow Notation: Double-headed curly arrows represent the movement of an electron pair (↷\curvearrowright). Single-headed 'fishhook' arrows represent the movement of a single electron (⇀\rightharpoonup).

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Radical Substitution: Occurs in alkanes (e.g., with Cl2Cl_2 or Br2Br_2 in UV light). It involves three stages: Initiation (homolytic fission), Propagation (regeneration of radicals), and Termination (combination of radicals).

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Nucleophilic Substitution (SN1S_N1 and SN2S_N2): Mechanisms for the reaction of haloalkanes. SN2S_N2 is a concerted one-step process favored by primary haloalkanes, while SN1S_N1 is a two-step process involving a carbocation intermediate (C+C^+), favored by tertiary haloalkanes.

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Electrophilic Addition: The characteristic reaction of alkenes. The high electron density of the π\pi bond attracts electrophiles (e.g., HBrHBr or Br2Br_2), leading to the addition of atoms across the double bond.

📐Formulae

A:B→UVA⋅+B⋅ (Homolytic Fission)A:B \xrightarrow{UV} A\cdot + B\cdot \text{ (Homolytic Fission)}

A:B→A++:B− (Heterolytic Fission)A:B \rightarrow A^+ + :B^- \text{ (Heterolytic Fission)}

Rate=k[R−X][Nu−] (Second-order, SN2)Rate = k[R-X][Nu^-] \text{ (Second-order, } S_N2)

Rate=k[R−X] (First-order, SN1)Rate = k[R-X] \text{ (First-order, } S_N1)

💡Examples

Problem 1:

Describe the initiation step for the reaction between methane (CH4CH_4) and chlorine (Cl2Cl_2) in the presence of ultraviolet (UV) light.

Solution:

Cl−Cl→UVCl⋅+Cl⋅Cl-Cl \xrightarrow{UV} Cl\cdot + Cl\cdot

Explanation:

The UV light provides the necessary energy to break the Cl−ClCl-Cl covalent bond homolytically. Each chlorine atom takes one electron from the shared pair, resulting in the formation of two highly reactive chlorine radicals.

Problem 2:

Identify the Lewis acid and Lewis base in the reaction: BF3+:NH3→F3B−NH3BF_3 + :NH_3 \rightarrow F_3B-NH_3

Solution:

Lewis Acid: BF3BF_3; Lewis Base: :NH3:NH_3

Explanation:

Boron trifluoride (BF3BF_3) has an incomplete octet and accepts an electron pair from the nitrogen atom in ammonia (:NH3:NH_3). Since BF3BF_3 accepts the pair, it is the Lewis acid, and :NH3:NH_3 is the Lewis base (donor).

Problem 3:

Outline the SN2S_N2 mechanism for the reaction between chloromethane (CH3ClCH_3Cl) and hydroxide ions (OH−OH^-).

Solution:

OH−+CH3Cl→[HO⋯CH3⋯Cl]‡→CH3OH+Cl−OH^- + CH_3Cl \rightarrow [HO \cdots CH_3 \cdots Cl]^{\ddagger} \rightarrow CH_3OH + Cl^-

Explanation:

The nucleophile (OH−OH^-) attacks the electron-deficient carbon atom from the opposite side of the leaving group (ClCl). A transition state is formed where the C−OHC-OH bond is partially formed and the C−ClC-Cl bond is partially broken. This is a one-step bimolecular process.