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

Grade 12IBChemistry

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

🔑Concepts

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A chemical reaction involves the breaking and forming of bonds, which can be understood as the movement of electron pairs.

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Lewis Acids and Bases: A Lewis acid is defined as an electron-pair acceptor (e.g., BF3BF_3, H+H^+), while a Lewis base is an electron-pair donor (e.g., NH3NH_3, H2OH_2O).

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Nucleophiles and Electrophiles: A nucleophile is an electron-rich species that donates an electron pair to form a new covalent bond (e.g., OH−OH^-, CN−CN^-). An electrophile is an electron-deficient species that accepts an electron pair (e.g., NO2+NO_2^+, CH3+CH_3^+).

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Homolytic Fission: This occurs when a covalent bond breaks such that each atom takes one of the shared electrons, resulting in the formation of free radicals. It is represented by 'fishhook' (single-headed) arrows: X−Y→X⋅+Y⋅X-Y \rightarrow X^{\cdot} + Y^{\cdot}.

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Heterolytic Fission: This occurs when a covalent bond breaks such that one atom takes both shared electrons, resulting in the formation of a cation and an anion: X−Y→X++:Y−X-Y \rightarrow X^+ + :Y^-.

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Curly Arrow Notation: Double-headed curly arrows are used to represent the movement of an electron pair from a source (lone pair or bond) to an electron-deficient site.

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Coordinate Covalent Bond: A type of bond formed when one atom provides both electrons for the shared pair in a Lewis acid-base reaction.

📐Formulae

A+:B→A−BA + :B \rightarrow A-B

X−X→UVX⋅+X⋅X-X \xrightarrow{UV} X^{\cdot} + X^{\cdot}

R−X→R++X−R-X \rightarrow R^+ + X^-

Nu:+R−X→Nu−R+X−Nu: + R-X \rightarrow Nu-R + X^-

💡Examples

Problem 1:

Identify the Lewis acid and Lewis base in the reaction between ammonia (NH3NH_3) and boron trifluoride (BF3BF_3), and represent the product formed.

Solution:

In the reaction NH3+BF3→H3N→BF3NH_3 + BF_3 \rightarrow H_3N \rightarrow BF_3:

  • NH3NH_3 is the Lewis base because the Nitrogen atom has a lone pair of electrons to donate.
  • BF3BF_3 is the Lewis acid because the Boron atom has an incomplete octet and can accept an electron pair.
  • The product contains a coordinate covalent bond (dative bond) represented as H3N→BF3H_3N \rightarrow BF_3.

Explanation:

The movement of the electron pair from the NN atom to the BB atom completes the octet for Boron and forms a stable adduct.

Problem 2:

Describe the homolytic fission of a chlorine molecule (Cl2Cl_2) and the species produced.

Solution:

Cl−Cl→UV lightCl⋅+Cl⋅Cl-Cl \xrightarrow{UV \text{ light}} Cl^{\cdot} + Cl^{\cdot} The bond breaks equally, and each chlorine atom retains one electron from the shared pair.

Explanation:

Homolytic fission typically requires energy in the form of UV light or heat. The resulting species, Cl⋅Cl^{\cdot}, are called free radicals and are highly reactive due to their unpaired electrons.

Problem 3:

Using curly arrow notation, show the mechanism for the reaction between a hydroxide ion (OH−OH^-) and a chloromethane (CH3ClCH_3Cl) molecule.

Solution:

The nucleophile OH−OH^- attacks the electron-deficient carbon atom: HO−↷C(H3)−Cl→HO−CH3+Cl−HO^- \curvearrowright C(H_3)-Cl \rightarrow HO-CH_3 + Cl^- The arrow starts at the lone pair of the oxygen in OH−OH^- and points to the CC atom. Simultaneously, a second arrow starts from the C−ClC-Cl bond and points to the ClCl atom.

Explanation:

This is a nucleophilic substitution reaction. The C−ClC-Cl bond is polar (Cδ+−Clδ−C^{\delta+} - Cl^{\delta-}), making the carbon an electrophilic center susceptible to attack by the nucleophilic hydroxide ion.