3.7.3Physical Chemistry
Nucleophilic Substitution — AQA AS Chemistry Revision Notes & Topical Questions
5 min read 5 parts Lesson 92 of 102
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Section 1 of 5Why Haloalkanes React
Why Haloalkanes React
Polar bond
- Polar bond Halogens are more electronegative than carbon, so the C–X bond is polar with δ+ on carbon and δ− on the halogen.
Nucleophile attack
- Nucleophile attack An electron-rich nucleophile is attracted to the δ+ carbon and donates a lone pair to form a new bond.
Leaving group
- Leaving group The C–X bond breaks heterolytically, and the halide ion leaves with both bonding electrons.
Net result
- Net result The halogen is substituted by the nucleophile, so this is nucleophilic substitution.
Section 2 of 5Reactions of Bromoethane
With aqueous hydroxide
- With aqueous hydroxide Warm aqueous NaOH gives ethanol: CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻.
With cyanide
- With cyanide Heating under reflux with ethanolic KCN gives propanenitrile: CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻.
With ammonia
- With ammonia Heating with excess ethanolic ammonia gives ethylamine: CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br.
Why excess ammonia
- Why excess ammonia An excess of ammonia reduces further substitution, which would give secondary and tertiary amines.
Cyanide advantage
- Cyanide advantage The nitrile has one more carbon than the haloalkane, so this reaction lengthens a carbon chain.
| Nucleophile | Conditions | Organic product |
|---|---|---|
| OH⁻ | Aqueous NaOH, warm | Alcohol |
| CN⁻ | Ethanolic KCN, reflux | Nitrile |
| NH₃ | Excess ethanolic NH₃, heat | Primary amine |
Section 3 of 5Mechanism Step by Step
1
Draw the haloalkane with the δ+ on the carbon and δ− on the halogen of the polar C–X bond.
2
Draw the nucleophile with a lone pair, for example :OH⁻ or :CN⁻, showing the negative charge.
3
A curly arrow goes from the lone pair of the nucleophile to the δ+ carbon atom.
4
A second curly arrow goes from the middle of the C–X bond to the halogen atom.
5
Show the products: the substituted organic molecule and the halide ion, X⁻.
Section 4 of 5Mechanism Marking Points
Mechanism Marking Points
Arrow shape
- Arrow shape Arrows must be curly with a full head, as half-headed fishhook arrows are used for radicals only.
Arrow start
- Arrow start The first arrow starts at the nucleophile's lone pair, and never at an atom.
Charges
- Charges The carbon carries δ+, not a full positive charge, in this mechanism.
Free ion
- Free ion Draw :OH⁻ or :CN⁻ as the free ion, not as NaOH or KCN with a bond to the metal.
Section 5 of 5Exam Checklist & Tips
3.7.3 – Key Points for the Exam
- Nucleophile An electron-pair donor that attacks an electron-deficient carbon atom.
- Nucleophilic substitution A reaction in which a nucleophile replaces an atom or group on a carbon.
- Leaving group The atom or group that leaves with the bonding pair, here a halide ion.
- Polar bond A covalent bond with unequal sharing of electrons, giving partial charges.
Common Mistakes
- Drawing the nucleophile's arrow towards the halogen instead of the δ+ carbon.
- Using ethanolic alkali to make an alcohol; aqueous alkali is needed for substitution.
- Showing the product without the halide ion as a leaving product.
- Using a fishhook arrow in an ionic mechanism.
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