Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 2 - 10.9 Nucleophilic substitution mechanisms-Study Notes - New Syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 2 – 10.9 Nucleophilic substitution mechanisms- Study Notes- New syllabus

Edexcel International A Level (IAL) Chemistry (YCH11) -Unit 2 – 10.9 Nucleophilic substitution mechanisms- Study Notes -International A Level (IAL) Chemistry (YCH11) – per latest Syllabus.

Key Concepts:

10.9 understand the mechanisms of the nucleophilic substitution reactions between primary halogenoalkanes and:

i aqueous potassium hydroxide

ii ammonia

SN1 and SN2 substitution mechanisms will be tested in Unit 4

Edexcel International A Level (IAL) Chemistry (YCH11) -Concise Summary Notes- All Topics

10.9 Nucleophilic Substitution Mechanisms of Primary Halogenoalkanes

Primary halogenoalkanes undergo nucleophilic substitution reactions via a specific mechanism known as the \( \mathrm{S_N2} \) mechanism. This mechanism involves a single step in which bond breaking and bond formation occur simultaneously. The nature of the nucleophile and reaction conditions determines the product formed.

SN2 Mechanism

The \( \mathrm{S_N2} \) mechanism is a one-step nucleophilic substitution in which the nucleophile attacks the carbon atom as the leaving group departs.

  • Occurs mainly in primary halogenoalkanes due to low steric hindrance.
  • Involves a single transition state (no intermediate).
  • Bond formation and bond breaking occur simultaneously.
  • The nucleophile attacks from the opposite side (backside attack).

(i) Reaction with Aqueous Potassium Hydroxide

Primary halogenoalkanes react with aqueous KOH to form alcohols via an \( \mathrm{S_N2} \) mechanism.

  • \( \mathrm{OH^-} \) acts as a strong nucleophile.
  • It attacks the \( \delta^+ \) carbon from the opposite side of the halogen.
  • The \( \mathrm{C-X} \) bond breaks heterolytically as the new \( \mathrm{C-OH} \) bond forms.
  • A halide ion is released.

Example: \( \mathrm{CH_3CH_2Br + OH^- \rightarrow CH_3CH_2OH + Br^-} \)

(ii) Reaction with Ammonia

Primary halogenoalkanes react with ammonia to form amines via an \( \mathrm{S_N2} \) mechanism.

  • \( \mathrm{NH_3} \) acts as a nucleophile due to its lone pair.
  • It attacks the \( \delta^+ \) carbon in a single step.
  • The \( \mathrm{C-X} \) bond breaks as the \( \mathrm{C-N} \) bond forms.
  • An intermediate ammonium ion is formed, which then loses a proton to give the amine.

Example: \( \mathrm{CH_3CH_2Br + NH_3 \rightarrow CH_3CH_2NH_2 + HBr} \)

Why Primary Halogenoalkanes Favour SN2

  • The carbon atom is less hindered, allowing easy access for the nucleophile.
  • There is minimal steric hindrance from surrounding alkyl groups.
  • Formation of unstable primary carbocations is avoided, so \( \mathrm{S_N1} \) is not favoured.

Key Features of the Mechanism

  • Single-step mechanism with no intermediate.
  • Involves simultaneous bond making and breaking.
  • Requires a strong nucleophile.
  • Rate depends on both the halogenoalkane and nucleophile concentration.

Example 1:

Explain why primary halogenoalkanes undergo nucleophilic substitution via an \( \mathrm{S_N2} \) mechanism rather than an \( \mathrm{S_N1} \) mechanism.

▶️ Answer/Explanation

Primary halogenoalkanes do not readily form carbocations because primary carbocations are highly unstable.

Therefore, the \( \mathrm{S_N1} \) mechanism, which involves carbocation formation, is not favoured.

Instead, the reaction occurs in a single step where the nucleophile attacks the carbon as the leaving group departs.

The low steric hindrance around the carbon allows this simultaneous process, favouring the \( \mathrm{S_N2} \) mechanism.

Example 2:

Describe the mechanism of the reaction between bromoethane and ammonia.

▶️ Answer/Explanation

The ammonia molecule, with a lone pair of electrons, acts as a nucleophile and attacks the \( \delta^+ \) carbon atom in bromoethane from the opposite side of the bromine atom.

As the \( \mathrm{C-N} \) bond forms, the \( \mathrm{C-Br} \) bond breaks heterolytically, releasing a bromide ion.

An intermediate ethylammonium ion is formed, which then loses a proton to produce ethylamine.

This occurs in a single step characteristic of the \( \mathrm{S_N2} \) mechanism.

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