Hoffmann Ammonolysis of Alkyl Halides
Hoffmann Ammonolysis is a method for preparing amines by the nucleophilic substitution reaction of an alkyl halide with an ethanolic solution of ammonia. The reaction involves the cleavage of the C-X bond by the ammonia molecule.
1. General Reaction
When an alkyl halide ($R-X$) is heated with ethanolic ammonia in a sealed tube at $373 \ K$, a mixture of amines is formed.
The sequence of formation is:
- Primary Amine ($1^\circ$)
- Secondary Amine ($2^\circ$)
- Tertiary Amine ($3^\circ$)
- Quaternary Ammonium Salt ($4^\circ$)
2. Detailed Mechanism
The reaction follows an $S_N2$ mechanism where ammonia acts as the nucleophile.
Step 1: Formation of Primary Amine
Ammonia attacks the alkyl halide, displacing the halide ion.
The free amine is generated from the ammonium salt by reaction with a strong base (like NaOH) or excess ammonia.
Step 2: Polyalkylation
The primary amine formed ($R-NH_2$) still has a lone pair on nitrogen and behaves as a nucleophile. It attacks another molecule of alkyl halide to form a secondary amine, then a tertiary amine, and finally a quaternary ammonium salt.
3. Reactivity Order
The reactivity of alkyl halides towards ammonolysis depends on the strength of the C-X bond.
Reactivity Order of Halides
$$ R-I > R-Br > R-Cl $$
Alkyl iodides are the most reactive because the C-I bond is the weakest and $I^-$ is the best leaving group.
4. Controlling the Product
Since the reaction typically yields a mixture, specific conditions are needed to favor one product.
To get Primary Amine
Use a large Excess of Ammonia. This increases the probability of $R-X$ colliding with $NH_3$ rather than with an amine product.
To get Quaternary Salt
Use an Excess of Alkyl Halide. This ensures exhaustive methylation/alkylation occurs until the quaternary stage is reached.
5. Limitations
- Aryl Halides: Do not undergo ammonolysis under normal conditions because the C-X bond has partial double bond character (low reactivity towards nucleophilic substitution).
- Tertiary Alkyl Halides: Tend to undergo Elimination rather than substitution with ammonia (a base) to form alkenes.
- Mixture Separation: Separating the mixture of $1^\circ, 2^\circ, 3^\circ$ amines is difficult (can be done using fractional distillation or Hinsberg's reagent).
Hoffmann Ammonolysis Quiz
Test your concepts on Amine synthesis. 10 MCQs with explanations.
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