Amines classification
Primary (1°): R-NH₂. Secondary (2°): R₂NH. Tertiary (3°): R₃N. Aliphatic vs aromatic. Quaternary ammonium salt: R₄N⁺X⁻.
-- NCERT, p. 2The trap that costs marks here: confusing Hofmann elimination product selectivity with Saytzeff (Zaitsev) selectivity. When you see a quaternary ammonium hydroxide undergoing elimination, the product is the least substituted alkene — the opposite of what you'd pick for an alkyl halide with a base.
Nomenclature and classification of amines
Amines are derivatives of ammonia where one, two, or three hydrogen atoms are replaced by alkyl or aryl groups (NCERT Class 12 Chemistry Chapter 13, page 2).
Classification:
IUPAC naming rules:
Common name system: alkyl group names + "amine" as one word (methylamine, diethylamine, triphenylamine).
Hofmann elimination — the nomenclature-adjacent reaction trap: Quaternary ammonium hydroxide (R₄N⁺OH⁻) heated → eliminates to form the least substituted alkene (Hofmann product). This is anti-Saytzeff. The bulky leaving group (NR₃) and base attack the less hindered β-hydrogen. Confusing this with standard E2 of alkyl halides (Saytzeff → most substituted alkene) is the high-frequency trap for this topic.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following is a tertiary amine?
The IUPAC name of (CH₃)₂CHNH₂ is:
The IUPAC name of CH₃–NH–C₂H₅ is:
Hofmann elimination of N,N,N-trimethylbutan-2-aminium hydroxide gives predominantly:
Which compound is named N,N-diethylethanamine?
An amine has the structure: C₆H₅–NH–CH₃. Its correct IUPAC name is:
In Hofmann elimination, why does the least substituted alkene form preferentially?
Consider the quaternary salt: (CH₃)₃N⁺–CH₂–CH(CH₃)₂ OH⁻. On heating, which alkene is the major Hofmann elimination product?
Given
- Substrate: tetraethylammonium hydroxide, (C₂H₅)₄N⁺ OH⁻ - Reaction: strong heating (pyrolysis)
Required
Identify the major elimination product.
Concept
Hofmann elimination: quaternary ammonium hydroxide → least substituted alkene + tertiary amine + water. Anti-Saytzeff selectivity due to steric bulk of the –NR₃ leaving group.
Formula / Rule
Hofmann rule: β-elimination from quaternary ammonium salts preferentially removes the β-H from the least substituted carbon → terminal alkene.
Substitution / Analysis
All four groups on nitrogen are identical (ethyl). Each ethyl group has β-hydrogens on the terminal CH₃. Elimination from any ethyl group gives the same product: ethene (CH₂=CH₂).
Calculation
(C₂H₅)₄N⁺ OH⁻ → CH₂=CH₂ + (C₂H₅)₃N + H₂O Since all groups are equivalent, there is no substitution-selectivity question — ethene is the sole alkene product.
Final answer
**Major product: ethene (CH₂=CH₂)**, with triethylamine as the amine by-product.
Common trap
Picking a higher alkene (butene) by imagining two ethyl groups combine. They don't — Hofmann elimination breaks one C–N bond and removes one β-H from the same alkyl chain. Each elimination event produces a two-carbon alkene from one ethyl group.
Similar NEET-style question
"Exhaustive methylation of propan-1-amine followed by treatment with Ag₂O and heating gives which alkene as the major product?" (Answer: propene — the only alkene possible from the propyl chain, which is also the least substituted terminal alkene consistent with Hofmann.) ---
Primary (1°): R-NH₂. Secondary (2°): R₂NH. Tertiary (3°): R₃N. Aliphatic vs aromatic. Quaternary ammonium salt: R₄N⁺X⁻.
-- NCERT, p. 2Aqueous basicity order: 2° ≈ 1° > 3° > NH3. Aromatic amines much weaker due to resonance delocalisation of lone pair.
| Symbol | Quantity | SI Unit |
|---|---|---|
| pKb | -log Kb | - |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Inorganic Exception
Student treats aniline as stronger base than methylamine. Aniline weaker due to lone-pair delocalization into ring.
Basicity ordering across aromatic and aliphatic amines.
Aliphatic amines (CH3NH2) > NH3 > aromatic amines (C6H5NH2). Aniline lone pair delocalised into ring → less available for protonation.
Category: Organic Reaction Conditions
Quaternary ammonium hydroxide (Hofmann) elimination favours LESS substituted alkene (anti-Saytzeff). Halide elimination (E2) follows Saytzeff (more substituted).
E2 elimination question with quaternary ammonium hydroxide vs alkyl halide.
Hofmann: bulky leaving group + base → least substituted alkene (Hofmann product). Saytzeff: alkyl halide + base → most substituted alkene (Zaitsev product).
Root cause: concept gap
Gas phase: 3° > 2° > 1° > NH₃. Aqueous: 2° > 1° > 3° > NH₃ (steric + solvation effects).
Root cause: concept gap
Aniline's lone pair delocalised into ring → less basic. Order: aliphatic (CH3NH2) > NH3 > aromatic (PhNH2).
Root cause: concept gap
Sandmeyer: ArN₂⁺ + CuCl/CuBr/CuCN → ArCl/ArBr/ArCN. Gattermann: Cu powder + HX. Hydrolysis: H₂O → ArOH.
Root cause: concept gap
Hofmann (quaternary R₄N⁺ + OH⁻): least substituted alkene. Saytzeff (alkyl halide + base): most substituted.
11 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
The highest number of helium atoms is in
The correct order of energies of molecular orbitals of N molecule, is 2
Which of the following reactions will NOT give primary amine as the product?
Which of the following sequence of reactions is suitable to synthesize chlorobenzene?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
uses gas phase order in aqueous
Doesn't account for solvation reversal
confuses sandmeyer gattermann
Mixes up Cu reagents
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