Organic Chemistry · Amines and Heterocycles
Reactions of Amines
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In 30 seconds
An amine carries a nitrogen atom with a lone pair of electrons, and nearly every reaction in this topic is that lone pair doing one of two jobs: grabbing a proton (acting as a base) or attacking an electron-poor carbon (acting as a nucleophile). Alkyl amines have conjugate-acid pKa values near 10–11, so they are strong bases and good nucleophiles
Amine reactions fall into a few families: alkylation Adding an alkyl group to nitrogen by S_N2 displacement Full entry → (adding alkyl groups to nitrogen), acylation Converting an amine into an amide with an acid chloride/anhydride Full entry → (converting the amine into an amide A carbonyl attached to nitrogen, RC(=O)NR2 Full entry →), condensation with aldehydes and ketones (imines and enamines), reaction with nitrous acid (diazonium ions), and oxidation. Learn where the electrons go and why each reaction stops, and you can predict unfamiliar products.
Why this matters
- Drug synthesis: Lidocaine, antihistamines, and countless medicines are built by alkylating or acylating amines.
- Peptide bonds: Nature links amino acids by acylating one amine with a neighboring carboxyl — the amine-to-amide conversion makes proteins.
- Vision: Rhodopsin holds its retinal chromophore as an imine (Schiff base) formed between an aldehyde and a lysine side chain.
The college version
Core Concepts
The lone pair: base and nucleophile in one
Basicity is reported as the pKa of the conjugate acid (pKaH): for methylamine, pKaH ≈ 10.6, so at physiological pH alkyl amines are mostly protonated ammonium ions. Nucleophilicity tracks basicity for unhindered amines; bulky amines (e.g., triethylamine) stay basic but are poor nucleophiles because the nitrogen is crowded.
Alkylation: to the quaternary ammonium salt
An amine displaces a halide from an alkyl halide:
RNH2 + CH3I ⟶ RNH2CH3 + I-
The product is still an amine and still nucleophilic, so it reacts again. With excess methyl iodide and base, all N–H hydrogens are replaced:
RNH2 + 3 CH3I excess, base⟶ RN+(CH3)3 I- + 3 HI
The quaternary ammonium salt Nitrogen with four alkyl groups, R4N+ Full entry → has no N–H bond and a permanent positive charge; the reaction stops there. This over-alkylation is why alkylation is a poor route to pure secondary or tertiary amines — reductive amination (Topic 6) is preferred.
Acylation: the off switch
An acid chloride converts an amine into an amide:
RNH2 + CH3C(=O)Cl ⟶ CH3C(=O)NHR + HCl
In words: the nitrogen lone pair attacks the carbonyl carbon, the C=O π electrons move onto oxygen, chloride leaves, and the HCl is neutralized by added base (aqueous hydroxide or pyridine — Schotten–Baumann conditions). The amide's lone pair is delocalized into the carbonyl, so amides are nearly nonbasic (pKaH ≈ −0.5). Acylation therefore turns the amine off — the standard way to protect an amino group before chemistry elsewhere in the molecule.
Condensation: imines and enamines
A primary amine condenses with an aldehyde or ketone to give an imine (Schiff base), losing water:
RNH2 + R'2C=O ⇌ R'2C=NR + H2O
Mechanism in words: nitrogen attacks the carbonyl carbon, a proton transfer turns C–OH into water, and water leaves as the C=N bond forms. A secondary amine gives an enamine A C=C–N compound from a secondary amine + carbonyl Full entry → (C=C–N) instead. Imine formation is reversible — the basis of reductive amination.
Nitrous acid and oxidation
Nitrous acid (from NaNO₂ + HCl) converts a primary amine into a diazonium ion A RN2+ group from amine + nitrous acid Full entry →:
RNH2 + HNO2 + H+ ⟶ RN2 + + 2 H2O
Aliphatic diazonium ions are unstable: N₂ leaves, and the resulting carbocation gives a messy mixture of alcohols, alkenes, and substitution products. Aromatic diazonium salts are stable when cold — the gateway to Sandmeyer chemistry (Topic 8). Quaternary ammonium hydroxides undergo the Hofmann elimination on heating; the bulky amine leaving group makes it favor the least substituted alkene — opposite to most E2 reactions.
How It Works / Step-by-Step Process
To predict the product of any amine reaction:
- Classify the amine: count the N–H bonds (primary, secondary, tertiary).
- Identify the reagent's role: alkyl halide (alkylation), acid chloride (acylation), aldehyde/ketone (condensation), nitrous acid (diazotization), peroxide (oxidation).
- Predict where the lone pair goes: onto carbon (attack), onto H⁺ (protonation), or nowhere (no N–H left).
- Check the stopping point: if the product still has an N–H, expect further reaction; acylation and quaternization stop the train.
- Name the product class: ammonium salt, amide, imine, enamine, diazonium salt, or alkene.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Amines | Amides | Amines are basic and nucleophilic (pKaH ~10); amides are nearly neutral (pKaH ≈ −0.5) |
| Alkylation | Acylation | Alkylation adds alkyl groups and over-reacts; acylation forms an amide and stops after one addition |
| Imine | Enamine | Imine: C=N with H on nitrogen (1° amine); enamine: C=C–N (2° amine) |
| "Tertiary amines form amides" | 1°/2° amine acylation | Acylation needs an N–H bond; tertiary amines just form ammonium salts |
| Hofmann product | Zaitsev product | Hofmann elimination favors the least substituted alkene; ordinary E2 favors the most substituted |
Eli explains
The same idea, in plain words
Explain it like I’m 10
An amine is like a kid holding out a sticky hand — the lone pair — ready to grab protons or carbon atoms. Give it small carbon pieces (alkyl groups) and it grabs them one after another until all four hands are full, then it stops. Put a "coat" on it (an acyl group) and the sticky hand gets covered, so it calms down. With nitrous acid, the amine trades its sticky hand for a "pop-off" handle that chemists can later replace with almost anything.
Worked example
Example 1: How much methyl iodide quaternizes an amine?
Benzylamine is converted to its quaternary ammonium salt with excess methyl iodide; each molecule needs three methyl groups. How many grams of methyl iodide (molar mass 141.94 g/mol) are needed for 2.5 mmol of amine?
Moles of methyl iodide:
n(CH3I) = 3 × n(amine) = 3 × 0.0025 mol = 0.0075 mol
Converting to mass:
m = n × M = 0.0075 mol × 141.94 gmol = 1.06 g
Units check: mol × g/mol = g. The answer, 1.06 g, also shows why alkylation is wasteful for making pure mono- or dialkylamines: three equivalents of alkylating agent are consumed, and the product is still a salt.
Example 2: Acylation kills basicity — prove it with numbers
Compare hexylamine and its acetylated product, N-hexylacetamide, at pH 7 (pKaH ≈ 10.6 for the amine, ≈ −0.5 for the amide). Henderson–Hasselbalch in the form:
log[B][BH+] = pH - pKaH
For hexylamine:
log[B][BH+] = 7.0 - 10.6 = -3.6 ⇒ [B][BH+] = 10-3.6 ≈ 2.5 × 10-4
Only ~0.025% of the amine is free base; 99.97% is protonated. For the amide:
log[B][BH+] = 7.0 - (-0.5) = 7.5 ⇒ [B][BH+] ≈ 3 × 107
The amide is essentially 100% neutral — exactly why acylation protects and deactivates amine nitrogens.
Key takeaways
- Alkylation over-reacts: excess CH₃I + base drives 1° amines to quaternary ammonium salts (3 equivalents per N–H₂).
- Acylation is the off switch: acid chloride → amide; amides are essentially nonbasic (pKaH ≈ −0.5).
- Primary amine + aldehyde/ketone → imine (C=N); secondary amine → enamine (C=C–N).
- Primary aliphatic amines + HNO₂ → unstable diazonium ions → messy mixtures; aromatic amines give stable cold diazonium salts.
- Hofmann elimination gives the least substituted alkene; Cope elimination does the same via a syn pathway.
- Amines are poor leaving groups; quaternization or protonation is required before nitrogen can leave.
Check yourself
5 review questions from the chapter. Try each one, then open the answer.
Why is it hard to stop methyl iodide alkylation at the secondary amine stage?
Show answer
The mono-alkylated product is still an amine with a lone pair and N–H bonds, so it stays nucleophilic and reacts again; excess methyl iodide and base drive it all the way to the quaternary ammonium salt.
Predict the product when cyclohexylamine is treated with acetyl chloride and aqueous NaOH.
Show answer
N-cyclohexylacetamide: the amine attacks the acid chloride's carbonyl carbon, chloride leaves, and the HCl is neutralized by hydroxide.
What structural feature tells an imine from an enamine, and which amine class makes each?
Show answer
An imine has a C=N bond with a hydrogen on nitrogen (primary amine); an enamine has a C=C–N unit (secondary amine).
What happens when a primary aliphatic amine meets nitrous acid, and why is the result usually a mixture?
Show answer
Nitrous acid (from NaNO₂ + HCl) makes an aliphatic diazonium ion, which loses N₂ to give a carbocation; the carbocation forms a mixture of alcohols, alkenes, and rearranged products.
Which alkene — more substituted or less substituted — is favored in a Hofmann elimination, and why?
Show answer
The less substituted alkene (Hofmann product), because the bulky trialkylamine leaving group forces base to abstract the most accessible β-hydrogen.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- alkylation
- Adding an alkyl group to nitrogen by S_N2 displacement
- quaternary ammonium salt
- Nitrogen with four alkyl groups, R4N+
- acylation
- Converting an amine into an amide with an acid chloride/anhydride
- amide
- A carbonyl attached to nitrogen, RC(=O)NR2
- imine (Schiff base)
- A C=N compound from a primary amine + aldehyde/ketone
- enamine
- A C=C–N compound from a secondary amine + carbonyl
- diazonium ion
- A RN2+ group from amine + nitrous acid
Sources & references
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