Amines as Nucleophiles
A concise revision guide to amines as nucleophiles: their reactions with haloalkanes to give secondary amines and with acyl chlorides to give N-substituted amides.
Before You Start
The lone pair that makes an amine a base also makes it a nucleophile. Check these ideas first.
- A nucleophile is an electron-pair donor that attacks an electron-deficient (δ+) carbon atom.
- The carbon bonded to the halogen in a haloalkane is δ+, and ammonia substitutes the halogen: Nucleophilic Substitution Reactions.
- The carbonyl carbon of an acyl chloride is strongly δ+ (O and Cl both withdraw electrons) and reacts rapidly with water, alcohols, ammonia and amines: Acyl Chlorides.
- Amines as bases, and the salts they form with HCl: Amines as Bases.
With Haloalkanes: Secondary and Tertiary Amines
A primary amine attacks a haloalkane exactly as ammonia does. The product has two carbon groups on nitrogen: a secondary amine.
| Reaction | Equation | Product class |
|---|---|---|
| butylamine + bromoethane | CH₃CH₂CH₂CH₂NH₂ + CH₃CH₂Br → CH₃CH₂CH₂CH₂NHCH₂CH₃ + HBr | secondary amine: N-ethylbutylamine, CH₃CH₂CH₂CH₂NHCH₂CH₃ |
| secondary amine + bromoethane | CH₃CH₂CH₂CH₂NHCH₂CH₃ + CH₃CH₂Br → CH₃CH₂CH₂CH₂N(CH₂CH₃)₂ + HBr | tertiary amine: N,N-diethylbutylamine, CH₃CH₂CH₂CH₂N(CH₂CH₃)₂ |
| tertiary amine + bromoethane | one more substitution | quaternary ammonium salt |
- Conditions are the same as for ammonia: in ethanol, heated under pressure.
- The HBr formed is taken up by excess amine as its salt, so the equation is often written with two moles of amine.
- Each product still has a lone pair on nitrogen, so the haloalkane is attacked again: a mixture of secondary and tertiary amines and a quaternary ammonium salt forms.
- The quaternary ammonium salt has four carbon groups on a positive nitrogen and no lone pair, so substitution stops there. Excess amine favours the secondary amine.
Butylamine with bromoethane to the secondary amine, then on to the tertiary amine. Equations are all that is needed: no mechanism.
Exam focus: Name the product carefully: the group from the haloalkane goes on the nitrogen and takes the prefix N-. butylamine with bromoethane gives N-ethylbutylamine.
Check: Amines with Haloalkanes
Predict and name the amine formed when a named amine attacks a haloalkane, state the conditions and explain why a mixture of products forms.
Quick Check: Following Repeated Substitution
Work out what each amine goes on to form when the alkylating reagent is in large excess.
With Acyl Chlorides: Secondary (N-substituted) amides
The amine lone pair attacks the strongly δ+ carbonyl carbon of an acyl chloride. The reaction is a nucleophilic addition-elimination (the name is all that is needed): the amine adds, then chloride is eliminated, and an amide forms.
| Reaction | Equation | Product |
|---|---|---|
| butylamine + ethanoyl chloride | CH₃COCl + CH₃CH₂CH₂CH₂NH₂ → CH₃CONHCH₂CH₂CH₂CH₃ + HCl | N-butylethanamide, CH₃CONHCH₂CH₂CH₂CH₃ (secondary (N-substituted) amide) |
| with excess amine | CH₃COCl + 2CH₃CH₂CH₂CH₂NH₂ → CH₃CONHCH₂CH₂CH₂CH₃ + CH₃CH₂CH₂CH₂NH₃⁺Cl⁻ | the HCl is taken by the second amine as a salt |
| phenylamine + ethanoyl chloride | CH₃COCl + C₆H₅NH₂ → CH₃CONHC₆H₅ + HCl | N-phenylethanamide, a white solid |
| ammonia + ethanoyl chloride | CH₃COCl + NH₃ → CH₃CONH₂ + HCl | ethanamide, a primary amide (Amides) |
- Conditions: room temperature, no catalyst. The reaction is rapid and vigorous, with white fumes of HCl unless excess amine is present.
- The product is a secondary amide: the group from the amine sits on the nitrogen and takes the prefix N-.
- The amide link –CONH– formed here is the link in polyamides (Condensation Polymers: Polyamides and Polyesters).
Ethanoyl chloride with butylamine and with phenylamine: the amide link is boxed in each product. The equation and the name are what the exam asks for.
Exam focus: Write the amide with the N bonded to the carbonyl carbon: CH₃CO–NH–R. butylamine + ethanoyl chloride gives N-butylethanamide; phenylamine gives N-phenylethanamide.
Quick Check: Reasoning About Amide Formation
Pick the one accurate statement in each set.
Worked Example: Naming the Amide Product
Name the amide formed when propanoyl chloride reacts with methylamine.
Step 2. The amine supplies the group on the nitrogen: methyl, written with the prefix N-.
Five reactant pairs and their amide products, with the three naming rules.
What the lone pair attacks in a haloalkane and in an acyl chloride: the δ+ carbon, the leaving group, the product class and the conditions.
Check: Amines with Acyl Chlorides
Write the equation for an amine or ammonia with an acyl chloride, name the amide formed and state what happens to the HCl.
Common Exam Points
- The nitrogen lone pair attacks a δ+ carbon: amines are nucleophiles as well as bases.
- Butylamine + bromoethane → N-ethylbutylamine + HBr in ethanol under pressure: a secondary amine, then a tertiary amine with more bromoethane.
- A mixture forms because every amine produced still has a lone pair; the series ends at a quaternary ammonium salt.
- Butylamine + ethanoyl chloride → N-butylethanamide + HCl at room temperature: nucleophilic addition-elimination.
- Phenylamine + ethanoyl chloride → N-phenylethanamide + HCl, a white solid; ammonia gives ethanamide.
- With excess amine the HCl is taken as the alkylammonium chloride: CH₃COCl + 2RNH₂ → CH₃CONHR + RNH₃⁺Cl⁻.
- Name the product from the acyl group and the N-substituent: a secondary amide such as N-methylpropanamide.
FAQs
Use these quick answers to check how the lone pair of butylamine attacks a haloalkane and an acyl chloride, the amines and amides formed and how to name them.
Why can an amine act as both a base and a nucleophile?
Both roles use the same lone pair on nitrogen. As a base it accepts a proton, H⁺. As a nucleophile it attacks an electron-deficient carbon, the δ+ carbon of a haloalkane or the carbonyl carbon of an acyl chloride, forming a new C–N bond.
What forms when butylamine reacts with bromoethane?
A secondary amine, N-ethylbutylamine, CH₃CH₂CH₂CH₂NHCH₂CH₃, and HBr, which excess amine takes up as a salt.
The new amine still has a lone pair, so it can attack more bromoethane to give a tertiary amine and finally a quaternary ammonium salt, so a mixture forms.
What is the product when an amine reacts with an acyl chloride?
A secondary amide. Ethanoyl chloride and butylamine give N-butylethanamide, CH₃CONHCH₂CH₂CH₂CH₃, plus HCl, in a vigorous reaction at room temperature. Phenylamine gives N-phenylethanamide, a white solid. The reaction is called nucleophilic addition-elimination.
Why is a second mole of amine needed with an acyl chloride?
The HCl released would protonate the amine, so one mole of amine is used up as its ammonium chloride salt: CH₃COCl + 2RNH₂ → CH₃CONHR + RNH₃⁺Cl⁻. Using excess amine ensures some free amine remains as the nucleophile and the yield of amide stays high.
How do I name the amide formed from an amine and an acyl chloride?
Name the acyl chloride part as the amide: ethanoyl chloride gives ethanamide. Then add the alkyl group that sits on the nitrogen as an N- prefix, so methylamine gives N-methylethanamide and phenylamine gives N-phenylethanamide. The small molecule lost is always HCl.
Copyright notice: This OLS revision content, including the explanations, layout, diagrams, tables and embedded learning structure, is authored for Online Learning System by Dr. Mohammed Al-Fatah. It may not be copied, reproduced, redistributed or adapted without written permission.
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