Did you know? Making a primary amine creates a molecule with another reactive nitrogen lone pair. In the bromoethane–ammonia reaction, that means the desired ethylamine can sometimes compete for further alkylation.
JC2 H2 Chemistry Tutor · Hougang · 2027 syllabus 9476. Twelve original learning chapters and a hypothetical mole ledger. The discussion is theoretical and is not a recipe for hazardous, pressurised organic reactions.
Twelve chapters · Worked mole ledger · Transfer problem · FAQs · Syllabus
Chemistry 234 · Connected JC tutorials: JC1 H1 | Limestone purity from CO₂ · JC1 H2 | AlCl₃ dimers and bridging chlorides · JC2 H3 | Neopentyl SN2 steric hindrance
Twelve chapters: substitution, primary amines and selectivity
- 1. 1. Identify the electrophilic carbon and its leaving group
- 2. 2. Explain why NH₃ can attack carbon with an electron pair
- 3. 3. Draw the immediate ethylammonium product with a positive charge
- 4. 4. Show how a second ammonia molecule gives ethylamine
- 5. 5. Classify ethylamine by nitrogen substituents, not carbon count
- 6. 6. Understand why the intended amine can still react
- 7. 7. Distinguish later secondary, tertiary and quaternary nitrogen products
- 8. 8. Explain what excess ammonia changes—and what it cannot promise
- 9. 9. Solve a hypothetical selective-substitution mole ledger
- 10. 10. Distinguish ammonia substitution from hydroxide substitution and elimination
- 11. 11. Diagnose six tempting errors in amine product prediction
- 12. 12. Apply the idea to 1-bromopropane and a changed limiting amount
1. Identify the electrophilic carbon and its leaving group
Bromoethane, CH₃–CH₂–Br, is a primary halogenoalkane. The carbon directly attached to bromine is a saturated CH₂ carbon connected to only one other carbon. Because bromine is more electronegative, the C–Br bond is polar and can undergo nucleophilic substitution under suitable conditions.
An organic reaction is not determined solely by seeing bromine in a formula. You must identify the reagent, the reactive atom and the type of bond change. In the 2027 H2 Chemistry syllabus, bromoethane is a named example for making a primary amine by reaction with ammonia in ethanol under heating and pressure. The exact conditions are syllabus information, not an invitation to perform a dangerous reaction independently.
This guide asks a subtler question beyond simply naming the primary amine: when the desired product itself has a nitrogen lone pair, why might later reactions compete with its formation?
2. Explain why NH₃ can attack carbon with an electron pair
Ammonia has a lone pair on nitrogen. As a Lewis base it donates an electron pair; in this substitution context it acts as a nucleophile toward the partially positive carbon of CH₃CH₂Br. The same electron-pair idea appears in a wide range of H2 organic mechanisms.
In the ideal SN2 drawing, a full curved arrow begins at the nitrogen lone pair and points to the bromine-bearing carbon, while a second full curved arrow from the C–Br bond points to bromine. Formation of the C–N bond and cleavage of C–Br occur together within the concerted substitution step.
A common drawing error starts the first arrow at the carbon as though that electron-deficient atom were donating electrons to nitrogen. Mechanistic arrows represent the movement of electron pairs, and the arrow should begin where those electrons already exist.
3. Draw the immediate ethylammonium product with a positive charge
The immediate organic species arising from a neutral ammonia nucleophile gaining a bond to carbon is CH₃CH₂NH₃⁺. It has four bonds at nitrogen: one N–C and three N–H. Bromide is the accompanying negative ion after C–Br cleavage.
This is not yet neutral ethylamine, whose formula is CH₃CH₂NH₂. A student who erases an N–H bond without showing its fate has omitted a necessary chemical change. Nitrogen bond counting gives the clue: a four-bonded ammonium nitrogen typically has positive formal charge.
The chemical name “ethylammonium” is distinct from “ethylamine”. The former is a protonated nitrogen species, while the latter has a lone pair available for further acid/base or nucleophilic chemistry.
4. Show how a second ammonia molecule gives ethylamine
Another ammonia molecule can accept a proton from ethylammonium, producing neutral ethylamine CH₃CH₂NH₂ and ammonium NH₄⁺. In the simple net accounting, bromide accompanies ammonium as NH₄Br. The combined school equation is CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br.
One ammonia molecule contributes nitrogen to the organic product. The other serves as the base in the proton transfer. This explains why the net coefficient of ammonia is two even though only one nitrogen atom becomes attached to the carbon chain.
Check conservation independently: carbon remains two; bromine from the original substrate ends in the ammonium bromide component; the total nitrogen count is two on each side; and all hydrogens are accounted for. Balanced net equations complement, but do not replace, a valid electron-pair mechanism.
5. Classify ethylamine by nitrogen substituents, not carbon count
Ethylamine is a primary amine because its nitrogen is directly bonded to one carbon-containing group and has two N–H bonds. The word primary does not mean its carbon chain contains one carbon atom. An ethyl group contains two carbon atoms, but it is still just one carbon-containing substituent on nitrogen.
A secondary amine has two carbon-containing substituents and one N–H bond, whereas a tertiary amine has three carbon substituents and a nitrogen lone pair. A quaternary ammonium species has four carbon substituents and carries positive charge, with no lone pair available on that nitrogen for the same type of further alkylation.
When questions ask for an amine class, mark the nitrogen and count its directly bonded carbon groups. Classifying from the alkyl chain’s length or from the number of molecules used during a synthesis will give the wrong answer.
6. Understand why the intended amine can still react
Neutral ethylamine has a nitrogen lone pair. Once some ethylamine exists in the mixture, it may also compete as a nucleophile for remaining bromoethane. A subsequent carbon–nitrogen bond-forming substitution can lead, after suitable proton transfer, to diethylamine (CH₃CH₂)₂NH.
This is a meaningful selectivity problem. The reaction can produce the primary amine as intended, but the primary amine is not chemically inert. Reacting species change over time, and the availability of bromoethane and competing nitrogen bases can influence which products accumulate.
Do not use this structural reasoning to invent a precise yield of diethylamine. A qualitative possibility is not a quantitative prediction; reaction rates, concentrations, medium and changing speciation would be needed to calculate product proportions.
7. Distinguish later secondary, tertiary and quaternary nitrogen products
If a secondary amine is alkylated again, a tertiary amine with three ethyl groups on nitrogen may be formed in a suitably chosen reaction model. Triethylamine, (CH₃CH₂)₃N, has no N–H hydrogen but still has a lone pair. A further alkylation may lead to tetraethylammonium, (CH₃CH₂)₄N⁺.
The final quaternary ammonium species cannot lose an N–H proton to become a neutral “quaternary amine”: there is no N–H hydrogen left to remove. The positive charge is a direct consequence of nitrogen having four carbon-bonded groups under the simple Lewis picture.
This sequence is an explanatory extension of the syllabus’s ethylamine formation example. It is not a guarantee that every alkylation stage will occur under one fixed set of conditions or in any particular proportion.
8. Explain what excess ammonia changes—and what it cannot promise
If free ammonia is abundant relative to the growing concentration of ethylamine, many available electrophilic halogenoalkane encounters may involve ammonia rather than the first organic amine product. Under suitable conditions that can favour the formation of primary amine over some further alkylation pathways.
But the phrase excess ammonia cannot guarantee a single perfectly pure product. Amine nucleophilicity and concentrations both matter, and a competing alkylation might still occur. Yield and selectivity are not the same quantity: a reagent can be available in excess without directing every reacting molecule toward one desired product.
A stronger answer explains the competition: NH₃ and CH₃CH₂NH₂ both contain nitrogen lone pairs, and their relative encounters and reaction probabilities depend on actual chemistry. The syllabus’s principal named product remains ethylamine; later alkylation is a bounded advanced consideration.
9. Solve a hypothetical selective-substitution mole ledger
Take an entirely constructed question with 0.0100 mol bromoethane and 0.0500 mol NH₃, and explicitly assume that only the net primary-amine-forming reaction occurs and that the limiting reagent reacts completely. From CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br, the 0.0100 mol bromoethane requires 0.0200 mol ammonia.
Ammonia is in excess by 0.0300 mol. The theoretical primary amine amount is 0.0100 mol. Using simple relative masses C=12.0, H=1.0 and N=14.0 gives Mᵣ(C₂H₇N)=45.0 and a theoretical ethylamine mass of 0.450 g. The same amount 0.0100 mol NH₄Br appears in the net ledger.
| Species | Initial amount | Ideal final amount |
|---|---|---|
| Bromoethane | 0.0100 mol | 0 |
| Ammonia | 0.0500 mol | 0.0300 mol |
| Ethylamine | 0 | 0.0100 mol |
| Ammonium bromide | 0 | 0.0100 mol |
The 0.450 g is a theoretical maximum under the intentionally simplified conditions, not a measured yield. If secondary amines, other products or incomplete conversion were relevant, the actual primary-amine amount would require additional data. Stating that limitation is part of doing the calculation correctly.
10. Distinguish ammonia substitution from hydroxide substitution and elimination
The same carbon–bromine bond can undergo different transformations depending on the reactant and conditions. A hydroxide-containing substitution can yield an alcohol, whereas an appropriate nitrogen nucleophile can yield an amine. The key outcome is the new bond that takes the place of the C–Br connection.
The H2 syllabus separately gives 2-bromopropane with ethanolic NaOH and heat as an elimination example. That reaction forms an alkene rather than simply replacing bromine with nitrogen or hydroxyl. Do not classify reaction type from the solvent name alone: read the substrate, reagent, conditions and mechanistic pathway.
This is a useful way of testing understanding with a reaction map. One starting halogenoalkane can be connected to multiple functional-group products, but the chemical changes must be justified by the stated conditions.
11. Diagnose six tempting errors in amine product prediction
- Drawing neutral ethylamine as the immediate product of NH₃ attack without an intermediate or deprotonation step.
- Counting the coefficient 2NH₃ as two nitrogen atoms incorporated into one ethylamine molecule.
- Calling ethylamine secondary because its ethyl group has two carbon atoms.
- Assuming the primary amine’s nitrogen can no longer act as a nucleophile.
- Treating tetraethylammonium as a neutral tertiary amine that can lose an N–H proton.
- Inventing an exact secondary-amine percentage from the presence of “excess ammonia” alone.
Each mistake has a different origin. Errors in nitrogen formal charge are repaired by bond counting; errors in naming are repaired by counting directly attached carbon groups; errors in selectivity need a competition model. A good tutor does not assign twenty extra questions until the source of the error is known.
The complete reasoning links molecular structure, electron-pair movement, acid–base follow-up and the possibility of further substitution. That is more durable than simply memorising “bromoethane + ammonia = ethylamine”.
12. Apply the idea to 1-bromopropane and a changed limiting amount
Consider the structurally related primary halogenoalkane 1-bromopropane, CH₃CH₂CH₂Br. By analogous nucleophilic substitution with ammonia, its initial primary-amine product can be propan-1-amine, CH₃CH₂CH₂NH₂. The mechanism and nitrogen bond-counting principles remain the same even though the chain has gained one carbon.
In a second invented mole test, suppose 0.0120 mol halogenoalkane undergoes the same completely selective primary-amination net stoichiometry but only 0.0200 mol ammonia is available. Since the net equation requires 2NH₃ per haloalkane, ammonia is limiting: maximum primary amine is 0.0100 mol, leaving 0.00200 mol halogenoalkane in the idealised ledger.
- Identify the carbon attached to the halogen.
- Draw electron donation from ammonia nitrogen and departure of halide.
- Show the resulting alkylammonium species and subsequent proton transfer.
- Count carbon substituents on nitrogen to name the amine class.
- Account for the extra ammonia base in the net equation.
- Qualify any purity or selectivity claims when further alkylation is possible.
The student who can complete those steps with a different substrate is ready to reason about primary-amine formation rather than simply reproducing one named 2027 syllabus equation.
Frequently asked questions
Why is ethylamine the named product of bromoethane and ammonia?
Nitrogen’s lone pair can displace bromide through substitution, and the first alkylammonium product can be deprotonated to yield ethylamine.
Why does the balanced overall equation use two NH₃?
One NH₃ supplies the nitrogen in ethylamine; another accepts the proton equivalent to give ammonium bromide.
Does a primary amine always stay primary after formation?
Not if suitable further alkylation occurs. Its nitrogen still has a lone pair that can react with additional halogenoalkane.
Why can excess ammonia help?
It can favour ammonia competing for the halogenoalkane rather than the newly formed amine, but cannot guarantee a unique pure product.
Does the H2 syllabus require practical preparation under pressure?
The 2027 9476 syllabus names ethanol, ammonia and heated pressure conditions for the reaction. That is course chemistry, not advice for unsupervised synthesis.
The small-group tutorial method
I begin by hiding the reaction-product name and asking the student to draw bromoethane, locate the electrophilic carbon and show where ammonia’s electron pair moves. We then distinguish ethylammonium from ethylamine, check the proton-transfer step and compare primary versus secondary amine structures. A new substrate and new mole values provide the exit test. The tutor corrects the first wrong charge, product or assumption rather than supplying only a finished equation.
The immutable eduKateSG Secondary 1 Mathematics Tutor Clementi reference is an unchanged model of guided teaching and feedback in a separate Mathematics programme. It is not a statement that this Chemistry class operates at that venue or has a current vacancy. Contact eduKateSG about current subject support.
Official curriculum and further links
Syllabus: 2027 Singapore-Cambridge H2 Chemistry 9476 Sections 11.3, 11.5 and 11.9 and SEAB 2027 official listing. Section 11.5(a)(i) names formation of primary amines from bromoethane with ammonia in ethanol heated under pressure. The further-alkylation discussion is a clearly labelled selectivity extension, not a fixed experimental yield claim.
Related: SN1/SN2 mechanisms · Ethylamine versus ethanamide basicity · Substitution and elimination conditions. Organic reagents and pressurised conditions require trained supervision; the article contains no practical synthesis procedure.
Original eduKateSG H2 Chemistry tutorial · 31 October 2026 · Structure, electron arrows, selectivity, quantitative limits.