Did you know that making one molecule of an N-substituted amide from ethanoyl chloride and a primary amine can consume two molecules of the amine, even though only one of them becomes part of the amide? The first amine molecule supplies the nitrogen that forms the new bond to the acyl carbon. The second can act as a base, accepting the proton associated with the hydrogen chloride produced in the overall reaction. Under the familiar simple reaction model without a separate added base, the extra amine ends up as an alkylammonium chloride salt.
For JC students searching for an H2 Chemistry tutor in Hougang, ethanoyl chloride methylamine amide reaction, why two moles amine are needed, N-methylethanamide, acyl chloride with primary amine, 9476 nucleophilic acyl substitution or A-Level organic synthesis stoichiometry, the key lesson is not to confuse the atoms incorporated into the desired product with all the reagent molecules consumed along the way.
Take ethanoyl chloride, CH₃COCl, and methylamine, CH₃NH₂. A familiar idealised equation is:
CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃Cl.
The main organic product is N-methylethanamide, CH₃CONHCH₃. The other product is methylammonium chloride, CH₃NH₃⁺Cl⁻, a salt formed through proton acceptance by another methylamine molecule.
The first amine’s nitrogen is now part of the amide C(=O)–N bond. The second amine’s nitrogen remains in an ammonium salt. Those are two different roles played by the same chemical reagent.
This fits the revised Singapore H2 Chemistry 9476 syllabus treatment of acyl chlorides reacting with primary amines to form amides and the acid–base behaviour of nitrogen compounds. We will balance the reaction, explain the chemical mechanism, calculate limiting reagents, compare ammonia and amines, and show when the second equivalent can be supplied by a different external base instead.
At eduKate, we aim to explain the first unstable concept in an organic pathway and then ask the learner to solve a changed molecule independently. Families reading from Hougang should verify current lesson availability, class venue and transport arrangements directly; a Hougang-targeted guide does not by itself establish a classroom in Hougang.
Hougang Chemistry Tutor Library · Official 2027 H2 Chemistry syllabus directory
Identify the two functional groups
Ethanoyl chloride, CH₃COCl, contains an acyl chloride functional group:
R–C(=O)–Cl.
The reactive centre is the carbonyl carbon, bonded to oxygen through a double bond and to chlorine through a single bond.
Methylamine, CH₃NH₂, is a primary amine. The nitrogen has a lone pair and can donate that electron pair to an electrophilic reaction centre.
It can also accept a proton, behaving as a Brønsted–Lowry base.
That dual behaviour is important.
A single reagent may participate as a nucleophile through electron-pair donation to carbon, and as a base through proton acceptance.
Those words describe different reaction roles. In our overall equation, different methylamine molecules can perform the different roles.
The resulting amide is not a simple ammonium salt; it contains a new C(=O)–N covalent linkage.
What is an amide?
An amide contains a carbonyl group directly bonded to nitrogen, as in:
R–C(=O)–N(R′)(R″),
where the nitrogen substituents can include hydrogen or organic groups.
The specific product in our example is:
CH₃–C(=O)–NH–CH₃.
This is N-methylethanamide.
The name tells us two things:
- Ethano- refers to the two-carbon acyl portion originally provided by ethanoyl chloride.
- N-methyl- tells us that a methyl group is attached to the nitrogen atom, not to the second carbon of the ethanoyl chain.
The product contains three carbon atoms total.
That carbon count is important. The ethanoyl fragment contributes two carbons, and the methylamine incorporated into the amide contributes one carbon.
The second methylamine molecule does not add a fourth carbon to the desired amide. Its carbon remains in the methylammonium chloride coproduct.
Why the acyl carbon is an electrophile
The C=O bond is polar because oxygen attracts electron density more strongly than carbon.
The carbonyl carbon is correspondingly electron-deficient and susceptible to attack by an appropriate nucleophile.
Methylamine’s nitrogen lone pair can interact with that carbonyl carbon and form a new bond.
The initial electron-pair attack changes the bonding at the carbonyl centre, producing a tetrahedral acyl-addition intermediate in the familiar mechanism.
The carbonyl group is then re-formed as the chloride-containing leaving group departs.
This is described as nucleophilic acyl substitution: one group attached to the acyl carbon is ultimately replaced by another.
It is not the same as SN2 displacement at a saturated alkyl carbon, and it is not an electrophilic addition across an alkene double bond.
The functional group and local bonding environment determine the mechanism.
The first amine becomes part of the amide
The molecule CH₃NH₂ has a nitrogen lone pair.
In the simplified mechanism, nitrogen uses that electron pair to form a bond with the electrophilic acyl carbon of CH₃COCl.
After nucleophilic addition, elimination of chloride and appropriate proton transfer, the desired amide framework forms:
CH₃CO–NH–CH₃.
One original N–H proton of methylamine is effectively lost during the transformation.
The resulting nitrogen in the amide is bonded to the carbonyl carbon, the methyl group and one remaining hydrogen.
Therefore one methylamine molecule is a structural building block of the desired product.
The key chemical transformation can be summarised as:
Acyl chloride C–Cl → amide C–N.
However, that summary is not the complete acid–base inventory. We still need to account for the displaced chloride and the proton removed from the reacting amine.
Where does the acid byproduct come from?
The simple formal net acylation can be written:
CH₃COCl + CH₃NH₂ → CH₃CONHCH₃ + HCl.
This equation is balanced and shows that HCl is generated formally when the C–Cl group is replaced by the amine-derived nitrogen group and a proton is released.
But methylamine is itself a base. If additional free methylamine is present, it can accept the HCl-derived proton:
CH₃NH₂ + HCl → CH₃NH₃⁺Cl⁻.
The product can be written as CH₃NH₃Cl, methylammonium chloride.
Add these two equations and cancel the formal HCl intermediate:
CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃Cl.
That is why two methylamine molecules appear in the familiar full reaction when the second amine is the acid scavenger.
One is incorporated into the amide; the other is protonated.
Is the second methylamine molecule always compulsory?
No. The two-equivalent equation describes the simple case where another methylamine molecule accepts the acid generated, with no separate external base.
If a suitable different base is present to remove the proton, the second equivalent of methylamine need not be the species that captures it.
In general form:
RCOCl + R′NH₂ + B → RCONHR′ + BH⁺ + Cl⁻,
where B is a suitable base, with the actual ionic form and charge balance depending on that base.
Under this model, one equivalent of the primary amine can be incorporated into the amide, while a different base accepts the proton.
Therefore it would be inaccurate to say:
“Every acyl chloride reaction requires precisely two molecules of every primary amine under all conditions.”
The chemically correct statement is:
“In the familiar reaction where the amine itself neutralises the generated HCl, two amine equivalents are included per acyl chloride; with an appropriate external base, the stoichiometric amine requirement can differ.”
This distinction is useful for exam questions that explicitly name a base in addition to the amine.
Balance atoms and charge in the full equation
Use the equation:
CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃Cl.
Check each species.
Ethanoyl chloride has formula C₂H₃OCl.
Methylamine is CH₅N.
N-methylethanamide is C₃H₇NO.
Methylammonium chloride is CH₆NCl.
The left-hand total is:
C: 2 + 2(1) = 4.
H: 3 + 2(5) = 13.
N: 2. O: 1. Cl: 1.
The right-hand total is:
C: 3 + 1 = 4.
H: 7 + 6 = 13.
N: 1 + 1 = 2. O: 1. Cl: 1.
The equation is balanced.
A student who omits the ammonium salt has failed to account for atoms and proton transfer unless the alternative byproduct HCl or a suitable added base is explicitly included.
An examination-friendly reaction table
| Starting materials | Desired covalent product | Other product in stated model | Main lesson |
|---|---|---|---|
| CH₃COCl + 2CH₃NH₂ | N-methylethanamide | Methylammonium chloride | One amine incorporated; one captures acid |
| CH₃COCl + 2NH₃ | Ethanamide | Ammonium chloride | Ammonia serves as nucleophile and base |
| CH₃COCl + CH₃NH₂ + suitable external base | N-methylethanamide | Protonated external base and chloride | Acid scavenger need not be another amine |
| CH₃COCl + CH₃CH₂OH | Ethyl ethanoate | HCl in simple overall equation | Alcohol gives ester, not amide |
The table distinguishes the incoming nucleophile from the functionality of the organic product.
Do not assume that any nitrogen-containing molecule gives the same amide or that oxygen- and nitrogen-containing nucleophiles are interchangeable.
Compare ammonia and methylamine
The simplest nitrogen nucleophile comparison is ammonia, NH₃.
Reacting ethanoyl chloride with sufficient ammonia can produce ethanamide, CH₃CONH₂.
The familiar overall equation is:
CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl.
The desired product contains a carbonyl carbon bonded to NH₂, because the incoming nucleophile supplied nitrogen and hydrogens without a methyl substituent.
When methylamine is used instead, the amide nitrogen carries a methyl substituent:
CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃Cl.
The molecular products have different carbon counts and names.
If a student writes ethanamide as the main product from methylamine, they have lost the incoming CH₃ group.
The correct method is to draw what the nucleophile actually contributes.
Why a primary amine is not the same as a primary amide
Methylamine, CH₃NH₂, is a primary amine because nitrogen is bonded to one carbon-containing group and two hydrogens.
N-methylethanamide, CH₃CONHCH₃, is an amide containing the C(=O)–N linkage.
Do not classify an amide as a simple amine just because it contains nitrogen bonded to carbon.
The amide nitrogen’s lone pair can interact through resonance with the adjacent carbonyl group. This makes ordinary amides much less basic than comparable simple alkylamines under familiar conditions.
That difference helps explain why additional methylamine, rather than the final N-methylethanamide product, is the useful base in the simple acid-capture equation.
The nitrogen environments differ before and after acylation.
This is a major transfer point between the H2 nitrogen compounds and carboxylic acid derivatives chapters.
What about a secondary amine?
A secondary amine has nitrogen bonded to two carbon-containing groups and still has an N–H bond.
Under suitable acylation conditions, it can form an N,N-disubstituted amide by replacing that N–H connection with an acyl-derived carbon–nitrogen bond.
For example, dimethylamine can give an N,N-dimethyl amide derivative under a suitable acyl chloride reaction.
The formal proton-loss and acid-scavenging issue remains relevant.
A tertiary amine lacks an N–H bond, so it cannot simply undergo the same straightforward conversion into a neutral N-acyl tertiary amide through loss of an N–H proton.
Tertiary amines can still act as bases or participate in other acyl chemistry, but that is a different mechanistic context.
In H2, focus first on the exact primary-amine or ammonia example supplied by the question rather than blindly extending one equation to every nitrogen reagent.
Worked mole example: acid chloride as limiting reagent
Suppose a hypothetical theoretical problem states:
n(CH₃COCl) = 0.0200 mol.
n(CH₃NH₂) = 0.0500 mol.
No separate proton-accepting base is included, and the model equation is:
CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃Cl.
Each mole of ethanoyl chloride requires two moles of methylamine.
To consume 0.0200 mol ethanoyl chloride completely, we need:
n(methylamine required) = 2 × 0.0200 = 0.0400 mol.
There are 0.0500 mol available, so methylamine is in excess.
Ethanoyl chloride is the limiting reagent.
The theoretical desired amide amount is:
n(N-methylethanamide) = 0.0200 mol.
The excess methylamine amount remaining under the simple stoichiometric model is:
0.0500 − 0.0400 = 0.0100 mol.
The key is not to treat the reactant mole ratio as 1:1 when the stated equation explicitly consumes an additional amine equivalent in acid neutralisation.
Calculate the theoretical amide mass
N-methylethanamide has formula C₃H₇NO.
Using approximate atomic masses C = 12, H = 1, N = 14 and O = 16:
Mr(C₃H₇NO) = 3(12) + 7(1) + 14 + 16 = 73.
From the previous limiting-reagent calculation:
n(amide) = 0.0200 mol.
Therefore:
m(amide) = 0.0200 × 73 = 1.46 g.
The theoretical methylammonium chloride coproduct amount is also 0.0200 mol under the balanced equation.
This is a purely hypothetical stoichiometric calculation, not a predicted practical yield or a preparation procedure. Actual reaction conversion, side reactions and recovery conditions would require separate evidence.
A student who calculates twice as much amide because there are two methylamine molecules in the equation has confused the amine’s two roles.
Only one methylamine molecule is incorporated into each amide molecule.
What if methylamine is limiting?
Now change the invented initial amounts to:
0.0200 mol ethanoyl chloride.
0.0300 mol methylamine.
Keep the same ideal full reaction model with methylamine acting as both nucleophile and acid scavenger.
The 2:1 methylamine-to-acyl chloride requirement means that 0.0300 mol methylamine can support at most:
0.0300/2 = 0.0150 mol of the stated overall reaction.
Thus the theoretical N-methylethanamide amount is 0.0150 mol.
Using Mr 73:
m(amide) = 0.0150 × 73 = 1.095 g ≈ 1.10 g.
Under the ideal model, ethanoyl chloride is left in excess. A real system would require care because unreacted acyl chloride can participate in other reactions, especially with moisture.
The purpose here is to practise limiting reagent identification using all balanced coefficients.
If a different base is present, the stoichiometry must be rewritten to reflect that new condition before doing the arithmetic.
Why a reaction can have a reasonable product yield but poor atom economy
In the full simple reaction equation, two methylamine molecules are consumed for every one amide formed.
One becomes part of the desired amide; the other becomes part of methylammonium chloride.
If the desired product is defined as N-methylethanamide only, the coproduct counts outside the atom-economy numerator.
Using the approximate formula masses:
Mr(CH₃COCl) = 78.5.
Mr(CH₃NH₂) = 31.0.
Mr(N-methylethanamide) = 73.0.
The total reactant formula mass is:
78.5 + 2(31.0) = 140.5.
So the ideal reaction atom economy for the desired amide is:
73.0/140.5 × 100% ≈ 52.0%.
The remaining reactant mass is accounted for by methylammonium chloride, with Mr 67.5.
Check:
73.0 + 67.5 = 140.5.
This calculation tells us about stoichiometric material incorporation, not the actual isolated yield of amide in a laboratory or plant.
A second atom-economy comparison
Suppose the desired amide can be made under conditions using one equivalent of methylamine and a different suitable external base.
We cannot automatically claim that route has exactly 100% atom economy, because the external base and its protonated coproduct also contribute to the material balance.
The new atom-economy calculation must include all stoichiometrically consumed reagents and the defined desired product, using the fully balanced equation for the specific base.
A base that is regenerated catalytically would be treated differently from a base consumed as a salt.
This is another reason the words “with a suitable external base” are not sufficient numerical data for an exact atom-economy percentage.
Chemistry students should resist inventing a green-chemistry score when the reaction and process boundary have not been fully specified.
Why a catalyst is not the same as an acid scavenger
A catalyst speeds a process through a reaction pathway and is regenerated overall.
An acid scavenger, such as an added base that becomes permanently protonated in the net equation, is consumed stoichiometrically.
If a base ends as BH⁺Cl⁻, it should not automatically be called a catalyst simply because it supports the formation of the amide.
It is possible for substances to play several roles in a chemical system, but the overall balanced reaction tells us whether they are regenerated or consumed under the model.
This matters in reaction conditions and industrial efficiency questions.
A student who omits a consumed base from a mass-balance calculation may underestimate the reactant input.
How the mechanism differs from ordinary alkyl substitution
The acyl chloride’s electrophilic centre is an sp² carbonyl carbon.
The amine attacks the C=O-associated centre and the reaction proceeds through addition–elimination in the familiar nucleophilic acyl substitution model.
An alkyl halide such as bromoethane instead reacts at a saturated sp³ carbon, where substitution can follow an SN1 or SN2 pathway depending on the substrate and conditions.
Both reaction families can involve a nucleophile and a leaving group, but the local bonding changes and transition structures differ.
Do not automatically draw an SN2 backside attack at a saturated carbon when the reagent is CH₃COCl.
The carbon bonded to chlorine here is also bonded through a double bond to oxygen. That identifies the acyl chloride functional group.
The best first step in a mechanism question is to mark that C=O bond before drawing any arrows.
Why amides are important in protein chemistry
Amide linkages occur in many organic compounds.
A peptide bond in a protein is an amide linkage formed between amino acid residues through a different biological or synthetic condensation context.
The structural feature –C(=O)–N– is shared.
That does not mean proteins are routinely produced by treating amino acids with ethanoyl chloride. Biological peptide formation involves organised cellular machinery and other activation chemistry.
The connection is functional-group identity, not an assertion of an identical practical method.
Recognising the amide bond helps students understand why peptides have different chemical and physical properties from free amines and carboxylic acids.
It also supports the 9476 polymers section, where condensation polymers and their hydrolysis are relevant.
Why the amide nitrogen is less basic
A simple alkylamine has a nitrogen lone pair that is relatively available to accept a proton under ordinary conditions.
In an amide, the nitrogen is adjacent to a carbonyl group. Its lone pair can interact with the carbonyl π system, creating resonance contributors that share electron density over the N–C–O framework.
That means the lone pair is less available for proton acceptance than in a comparable simple alkylamine.
Consequently amides are generally much less basic than their corresponding amine precursors.
This reinforces why an extra molecule of free methylamine can neutralise the HCl byproduct in the simple equation.
The reaction product’s nitrogen does not behave just like an unreacted amine molecule, because its electron environment has changed.
A student who labels both the amide and methylamine as equally basic nitrogen compounds has missed the effect of carbonyl conjugation.
Why HCl capture affects the reaction mixture
Hydrogen chloride produced during acylation can protonate a free primary amine:
CH₃NH₂ + H⁺ → CH₃NH₃⁺.
The protonated methylammonium species has no available neutral amine lone pair in the same way as unprotonated methylamine for the desired nucleophilic attack.
Thus acid generation and acid–base capture can affect the mixture of reactive amine species.
That is one reason the amount and role of added base matter in acylation systems.
The exact chemical equilibria depend on solvent, reagent composition and conditions; a simple 2:1 equation captures the net stoichiometry when an additional free amine equivalent accepts the acid.
It does not mean every molecule in the reaction passes through one rigid time sequence or that the acid must first accumulate as pure gaseous HCl.
The mechanism and net equation are different descriptions of the same overall chemistry.
A corrected reaction-pathway table
| Student’s proposed answer | What needs fixing | Better reasoning |
|---|---|---|
| CH₃COCl + CH₃NH₂ → N-methylethanamide with no other product | Atoms and proton balance incomplete | Account for chloride and a proton as HCl or a captured salt |
| Two methylamine molecules make one amide with two N atoms | Wrong desired product composition | One amine supplies product nitrogen, the other captures acid |
| CH₃COCl + methylamine forms ethanamide | Lost N-methyl substituent | Incoming CH₃NH₂ carries a methyl group onto amide nitrogen |
| Methylamine is only a nucleophile | Ignored its proton-accepting ability | Can another molecule neutralise HCl? |
| Any acyl chloride always needs exactly two amine equivalents | Ignored separately supplied external base | Which species accepts the generated proton? |
| Acyl chloride + amine is SN2 at an alkyl carbon | Misidentified functional group | Where is the C=O and electrophilic acyl carbon? |
| All nitrogen groups have equal basicity | Ignored amide resonance | Where is the nitrogen lone pair delocalised? |
A student-friendly six-step method
- Identify the acyl chloride group R–C(=O)–Cl.
- Identify the nucleophile, including the organic groups attached to its N atom.
- Draw the covalent amide product, retaining the acyl carbonyl and attaching the nucleophile-derived nitrogen.
- Account for chloride and the N–H proton, using HCl or the appropriate ammonium salt depending on the stated base.
- Balance the full equation, including any additional amine or external base acting as acid scavenger.
- Only then calculate moles, mass, limiting reagent or atom economy from the final stoichiometry.
This method is more reliable than memorising the phrase “acyl chloride plus amine gives amide” without accounting for the whole reaction.
Six changed practice questions
- Name the main amide formed when ethanoyl chloride reacts with methylamine under suitable conditions.
- Write the familiar full equation when methylamine itself neutralises the released acid.
- Explain why the second methylamine molecule does not become part of the amide.
- What main amide forms if the nucleophile is ammonia rather than methylamine?
- In the 2:1 model, how much methylamine is needed to consume 0.0100 mol ethanoyl chloride completely?
- If 0.0200 mol ethanoyl chloride reacts with 0.0500 mol methylamine under the ideal 2:1 equation, identify the limiting reagent and the theoretical amide mass using Mr = 73.
Answers
1. N-methylethanamide, CH₃CONHCH₃. 2. CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃Cl. 3. It accepts the proton associated with the HCl byproduct and becomes methylammonium chloride. 4. Ethanamide, CH₃CONH₂. 5. 0.0200 mol methylamine. 6. Ethanoyl chloride limits the reaction; 0.0200 mol amide forms theoretically, giving 1.46 g.
If the learner gets Question 1 right but Question 5 wrong, the weak link is full reaction stoichiometry, not naming.
A seven-session JC1-to-JC2 revision route
- Session 1: distinguish amines, amides, acyl chlorides and alkyl halides from structural formulae.
- Session 2: identify the electrophilic carbonyl carbon and the amine lone-pair nucleophile.
- Session 3: construct the N-substituted amide product, counting its carbon atoms carefully.
- Session 4: explain acid generation and why an extra amine can form an ammonium chloride salt.
- Session 5: balance the full equation and repeat using ammonia instead of methylamine.
- Session 6: calculate limiting reactants, theoretical product masses and reaction atom economy.
- Session 7: solve a changed acylation problem with an explicitly supplied external base and explain how the stoichiometry differs.
The final session tests whether the student understands why two amine equivalents appear instead of memorising that coefficient as a universal law.
What parents choosing an H2 Chemistry tutor near Hougang can ask
Bring a marked acyl chloride or amine reaction question and ask whether the student has confused the electrophilic reaction centre, amide product naming, the acid byproduct, proton capture or limiting-reagent coefficients.
A good three-student tutorial should finish with a changed amine structure so each learner independently identifies the desired amide and the separate salt coproduct. One-to-one tuition should apply the same transfer test.
Confirm current H2 9476 familiarity, actual teaching venue, timing and availability. A Hougang-targeted educational page does not establish that the class is physically located in Hougang.
Why this matters beyond the A Levels
Amide-bond formation is important in pharmaceuticals, materials science and the chemistry of proteins and polymers.
Chemists also need to plan how a reaction’s acid byproducts will be managed and what happens to the stoichiometric base or other additives.
The useful habit is to distinguish molecules incorporated into a product from reagents consumed in support of the transformation. That matters for reaction mass balance, atom economy, cost and environmental assessment.
The underlying skill can support university Chemistry and chemical engineering without promising any examination grade or admissions outcome.
Frequently asked questions
Why do two methylamine molecules appear in the full reaction?
One forms the amide C–N bond, while the other accepts HCl-derived acid and becomes methylammonium chloride.
What does ethanoyl chloride and methylamine produce?
N-methylethanamide as the desired organic product, with methylammonium chloride in the familiar two-equivalent amine reaction model.
Must every acyl chloride reaction use two equivalents of amine?
No. An appropriate separately added base can accept the proton instead, changing the reagent stoichiometry.
Does ammonia give the same amide as methylamine?
No. Ammonia gives ethanamide, while methylamine supplies an N-methyl group and gives N-methylethanamide.
Is the reaction SN2 substitution?
The acyl chloride reacts by nucleophilic acyl substitution at the carbonyl carbon, commonly described through an addition–elimination pathway, not ordinary SN2 at an sp³ alkyl carbon.
Is methylammonium chloride part of the desired amide structure?
No. It is a separate ionic coproduct in the simple model.
Is the acyl chloride–primary amine reaction in H2 9476?
Yes. The revised syllabus includes formation of amides from acyl chlorides and reactions with primary amines, together with nitrogen compound chemistry.
Continue learning
Why ethanoyl chloride esterification differs from acid esterification · JC2 amines, amides and amino acid chemistry · Why amides are less basic than amines · How nucleophiles change products in halogenoalkane substitution · Polyamides and condensation polymer hydrolysis · Hougang Chemistry Tutor Library · eduKate Singapore JC Chemistry enquiries.
Safety and model note: Acyl chlorides are corrosive and moisture-sensitive; methylamine and related reagents can be hazardous. Equations and numerical calculations are theoretical educational descriptions, not instructions for reagent preparation or unsupervised synthesis. The two-amine stoichiometry assumes the amine itself acts as acid scavenger and no different base replaces that role.