← Explore exam analytics

Amines · TOPIC LESSON

Amine Synthesis and Reactions

Use the nitrogen lone pair to connect basicity, nucleophilicity and selective synthesis.

JEE Main · JEE Advanced · NEET UG49 min study guide12 indexed sections

BY THE END, YOU CAN

  • Classify amines by nitrogen substitution and distinguish amines, amides and ammonium salts.
  • Compare basicity with conjugate-acid pKa while accounting for resonance and solvent.
  • Choose amine syntheses with explicit carbon-count and overalkylation checks.
  • Explain Gabriel substitution, amide reduction and Hofmann rearrangement as distinct routes.
  • Predict acylation, nitrous-acid response and aromatic substitution with appropriate limits.

BEFORE YOU START

  • Acid–base equilibria, pKa and resonance.
  • SN2 substitution and nucleophilic acyl substitution.
  • Carbonyl structures and organic carbon accounting.

Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.

CONNECT THE CONCEPT TO THE EXAM

Selected exam questions & worked solutions

These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.

No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.

01 / UNDERSTAND & APPLY

1. Count carbon groups attached directly to nitrogen

A primary amine RNH₂ has one carbon group bonded to nitrogen, a secondary amine R₂NH has two, and a tertiary amine R₃N has three. The labels do not refer to how substituted the neighboring carbon is. Thus tert-butylamine is a primary amine because nitrogen is attached to only one carbon group.

A quaternary ammonium ion R₄N⁺ has four carbon substituents and no nitrogen lone pair available for ordinary proton acceptance. It is not a neutral tertiary amine with an extra hidden hydrogen. A protonated tertiary amine R₃NH⁺ is also positively charged but contains an N–H bond and can lose a proton to regenerate R₃N.

An arylamine has nitrogen directly bonded to an aromatic ring, as in aniline PhNH₂. Benzylamine PhCH₂NH₂ instead has a methylene group separating nitrogen from the ring. That interruption prevents the nitrogen lone pair from being directly conjugated with the aromatic π system in the same way.

An amide RCONH₂ is a different functional group from an amine because nitrogen is directly attached to a carbonyl carbon. Lone-pair delocalization gives the C–N bond partial double-bond character and changes geometry, basicity and reactivity. Ordinary pyramidal amines often undergo rapid nitrogen inversion, so a three-different-substituent nitrogen is not automatically an isolable stable stereocenter.

ORIGINAL WORKED EXAMPLE

Classify by nitrogen connectivity

Classify (CH₃)₃CNH₂, (CH₃)₂NH and CH₃CONH₂.

  1. In (CH₃)₃CNH₂, nitrogen is attached to one carbon group, so the compound is a primary amine.
  2. In (CH₃)₂NH, nitrogen has two methyl substituents, so it is a secondary amine.
  3. CH₃CONH₂ has nitrogen attached to a carbonyl carbon and is an amide, not an ordinary primary amine.
  4. The substitution of the carbon next to nitrogen does not determine the amine class.

Answer Primary amine; secondary amine; amide, respectively.

02 / UNDERSTAND & APPLY

2. Compare the conjugate acids, not just the drawn lone pairs

For an amine base B in water, B+H₂O⇌BH⁺+OH⁻ defines K_b. Its conjugate acid satisfies BH⁺⇌B+H⁺ with K_a. At a common temperature, K_aK_b=K_w, so pK_a(BH⁺)+pK_b(B)=pK_w. At 25 °C the usual dilute-water approximation uses pK_w≈14.00.

A larger conjugate-acid pK_a means a stronger base B, because BH⁺ is less willing to give its proton back. Conversely, a smaller pK_b means a stronger base. Many ranking errors arise from using the direction of pK_a for the acid while forgetting that the desired comparison is between its conjugate bases.

Basicity is an equilibrium property, whereas nucleophilicity describes a reaction rate toward an electrophile under specified conditions. A bulky amine can be a useful base yet a poor nucleophile at a crowded carbon center. Solvent and steric effects can alter nucleophilicity rankings without an identical change in thermodynamic basicity.

Protonation consumes the lone pair in a new N–H bond and gives an ammonium ion. In an electron-pair mechanism, the arrow begins at nitrogen’s lone pair and ends at the proton; the acid H–A bond pair goes to A. A positively charged ammonium nitrogen has no additional ordinary lone pair to donate until deprotonation occurs.

K_b(B)K_a(BH⁺)=K_w; pK_b=pK_w−pK_a(BH⁺)

Use values for the same solvent and temperature.

pH=pK_a(BH⁺)+log([B]/[BH⁺])

Dilute buffer approximation with both base and conjugate acid present.

ORIGINAL WORKED EXAMPLE

Use two hypothetical conjugate-acid pKa values

At 25 °C in water, base B₁ has pKa(B₁H⁺)=10.2 and B₂ has pKa(B₂H⁺)=5.2. Which is stronger, and by what factor in Kb?

  1. pKb(B₁)=14.0−10.2=3.8; pKb(B₂)=14.0−5.2=8.8.
  2. B₁ has the smaller pKb and is the stronger base.
  3. The pKb difference is 5.0, so Kb(B₁)/Kb(B₂)=10⁵.
  4. The result compares equilibrium constants in the same solvent, not reaction rates toward every electrophile.

Answer B₁ is stronger by a factor of 10⁵ in Kb.

03 / UNDERSTAND & APPLY

3. Resonance, induction and solvation can compete

Alkyl groups can increase electron availability at nitrogen through their electronic effects, helping many alkylamines be stronger aqueous bases than ammonia. However, solvation of the neutral amine and its ammonium ion, steric crowding and hydrogen bonding also affect the equilibrium. A universal aqueous order tertiary>secondary>primary is therefore unreliable.

In aniline, the nitrogen lone pair can delocalize into the aromatic ring. Protonation at nitrogen removes that lone pair from conjugation, so the neutral base receives resonance stabilization that is not retained in the same form by anilinium. This makes aniline substantially less basic than a comparable unconjugated alkylamine.

Electron-withdrawing substituents on an aniline ring generally weaken basicity, while suitable electron-donating substituents can strengthen it. Position matters because resonance communication differs at ortho, meta and para positions. Ortho steric and solvation effects can compete with the simple electronic prediction, so exact close rankings may need measured data.

Amide nitrogen is much less basic than ordinary amine nitrogen because its lone pair is delocalized into C=O. Under sufficiently acidic conditions, amides can be protonated, commonly at oxygen; it is too strong to say an amide can never accept a proton. The key contrast is that ordinary amine-like nitrogen protonation and nucleophilicity are strongly suppressed.

Pyridine and pyrrole illustrate another distinction. Pyridine’s basic lone pair is outside the aromatic π sextet, whereas pyrrole’s nitrogen lone pair contributes to aromaticity. Removing the latter from the aromatic system by nitrogen protonation is unfavorable in a simple comparison. Actual preferred protonation sites and strongly acidic behavior require more detailed structural analysis.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Aniline versus benzylamine connectivityAniline has NH₂ bonded directly to an aromatic six-membered ring, allowing nitrogen lone-pair conjugation with the ring. Benzylamine has NH₂–CH₂–ring connectivity; the sp³ methylene interrupts direct conjugation. Under ordinary comparable aqueous conditions benzylamine is the stronger base. Dots identify the nitrogen lone pair and inner circles denote aromatic rings.Direct conjugation versus a CH₂ spacerAniline and benzylamine have different lone-pair accessNH₂Aniline: N is directly bonded to the aromatic ring.Lone pair can conjugate with the ring π system.CH₂NH₂sp³ spacerBenzylamine: CH₂ interrupts direct conjugation.Usually the stronger aqueous base of this pair.Dots mark a lone pair; circles mark aromatic rings.Compare the same solvent and protonation conditions.
Connectivity comparison, not an orbital-density map. Basicity also depends on solvation and other substituents; the stated ranking is for this specific pair.

ORIGINAL WORKED EXAMPLE

Spot interrupted versus direct conjugation

Which is normally more basic in water, aniline PhNH₂ or benzylamine PhCH₂NH₂? Explain the structural difference.

  1. Aniline has nitrogen directly attached to the aromatic ring, permitting lone-pair conjugation.
  2. Benzylamine has an intervening sp³ CH₂ group that interrupts direct lone-pair overlap with the ring π system.
  3. The benzylamine lone pair is therefore more available for protonation in this comparison.
  4. Other substituents and solvent can change quantitative values, but the stated pair illustrates the resonance distinction.

Answer Benzylamine is normally the stronger aqueous base.

04 / UNDERSTAND & APPLY

4. Protonation changes charge and often changes phase preference

An amine plus a suitable acid forms an ammonium salt, for example RNH₂+HCl→RNH₃⁺Cl⁻. This is acid–base chemistry, not replacement of a carbon substituent on nitrogen. Treatment with a sufficiently strong base can regenerate the neutral amine from its conjugate acid.

Charged ammonium forms often have greater affinity for an aqueous phase than the corresponding neutral organic amine. This principle supports acid–base extraction, but actual distribution depends on molecular size, counterion, solvent and concentration. Protonation does not guarantee that every ammonium salt is infinitely water-soluble.

For the dilute equilibrium BH⁺⇌B+H⁺, [B]/[BH⁺]=10^(pH−pKa). The protonated fraction is 1/[1+10^(pH−pKa)] when these are the only relevant forms. If pH is two units below pKa, roughly 99% is protonated; if two units above, roughly 1% is protonated.

A distribution calculation can require both acid–base speciation and a partition coefficient for the neutral form. The fraction protonated in one phase alone does not establish the total fraction extracted into that phase. Distinguish chemical speciation from mass distribution between phases.

ORIGINAL WORKED EXAMPLE

Find a protonated fraction

A simple amine has conjugate-acid pKa=9.00. At pH 7.00, estimate the fraction protonated assuming only B and BH⁺ and ideal dilute behavior.

  1. [B]/[BH⁺]=10^(7−9)=0.0100.
  2. Protonated fraction is 1/(1+0.0100)=0.9901.
  3. Thus about 99.0% of the amine species in that phase is protonated.
  4. This does not by itself give an extraction yield between two solvents.

Answer Approximately 99.0% protonated.

05 / UNDERSTAND & APPLY

5. Alkylation can keep going beyond the desired amine

Ammonia can attack a suitable alkyl halide by SN2 substitution, initially producing an alkylammonium salt; deprotonation gives a primary amine. But that primary amine is itself a nucleophile and can react again. Further alkylation can generate secondary and tertiary amines and finally a quaternary ammonium salt.

Using excess ammonia can favor primary-amine formation in some cases, but it does not turn a potentially multiple-alkylation process into a universally selective reaction. A synthesis answer should acknowledge the mixture risk and choose a more selective route when a pure primary product is required.

Because the route uses SN2 chemistry, methyl and suitable primary alkyl electrophiles are favorable. A tertiary alkyl halide does not provide a clean analogous substitution because steric hindrance and elimination intervene. An unactivated aryl halide also does not undergo ordinary backside SN2 at its sp² carbon.

Reductive amination offers a complementary route. A carbonyl compound reacts with ammonia or an amine to form an imine or iminium-related intermediate, which is then selectively reduced. The carbonyl carbon remains in the skeleton and becomes bonded to nitrogen. Reagent choice must favor the desired imine/iminium reduction rather than simply reducing all starting carbonyl to alcohol.

ORIGINAL WORKED EXAMPLE

Read a reductive-amination skeleton

Predict the amine connectivity from cyclohexanone and methylamine under suitable reductive-amination conditions.

  1. The carbonyl carbon is already a member of the cyclohexane ring.
  2. Condensation with methylamine establishes a C–N bond to a nitrogen bearing CH₃.
  3. Reduction converts the imine/iminium C=N linkage into a C–N single bond with an added hydrogen at the former carbonyl carbon.
  4. The product is N-methylcyclohexanamine, a secondary amine; the ring is not expanded.

Answer N-methylcyclohexanamine, with cyclohexyl and methyl groups on nitrogen.

06 / UNDERSTAND & APPLY

6. Gabriel synthesis controls nitrogen alkylation through a masked amine

In the Gabriel route, deprotonated phthalimide provides a nitrogen nucleophile that undergoes alkylation with a suitable SN2 electrophile. The resulting N-alkyl phthalimide is then cleaved to release a primary amine RNH₂. Nitrogen’s temporary attachment to the imide framework prevents the same uncontrolled sequence of ordinary amine overalkylation.

The carbon skeleton R comes from the alkyl electrophile. No extra carbon from the phthalimide protecting framework remains in the released amine. Hydrolysis or another appropriate cleavage removes that framework after alkylation; omitting the cleavage step leaves an imide rather than the desired free amine.

The electrophile restrictions remain important: a suitable primary alkyl halide is a standard choice, while tertiary halides favor elimination and unactivated aryl halides do not undergo the required ordinary SN2 displacement. Gabriel synthesis is therefore not a general method for preparing aniline directly from chlorobenzene.

This route primarily supplies primary amines in its standard form. It should not be invoked without modification to make an arbitrary secondary or tertiary amine. The route’s value is selective installation of one R group on a nitrogen that is later unmasked.

ORIGINAL WORKED EXAMPLE

Select a Gabriel-compatible target

Would a standard Gabriel route from 1-bromopropane or from bromobenzene be the straightforward route to the corresponding primary amine?

  1. 1-Bromopropane is a primary alkyl halide compatible with the SN2 alkylation step.
  2. After phthalimide alkylation and cleavage, its three-carbon group gives propan-1-amine.
  3. Bromobenzene has an aryl sp² C–Br bond and does not undergo the ordinary SN2 step required here.
  4. The failure is mechanistic, not because aniline is not a primary amine.

Answer The 1-bromopropane route is appropriate; ordinary Gabriel conditions do not convert bromobenzene to aniline.

07 / UNDERSTAND & APPLY

7. Nitro reduction installs an arylamine without changing the ring skeleton

An aromatic nitro group can be reduced to an amino group while retaining its bond to the same ring carbon: ArNO₂ becomes ArNH₂. This is a major route to anilines because an appropriate aromatic precursor can first be nitrated, then the nitro group reduced. Reduction does not relocate the substituent; the position established in the nitration or starting substrate remains part of the product connectivity.

The formal acidic half-reaction is ArNO₂+6H⁺+6e⁻→ArNH₂+2H₂O. In a strongly acidic reaction mixture the amine is normally present largely as ArNH₃⁺, and a separate basic workup can release the neutral arylamine. A product written as an anilinium salt and a product written as aniline can therefore represent different acid–base states after the same reduction.

Catalytic hydrogenation, for example H₂ with a suitable Pt catalyst, can reduce a nitro group, with the net balance ArNO₂+3H₂→ArNH₂+2H₂O. Other reducible groups may also react: alkenes, carbonyl groups or hydrogenolysis-sensitive substituents require a compatibility check. Do not assume that hydrogenation always changes only NO₂.

Metal/acid combinations such as Fe/HCl or Sn/HCl, and suitable SnCl₂/acid reductions, provide alternative nitro-to-amine routes. SnCl₂ can offer useful chemoselectivity in an appropriate multifunctional substrate, but acidic conditions themselves can affect acetals or other acid-sensitive groups. These are reagent-family distinctions, not universal guarantees for every substituted nitroarene. Neutral or alkaline partial-reduction conditions can lead to other nitrogen-containing products, so the medium is meaningful information.

ORIGINAL WORKED EXAMPLE

Track a ring substituent and the workup

Predict the principal connectivity obtained by a suitable complete Fe/HCl reduction of 1-methyl-3-nitrobenzene, followed by a basic workup. Does the methyl group become a methyleneamine side chain?

  1. The nitro nitrogen stays attached to its original aromatic carbon.
  2. Complete reduction changes NO₂ to the anilinium/amine functionality; it does not insert carbon.
  3. The methyl substituent remains at the meta position relative to nitrogen.
  4. The acid mixture favours the ammonium salt; the stated basic workup gives the neutral amine.

Answer 3-Methylaniline, m-CH₃C₆H₄NH₂. It is an arylamine, not methylbenzylamine.

08 / UNDERSTAND & APPLY

8. Distinguish carbon-retaining reduction from one-carbon-shorter rearrangement

Reduction of a nitrile R–C≡N gives R–CH₂NH₂ under suitable conditions. The nitrile carbon is retained and becomes the carbon attached to NH₂. If the nitrile was prepared by substituting an alkyl halide with cyanide, that earlier substitution added one carbon, so the overall sequence lengthens the original alkyl skeleton.

Reduction of an amide R–C(=O)NH₂ with a suitable strong hydride gives R–CH₂NH₂, retaining the carbonyl carbon. N-substituted amides can analogously give substituted amines. The carbonyl oxygen is removed in the reduction process; the product is not the carbonyl-containing starting amide with an extra hydrogen label.

Hofmann rearrangement of a primary amide instead gives RNH₂ with one fewer carbon than RCONH₂. In the textbook mechanism, N-halogenation and deprotonation are followed by migration of R from the carbonyl carbon to nitrogen, producing an isocyanate intermediate R–N=C=O. Hydrolysis and decarboxylation then release the amine.

The migrating R group remains intact in the elementary skeletal description, and the original carbonyl carbon is lost as inorganic carbon dioxide/carbonate according to the medium. This is not the same reaction as Hofmann elimination of a quaternary ammonium compound, which forms an alkene. Similar names do not imply similar carbon accounting.

Choosing between these routes depends on the target carbon number. Starting from a four-carbon amide, reduction gives a four-carbon amine while Hofmann rearrangement gives a three-carbon amine. A carbon-count audit often resolves a route-selection question before any detailed mechanism is drawn.

Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.

Amide reduction and Hofmann rearrangement have different carbon countsButanamide CH₃CH₂CH₂CONH₂ has its carbonyl carbon circled. Suitable strong-hydride reduction retains this carbon as CH₂ and gives four-carbon butan-1-amine. Hofmann rearrangement instead transfers the propyl group to nitrogen, passes through propyl isocyanate CH₃CH₂CH₂N=C=O, then hydrolysis and decarboxylation release three-carbon propan-1-amine and inorganic carbon from the original carbonyl carbon. The two routes are alternatives from the original butanamide; the Route B label explicitly resets the starting material.Keep or remove the carbonyl carbon?Two different routes from butanamideOCCH₃CH₂CH₂NH₂Circled carbon is the original carbonyl carbon.Route A: suitable strong-hydride reductionCH₃CH₂CH₂–CH₂–NH₂Butan-1-amine: all four carbons retained.Route B: restart from butanamideCH₃CH₂CH₂–N=C=OIsocyanate: propyl group has migrated to N.CH₃CH₂CH₂NH₂ + inorganic carbonPropan-1-amine: one fewer carbon in the amine.Original carbonyl carbon leaves as CO₂/carbonate.Hydrolysis/decarboxylation follows rearrangement.Hofmann rearrangement of the starting amide
Conceptual reaction map rather than a full elementary mechanism or handling procedure. The inorganic carbon form depends on medium; this rearrangement is distinct from Hofmann elimination. Route B restarts from butanamide, not from the Route A amine product.

ORIGINAL WORKED EXAMPLE

Two outcomes from one amide

Compare products of suitable strong-hydride reduction and Hofmann rearrangement of butanamide CH₃CH₂CH₂CONH₂.

  1. Hydride reduction retains the carbonyl carbon as CH₂, giving CH₃CH₂CH₂CH₂NH₂, butan-1-amine.
  2. Hofmann rearrangement transfers the propyl group to nitrogen and loses the carbonyl carbon.
  3. Its product is CH₃CH₂CH₂NH₂, propan-1-amine.
  4. The carbon counts are four and three respectively, even though both products are primary amines.

Answer Reduction: butan-1-amine. Hofmann rearrangement: propan-1-amine.

09 / UNDERSTAND & APPLY

9. Acylation changes basicity; exhaustive alkylation can enable elimination

Primary and secondary amines can react with suitable acid chlorides or anhydrides to form amides by nucleophilic acyl substitution. Nitrogen attacks the carbonyl carbon, a tetrahedral intermediate forms, and a leaving group is expelled as C=O is restored. Proton transfer or a base handles the acid generated during acylation.

A tertiary amine lacks an N–H proton needed to form the analogous neutral amide while retaining all three carbon groups. It may act as a base or participate in an acylammonium intermediate in appropriate chemistry, but one should not draw an ordinary neutral four-carbon-substituted amide nitrogen without addressing charge and bond changes.

Acylation of aniline to an anilide reduces nitrogen lone-pair donation into the aromatic ring compared with free NH₂ and can moderate excessive ring activation. The acyl group can later be removed under appropriate hydrolysis conditions. As with any protection strategy, installation, compatibility and removal must all be considered.

Exhaustive alkylation can convert an amine into a quaternary ammonium species that undergoes elimination under suitable conditions. Hofmann elimination often favors the less substituted accessible alkene, reflecting steric and mechanistic factors. Beta-hydrogen availability and geometry still matter; the product is not chosen by a universal “least substituted” slogan alone.

ORIGINAL WORKED EXAMPLE

Acylate a secondary amine

Predict the neutral organic product from dimethylamine and ethanoyl chloride with suitable acid-scavenging conditions.

  1. Dimethylamine is (CH₃)₂NH and has one N–H bond.
  2. Its nitrogen attacks the ethanoyl carbonyl; chloride leaves after tetrahedral-intermediate formation.
  3. Removal of the nitrogen proton gives CH₃C(=O)N(CH₃)₂.
  4. The product is an amide with much weaker nitrogen basicity than the starting amine.

Answer N,N-Dimethylethanamide, CH₃CON(CH₃)₂.

10 / UNDERSTAND & APPLY

10. Amine class and aromatic attachment change the reaction family

Primary aromatic amines can form aryl diazonium ions with nitrosating conditions in cold acid. These diazonium intermediates support substitution and azo-coupling chemistry treated in a dedicated lesson. Primary aliphatic amines form much less stable aliphatic diazonium intermediates that can lose nitrogen and produce mixtures or rearranged products rather than a generally isolable stable salt.

Secondary amines can form N-nitrosamines under suitable nitrosating conditions, while tertiary amines respond differently because they have no N–H bond. Ring nitrosation can occur for appropriate tertiary arylamines. These distinctions are useful for conceptual classification but should not be reduced to a single universal gas/no-gas test for every substrate.

Free aniline strongly activates an aromatic ring toward electrophilic substitution and directs attack mainly to ortho and para positions. Under strongly acidic conditions, protonation forms anilinium, whose electronic effects are different and whose ring is deactivated. Therefore reagent acidity can change the directing group actually present during a reaction.

Aniline can undergo extensive bromination under conditions where a less activated aromatic substrate would react more slowly. Acylation can moderate the activation and help control substitution before hydrolysis regenerates the amine. Direct Friedel–Crafts chemistry with unprotected amines is often problematic because nitrogen binds Lewis-acid catalysts or becomes protonated.

Test reactions involving nitrosating agents or isocyanide formation belong to supervised analytical chemistry and require appropriate safety controls. This lesson teaches product logic and limitations, not preparation or handling instructions for hazardous diagnostic reagents.

11 / UNDERSTAND & APPLY

11. Product-based distinctions: carbylamine and Hinsberg logic

The carbylamine reaction identifies the primary-amine structural pattern RNH₂, including suitable aliphatic and aromatic primary amines. Its conceptual net equation with chloroform and base is RNH₂+CHCl₃+3KOH→RNC+3KCl+3H₂O. The isocyanide product has R attached to nitrogen, R–N≡C, commonly represented with formal N⁺/C⁻ charges; it is not the nitrile R–C≡N.

Secondary and tertiary amines do not give the ordinary primary-amine carbylamine conversion. The customary reaction conditions involve base in an alcoholic medium and heating, but no laboratory handling procedure is implied here. Isocyanides and the reagents can be hazardous; an odour description in a textbook is not permission to smell a reaction mixture. The structural product and validated analytical evidence, not an improvised odour test, are the meaningful concepts.

The Hinsberg distinction uses reaction with a sulfonyl chloride such as PhSO₂Cl. A primary amine gives PhSO₂NHR, which retains an N–H bond. The electron-withdrawing sulfonyl group makes that hydrogen sufficiently acidic for deprotonation under the suitable basic test conditions; the resulting anion often provides aqueous-base solubility. Acidification can restore the neutral sulfonamide and reduce its solubility.

A secondary amine gives PhSO₂NR₂, which has no N–H bond to remove. It therefore lacks that particular base-solubilizing deprotonation pathway. A tertiary amine has no N–H bond in the starting amine and does not form the corresponding neutral sulfonamide by the ordinary substitution sequence; it can instead be protonated by acid to an ammonium salt.

These classical solubility distinctions assume a suitable simple substrate and a completed reaction. Very small molecules, additional acidic/basic groups, intrinsic solubility, steric hindrance and slow sulfonylation can complicate observations. Failure to see a precipitate alone is not an infallible classification. Start by drawing the sulfonamide product and counting its remaining N–H bonds, then apply the stated pH and solubility assumptions.

ORIGINAL PRACTICE · 1

Compare the Hinsberg products of ethanamine and diethylamine under the usual simple-substrate model.

Reveal answer & reasoning

Ethanamine gives PhSO₂NHCH₂CH₃; diethylamine gives PhSO₂N(CH₂CH₃)₂.

The primary-amine product retains an acidic N–H bond and can form a base-soluble anion. The secondary-amine product has no N–H bond and cannot dissolve by that same deprotonation mechanism.

ORIGINAL PRACTICE · 2

What is the carbylamine product connectivity from aniline in the conceptual primary-amine reaction?

Reveal answer & reasoning

Phenyl isocyanide, C₆H₅–N≡C.

The original phenyl–nitrogen bond is retained. The product is not benzonitrile, C₆H₅–C≡N; those structures connect the phenyl group to different atoms.

12 / UNDERSTAND & APPLY

12. Eight solved synthesis and basicity checks

For each route, count carbon atoms and nitrogen substituents separately. For each basicity comparison, specify the solvent or use supplied conjugate-acid data.

  • Amine degree counts groups on nitrogen, not on the neighboring carbon.
  • A higher conjugate-acid pKa means a stronger base.
  • Do not impose one gas-phase or aqueous basicity order on every solvent.
  • Gabriel synthesis still requires a suitable SN2 electrophile.
  • Amide reduction retains the carbonyl carbon; Hofmann rearrangement removes it.
  • Hofmann rearrangement and Hofmann elimination are different reactions.
  • Protonation and acylation change aniline’s ring reactivity.

ORIGINAL PRACTICE · 1

Is (CH₃CH₂)₃N primary, secondary or tertiary?

Reveal answer & reasoning

Tertiary amine.

Nitrogen is directly bonded to three ethyl groups and has no N–H bond in the neutral molecule.

ORIGINAL PRACTICE · 2

At 25 °C, an amine conjugate acid has pKa=10.5. Find pKb in water.

Reveal answer & reasoning

3.5.

pKb=pKw−pKa≈14.0−10.5=3.5 under the stated dilute-water convention.

ORIGINAL PRACTICE · 3

For an amine with pKa(BH⁺)=8.0 at pH 8.0, what is [B]/[BH⁺]?

Reveal answer & reasoning

1.

The ratio is 10^(pH−pKa)=10⁰=1, so the two forms have equal concentrations in the stated model.

ORIGINAL PRACTICE · 4

Why is an amide nitrogen ordinarily less basic than an alkylamine nitrogen?

Reveal answer & reasoning

Its lone pair is delocalized into the carbonyl group.

Nitrogen protonation would remove that delocalization advantage; strong-acid oxygen protonation is a separate possibility.

ORIGINAL PRACTICE · 5

What amine results from reduction of CH₃CH₂C≡N?

Reveal answer & reasoning

Propan-1-amine, CH₃CH₂CH₂NH₂.

The nitrile already contains three carbons, including its C≡N carbon. Reduction converts that carbon to CH₂ while retaining the chain.

ORIGINAL PRACTICE · 6

What product comes from Hofmann rearrangement of benzamide PhCONH₂?

Reveal answer & reasoning

Aniline, PhNH₂.

The phenyl group migrates to nitrogen and the original carbonyl carbon is lost; this differs from amide reduction, which would give benzylamine.

ORIGINAL PRACTICE · 7

Why can direct alkylation of ammonia give more than a primary amine?

Reveal answer & reasoning

The primary product remains nucleophilic and can be alkylated again.

Successive substitutions can form secondary, tertiary and quaternary ammonium products unless conditions and route design control them.

ORIGINAL PRACTICE · 8

Does protonating aniline preserve the free-NH₂ group’s strong ortho/para ring activation?

Reveal answer & reasoning

No.

Anilinium lacks the freely donating nitrogen lone pair and has different, deactivating electronic effects. The protonation state must be considered during electrophilic substitution.

KEEP THE BIG IDEAS

Quick revision

  • Nitrogen lone-pair availability links basicity and nucleophilic chemistry, but the two properties are not identical.
  • Resonance and solvation explain why simple alkyl-count rankings can fail.
  • Synthesis choices must control overalkylation and carbon number.
  • Gabriel unmasking gives a primary amine from a compatible alkyl electrophile.
  • Reduction and Hofmann rearrangement have different carbon-retention outcomes.
  • Aromatic versus aliphatic attachment and protonation state determine important reaction differences.
Back to the top ↑

READ WITH CONTEXT

References & scope

The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.

Explore this topic in context

Analytics use a separately reviewed, incomplete historical corpus. Opening a topic does not change its counts or denominator.

KEEP EXPLORING

Your next idea is waiting.

Back to Chemistry index →Browse all topic notes