9. Amines

What you will learn

• Draw and classify amines with explicit lone pairs.
• Name carbon-chain and nitrogen substituents unambiguously.
• Choose carbon-retaining, carbon-extending or carbon-shortening preparations.
• Explain basicity with resonance and solvent effects.
• Compare class-dependent reactions and selective protection.
• Use aromatic diazonium intermediates for positional planning and coupling.

Big idea: the nitrogen lone pair connects structure with both basicity and bond formation.

The same formal functional-group label can hide important differences. An aliphatic lone pair, an aromatic delocalised pair and a protonated nitrogen do not have the same availability or reactivity. Begin by drawing the bonds and locating the electron pair.

The later sections develop diazonium chemistry as a planning network. Track the ring position and the nitrogen atoms separately. That distinguishes replacement, removal and coupling instead of treating them as a list of disconnected named reactions.

Gabriel and Hofmann routes became useful because they solved different synthetic problems: controlling nitrogen substitution and changing the carbon skeleton. Named reactions are most valuable when their structural purpose is remembered, rather than only the reagent names.

An amine’s lone pair is central to both its shape and its chemical behaviour.

Amines can be viewed as ammonia derivatives in which one or more hydrogens are replaced by carbon groups. Ordinary neutral amine nitrogen commonly has three sigma bonds and one lone pair. The bonded-atom arrangement is approximately pyramidal in a simple aliphatic amine.

The lone pair can accept a proton or form a bond to an electrophile. Its availability depends on attached groups and the surrounding medium. A free lone pair on an aliphatic amine is not equivalent to a lone pair strongly delocalised next to a carbonyl in an amide.

A pyramidal drawing does not imply that every simple amine has a stable isolable nitrogen-centred enantiomer. Rapid pyramidal inversion commonly interconverts the arrangements of ordinary amines. Structural constraints or different nitrogen bonding can change that situation, but a static screenshot alone does not prove configurational stability.

Why should an exam drawing of a simple amine include a lone pair?

Three bonds alone omit the electronic feature that explains proton acceptance and nucleophilic behaviour. The lone pair occupies an electron domain, not a fourth atom position.

Result: Show the lone pair explicitly in a Lewis or mechanism sketch; use the 3D model to understand the pyramid.

1. What local bonded geometry is usual for a simple neutral aliphatic amine?
2. Does a static pyramidal drawing prove a simple amine’s nitrogen configuration is permanently fixed?

Answers

1. Approximately trigonal pyramidal.
2. No. Rapid nitrogen inversion commonly occurs.

Primary, secondary and tertiary amine labels count carbon groups attached directly to nitrogen.

A primary amine has one carbon group and two N–H bonds. A secondary amine has two carbon groups and one N–H bond. A tertiary amine has three carbon groups and no N–H bond. These labels differ from the alcohol classification, which counts carbon neighbours at the hydroxyl-bearing carbon.

An arylamine has nitrogen directly attached to an aromatic carbon. A benzylamine-type structure has nitrogen attached to a saturated side-chain carbon instead. That difference changes the resonance connection of the nitrogen lone pair. A molecule containing a ring is not automatically an arylamine.

Classify $\mathrm{C_6H_5CH_2NH_2}$ by nitrogen substitution and by ring attachment.

Only one carbon group is attached to N, so it is primary. A methylene group separates nitrogen from the aromatic ring.

Result: It is a primary benzylic amine, not an aniline-type directly attached arylamine.

1. How many carbon groups attach directly to nitrogen in a secondary amine?
2. Does absence of N–H bonds automatically mean a compound is not an amine?

Answers

1. Two.
2. No. Tertiary amines have no N–H bond.

Naming an amine requires you to distinguish carbon-chain positions from substitutions on the nitrogen itself.

For a simple primary amine, select a suitable parent and use an -amine suffix with a locant when needed. Substituents attached to nitrogen can be marked with N-. This prevents a nitrogen methyl group from being mistaken for a carbon-chain branch.

Common names often list alkyl groups followed by amine. Established aromatic names such as aniline are widely recognised. In a systematic answer, preserve the actual connection: a nitrogen directly on a ring and a nitrogen on a side chain are different structures.

Name $\mathrm{CH_3CH_2NHCH_3}$.

Take the ethyl group as the ethanamine parent. The extra methyl is attached to nitrogen, not to a parent-chain carbon.

Result: N-Methylethanamine; ethylmethylamine is a recognised common description.

Write N,N-dimethylpropan-1-amine in condensed form.

Start with propan-1-amine. Replace both nitrogen hydrogens with methyl groups. Check that neutral nitrogen has three carbon bonds and one lone pair.

Result: $\mathrm{CH_3CH_2CH_2N(CH_3)_2}$.

A good final check counts all carbons and all groups directly attached to N. Then read the name back into a new sketch. If it gives a different branch position or a different amine class, the original naming step needs revision.

1. What does N,N-dimethyl tell you?
2. Why is benzylamine not the same as aniline?

Answers

1. Two methyl groups are attached to the named nitrogen.
2. A CH₂ group separates N from the ring in benzylamine.

The best amine preparation depends on whether you need to retain, extend or shorten the starting carbon skeleton.

A suitable reducing system converts a nitro group to a primary amino group. Aryl nitro compounds are important precursors to arylamines. Catalytic hydrogenation gives a useful overall stoichiometric representation.

$$\mathrm{C_{6}H_{5}NO_{2}} + 3 \mathrm{H_{2}} \xrightarrow{\mathrm{catalyst}} \mathrm{C_{6}H_{5}NH_{2}} + 2 \mathrm{H_{2}O}$$

Ammonia can substitute at a suitable alkyl halide. The first amine product can itself react further, so a mixture of substitution classes is possible. Excess ammonia can favour primary amine formation but does not turn the reaction into an automatically perfect single-product method.

$$\mathrm{CH_{3}CH_{2}Br} + 2 \mathrm{NH_{3}} \longrightarrow \mathrm{CH_{3}CH_{2}NH_{2}} + \mathrm{NH_{4}Br}$$

Nitrile reduction retains its nitrile carbon as a carbon adjacent to nitrogen in the primary amine. When the nitrile was prepared from a haloalkane and cyanide, the overall sequence extends that original skeleton by one carbon. Amide reduction with a suitable hydride system also retains the carbonyl carbon, changing it to a reduced carbon centre after work-up.

$$\mathrm{CH_{3}CH_{2}CN} + 2 \mathrm{H_{2}} \xrightarrow{\mathrm{catalyst}} \mathrm{CH_{3}CH_{2}CH_{2}NH_{2}}$$

A deprotonated phthalimide system can undergo suitable alkyl substitution, then release a primary amine on cleavage. The approach is useful for primary aliphatic amines because the protected nitrogen prevents the ordinary repeated alkylation sequence. Ordinary aryl halides do not provide the same SN2 route, so it is not a general method for preparing aniline.

An appropriate primary amide can lose its carbonyl carbon in the rearrangement sequence, giving a primary amine with one fewer carbon. This is different from amide reduction, which retains that carbon. Recognising that distinction is a high-value synthesis-planning skill.

$$\mathrm{CH_{3}CH_{2}CH_{2}CONH_{2}} + \mathrm{Br_{2}} + 4 \mathrm{NaOH} \longrightarrow \mathrm{CH_{3}CH_{2}CH_{2}NH_{2}} + \mathrm{Na_{2}CO_{3}} + 2 \mathrm{NaBr} + 2 \mathrm{H_{2}O}$$

Compare reduction and Hofmann degradation of butanamide.

Track the carbonyl carbon. Reduction retains it in the organic skeleton; degradation transfers it out of the organic product.

Result: Reduction gives a four-carbon amine, while Hofmann degradation gives a three-carbon amine. The word “amide” alone does not specify the carbon count of the product.

1. Why can ordinary ammonolysis produce several amine classes?
2. Why is the common Gabriel route unsuitable for an ordinary aryl halide?

Answers

1. The first amine product can be alkylated further.
2. That aromatic carbon does not provide the required ordinary alkyl SN2 pathway.

Nitrogen substitution changes both the shape of an amine and the intermolecular bonds it can make.

Primary and secondary amines can donate N–H hydrogen bonds and accept at a suitable lone pair. Tertiary amines lack N–H donors but can accept hydrogen bonds in their neutral form. The ability to self-associate therefore differs among amine classes, even when their molecular masses are similar.

Compared with an appropriate alcohol, an amine often has weaker hydrogen-bonding association. Comparisons must still control size, shape, branching and the number of functional groups. There is no universal boiling-point ordering that works for arbitrary molecules selected only by their primary/secondary/tertiary labels.

Small amines can interact well with water. Growing nonpolar carbon portions can reduce water solubility. Protonation often produces an ammonium salt with very different interactions and solubility from the neutral amine. The medium’s pH can therefore change which form dominates.

A tertiary amine has no N–H bond. Does that prevent it from interacting with water through hydrogen bonding?

Water can donate a hydrogen bond to the nitrogen lone pair. The absence of an amine donor does not remove its acceptor role.

Result: No. A neutral tertiary amine can accept a hydrogen bond from water.

In a property explanation, specify whether you mean a neutral amine or its protonated form. An ammonium centre does not retain the same free lone-pair acceptor behaviour as the corresponding neutral amine.

1. Which ordinary amine classes have an N–H donor?
2. Why can acid addition change amine solubility?

Answers

1. Primary and secondary amines.
2. It converts a neutral amine into an ionic ammonium form.

The lone pair can accept a proton or make a new bond. Resonance, solvation and steric access control how readily it does either job.

A base accepts a proton under the Brønsted description. For an amine in water, the equilibrium produces an ammonium ion and hydroxide. A larger base-ionisation constant means stronger basicity in the same solvent and temperature convention; a smaller pKb means the same thing on the logarithmic scale.

$$\mathrm{CH_{3}CH_{2}NH_{2}} + \mathrm{H_{2}O} \rightleftharpoons \mathrm{CH_{3}CH_{2}NH_{3}}^{+} + \mathrm{OH}^{-}$$

Compare hypothetical amines with $pK_b=4.20$ and $5.10$.

The lower pKb corresponds to a larger Kb. Compare the constants rather than interpreting a larger p-number as a stronger base.

Result: $K_b$ values are approximately 0.000063096 and 0.000007943. The first is about 7.943 times larger in this stated convention.

Alkyl groups can increase electron density at nitrogen, but aqueous basicity also depends on how well the resulting ion is solvated and how steric crowding affects its environment. A simple gas-phase electron-donation ordering cannot be transferred automatically to water. Tertiary is not a universal synonym for strongest aqueous amine base.

In aniline, the nitrogen lone pair can interact with the aromatic system, making it less available for protonation than a simple aliphatic comparison. Electron-withdrawing ring groups often reduce basicity further. Electron-donating groups can act oppositely, but position, resonance and steric effects must be considered.

Alkylation adds carbon groups through nitrogen nucleophilicity and can continue to a quaternary ammonium form. Acylation of a primary or secondary amine gives an amide under suitable conditions. A tertiary amine lacks the N–H needed for ordinary neutral amide formation by the same simple replacement route.

$$\mathrm{CH_{3}CH_{2}NH_{2}} + \mathrm{CH_{3}COCl} \longrightarrow \mathrm{CH_{3}CONHCH_{2}CH_{3}} + \mathrm{HCl}$$

A primary amine can give an isocyanide under the characteristic chloroform/alkali conditions. Isocyanide connectivity places nitrogen between the organic group and the terminal carbon, unlike a nitrile. The reaction is a structural-class test, not a route for untrained experimental use.

$$\mathrm{CH_{3}CH_{2}NH_{2}} + \mathrm{CHCl_{3}} + 3 \mathrm{KOH} \longrightarrow \mathrm{CH_{3}CH_{2}NC} + 3 \mathrm{KCl} + 3 \mathrm{H_{2}O}$$

Primary aliphatic amines give unstable diazonium species that release nitrogen, often producing alcohols and potentially rearranged products. Primary aromatic amines can give diazonium salts that persist under suitable cold acidic conditions. Secondary amines can form N-nitroso compounds. The amine class therefore changes the pathway, not merely the final product name.

$$\mathrm{CH_{3}CH_{2}NH_{2}} + \mathrm{HNO_{2}} \longrightarrow \mathrm{CH_{3}CH_{2}OH} + \mathrm{N_{2}} + \mathrm{H_{2}O}$$

In the classical Hinsberg comparison, a primary amine can form a sulfonamide retaining an N–H bond that can be deprotonated in base. A secondary amine gives a sulfonamide without that N–H site. A tertiary amine does not form the same sulfonamide and can instead be recognised through its acid–base behaviour. Interpret the complete sequence and solubility observations, not one isolated step.

An amino group strongly activates an aromatic ring and commonly directs ortho/para. Bromine water can produce extensive substitution. Strong acid protonates aniline and changes its directing influence, so unprotected nitration can be complicated. Acylating the amino group can moderate activation and support better control, followed by deprotection when appropriate.

$$\mathrm{C_{6}H_{5}NH_{2}} + 3 \mathrm{Br_{2}} \longrightarrow \mathrm{C_{6}H_{2}Br_{3}NH_{2}} + 3 \mathrm{HBr}$$

Heating aniline with an appropriate sulfuric-acid system can ultimately produce a sulfonated product such as sulfanilic acid. The zwitterionic description helps explain some of its properties: one part of a molecule can carry positive charge while another carries negative charge.

Why can protecting an amino group help a ring-substitution plan?

The unprotected group can overactivate the ring or become protonated in the reagent mixture. Acyl protection changes its electronic effect and reaction compatibility.

Result: Protection is a way of controlling selectivity, not simply hiding a group in the written formula.

1. Which ordinary amine class gives the carbylamine test?
2. Why does aniline’s lone pair behave differently from an alkylamine lone pair?
3. Does a larger pKb indicate a stronger base?

Answers

1. Primary amines.
2. It can be delocalised into the aromatic electronic system.
3. No. In the same convention, a smaller pKb indicates a stronger base.

A diazonium group is a temporary transformation point on an aromatic ring, allowing one well-positioned group to be replaced by several alternatives.

Diazotisation converts a suitable primary aromatic amine into a diazonium salt using nitrous acid generated in the acidic mixture. The usual school-level preparation uses a cold aqueous acidic range near the freezing point of water, often specified as about 273–278 K. Keeping the conditions stated is part of identifying the intended intermediate.

$$\mathrm{C_{6}H_{5}NH_{2}} + \mathrm{HNO_{2}} + \mathrm{HCl} \longrightarrow \mathrm{C_{6}H_{5}N_{2}Cl} + 2 \mathrm{H_{2}O}$$

The aromatic ring helps the diazonium species behave differently from an ordinary aliphatic diazonium intermediate. The resulting salt is still a reactive intermediate, not an indefinitely stable everyday solid. Preparation, storage and reaction conditions cannot be inferred from the formula alone.

Why is diazotisation useful after placing an amino group at a desired ring position?

The amino group marks a carbon position. Its diazonium derivative can then be transformed at that position into several other groups.

Result: It links positional planning with a range of functional-group replacements.

The reaction is specific to the appropriate primary aromatic amine pathway. Secondary and tertiary amines do not produce the same diazonium salt by simply following the same written equation.

1. What is the required amine class in the familiar aromatic diazotisation route?
2. Why is the cold acidic condition part of the answer?

Answers

1. A suitable primary aromatic amine.
2. It identifies conditions under which the intended diazonium intermediate can be formed and used rather than rapidly changing by other routes.

Diazonium salts are useful because they can react readily; their physical behaviour must be described together with that reactivity.

Benzenediazonium chloride is commonly discussed as a water-soluble ionic species in cold solution. Warming can promote hydrolysis and other decomposition pathways. Some other counterions give salts with different solubility and thermal behaviour, so “all diazonium salts behave identically” is incorrect.

A solid-state formula and a solution intermediate are not equivalent safety or stability descriptions. In particular, a textbook representation does not imply that a dry sample is suitable to isolate or handle casually. For school chemistry, the important point is to connect the counterion and conditions with the intended chemical pathway.

A diazonium conversion works in cold aqueous solution but gives a different product after warming. Is that automatically an experimental contradiction?

Temperature can change which reaction occurs readily and how long an intermediate persists. The original cold-solution statement did not claim stability at every temperature.

Result: No. The reaction conditions and timescale are essential parts of the description.

When comparing physical properties, separate solubility from chemical persistence. A compound can dissolve readily and still react rapidly once dissolved.

1. Can changing the counterion change a diazonium salt’s solubility?
2. Why must temperature accompany a statement about a diazonium solution?

Answers

1. Yes.
2. It affects the lifetime of the intermediate and the reactions that compete.

Diazonium reactions divide neatly into those that expel nitrogen and those that retain the two nitrogen atoms in a new linkage.

Sandmeyer-type conversions replace the diazonium group using suitable copper(I) reagent systems to give selected chlorido, bromido or cyanido substitution products. Iodide replacement and hydrolysis provide additional routes. Match the reagent with the desired substituent rather than treating every conversion as one universal reaction.

$$\mathrm{C_{6}H_{5}N_{2}Cl} \xrightarrow{\mathrm{CuCl/HCl}} \mathrm{C_{6}H_{5}Cl} + \mathrm{N_{2}}$$

$$\mathrm{C_{6}H_{5}N_{2}}^{+} + \mathrm{CN}^{-} \xrightarrow{\mathrm{CuCN}} \mathrm{C_{6}H_{5}CN} + \mathrm{N_{2}}$$

$$\mathrm{C_{6}H_{5}N_{2}}^{+} + \mathrm{I}^{-} \longrightarrow \mathrm{C_{6}H_{5}I} + \mathrm{N_{2}}$$

A suitable tetrafluoroborate diazonium route can produce an aryl fluoride in the Balz–Schiemann transformation. Replacing diazonium by hydrogen can use a reducing system such as hypophosphorous acid. The nitrogen gas leaving group makes these transformations chemically distinct from an ordinary aryl-halide substitution.

The Gattermann diazonium-replacement comparison uses copper powder with an appropriate hydrogen halide, rather than the copper(I)-salt formulation identifying the familiar Sandmeyer routes. Keep the distinguishing reagent labels when an exam asks for the named method.

$$\mathrm{C_{6}H_{5}N_{2}BF_{4}} \xrightarrow{\Delta} \mathrm{C_{6}H_{5}F} + \mathrm{BF_{3}} + \mathrm{N_{2}}$$

$$\mathrm{C_{6}H_{5}N_{2}}^{+} + \mathrm{H_{3}PO_{2}} + \mathrm{H_{2}O} \longrightarrow \mathrm{C_{6}H_{6}} + \mathrm{H_{3}PO_{3}} + \mathrm{N_{2}} + \mathrm{H}^{+}$$

An activated aromatic partner can couple with an aromatic diazonium species to form an azo linkage, N=N. Phenolic and amino-substituted rings are familiar partners. The medium controls the reactive form and available ring position; para coupling is commonly favoured when that position is available in the simple examples.

$$\mathrm{C_{6}H_{5}N_{2}}^{+} + \mathrm{C_{6}H_{5}OH} \longrightarrow \mathrm{C_{6}H_{5}N=NC_{6}H_{4}OH} + \mathrm{H}^{+}$$

The extended conjugated structure can absorb visible light, making azo compounds important colour systems. The colour is connected with electronic structure, not simply with the presence of any nitrogen atom.

How can an observed product show whether the diazonium nitrogens were retained?

Look for an azo N=N connection between carbon groups. Ordinary replacement products instead remove the diazonium nitrogens as molecular nitrogen.

Result: A new azo linkage indicates nitrogen retention; the aryl substitution routes shown above release nitrogen gas.

1. Which reagent class can replace a diazonium group with a nitrile group in the named copper route?
2. What new connection defines azo coupling?

Answers

1. A suitable copper(I) cyanide system.
2. An N=N linkage joining the diazonium-derived group with the activated coupling partner.

An aromatic amine can be a planning tool even when no nitrogen remains in the final target.

Amino groups can help place substituents through their directing effect, then be transformed into diazonium groups. Replacement can install halogen, hydroxyl, nitrile or hydrogen. This lets a synthesis use one group for positional control and later replace it with another.

The final hydrogen replacement is especially instructive: a group can be introduced temporarily to influence where another reaction occurs, then removed. A multistep plan is therefore more than listing reagents. It tracks positions, carbon count, functional-group compatibility and any temporary protection.

Plan how an already positioned aromatic amino group could lead to an aromatic carboxylic acid at that same carbon.

Diazotise the suitable primary aromatic amine. Use an appropriate cyanide replacement to make the aryl nitrile. Hydrolyse the nitrile under suitable conditions. The nitrile carbon becomes the acid carbon.

Result: The ring position is retained while the new carboxyl carbon comes from the cyanide source. Count that carbon explicitly.

A reliable exam scheme writes the structure after each step. This reveals whether a later reagent would attack another group or whether a proposed reaction changes an unintended ring position. Conditions such as acidic work-up or controlled temperature should be written where they decide the actual species.

1. Why can diazonium chemistry improve aromatic positional control?
2. What carbon-count change occurs on converting aryl diazonium to nitrile and then acid?

Answers

1. It transforms a functional group already located at the intended carbon.
2. The ring gains a carboxyl carbon supplied through cyanide.

Try these original questions without looking at the answers. Show equations, conditions and units when relevant.

9.1 Classify $\mathrm{CH_3NHCH_2CH_3}$ and $\mathrm{(CH_3)_3N}$.

9.2 Explain the lone pair’s two roles in a simple amine.

9.3 Name $\mathrm{CH_3CH_2CH_2NHCH_3}$.

9.4 Compare carbon counts from propanamide reduction and Hofmann degradation.

9.5 Why is excess ammonia useful but not absolute proof of one product in alkyl-halide ammonolysis?

9.6 Why is aniline commonly a weaker base than a simple alkylamine in a suitable comparison?

9.7 Compare model bases with $pK_b=3.80$ and $4.80$.

9.8 Why does tertiary not automatically mean strongest aqueous amine base?

9.9 State the class tested by the carbylamine reaction and the product class.

9.10 What structural feature lets an ordinary primary-amine sulfonamide dissolve by deprotonation in a Hinsberg comparison?

9.11 Why can acyl protection help control aniline ring substitution?

9.12 Give the intended conditions for the familiar aromatic diazotisation intermediate.

9.13 Distinguish diazonium replacement from azo coupling by the nitrogen atoms.

9.14 Plan a route from a positioned aryl amino group to an aryl iodide, and state why the ring position is retained.

Answers

9.1 Secondary and tertiary amines, respectively.

9.2 It contributes to the electron-domain picture of shape and can act as a proton-accepting or nucleophilic pair.

9.3 N-Methylpropan-1-amine.

9.4 Reduction retains a three-carbon amine skeleton; degradation gives a two-carbon amine skeleton.

9.5 It can favour the first substitution while the product amine still remains capable of further alkylation.

9.6 Its lone pair is delocalised into the aromatic system and is less available for protonation.

9.7 The first has $10$ times larger Kb in the same convention.

9.8 Solvation and steric effects combine with electron donation, so class alone is insufficient.

9.9 Primary amines; isocyanides.

9.10 It retains an N–H bond whose proton can be removed under the relevant basic conditions.

9.11 It moderates the amino group’s electronic and chemical behaviour, reducing uncontrolled activation or protonation effects.

9.12 A suitable primary aromatic amine with nitrous acid generated in cold aqueous acid, commonly near 273–278 K in the school-level description.

9.13 Replacement releases them as nitrogen gas; coupling retains them in an N=N linkage.

9.14 Form the corresponding diazonium species under suitable cold acidic conditions, then replace the diazonium group with iodide. The transformation occurs at the carbon already bearing the amino-derived group.

Quick-check answers appear directly after each topic. The selected explanations below show how to check the reasoning, rather than memorise a final line.

What structural feature lets an ordinary primary-amine sulfonamide dissolve by deprotonation in a Hinsberg comparison?

It retains an N–H bond whose proton can be removed under the relevant basic conditions.

Why can acyl protection help control aniline ring substitution?

It moderates the amino group’s electronic and chemical behaviour, reducing uncontrolled activation or protonation effects.

Give the intended conditions for the familiar aromatic diazotisation intermediate.

A suitable primary aromatic amine with nitrous acid generated in cold aqueous acid, commonly near 273–278 K in the school-level description.

Distinguish diazonium replacement from azo coupling by the nitrogen atoms.

Replacement releases them as nitrogen gas; coupling retains them in an N=N linkage.

Plan a route from a positioned aryl amino group to an aryl iodide, and state why the ring position is retained.

Form the corresponding diazonium species under suitable cold acidic conditions, then replace the diazonium group with iodide. The transformation occurs at the carbon already bearing the amino-derived group.