Amines — JEE Mains Chemistry Formulas and Traps

Structure, Physical Properties and Basicity

Basic strength of aliphatic amines and amides

Base dissociation of an amine and its pKb

\mathrm{RNH_2 + H_2O \rightleftharpoons RNH_3^+ + OH^-} \qquad K_b = \dfrac{[\mathrm{RNH_3^+}][\mathrm{OH^-}]}{[\mathrm{RNH_2}]} \qquad pK_b = -\log K_b

Basic strength of aryl amines and nitrogen heterocycles

A base and its conjugate acid at 298 K

pK_a(\mathrm{BH^+}) + pK_b(\mathrm{B}) = 14

Structure and physical properties of amines

Property	What is observed	Reason
Shape at nitrogen	Pyramidal, C–N–C about 108° in $\mathrm{(CH_3)_3N}$	$sp^3$ nitrogen with one lone pair
Physical state	Lower aliphatic amines are gases with a fishy smell; 1° amines with three or more carbons are liquids	Molar mass and hydrogen bonding rise together
Boiling point of isomers	1° > 2° > 3°	Two N–H, one N–H, then no N–H for intermolecular hydrogen bonds
Amine against alcohol	Alcohol boils higher at similar molar mass	O–H is more polar than N–H, so its hydrogen bonds are stronger
Solubility in water	Lower amines dissolve; higher amines and aniline barely dissolve	Hydrogen bonds to water, outweighed by a large hydrophobic part
Aniline on storage	Colourless when pure, turns brown on standing	Atmospheric oxidation of the activated ring

Count the N–H bonds first: they decide hydrogen bonding, and hydrogen bonding decides boiling point.

Common traps

Primary amines associate more than secondary amines

A primary amine has two N–H bonds and a secondary amine has one, so primary amines form more intermolecular hydrogen bonds and boil higher than their secondary isomers.

Aniline darkens by oxidation, not reduction

The electron-rich ring of an aryl amine is oxidised by air, giving coloured products. A statement that arylamines colour on storage by atmospheric reduction is false.

Tertiary is not the strongest base in water

In the gas phase a tertiary amine is the strongest, but in water it loses out on solvation and crowding. Trimethylamine is weaker than methylamine in water; triethylamine is weaker than diethylamine.

The methyl and ethyl orders are different

In water the methyl series runs 2° > 1° > 3°, while the ethyl series runs 2° > 3° > 1°. Swapping the two is the most common slip in these questions.

An amide is not an amine

Acetamide has a nitrogen with a lone pair, but the lone pair is shared with the carbonyl group. Acetamide is a weaker base than aniline, not a stronger one.

Benzylamine is not an aryl amine for basicity

In benzylamine the nitrogen sits on a

\mathrm{CH_2}

group, not on the ring, so its lone pair is not delocalised. It is about as basic as an aliphatic amine and far stronger than aniline.

Pyridine and pyrrole are not equally basic

Pyridine's lone pair lies in the ring plane, outside the aromatic π system, so it can take a proton. Pyrrole's lone pair is part of the aromatic sextet; protonating it destroys aromaticity, so pyrrole is almost non-basic.

Preparation: Reduction, Ammonolysis and Gabriel Synthesis

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Ammonolysis of alkyl halides

Successive alkylation in ammonolysis

\mathrm{NH_3 \xrightarrow{RX} RNH_2 \xrightarrow{RX} R_2NH \xrightarrow{RX} R_3N \xrightarrow{RX} R_4N^+X^-}

Gabriel phthalimide synthesis of primary amines

Gabriel phthalimide synthesis

\text{phthalimide} \xrightarrow{\mathrm{KOH}} \text{K-phthalimide} \xrightarrow{\mathrm{RX}} \text{N-alkylphthalimide} \xrightarrow{\mathrm{NaOH(aq)}} \mathrm{RNH_2} + \text{phthalate}

Reduction routes to amines: nitro compounds, nitriles and amides

Starting compound	Reagent	Product	Carbon count
$\mathrm{C_6H_5NO_2}$	$\mathrm{Sn/HCl}$ , $\mathrm{Fe/HCl}$ or $\mathrm{H_2/Pd}$	$\mathrm{C_6H_5NH_2}$	Unchanged
$\mathrm{RC{\equiv}N}$	$\mathrm{LiAlH_4}$ , $\mathrm{H_2/Ni}$ or $\mathrm{Na(Hg)/C_2H_5OH}$	$\mathrm{RCH_2NH_2}$	One more than the halide RX
$\mathrm{RCONH_2}$	$\mathrm{LiAlH_4}$ , then $\mathrm{H_2O}$	$\mathrm{RCH_2NH_2}$	Unchanged
$\mathrm{RC{\equiv}N}$	$\mathrm{SnCl_2/HCl}$ , then $\mathrm{H_3O^+}$	$\mathrm{RCHO}$ , not an amine	Unchanged
$\mathrm{RCONH_2}$	$\mathrm{Br_2}$ and NaOH (Hofmann)	$\mathrm{RNH_2}$	One fewer

Reduction keeps every carbon; only the Hofmann route drops one.

Common traps

Nitrobenzene needs acid or a catalyst to reach aniline

A metal reduces nitrobenzene to aniline only in acid. In neutral solution the reduction stops at intermediate stages; for example

\mathrm{Zn/NH_4Cl}

gives N-phenylhydroxylamine.

Two amide reactions, two carbon counts

\mathrm{LiAlH_4}

reduces

\mathrm{RCONH_2}

to

\mathrm{RCH_2NH_2}

and keeps every carbon.

\mathrm{Br_2}

with NaOH turns the same amide into

\mathrm{RNH_2}

, one carbon shorter.

Ammonolysis rarely gives one amine

Each amine formed is a better nucleophile than ammonia and reacts again. Expect a mixture of 1°, 2° and 3° amines and the quaternary salt; only a large excess of ammonia makes the primary amine dominant.

Ammonolysis breaks a C–X bond

Ammonolysis means ammonia replacing a halogen on an alkyl or benzyl carbon. Acylation of an amine, reduction of a nitrile or protonation of aniline by HCl is not ammonolysis.

Gabriel gives no aryl amines

4-Methoxyaniline, aniline or any amine with

\mathrm{NH_2}

on a ring carbon cannot be made by Gabriel synthesis, because an aryl halide does not undergo SN2.

Gabriel gives no secondary amines

After one alkylation the nitrogen of the phthalimide carries no hydrogen, so only one alkyl group can be attached. The product after hydrolysis is always a primary amine.

Hofmann Bromamide Degradation

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Hofmann bromamide degradation: equation, intermediates and scope

Hofmann bromamide degradation

\mathrm{RCONH_2 + Br_2 + 4NaOH \to RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O}

Hofmann degradation in multistep sequences

Halide to amine with the same carbon count

\mathrm{RX \xrightarrow{Mg,\ ether} RMgX \xrightarrow{CO_2,\ H_3O^+} RCOOH \xrightarrow{NH_3,\ \Delta} RCONH_2 \xrightarrow{Br_2,\ NaOH} RNH_2}

Common traps

Benzamide gives aniline, not benzylamine

Hofmann degradation removes the carbonyl carbon.

\mathrm{C_6H_5CONH_2}

gives

\mathrm{C_6H_5NH_2}

. Benzylamine would come from reducing benzamide with

\mathrm{LiAlH_4}

.

Aryl groups migrate too

The group that moves from carbon to nitrogen can be alkyl or aryl. A statement that only an alkyl group migrates is false; benzamide reacts perfectly well.

One bromine, four hydroxides

The balanced equation uses one

\mathrm{Br_2}

and four NaOH per amide. Two hydroxides neutralise the two HBr equivalents and two more trap

\mathrm{CO_2}

as carbonate.

The Grignard route does not lengthen the amine

\mathrm{CO_2}

adds one carbon, but the Hofmann step removes it again. Starting from bromobenzene you end at aniline, not benzylamine.

Isocyanides and nitriles hydrolyse differently

Acid hydrolysis of

\mathrm{R{-}NC}

gives

\mathrm{RNH_2}

and formic acid; acid hydrolysis of

\mathrm{R{-}CN}

gives

\mathrm{RCOOH}

. The atom bonded to R decides.

Acylation, Nitrous Acid and Hofmann Elimination

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Acylation of amines: products, stoichiometry and yield

Acetylation of an amine and the mass it adds

\mathrm{RNH_2 + (CH_3CO)_2O \to RNHCOCH_3 + CH_3COOH} \qquad \Delta M = +42\ \mathrm{g\,mol^{-1}}\ \text{per acetyl group}

Amine salts, nitrous acid on primary aliphatic amines, and Hofmann elimination

Nitrous acid on a primary aliphatic amine

\mathrm{RNH_2 \xrightarrow{NaNO_2,\ HCl,\ cold} [RN_2^+Cl^-] \xrightarrow{H_2O} ROH + N_2 + HCl} \qquad n(\mathrm{N_2}) = n(\mathrm{RNH_2})

Common traps

Acylation is one to one per nitrogen

One mole of aniline gives one mole of acetanilide, whatever the excess of anhydride. The excess reagent does not add a second acetyl group to the amide nitrogen under normal conditions.

The better nucleophile is acylated first

In a molecule with an alkyl

\mathrm{NH_2}

and an amide or aryl nitrogen, one equivalent of anhydride acylates the alkyl

\mathrm{NH_2}

. In 4-aminophenol it acylates the

\mathrm{NH_2}

, not the OH.

Aliphatic and aromatic primary amines differ with nitrous acid

A primary aliphatic amine loses

\mathrm{N_2}

at once and gives an alcohol. A primary aromatic amine at 273–278 K gives a diazonium salt that stays in solution; it loses

\mathrm{N_2}

only on warming.

Hofmann elimination is not Saytzeff elimination

Dehydrohalogenation of an alkyl halide usually gives the more substituted alkene. Elimination from a quaternary ammonium salt gives the less substituted alkene as the major product.

Carbylamine and Hinsberg Tests

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Carbylamine (isocyanide) test for primary amines

Carbylamine reaction

\mathrm{RNH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R{-}N{\equiv}C + 3KCl + 3H_2O}

Hinsberg test: benzenesulphonyl chloride with primary, secondary and tertiary amines

Amine class	Example	Product with the reagent	Behaviour in alkali
Primary aliphatic	$\mathrm{C_2H_5NH_2}$	$\mathrm{C_6H_5SO_2NHC_2H_5}$	Dissolves: the N–H is acidic
Primary aromatic	$\mathrm{C_6H_5NH_2}$	$\mathrm{C_6H_5SO_2NHC_6H_5}$	Dissolves: the N–H is acidic
Secondary aliphatic	$\mathrm{(C_2H_5)_2NH}$	$\mathrm{C_6H_5SO_2N(C_2H_5)_2}$	Insoluble solid: no N–H left
Secondary aromatic	$\mathrm{C_6H_5NHCH_3}$	$\mathrm{C_6H_5SO_2N(CH_3)C_6H_5}$	Insoluble solid: no N–H left
Tertiary	$\mathrm{(C_2H_5)_3N}$ or $\mathrm{C_6H_5N(CH_3)_2}$	No reaction	The amine is unchanged

Reacts and dissolves: primary. Reacts and stays solid: secondary. No reaction: tertiary.

Identifying an amine from its test results

Test	Primary aliphatic	Primary aromatic	Secondary	Tertiary
$\mathrm{CHCl_3}$ + alcoholic KOH, heat	Foul-smelling isocyanide	Foul-smelling isocyanide	No isocyanide	No isocyanide
Hinsberg's reagent, then alkali	Sulphonamide, dissolves	Sulphonamide, dissolves	Sulphonamide, insoluble solid	No reaction
$\mathrm{NaNO_2}$ + HCl, cold	$\mathrm{N_2}$ gas and an alcohol	Diazonium salt, no gas at 273–278 K	N-Nitrosamine, yellow oil	Aliphatic: soluble salt; aromatic: p-nitroso compound
Diazotise, then β-naphthol in NaOH	No dye	Orange-red azo dye	No dye	No dye
Dilute HCl	Dissolves as a salt	Dissolves as a salt	Dissolves as a salt	Dissolves as a salt

The dye test is the one that separates primary aromatic from primary aliphatic amines.

Common traps

The carbylamine test covers aromatic amines too

Aniline gives phenyl isocyanide just as ethanamine gives ethyl isocyanide. The test separates primary from secondary and tertiary amines, not aliphatic from aromatic.

Isocyanide from a halide needs AgCN

KCN with an alkyl halide gives a nitrile, R–C≡N. AgCN gives the isocyanide, R–N≡C. An aryl halide gives neither under ordinary conditions.

A secondary amine's product does not dissolve in alkali

The sulphonamide from a secondary amine has no hydrogen on nitrogen, so alkali has nothing to remove. It stays as an insoluble solid. Only a primary amine's product dissolves.

A clear solution means a primary amine

When a primary amine reacts with the reagent in alkali, the sulphonamide dissolves as its salt and the solution stays clear. A precipitate means a secondary amine; an unchanged amine means a tertiary one.

Dissolving in acid does not identify the class

Primary, secondary and tertiary amines all dissolve in dilute mineral acid as salts. Solubility in acid shows only that the compound is a base.

Benzylamine behaves as an aliphatic amine in the tests

Benzylamine gives the carbylamine test and an alkali-soluble Hinsberg product, but with nitrous acid it gives nitrogen gas and benzyl alcohol, not a stable diazonium salt, so it gives no azo dye.

Electrophilic Substitution in Aniline

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Bromination of aniline and protection by acetylation

Bromination of aniline with bromine water

\mathrm{C_6H_5NH_2 + 3Br_2 \xrightarrow{H_2O} 2,4,6{-}Br_3C_6H_2NH_2\downarrow + 3HBr}

Nitration of aniline: the anilinium ion and meta product

Product distribution in the direct nitration of aniline

\mathrm{C_6H_5NH_2 \xrightarrow{HNO_3,\ H_2SO_4,\ 288\ K}}\ p\ (51\%) + m\ (47\%) + o\ (2\%)

Friedel–Crafts failure, sulphonation and oxidation of aniline

Reaction of aniline	Reagent and conditions	Result	Reason
Friedel–Crafts alkylation or acylation	RCl or RCOCl with anhydrous $\mathrm{AlCl_3}$	No ring substitution; N– $\mathrm{AlCl_3}$ complex	Nitrogen is a Lewis base; the complex deactivates the ring
Sulphonation	Conc. $\mathrm{H_2SO_4}$ , then 453–473 K	Sulphanilic acid, $\mathrm{H_3N^+C_6H_4SO_3^-}$	Anilinium hydrogensulphate rearranges on heating
Oxidation	Acidified $\mathrm{K_2Cr_2O_7}$	p-Benzoquinone	The ring is very electron-rich
Nitration	$\mathrm{HNO_3/H_2SO_4}$ , 288 K	Mixture of para, meta and ortho nitroanilines	Partial protonation to the anilinium ion
Bromination	Bromine water	2,4,6-Tribromoaniline	Strong activation by $\mathrm{NH_2}$

Wherever aniline meets an acid, think of the protonated or complexed nitrogen first.

Common traps

Bromine water cannot stop at one bromine

The amino group activates all three ortho and para positions. With bromine water they are all substituted at once; monobromination needs the amine protected as its acetyl derivative.

The protecting group must come off

Bromination of acetanilide gives 4-bromoacetanilide. The last step, hydrolysis, is what turns it into 4-bromoaniline; a sequence without it ends at the amide.

NH₂ is not meta-directing

The amino group directs ortho and para. The meta product in nitration comes from the anilinium ion formed in the acid, whose positive nitrogen directs meta.

Para is still slightly ahead of meta

Direct nitration gives about 51% para and 47% meta, with only 2% ortho. A statement that meta exceeds ortho is true; one that meta is the only product is false.

Friedel–Crafts on aniline gives no ring product at all

The

\mathrm{AlCl_3}

is tied up by the nitrogen. Do not predict ortho, para or meta alkylanilines; the ring is not alkylated or acylated.

A strongly activated ring is also easily oxidised

Aniline darkens in air and is oxidised to p-benzoquinone by dichromate. Oxidising agents in a sequence can destroy the amine before any intended step.

Diazonium Salts: Stability and Replacement Reactions

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Diazotisation and the stability of diazonium salts

Diazotisation of aniline

\mathrm{C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273{-}278\ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O}

Synthesis planning with diazonium salts: the amino group as a temporary director

Removing the amino group after it has directed

\mathrm{ArNH_2 \xrightarrow{NaNO_2,\ HCl,\ 273\ K} ArN_2^+Cl^- \xrightarrow{H_3PO_2,\ H_2O} ArH + N_2}

Replacement reactions of arenediazonium salts

Reagent	Product from ArN₂⁺	Name or note
$\mathrm{CuCl/HCl}$	ArCl	Sandmeyer
$\mathrm{CuBr/HBr}$	ArBr	Sandmeyer
$\mathrm{CuCN/KCN}$	ArCN	Sandmeyer
Cu powder with HCl or HBr	ArCl or ArBr	Gattermann
KI	ArI	No copper needed
$\mathrm{HBF_4}$ , then heat	ArF, with $\mathrm{BF_3}$ and $\mathrm{N_2}$	Balz–Schiemann
$\mathrm{H_2O}$ , warm	ArOH	Phenol and $\mathrm{N_2}$
$\mathrm{H_3PO_2}$ and $\mathrm{H_2O}$	ArH	Reductive removal; $\mathrm{H_3PO_3}$ forms
$\mathrm{CH_3CH_2OH}$	ArH	Ethanol is oxidised to ethanal

Sandmeyer delivers Cl, Br and CN only; F and I come by other reagents.

Common traps

Electron-withdrawing groups destabilise the diazonium salt

A para nitro or cyano group pulls electrons away from a ring that is already carrying a positive charge, so the salt becomes less stable. A para methoxy group does the opposite.

Only a nitrogen on the ring gives a stable salt

Diazotisation at 273–278 K gives a usable salt only from a primary aromatic amine. Aliphatic amines, benzylamine included, lose nitrogen at once.

Ethanol reduces, it does not make an ether

Benzenediazonium chloride with ethanol gives benzene and ethanal. A reaction claiming phenetole,

\mathrm{C_6H_5OC_2H_5}

, as the product is wrong.

Fluoro- and iodobenzene are not Sandmeyer products

Copper(I) salts deliver chloride, bromide and cyanide. Iodobenzene comes from KI; fluorobenzene comes from the fluoroborate on heating.

The director is whatever is on the ring at that step

The same nitrogen directs meta as

\mathrm{NO_2}

and ortho-para as

\mathrm{NH_2}

. A sequence that reduces before brominating puts the bromine in a different place from one that brominates first.

Removing NH₂ leaves the pattern it created

After

\mathrm{H_3PO_2}

removes the diazo group, the substituents stay where the amino group sent them. Two bromines placed ortho and para to an

\mathrm{NH_2}

are meta to each other once the

\mathrm{NH_2}

is gone.

Coupling Reactions and Azo Dyes

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Coupling of diazonium salts with phenols and aryl amines

Coupling of benzenediazonium chloride with phenol

\mathrm{C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{OH^-} C_6H_5{-}N{=}N{-}C_6H_4{-}OH\ (para) + HCl}

Azo dye stoichiometry and the Griess–Ilosvay test

Mass of dye from the diazotised amine

n(\text{dye}) = n(\mathrm{ArNH_2}) \qquad m(\text{dye}) = n(\text{dye}) \times M(\text{dye}) \qquad \%\mathrm{N} = \dfrac{\text{mass of N}}{M} \times 100

Common traps

Coupling keeps the nitrogen

In replacement reactions

\mathrm{N_2}

leaves; in coupling both nitrogens stay in the product as the

\mathrm{{-}N{=}N{-}}

bridge. An azo dye always contains the two nitrogens of the diazonium ion.

Only strongly activated rings couple

The diazonium ion is a weak electrophile. Benzene, toluene or chlorobenzene do not couple; phenols, naphthols and aryl amines do.

Ortho methyls slow coupling on dimethylanilines

Two methyl groups beside

\mathrm{N(CH_3)_2}

push it out of the ring plane. Its lone pair no longer feeds the ring, so the ring is less activated and couples more slowly.

When aniline plays both roles, it is used twice

To make aniline yellow from aniline alone, one aniline is diazotised and a second is the coupling partner. The moles of dye are half the moles of aniline used in total, but equal to the moles of diazonium salt.

The Griess–Ilosvay colour is red

The azo dye from diazotised sulphanilic acid and 1-naphthylamine is red. The test detects nitrite; it uses the amine chemistry but is not a test for amines.

Amines formulas

Structure, Physical Properties and Basicity

Basic strength of aliphatic amines and amides

Basic strength of aryl amines and nitrogen heterocycles

Structure and physical properties of amines

Common traps

Preparation: Reduction, Ammonolysis and Gabriel Synthesis

Ammonolysis of alkyl halides

Gabriel phthalimide synthesis of primary amines

Reduction routes to amines: nitro compounds, nitriles and amides

Common traps

Hofmann Bromamide Degradation

Hofmann bromamide degradation: equation, intermediates and scope

Hofmann degradation in multistep sequences

Common traps

Acylation, Nitrous Acid and Hofmann Elimination

Acylation of amines: products, stoichiometry and yield

Amine salts, nitrous acid on primary aliphatic amines, and Hofmann elimination

Common traps

Carbylamine and Hinsberg Tests

Carbylamine (isocyanide) test for primary amines

Hinsberg test: benzenesulphonyl chloride with primary, secondary and tertiary amines

Identifying an amine from its test results

Common traps

Electrophilic Substitution in Aniline

Bromination of aniline and protection by acetylation

Nitration of aniline: the anilinium ion and meta product

Friedel–Crafts failure, sulphonation and oxidation of aniline

Common traps

Diazonium Salts: Stability and Replacement Reactions

Diazotisation and the stability of diazonium salts

Synthesis planning with diazonium salts: the amino group as a temporary director

Replacement reactions of arenediazonium salts

Common traps

Coupling Reactions and Azo Dyes

Coupling of diazonium salts with phenols and aryl amines

Azo dye stoichiometry and the Griess–Ilosvay test

Common traps

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