JEE Main & Advanced

Amines

Amines for JEE Main & Advanced

1
Module 1

Structure, Preparation, and Basicity

Classification, Nomenclature, and PreparationTopic 1

Amines: Organic derivatives of $NH_3$ where one or more H is replaced by alkyl/aryl groups.

Classification (based on number of R groups on N):

TypeStructureExample
Primary (1°)$R-NH_2$ (1 H replaced)Methylamine ($CH_3NH_2$), aniline ($C_6H_5NH_2$)
Secondary (2°)$R_2NH$ (2 H replaced)Dimethylamine ($(CH_3)_2NH$), N-methylaniline
Tertiary (3°)$R_3N$ (3 H replaced)Trimethylamine ($(CH_3)_3N$), N,N-dimethylaniline
Quaternary ammonium$R_4N^+X^-$$(CH_3)_4N^+Cl^-$

Also:

  • Aliphatic ($R$ = alkyl): methylamine, ethylamine
  • Aromatic ($R$ = aryl): aniline ($C_6H_5NH_2$)
  • Mixed: N-methylaniline ($C_6H_5NHCH_3$)

Structure: N is $sp^3$ hybridized; pyramidal (like $NH_3$); lone pair on N.

  • Bond angle ~$107°$ (slightly less than tetrahedral due to lp-bp repulsion)

Nomenclature:

CompoundCommonIUPAC
$CH_3NH_2$MethylamineMethanamine
$(CH_3)_2NH$DimethylamineN-methylmethanamine
$(CH_3)_3N$TrimethylamineN,N-dimethylmethanamine
$C_6H_5NH_2$AnilineBenzenamine (or aniline accepted)
$C_6H_5NHCH_3$N-methylanilineN-methylbenzenamine

Suffix: replace last 'e' of alkane with -amine. Substituents on N: prefix N-.

Preparation:

1. Reduction of Nitro Compounds: $RNO_2 + 6[H] \xrightarrow{Sn/HCl \text{ or } Fe/HCl \text{ or } H_2/Pt} RNH_2 + 2H_2O$

  • For aniline: $C_6H_5NO_2 + 6[H] \to C_6H_5NH_2 + 2H_2O$
  • Common reagents: $Sn + HCl$, $Fe + HCl$, $H_2/Ni$, $Zn/HCl$

2. Reduction of Nitriles ($RCN$): $RCN + 4[H] \xrightarrow{Na/C_2H_5OH \text{ or } LiAlH_4 \text{ or } H_2/Ni} RCH_2NH_2$

  • Adds one C; 1° amine
  • $CH_3CN \to CH_3CH_2NH_2$ (ethylamine, from methyl cyanide)

3. Reduction of Amides: $RCONH_2 + 4[H] \xrightarrow{LiAlH_4} RCH_2NH_2 + H_2O$

  • Note: same number of C; 1° amine

4. Ammonolysis of Alkyl Halides (Hofmann's Method): $RX + NH_3 \to RNH_2 + HX$ (further alkylation possible)

  • Gives mix of $1°, 2°, 3°$ amines and quaternary salt
  • Order of reactivity: $RI > RBr > RCl$
  • Industrial way for simple amines

Problems: over-alkylation. Solution: excess NH₃ for $1°$ amine; excess RX for higher amines.

5. Hofmann Bromamide Degradation: $RCONH_2 + Br_2 + 4NaOH \to RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O$

  • Amide → $1°$ amine with one fewer C
  • Specific for $1°$ amine; no other isomers
  • e.g., $CH_3CH_2CONH_2 \to CH_3CH_2NH_2$

6. Gabriel Phthalimide Synthesis: For pure $1°$ amines only.

  1. Potassium phthalimide + RX → N-alkyl phthalimide
  2. Hydrolyze (or hydrazine — Ing-Manske) → $RNH_2$ + phthalic acid

$\text{Phthalimide}-K + RX \to \text{Phthalimide}-R \xrightarrow{NaOH \text{ or } NH_2NH_2} RNH_2 + \text{phthalate}$

Limitations: Doesn't work with aryl halides (aryl C-X bond is unreactive in $S_N2$); doesn't give aryl amines.

Physical Properties:

  • Lower amines: gases or liquids; pungent (fishy) smell
  • BP: $RNH_2$ > $R_2NH$ > $R_3N$ (more H-bonding with more N-H bonds)
  • BP order: $RNH_2 \approx ROH > R_3N \approx RH$ (alkanes)
  • Solubility: lower amines very soluble in water (H-bond with water); decreases with chain length
  • Aniline: oily liquid; turns brown on air exposure (auto-oxidation)
Worked Examples
1

Predict products and identify if pure $1°$ amine: (a) $CH_3Br + NH_3$ (excess NH₃) (b) Gabriel phthalimide synthesis with $CH_3CH_2Br$.

Show solution

(a) With excess NH₃: $CH_3Br + NH_3 \to CH_3NH_2 + HBr$ (mostly $1°$, but some $2°, 3°$ also possible). (b) Gabriel: $CH_3CH_2Br$ + K-phthalimide → N-ethylphthalimide → hydrolyze → pure ethylamine ($CH_3CH_2NH_2$).

Final Answer: Gabriel gives pure $1°$ amine; ammonolysis gives mix.

2

Why doesn't Gabriel synthesis work for aniline?

Show solution

Gabriel synthesis requires reaction of phthalimide-K with alkyl halide via $S_N2$. Aryl halides ($C_6H_5X$) don't undergo $S_N2$ due to:

  • Partial double-bond character in C-Cl (resonance)
  • $sp^2$ C resistant to nucleophilic attack
  • Steric hindrance from ring planar arrangement

Hence Gabriel can't make aryl amines.

Final Answer: Aryl halides don't undergo $S_N2$ — Gabriel only makes aliphatic primary amines.

✎ Self-Check — 5 questions0 / 5
Q1.

$(CH_3)_3N$ is:

Q2.

Reduction of nitrobenzene gives:

Q3.

Gabriel synthesis cannot make:

Q4.

Hofmann bromamide degradation gives:

Q5.

Reduction of $CH_3CN$ gives:

Basicity of Amines (Gas vs Aqueous Phase)Topic 2

Basicity: Ability to donate lone pair to H⁺. Lone pair on N is key.

Comparison of Basicity:

  • Amines are weaker bases than NaOH, KOH
  • Amines are stronger bases than NH₃ (in general)
  • $pK_b$ (or $pK_a$ of conjugate acid) measures basicity

Factors Affecting Basicity:

1. Inductive Effect (+I from alkyl): Alkyl groups donate electrons through sigma bonds → increases electron density on N → makes N more basic. Predicted order: $3° > 2° > 1° > NH_3$.

2. Resonance Effect: In aromatic amines, lone pair on N can delocalize into ring; reduces availability → less basic.

  • Aniline < methylamine in basicity.

3. Steric Effect: Bulky groups around N prevent approach of H⁺ → decreases basicity. $(CH_3)_3N$ less basic than $(CH_3)_2NH$ in some scenarios.

4. Solvation Effect (Aqueous): $1°$ amine cation ($RNH_3^+$) has more H's available for H-bonding with water → better solvated → more stable → more basic effective. $3°$ amine cation ($R_3NH^+$) has only one N-H; less solvated → less stable → less basic in water.

Order of Basicity:

In Gas Phase (no solvation; only +I effect): $3° > 2° > 1° > NH_3$ $(CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3$

In Aqueous Solution (combines +I and solvation): Generally: $2° > 1° > 3° > NH_3$ $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$

(Sometimes $1° > 2° > 3°$ if bulky groups cause more steric and less solvation.)

$pK_b$ values (aqueous, 25°C):

Amine$pK_b$
$NH_3$$4.75$
$CH_3NH_2$$3.36$
$(CH_3)_2NH$$3.27$ (most basic)
$(CH_3)_3N$$4.20$
Aniline$9.42$ (very weak)
Pyridine$8.85$
Picoline$\sim 8.0$

Basicity of Aromatic Amines:

Aniline < methylamine because:

  1. Lone pair on N delocalized into benzene ring via resonance (mesomeric effect)
  2. Lone pair less available for protonation
  3. Conjugate acid (anilinium ion) loses this resonance stabilization → less stable → less basic

Substituent Effects on Aniline:

Substituent (at p-position)Effect on Aniline Basicity
$-NO_2$ (p)Strongly decreases (EW)
$-CN$ (p)Decreases
$-Cl$, $-Br$ (p)Decreases slightly (EW dominates over +R)
$-CH_3$ (p)Increases (ED)
$-OCH_3$ (p)Increases (ED, +R)
$-NH_2$ (p)Increases (ED, +R)
$-OH$ (p)Increases

Effects in different positions:

  • p-toluidine ($p$-$CH_3$): more basic than aniline (+I)
  • p-nitroaniline: much less basic than aniline (-R, -I)
  • o-nitroaniline: weakest of three nitroanilines (ortho effect, H-bonding internal)
Worked Examples
1

Arrange in increasing basicity (aqueous): $NH_3$, $CH_3NH_2$, $(CH_3)_2NH$, $(CH_3)_3N$.

Show solution

In aqueous solution, considering both +I and solvation:

  • $NH_3$: only H, full solvation of $NH_4^+$, but no +I (least basic by +I)
  • $CH_3NH_2$: $1°$ amine, +I from CH₃, good solvation of $CH_3NH_3^+$
  • $(CH_3)_2NH$: more +I, still 2 H's for solvation
  • $(CH_3)_3N$: most +I but only 1 N-H for solvation; poor stabilization of cation

In aqueous: $NH_3 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH$.

Final Answer: $NH_3 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH$.

2

Arrange in increasing basicity: aniline, p-nitroaniline, p-toluidine, p-methoxyaniline.

Show solution

EW groups decrease basicity; ED groups increase.

  • p-nitroaniline: $-NO_2$ strongly EW → very weak base
  • aniline: standard
  • p-toluidine: $-CH_3$ mild ED → slightly more basic
  • p-methoxyaniline: $-OCH_3$ strong ED (+R) → most basic

Order: p-nitroaniline < aniline < p-toluidine < p-methoxyaniline.

Final Answer: p-nitroaniline < aniline < p-toluidine < p-methoxyaniline.

✎ Self-Check — 5 questions0 / 5
Q1.

Most basic in gas phase:

Q2.

Most basic in aqueous solution:

Q3.

Aniline is less basic than methylamine because:

Q4.

Most basic among:

Q5.

Decrease in basicity from gas to aqueous of $(CH_3)_3N$ is due to:

2
Module 2

Reactions of Amines

Reactions of Aliphatic Amines and TestsTopic 1

Reactions of Amines:

1. With Mineral Acids (Salt formation): $RNH_2 + HCl \to RNH_3^+Cl^-$ (alkyl ammonium chloride)

  • Salts soluble in water; amines extracted by acidic workup

2. Alkylation (Hofmann's Method): $RNH_2 + R'X \to RR'NH + HX$ (→ further alkylation → $R_3N$ → $R_4N^+X^-$ quaternary salt)

  • Gives mixture; isolated by fractional distillation/separation

3. Acylation (with acid chlorides or anhydrides): $RNH_2 + R'COCl \to R'CONHR + HCl$ (amide; "protected" amine) $RNH_2 + (R'CO)_2O \to R'CONHR + R'COOH$

  • $3°$ amines don't acylate (no N-H)
  • Acetylation of aniline → acetanilide ($C_6H_5NHCOCH_3$) — used in some old drugs

4. Carbylamine Reaction (Isocyanide Test) — for $1°$ amines: $RNH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} RNC + 3KCl + 3H_2O$

  • Gives isocyanide with very offensive smell
  • Specific test for $1°$ amines (both aliphatic and aromatic)
  • $2°$ and $3°$ amines don't give this

5. Hinsberg Test (with Benzenesulfonyl Chloride):

  • Distinguishes $1°, 2°, 3°$ amines.
AmineWith $C_6H_5SO_2Cl$ (Benzenesulfonyl chloride)+ KOH (test)
$1°$ ($RNH_2$)$C_6H_5SO_2NHR$ (sulfonamide; one N-H still)Soluble in KOH (acidic NH due to strong EW $SO_2$)
$2°$ ($R_2NH$)$C_6H_5SO_2NR_2$ (no N-H)Insoluble in KOH (precipitate)
$3°$ ($R_3N$)Doesn't react (no N-H to displace)Amine recovered

6. Reaction with Nitrous Acid (Diazotization):

(a) $1°$ Aliphatic amines + HNO₂ (cold) → unstable diazonium → loses $N_2$ → alcohol/alkene mix: $RNH_2 + HNO_2 \xrightarrow{0-5°C} ROH + N_2 + H_2O$ (Diazonium too unstable at any T)

(b) $1°$ Aromatic amines + HNO₂ at 0-5°C → stable arenediazonium salt: $ArNH_2 + HNO_2 + HCl \xrightarrow{0-5°C} ArN_2^+Cl^- + 2H_2O$

  • Stable below $5°C$ (called diazotization)

(c) $2°$ amines + HNO₂ → N-nitrosoamines (yellow oily; carcinogenic): $R_2NH + HNO_2 \to R_2N-NO + H_2O$

(d) $3°$ amines + HNO₂: only protonates (no reaction at N-H since none).

7. Oxidation:

  • Aniline + chromic acid → mostly degradation products; benzoquinone with some oxidants
  • Aliphatic amines can give various products; not commonly used
Worked Examples
1

Distinguish among $CH_3CH_2NH_2$ (1°), $(CH_3)_2NH$ ($2°$), $(CH_3)_3N$ ($3°$).

Show solution

Test 1: Carbylamine reaction (with $CHCl_3$ + alc. KOH):

  • Only $1°$ amine gives offensive isocyanide smell → $CH_3CH_2NH_2$ positive.
  • $(CH_3)_2NH$ and $(CH_3)_3N$ negative.

Test 2: Hinsberg test ($C_6H_5SO_2Cl$, then KOH):

  • $(CH_3)_2NH$ gives a sulfonamide ($C_6H_5SO_2N(CH_3)_2$) insoluble in KOH.
  • $(CH_3)_3N$ doesn't react; can be recovered as amine.

Final Answer: Carbylamine + Hinsberg identifies all three.

2

Predict products: (a) $C_6H_5NH_2 + (CH_3CO)_2O$ (b) $C_6H_5NH_2 + Br_2/H_2O$

Show solution

(a) Acylation: $C_6H_5NH_2 + (CH_3CO)_2O \to C_6H_5NHCOCH_3$ (acetanilide) $+ CH_3COOH$.

(b) Aniline is highly activated for EAS by $-NH_2$ (+R, strong activator). $Br_2$/water gives tribrominated product: $C_6H_5NH_2 + 3Br_2 \to 2,4,6$-tribromoaniline $+ 3HBr$ (white ppt).

Final Answer: (a) Acetanilide; (b) 2,4,6-tribromoaniline.

✎ Self-Check — 5 questions0 / 5
Q1.

Hinsberg reagent:

Q2.

Carbylamine test confirms:

Q3.

With Hinsberg + KOH:

Q4.

Aniline + cold HNO₂/HCl:

Q5.

$RNH_2 + (CH_3CO)_2O$ gives:

Diazonium Salts and Aromatic AminesTopic 2

Aromatic Diazonium Salts ($ArN_2^+ X^-$): Extremely useful synthetic intermediates.

Preparation (Diazotization): $ArNH_2 + HNO_2 + HCl \xrightarrow{0-5°C} ArN_2^+Cl^- + 2H_2O$ (or $HNO_2$ generated in situ from $NaNO_2 + HCl$)

Conditions: low T (0-5°C); diazonium salt decomposes above 10°C.

Reactions of Diazonium Salts (extremely versatile):

A. Replacement of $N_2^+$ by Other Groups:

1. Sandmeyer Reaction: $ArN_2^+Cl^- + CuCl \to ArCl + N_2 + CuCl$ $ArN_2^+Br^- + CuBr \to ArBr + N_2 + CuBr$ $ArN_2^+Cl^- + CuCN \to ArCN + N_2$

  • Replaces $N_2^+$ with Cl, Br, or CN

2. Gattermann Reaction: Variant with Cu/HCl or Cu/HBr (cheaper than Sandmeyer's Cu(I) salts): $ArN_2^+Cl^- + Cu/HCl \to ArCl + N_2 + CuCl$

3. Replacement by -I: With KI: $ArN_2^+Cl^- + KI \to ArI + N_2 + KCl$ (no catalyst needed; smooth)

4. Replacement by -F (Schiemann): $ArN_2^+BF_4^- \xrightarrow{\Delta} ArF + N_2 + BF_3$

5. Replacement by -OH (Hydrolysis): $ArN_2^+ + H_2O \xrightarrow{warm} ArOH + N_2 + H^+$ (gives phenol; warm dilute acid solution)

6. Replacement by -H (Deamination): $ArN_2^+ + H_3PO_2 \to ArH + N_2 + H_3PO_3$ (or hypophosphorous acid; useful for removing $-NH_2$ from ring)

7. Replacement by -NO₂: With Cu/NaNO₂.

B. Azo Coupling Reactions (Retention of $N_2$):

Diazonium ions are weak electrophiles; couple with strongly activated aromatics (phenols, anilines):

$ArN_2^+ + Ar'H \to Ar-N=N-Ar' + H^+$

Conditions:

  • With phenols: alkaline solution (pH 9-10): the phenoxide $Ar'O^-$ is highly activated; couples at para position.
  • With aromatic amines: acidic to neutral (pH 5-7): amine itself activated.

Examples:

  • Aniline + diazonium → orange dye (methyl orange in industry made from diazotized sulfanilic acid + N,N-dimethylaniline)
  • Phenol + diazonium → orange-red dye (p-hydroxyazobenzene)

Reactivity: Used in azo dyes industry (clothes, paper, food coloring).

Importance of $ArN_2^+$: Convert $-NH_2$ to many functional groups (-OH, -X, -CN, -NO₂, -H, -Ar) — making it a versatile route in aromatic synthesis.

Aromatic Amines Reactions Summary:

Aniline ($C_6H_5NH_2$) Properties:

  1. Lone pair on N donates to ring → strong activator, o/p director (for EAS).
  2. Less basic than aliphatic amines (resonance lock-up of lp).
  3. Cannot undergo Friedel-Crafts directly (Lewis acid forms complex with lone pair on N, deactivating ring).
  • Solution: protect $-NH_2$ as $-NHCOCH_3$ (acetanilide) first.

EAS on Aniline:

  • Bromination ($Br_2$/water): 2,4,6-tribromoaniline (uncontrolled; aniline very activated).
  • Mono-bromination requires acetanilide protection, then deprotection.
  • Nitration:
  • Direct $HNO_3 + H_2SO_4$: aniline gets protonated ($-NH_3^+$, m-director) → m-nitroaniline major (~50%) + o, p
  • To get p-nitroaniline: protect as acetanilide → nitrate → mostly para → hydrolyze.

Coupling reactions:

  • Aniline + nitrous acid → diazonium salt (then various reactions above)
  • Diazonium coupling with phenol/amine → azo dye
Worked Examples
1

Convert aniline to: (a) chlorobenzene (b) phenol (c) iodobenzene (d) fluorobenzene

Show solution

All via diazonium salt intermediate: $C_6H_5NH_2 \xrightarrow{HNO_2/HCl, 0-5°C} C_6H_5N_2^+Cl^-$ (a) Sandmeyer: + CuCl → $C_6H_5Cl + N_2$ (b) Hydrolyze: + $H_2O$/warm → $C_6H_5OH + N_2 + H^+$ (c) Direct: + KI → $C_6H_5I + N_2 + KCl$ (d) Schiemann: + $HBF_4$ → $C_6H_5N_2^+BF_4^-$ → $\Delta$ → $C_6H_5F + N_2 + BF_3$.

Final Answer: All via diazotization, then specific reagent.

2

Why does coupling of $C_6H_5N_2^+$ work with phenol but only at right pH?

Show solution

Coupling is EAS where $ArN_2^+$ is electrophile. Phenol's reactivity depends on pH:

  • In alkaline (pH 9-10): Phenol becomes $C_6H_5O^-$ (phenoxide). Strongly activated for EAS by $-O^-$ (strong +R donor). Couples at para position to give $p$-hydroxy azobenzene (orange dye).
  • In strongly acidic (low pH): Phenol is mostly $C_6H_5OH$; not as activated. $ArN_2^+$ also less reactive.
  • In strongly alkaline: $ArN_2^+$ converts to $ArN=N-OH$ (covalent diazohydroxide, not electrophilic).

Optimal: slightly alkaline (pH 9-10).

Final Answer: Need pH where phenoxide forms (good Nu) but diazonium remains ionic (still electrophile).

✎ Self-Check — 5 questions0 / 5
Q1.

Sandmeyer reaction:

Q2.

Schiemann reaction gives:

Q3.

Coupling of $ArN_2^+$ with phenol gives:

Q4.

Aniline + Friedel-Crafts:

Q5.

Convert $C_6H_5NH_2$ to $C_6H_5OH$:

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