Intuition The big pictureAn amine is NH₃ with one or more H replaced by carbon groups . Everything about its chemistry follows from ONE fact: the nitrogen has a lone pair . That lone pair can either:
grab a proton → this makes it a base (basicity).act as a nucleophile → attacks carbons/carbonyls (Hofmann, carbylamine, Hinsberg).
So if you remember "lone pair availability," you can predict almost everything. The whole subtopic is just "how available is the lone pair, and what does it attack?"
A base donates a lone pair to H⁺. For amines:
R - N H 2 + H 2 O ⇌ R - N H 3 + + O H − R\text{-}NH_2 + H_2O \rightleftharpoons R\text{-}NH_3^+ + OH^- R - N H 2 + H 2 O ⇌ R - N H 3 + + O H − K b K_b K b p K b = − log K b pK_b = -\log K_b p K b = − log K b Smaller p K b pK_b p K b
The stronger base is the one whose conjugate acid (R - N H 3 + R\text{-}NH_3^+ R - N H 3 + more stable (less willing to give H⁺ back). So we always ask two questions:
Is the lone pair more available to grab H⁺? (electronic effect)
Is the resulting cation more stabilised ? (solvation + induction)
Intuition Two cooperating effects
+I (inductive electron donation): an alkyl group pushes electron density toward N, making the lone pair richer → grabs H⁺ better.Solvation of the cation: once protonated, R - N H 3 + R\text{-}NH_3^+ R - N H 3 +
These two effects fight when we go to tertiary amines:
Amine
N–H bonds in cation
+I effect
Net basicity (water)
NH₃
4
none
weakest of aliphatic series
1° RNH₂
3
one R
strong
2° R₂NH
2
two R
usually strongest
3° R₃N
1
three R
drops (poor solvation)
Intuition No solvent = no solvation rescueIn the gas phase there is no water to stabilise the cation . Only the +I effect remains. More alkyl groups = more electron donation = stronger base. So:
Gas phase: ( C H 3 ) 3 N > ( C H 3 ) 2 N H > C H 3 N H 2 > N H 3 \text{Gas phase: } (CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3 Gas phase: ( C H 3 ) 3 N > ( C H 3 ) 2 N H > C H 3 N H 2 > N H 3 pure inductive order . The "anomaly" in water (2°>3°) only exists because solvation matters there.
Common mistake Steel-man: "More alkyl groups always = more basic"Why it feels right: +I clearly donates electrons, so more R should always help. The flaw: in water you also lose H-bond stabilisation of the cation. The tertiary cation has only one N–H to H-bond. Fix: in water , basicity = (+I) − (solvation loss), so 2° often wins. In gas phase , no solvation term, so strictly 3°>2°>1°>NH₃.
Intuition Lone pair is "busy"In aniline the N lone pair is delocalised into the benzene ring (resonance). A lone pair tied up in resonance is not available to grab H⁺.
Also, protonation destroys that resonance, so the conjugate acid is destabilised relative to the free base → base is weak.
C 6 H 5 N ¨ H 2 lone pair into ring p K b ≈ 9.4 ( very weak ) \underset{\text{lone pair into ring}}{C_6H_5\ddot N H_2}\quad pK_b \approx 9.4 \;(\text{very weak}) lone pair into ring C 6 H 5 N ¨ H 2 p K b ≈ 9.4 ( very weak )
Definition Hofmann eliminationConvert an amine to a quaternary ammonium hydroxide , then heat. It eliminates to give an alkene + amine + water. The "trick": make the leaving group bulky and positively charged.
Exhaustive methylation: treat amine with excess C H 3 I CH_3I C H 3 I R - N + ( C H 3 ) 3 I − R\text{-}N^+(CH_3)_3\,I^- R - N + ( C H 3 ) 3 I − Convert to hydroxide: with moist A g 2 O Ag_2O A g 2 O R - N + ( C H 3 ) 3 O H − R\text{-}N^+(CH_3)_3\,OH^- R - N + ( C H 3 ) 3 O H − Heat (Δ): O H − OH^- O H − β-hydrogen , kicks out neutral N ( C H 3 ) 3 N(CH_3)_3 N ( C H 3 ) 3 alkene .
C H 3 C H 2 C H 2 − N + ( C H 3 ) 3 O H − → Δ C H 3 C H = C H 2 + N ( C H 3 ) 3 + H 2 O CH_3CH_2CH_2\!-\!N^+(CH_3)_3\,OH^- \xrightarrow{\Delta} CH_3CH=CH_2 + N(CH_3)_3 + H_2O C H 3 C H 2 C H 2 − N + ( C H 3 ) 3 O H − Δ C H 3 C H = C H 2 + N ( C H 3 ) 3 + H 2 O
Intuition Bulky base + steric crowdingThe leaving group N + ( C H 3 ) 3 N^+(CH_3)_3 N + ( C H 3 ) 3 huge . The base must approach a β-H, but the H on the less hindered (less substituted) carbon is easier to reach. So the proton removed is the least sterically hindered one → the least substituted (Hofmann) alkene dominates.
Contrast: ordinary E2 (small leaving group, e.g. Br) gives the more substituted Zaitsev alkene.
Common mistake Steel-man: "Elimination always gives the more stable (Zaitsev) alkene"Why it feels right: more substituted alkenes are thermodynamically more stable, and Zaitsev is the default rule. Flaw: Hofmann's huge charged leaving group makes the transition state sterically controlled , not stability controlled. Fix: bulky base / bulky leaving group ⇒ Hofmann (less substituted) product.
Definition Carbylamine reactionA test for primary amines only (1° aliphatic OR aromatic). Heat the amine with chloroform (CHCl₃) and alcoholic KOH → produces a carbylamine (isocyanide) with a foul, offensive smell.
R - N H 2 + C H C l 3 + 3 K O H → Δ R - N ≡ C + 3 K C l + 3 H 2 O R\text{-}NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R\text{-}N\!\!\equiv\!\!C + 3KCl + 3H_2O R - N H 2 + C H C l 3 + 3 K O H Δ R - N ≡ C + 3 K C l + 3 H 2 O
Intuition WHY only 1° aminesKOH + CHCl₃ generates dichlorocarbene : C C l 2 :CCl_2 : C C l 2 two N–H bonds , which is exactly what's needed to lose both H's and both Cl's to form the R - N ≡ C R\text{-}N\equiv C R - N ≡ C no isocyanide, no foul smell.
Use: A positive (foul-smelling) test = primary amine present.
Definition Hinsberg reagentBenzenesulfonyl chloride , C 6 H 5 S O 2 C l C_6H_5SO_2Cl C 6 H 5 S O 2 C l
Intuition HOW the logic worksThe amine N attacks the S of the sulfonyl chloride (nucleophile attacks electrophile). What happens next depends on how many N–H bonds remain on the sulfonamide product:
Amine
Product
Key feature
Behaviour with KOH/NaOH
1° R - N H - S O 2 C 6 H 5 R\text{-}NH\text{-}SO_2C_6H_5 R - N H - S O 2 C 6 H 5 still has one acidic N–H (acidified by SO₂)
Dissolves in alkali (forms salt) → clear solution
2° R 2 N - S O 2 C 6 H 5 R_2N\text{-}SO_2C_6H_5 R 2 N - S O 2 C 6 H 5 no N–H leftInsoluble in alkali → precipitate persists
3° no reaction
no N–H to react
does not react (amine stays free)
Intuition WHY the 1° sulfonamide N–H is acidicThe strongly electron-withdrawing S O 2 SO_2 S O 2 R - N − - S O 2 C 6 H 5 R\text{-}N^-\text{-}SO_2C_6H_5 R - N − - S O 2 C 6 H 5 soluble salt . The 2° product has no N–H at all , so it can't form a salt → stays as a solid.
Common mistake Steel-man: "Tertiary amine gives a precipitate too"Why it feels right: all amines have a lone pair and seem like they should react. Flaw: tertiary amine has no N–H , so after attacking S it would form a charged species with no proton to lose; it reverts. Fix: 3° amine does not react with Hinsberg reagent (stays as oily free amine).
Recall Quick self-test (cover answers)
Why does 2° amine beat 3° in water? → solvation loss of 3° cation outweighs +I.
Why does order reverse in gas phase? → no solvation, only +I matters.
Why is aniline weak? → lone pair delocalised into ring.
Hofmann gives which alkene? → least substituted.
Carbylamine tests for? → primary amines.
Hinsberg: which amine dissolves in NaOH? → primary (acidic N–H).
Why is a 2° aliphatic amine more basic than NH₃ in water? +I effect of two alkyl groups raises lone-pair density AND the cation is well solvated/H-bonded.
Why does (CH₃)₃N fall below (CH₃)₂NH in water? Its cation has only one N–H, so poor solvation/H-bonding outweighs the larger +I effect.
What is the gas-phase basicity order of methylamines? (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃ — pure inductive order (no solvation).
Why is aniline a much weaker base than alkyl amines? Its N lone pair is delocalised into the benzene ring, so it's less available to bind H⁺; protonation also destroys resonance.
Smaller pK_b means? Stronger base (larger K_b).
What are the 3 steps of Hofmann elimination? (1) exhaustive methylation with excess CH₃I, (2) moist Ag₂O → quaternary ammonium hydroxide, (3) heat → alkene.
Which alkene does Hofmann elimination give and why? Least substituted (Hofmann) alkene; the bulky N⁺(CH₃)₃ leaving group makes base abstract the least hindered β-H.
Carbylamine test detects which amines and using what reagents? Primary amines (aliphatic & aromatic); CHCl₃ + alcoholic KOH, heat → foul-smelling isocyanide.
Reactive intermediate in the carbylamine reaction? Dichlorocarbene :CCl₂.
Hinsberg reagent is? Benzenesulfonyl chloride, C₆H₅SO₂Cl.
In Hinsberg test, why does the 1° product dissolve in NaOH? Its N–H is acidic (SO₂ withdraws electrons), forming a soluble sodium salt.
In Hinsberg test, behaviour of 2° amine? Forms sulfonamide with no N–H → insoluble in alkali (precipitate).
In Hinsberg test, behaviour of 3° amine? No N–H, so no reaction with the reagent.
Recall Feynman: explain to a 12-year-oldThe nitrogen has a little pair of "spare hands" (the lone pair). With those hands it can grab a hydrogen (that's being a base) or grab onto other molecules (that's reacting). Carbon groups push extra food to the hands so they grab better — but in water , after grabbing, water needs to hug the molecule to keep it happy; a too-crowded nitrogen can't be hugged well, so it's actually a bit worse. In empty space (gas) there's no one to hug, so the most-fed nitrogen always wins. Hofmann: we put a giant backpack on nitrogen and heat it; the backpack is so big the kicking-out happens at the roomiest spot, giving the simplest alkene. Carbylamine makes a stink only if nitrogen has two free hands (primary). Hinsberg sorts amines by counting how many hydrogens nitrogen still has after shaking hands with the reagent.
Basicity in water: "2-1-3-ammonia-aniline " → 2°>1°>3°>NH₃≫PhNH₂.Gas phase = pure greed: more alkyl = more base (3>2>1>NH₃). "No water, no manners — just inductive greed."Hofmann = Huge group → Humble (least) alkene. Carbylamine = CarbylAmine needs two N–H = primary only; CHCl₃ + alc. KOH = "Chloroform stinks." Hinsberg dissolve/don't: "1° D issolves (acidic H), 2° stays a D ot (precipitate), 3° D oesn't react."
Worked example Worked: predict Hofmann product of 2-aminobutaneQuaternise → C H 3 C H 2 C H ( N + M e 3 ) C H 3 O H − CH_3CH_2CH(N^+Me_3)CH_3\,OH^- C H 3 C H 2 C H ( N + M e 3 ) C H 3 O H −
β-H's available: on the CH₂ (toward ethyl) and on the terminal CH₃.
Why this step? Bulky N + M e 3 N^+Me_3 N + M e 3 less hindered β-H (the CH₃ side) → gives but-1-ene (least substituted), not but-2-ene.
Answer: but-1-ene major.
Worked example Worked: which is more basic,( C H 3 ) 2 N H (CH_3)_2NH ( C H 3 ) 2 N H
( C H 3 ) 2 N H (CH_3)_2NH ( C H 3 ) 2 N H Aniline: lone pair delocalised into ring → unavailable.
Why this step? Available lone pair = stronger base.
Answer: dimethylamine is far more basic (p K b ≈ 3.3 pK_b\approx3.3 p K b ≈ 3.3 ≈ 9.4 \approx9.4 ≈ 9.4
Water order 2 gt 1 gt 3 gt NH3
Gas phase 3 gt 2 gt 1 gt NH3
Intuition Hinglish mein samjhoDekho, amine ka pura khel ek hi cheez pe hai — nitrogen ke paas ek lone pair hota hai. Yeh lone pair agar H⁺ pakad le toh wo base ban jaata hai; agar kisi carbon ya carbonyl pe attack kare toh reaction hoti hai. Basicity samajhne ke liye sirf yeh poochho: lone pair kitna "free/available" hai, aur protonate hone ke baad jo cation banta hai wo kitna stable hai?
Water mein order hota hai: 2° > 1° > 3° > NH₃ ≫ aniline. Alkyl group +I effect se electron push karta hai (lone pair richer), par 3° amine ka cation crowded hota hai isliye paani usse achhe se H-bond/solvate nahi kar paata — isliye 3° neeche gir jaata hai. Gas phase mein paani hai hi nahi, toh sirf +I bachta hai, aur order ulta ho jaata hai: 3° > 2° > 1° > NH₃. Aniline weak isliye hai kyunki uska lone pair ring ke andar resonance mein chala jaata hai — busy lone pair H⁺ nahi pakad pata.
Hofmann elimination: amine ko excess CH₃I se quaternary banao, phir moist Ag₂O se hydroxide, phir heat. Leaving group N + ( C H 3 ) 3 N^+(CH_3)_3 N + ( C H 3 ) 3 bada/bulky hota hai, isliye base sabse kam hindered β-H uthata hai aur least substituted (Hofmann) alkene banta hai — Zaitsev ka ulta. Carbylamine test sirf primary amine ke liye hai: CHCl₃ + alcoholic KOH ke saath garam karo, badboo wala isocyanide banta hai (intermediate dichlorocarbene). Hinsberg test se 1°, 2°, 3° pehchaante hain: benzenesulfonyl chloride se 1° ka product NaOH mein ghul jaata hai (acidic N–H), 2° ka precipitate rehta hai (no N–H), aur 3° react hi nahi karta. Exam mein yeh teen tests aur dono basicity orders bahut aate hain — ratta mat maaro, "lone pair availability + solvation" yaad rakho, sab derive ho jaayega.
