4.4.1
Amine ek NH₃ hota hai jisme ek ya zyada H ki jagah carbon groups aate hain . Iske chemistry ka sab kuch EK fact se aata hai: nitrogen ke paas ek lone pair hai. Woh lone pair ya toh:
proton pakad sakta hai → isse woh base ban jaata hai (basicity).nucleophile ki tarah kaam kar sakta hai → carbons/carbonyls par attack karta hai (Hofmann, carbylamine, Hinsberg).
Toh agar tum "lone pair availability" yaad rakho, toh almost sab kuch predict kar sakte ho. Yeh poora subtopic bas "lone pair kitna available hai, aur woh kya attack karta hai?" hai.
Ek base lone pair H⁺ ko donate karta hai. Amines ke liye:
R - N H 2 + H 2 O ⇌ R - N H 3 + + O H − K b p K b = − log K b Chhota p K b
Zyada strong base woh hota hai jiska conjugate acid (R - N H 3 + zyada stable ho (H⁺ wapas dene ko taiyar na ho). Toh hum hamesha do sawaal poochhte hain:
Kya lone pair zyada available hai H⁺ pakadne ke liye? (electronic effect)
Kya resulting cation zyada stabilised hai? (solvation + induction)
Intuition Do cooperating effects
+I (inductive electron donation): ek alkyl group electron density N ki taraf push karta hai, lone pair ko aur rich banata hai → H⁺ behtar pakadta hai.Cation ki solvation: ek baar protonated hone ke baad, R - N H 3 +
Yeh dono effects ladte hain jab hum tertiary amines tak jaate hain:
Amine
Cation mein N–H bonds
+I effect
Net basicity (water)
NH₃
4
koi nahi
aliphatic series mein sabse weak
1° RNH₂
3
ek R
strong
2° R₂NH
2
do R
usually strongest
3° R₃N
1
teen R
girta hai (poor solvation)
Intuition Koi solvent nahi = koi solvation rescue nahiGas phase mein cation ko stabilise karne ke liye koi paani nahi hota . Sirf +I effect bachta hai. Zyada alkyl groups = zyada electron donation = zyada strong base. Toh:
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 hai. Paani mein "anomaly" (2°>3°) sirf isliye hai kyunki wahan solvation matter karta hai.
Common mistake Steel-man: "Zyada alkyl groups hamesha = zyada basic"Kyun sahi lagta hai: +I clearly electrons donate karta hai, toh zyada R hamesha help karna chahiye. Flaw: paani mein cation ki H-bond stabilisation bhi kho jaati hai. Tertiary cation ke paas sirf ek N–H hota hai H-bond ke liye. Fix: paani mein , basicity = (+I) − (solvation loss), toh 2° aksar jeetta hai. Gas phase mein , koi solvation term nahi, toh strictly 3°>2°>1°>NH₃.
Intuition Lone pair "busy" haiAniline mein N lone pair benzene ring mein delocalised ho jaata hai (resonance). Resonance mein laga hua lone pair H⁺ pakadne ke liye available nahi hota.
Saath hi, protonation us resonance ko khatam kar deta hai, toh conjugate acid free base ke relative destabilised ho jaata hai → base weak ho jaata hai.
lone pair into ring C 6 H 5 N ¨ H 2 p K b ≈ 9.4 ( very weak )
Definition Hofmann eliminationEk amine ko quaternary ammonium hydroxide mein convert karo, phir heat karo. Yeh eliminate karke alkene + amine + water deta hai. "Trick": leaving group ko bulky aur positively charged banao.
Exhaustive methylation: amine ko excess C H 3 I R - N + ( C H 3 ) 3 I − Hydroxide mein convert karo: moist A g 2 O R - N + ( C H 3 ) 3 O H − Heat karo (Δ): O H − β-hydrogen remove karta hai, neutral 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
Intuition Bulky base + steric crowdingLeaving group N + ( C H 3 ) 3 bahut bada hota hai. Base ko ek β-H ke paas jaana padta hai, lekin kam hindered (less substituted) carbon par H tak pahunchna asaan hota hai. Toh jo proton remove hota hai woh least sterically hindered hota hai → least substituted (Hofmann) alkene dominate karta hai.
Compare karo: ordinary E2 (chhota leaving group, e.g. Br) zyada substituted Zaitsev alkene deta hai.
Common mistake Steel-man: "Elimination hamesha zyada stable (Zaitsev) alkene deta hai"Kyun sahi lagta hai: zyada substituted alkenes thermodynamically zyada stable hote hain, aur Zaitsev default rule hai. Flaw: Hofmann ka bada charged leaving group transition state ko sterically controlled banata hai, stability controlled nahi. Fix: bulky base / bulky leaving group ⇒ Hofmann (less substituted) product.
Definition Carbylamine reactionYeh sirf primary amines ka test hai (1° aliphatic YA aromatic). Amine ko chloroform (CHCl₃) aur alcoholic KOH ke saath heat karo → ek carbylamine (isocyanide) banata hai jisme bahut buri, offensive smell hoti hai.
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 Sirf 1° amines mein KYUNKOH + CHCl₃ dichlorocarbene : C C l 2 do N–H bonds hote hain, jo bilkul wahi hai jo dono H's aur dono Cl's kho kar R - N ≡ C koi isocyanide nahi, koi buri smell nahi.
Use: Ek positive (buri smell wala) test = primary amine present hai.
Definition Hinsberg reagentBenzenesulfonyl chloride , C 6 H 5 S O 2 C l
Intuition Logic KAISE kaam karta haiAmine ka N, sulfonyl chloride ke S par attack karta hai (nucleophile attacks electrophile). Aage kya hota hai yeh depend karta hai ki sulfonamide product par kitne N–H bonds bache hain :
Amine
Product
Key feature
KOH/NaOH ke saath behaviour
1° R - N H - S O 2 C 6 H 5 abhi bhi ek acidic N–H hai (SO₂ se acidified)
alkali mein Dissolve ho jaata hai (salt banta hai) → clear solution
2° R 2 N - S O 2 C 6 H 5 koi N–H nahi bachaalkali mein Insoluble → precipitate rehta hai
3° koi reaction nahi
react karne ke liye koi N–H nahi
react nahi karta (amine free rehta hai)
Intuition 1° sulfonamide N–H acidic KYUN hota haiStrongly electron-withdrawing S O 2 R - N − - S O 2 C 6 H 5 soluble salt . 2° product mein bilkul N–H nahi hota, toh salt nahi ban sakta → solid reh jaata hai.
Common mistake Steel-man: "Tertiary amine bhi precipitate deta hai"Kyun sahi lagta hai: sabhi amines ke paas lone pair hoti hai aur lagta hai ki unhe react karna chahiye. Flaw: tertiary amine ke paas koi N–H nahi hota, toh S par attack karne ke baad ek charged species ban jaati hai jiske paas kho ne ke liye koi proton nahi; woh revert ho jaata hai. Fix: 3° amine Hinsberg reagent ke saath react nahi karta (oily free amine ki tarah rehta hai).
Recall Quick self-test (answers chhupao)
Paani mein 2° amine, 3° ko KYUN beat karta hai? → solvation loss of 3° cation, +I se zyada matter karta hai.
Gas phase mein order reverse KYUN ho jaata hai? → koi solvation nahi, sirf +I matter karta hai.
Aniline weak KYUN hai? → lone pair ring mein delocalised hai.
Hofmann kaun sa alkene deta hai? → least substituted.
Carbylamine test kiske liye hai? → primary amines.
Hinsberg: kaun sa amine NaOH mein dissolve hota hai? → primary (acidic N–H).
Paani mein ek 2° aliphatic amine, NH₃ se zyada basic KYUN hota hai? Do alkyl groups ka +I effect lone-pair density badhata hai AUR cation achhi tarah solvated/H-bonded hota hai.
Paani mein (CH₃)₃N, (CH₃)₂NH se neeche KYUN jaata hai? Iske cation ke paas sirf ek N–H hota hai, toh poor solvation/H-bonding, bade +I effect se zyada ho jaata hai.
Methylamines ka gas-phase basicity order kya hai? (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃ — pure inductive order (koi solvation nahi).
Aniline alkyl amines se bahut weak base KYUN hai? Iska N lone pair benzene ring mein delocalised hai, toh H⁺ bind karne ke liye kam available hai; protonation resonance bhi khatam kar deta hai.
Chhota pK_b matlab? Zyada strong base (bada K_b).
Hofmann elimination ke 3 steps kya hain? (1) excess CH₃I ke saath exhaustive methylation, (2) moist Ag₂O → quaternary ammonium hydroxide, (3) heat → alkene.
Hofmann elimination kaun sa alkene deta hai aur kyun? Least substituted (Hofmann) alkene; bulky N⁺(CH₃)₃ leaving group base ko least hindered β-H abstract karne par majboor karta hai.
Carbylamine test kaun se amines detect karta hai aur kaun se reagents use hote hain? Primary amines (aliphatic & aromatic); CHCl₃ + alcoholic KOH, heat → foul-smelling isocyanide.
Carbylamine reaction mein reactive intermediate kya hai? Dichlorocarbene :CCl₂.
Hinsberg reagent kya hai? Benzenesulfonyl chloride, C₆H₅SO₂Cl.
Hinsberg test mein 1° product NaOH mein KYUN dissolve hota hai? Iska N–H acidic hota hai (SO₂ electrons withdraw karta hai), ek soluble sodium salt banta hai.
Hinsberg test mein 2° amine ka kya behaviour hai? Sulfonamide banata hai jisme koi N–H nahi → alkali mein insoluble (precipitate).
Hinsberg test mein 3° amine ka kya behaviour hai? Koi N–H nahi, toh reagent ke saath koi reaction nahi.
Recall Feynman: ek 12-saal ke bachche ko samjhaoNitrogen ke paas ek chhoti si "spare hands" ki jodi hoti hai (lone pair). Un haathon se woh ek hydrogen pakad sakta hai (yeh base hona hai) ya doosre molecules par pakad sakta hai (yeh react karna hai). Carbon groups un haathon ko extra khaana push karte hain toh woh behtar pakad paate hain — lekin paani mein , pakadne ke baad, paani ko molecule ko khush rakhne ke liye usse hug karna padta hai; ek bahut bheed waala nitrogen ko achhi tarah hug nahi kiya ja sakta, toh woh actually thoda worse hota hai. Khaali jagah (gas) mein hugger koi nahi hota, toh sabse zyada khaya hua nitrogen hamesha jeetta hai. Hofmann: hum nitrogen par ek giant backpack daalte hain aur heat karte hain; backpack itna bada hota hai ki kicking-out sabse spacious jagah hoti hai, sabse simple alkene deta hai. Carbylamine sirf tab smell maarta hai jab nitrogen ke paas do free hands hoon (primary). Hinsberg amines ko sort karta hai yeh ginte hue ki nitrogen ke paas reagent se haath milaane ke baad kitne hydrogens bache hain.
Paani mein Basicity: "2-1-3-ammonia-aniline " → 2°>1°>3°>NH₃≫PhNH₂.Gas phase = pure greed: zyada alkyl = zyada base (3>2>1>NH₃). "Koi paani nahi, koi manners nahi — sirf inductive greed."Hofmann = Huge group → Humble (least) alkene. Carbylamine = CarbylAmine ko do N–H chahiye = sirf primary; CHCl₃ + alc. KOH = "Chloroform stinks." Hinsberg dissolve/don't: "1° D issolve karta hai (acidic H), 2° ek D ot ki tarah rehta hai (precipitate), 3° react hi D on't karta."
Worked example Worked: 2-aminobutane ka Hofmann product predict karoQuaternise karo → C H 3 C H 2 C H ( N + M e 3 ) C H 3 O H −
Available β-H's: CH₂ par (ethyl ki taraf) aur terminal CH₃ par.
Yeh step kyun? Bulky N + M e 3 less hindered β-H prefer karta hai (CH₃ side) → but-1-ene deta hai (least substituted), but-2-ene nahi.
Answer: but-1-ene major.
Water order 2 gt 1 gt 3 gt NH3
Gas phase 3 gt 2 gt 1 gt NH3
