Class 12 Chemistry · Chapter 9
Short answer:
Ye chapter Class 12 Chemistry Part II ka Chapter 9 hai. NCERT ke exercise me isme kul 18 questions hain — IUPAC naming, classification (1°/2°/3°), Hinsberg test se distinguish karna, basicity ordering (aliphatic amine vs NH3 vs aromatic amine), preparation methods (nitro compound reduction, Gabriel phthalimide synthesis, ammonolysis, Hofmann bromamide degradation), aur diazonium salts ki preparation + synthetic importance cover hote hain. Board exam me structure-based reasoning aur mechanism-in-words dono type ke questions aate hain — is chapter ki khaasiyat hai ki concept clear hone chahiye, sirf reaction ratna kaafi nahi hai.
Amines ammonia ke derivatives hain jinme ek, do ya teen H atoms alkyl/aryl groups se replace ho jaate hain — aur inki chemistry ka core idea hai nitrogen ka lone pair kitna 'available' hai H+ lene ke liye. Ye chapter proteins, alkaloids, drugs, dyes aur polymers banane ke industrial process me amines ka role dikhata hai, aur diazonium salts ke through amines ko organic synthesis ka ek powerful intermediate banata hai — jisse aap benzene ring pe bahut saare functional groups introduce kar sakte ho jo directly substitution se possible nahi hote.
Chapter 9 Summary — 5 Minute Revision
1. Classification aur Nomenclature
Ammonia (NH3) ke H atoms ko alkyl/aryl group se replace karke amines banti hain:
- 1° (primary) amine: R-NH2 — ek H replace, jaise CH3NH2 (methanamine)
- 2° (secondary) amine: R-NH-R' — do H replace, jaise (CH3)2NH (N-methylmethanamine)
- 3° (tertiary) amine: R-N(R')-R'' — teenon H replace, jaise (CH3)3N (N,N-dimethylmethanamine)
Zaroori hai: classification 1°/2°/3° yahan carbon attached hone ki tarah nahi hai (jaise alcohols me) — yaha nitrogen pe kitne H replace hue hain us basis pe hai. Isliye tert-butylamine (CH3)3C-NH2 ek primary amine hai, kyunki N pe sirf ek substituent hai, baaki do H hain.
IUPAC naming me suffix "-amine" use hota hai (methanamine, ethanamine). Secondary/tertiary amines me bade substituent ko parent chain lete hain aur chhote substituent ko N- prefix ke saath likhte hain — jaise N-methylethanamine, N,N-dimethylmethanamine.
2. Structure of Amines
Nitrogen sp3 hybridised hota hai — teen bonds (ya bonds+H) aur ek lone pair jo pyramidal geometry banata hai. Aromatic amines (jaise aniline) me lone pair benzene ring ke π system ke saath resonance me hissa leta hai — isse N-C(aryl) bond me partial double bond character aata hai aur nitrogen thoda planar hone ki taraf jaata hai.
3. Methods of Preparation
- Reduction of nitro compounds: R-NO2 + 6[H] (H2/Pd/Ni ya Fe+HCl ya Sn+HCl) → R-NH2 + 2H2O. Nitrobenzene se aniline banane ka ye industrial route hai.
- Ammonolysis of alkyl halides: R-X + excess NH3 → R-NH2 + HX (nucleophilic substitution). Problem: excess NH3 ke bawajood mixture of 1°, 2°, 3° amine aur quaternary ammonium salt ban jaata hai — separation mushkil hai.
- Gabriel phthalimide synthesis: Sirf pure 1° aliphatic amine banane ka clean method. Phthalimide ke N-H ko KOH se deprotonate karke potassium salt banate hain, phir R-X se alkylate karte hain, aur end me hydrolysis (ya hydrazine se) se free 1° amine nikalta hai phthalhydrazide ke saath. Limitation: aromatic amine (jaise aniline) is method se nahi ban sakti kyunki aryl halides SN2 substitution nahi dete (halide directly ring se attached hone ke karan nucleophilic substitution resist karta hai).
- Reduction of nitriles: R-C≡N + 4[H] (LiAlH4 ya H2/Ni) → R-CH2-NH2 (1° amine, ek carbon zyada).
- Reduction of amides: R-CO-NH2 + 4[H] (LiAlH4) → R-CH2-NH2.
- Hofmann bromamide degradation: R-CO-NH2 + Br2 + 4NaOH → R-NH2 + 2NaBr + Na2CO3 + 2H2O. Ye reaction ek carbon kam karta hai (amide se ek carbon kam wala 1° amine banta hai) — synthetically useful jab ek carbon chhota amine chahiye.
4. Physical Properties
1° aur 2° amines N-H bond ke through H-bonding kar sakte hain isliye inka boiling point corresponding alcohol se kam par corresponding alkane se zyada hota hai (N, O se kam electronegative hai isliye H-bonding weaker). 3° amine me N-H nahi hota isliye H-bonding nahi hoti — inka b.p. sabse kam hota hai apne isomers me. Lower amines paani me ghulnashil hain (H-bonding with water) — chain lambi hone par solubility ghatti hai.
5. Basic Character of Amines — sabse important concept
Amine ki basicity depend karti hai nitrogen ke lone pair ki H+ lene ki "availability" pe — jitna lone pair free hoga, utni zyada basicity.
- Aliphatic amine vs ammonia: Alkyl group +I (electron-donating) effect deta hai, jisse N pe electron density badhti hai aur lone pair zyada available hota hai. Isliye aliphatic amines NH3 se zyada basic hote hain (gas phase/theoretical order R2NH > RNH2 > R3N > NH3 expected hai, par aqueous solution me steric hindrance aur solvation effects ke karan actual order thoda alag hota hai — NCERT level pe general trend "aliphatic amine > NH3" hi expect kiya jaata hai).
- Aromatic amine vs aliphatic amine: Aniline ki basicity alcohol ke amine se kam hoti hai kyunki nitrogen ka lone pair benzene ring ke saath resonance me delocalise ho jaata hai — isliye wo H+ lene ke liye utna available nahi rehta. Isliye order hai: aliphatic amine > NH3 > aromatic amine (aniline).
- Substituent effect on aromatic amines: Ring pe electron-donating group (-CH3, -OCH3) basicity badhata hai (para position pe zyada effective). Electron-withdrawing group (-NO2, -Cl) basicity ghatata hai kyunki ye lone pair ko aur zyada delocalise/withdraw karta hai — isliye p-nitroaniline, aniline se kam basic hai.
6. Chemical Reactions
- Acylation: R-NH2 + (CH3CO)2O → R-NH-CO-CH3 + CH3COOH (amide banta hai — N-H protection ke liye use hota hai).
- Carbylamine reaction (test for 1° amine): R-NH2 + CHCl3 + 3KOH (alc.) → R-NC (foul smell isocyanide) + 3KCl + 3H2O. Sirf 1° amine (aliphatic ya aromatic) ye test deta hai — 2°/3° nahi dete.
- Reaction with HNO2 (nitrous acid): 1° aliphatic amine → alcohol + N2 + H2O (unstable diazonium banta hai jo turant decompose ho jaata hai). 1° aromatic amine (0-5°C) → stable aryl diazonium salt. 2° amine → N-nitrosamine (yellow oily). 3° amine → sirf salt banta hai (aliphatic) ya ring pe nitrosation (aromatic, para position).
- Hinsberg test (distinguish 1°/2°/3°): Amine + C6H5SO2Cl (benzenesulphonyl chloride) + KOH:
- 1° amine → sulphonamide banta hai jiska N-H acidic hai (SO2 group ke karan) → KOH me ghul jaata hai (clear solution).
- 2° amine → sulphonamide banta hai par N pe koi H nahi bachta (dono H replace) → KOH me ghulta nahi (insoluble precipitate).
- 3° amine → sulphonyl chloride ke saath react hi nahi karta (koi N-H nahi hai) → mixture insoluble rehta hai, acidify karne par amine layer/liquid separate ho jaata hai.
- Electrophilic substitution in aniline: -NH2 group strong o,p-directing aur ring-activating hai (resonance se ring pe electron density badhti hai). Bromination without catalyst hi ho jaata hai aur tribromo product (2,4,6-tribromoaniline) directly banta hai — mono-bromo ke liye pehle acetylation (protection) karna padta hai taaki -NH2 ki activating power kam ho jaaye.
7. Diazonium Salts — synthesis ka powerhouse
Preparation (diazotisation): Aromatic 1° amine + NaNO2 + HCl at 273-278 K (0-5°C) → Aryldiazonium chloride (Ar-N2+Cl-). Temperature 5°C se upar jaane par diazonium salt hydrolyse hokar phenol bana deta hai — isliye ice-cold condition maintain karna zaroori hai, aur salt ko turant use karna padta hai (storage stable nahi hai, decompose ho jaata hai).
Synthetic importance — diazonium salt se bahut saare functional groups benzene ring pe introduce ho sakte hain jo direct electrophilic substitution se possible nahi hote:
- Ar-N2+ + H3PO2 (ya ethanol) → Ar-H (N2 group remove — reductive deamination)
- Ar-N2+ + H2O (warm) → Ar-OH (phenol)
- Ar-N2+ + CuCN (Sandmeyer) → Ar-CN (nitrile)
- Ar-N2+ + CuCl/CuBr (Sandmeyer) → Ar-Cl / Ar-Br
- Ar-N2+ + KI → Ar-I
- Ar-N2+ + HBF4 then heat (Balz-Schiemann) → Ar-F
- Coupling reaction: Ar-N2+ + phenol/aromatic amine (alkaline/weakly acidic medium) → azo compound (Ar-N=N-Ar') — coloured dyes banane me use hota hai (jaise p-hydroxyazobenzene).
Poore Class 12 Chemistry ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q18)
Q1. Classify the following amines as primary, secondary or tertiary: (i) (CH3)2CHNH2 (ii) CH3(CH2)2NH2 (iii) CH3NHCH(CH3)2 (iv) (CH3)3CNH2 (v) C6H5NHC6H5
(i) Primary (1°) — isopropylamine
(ii) Primary (1°) — propan-1-amine
(iii) Secondary (2°) — N-methylisopropylamine (N pe do alkyl groups)
(iv) Primary (1°) — tert-butylamine. Yahan carbon quaternary hai par nitrogen pe sirf ek substituent hai, isliye classification amine me carbon ki nahi, N pe attached groups ki basis pe hoti hai
(v) Secondary (2°) — diphenylamine
Q2. Write IUPAC names of: (i) CH3(CH2)4NH2 (ii) (CH3)2CHCH(CH3)NH2 (iii) trimethylamine.
(i) Pentan-1-amine
(ii) 3-Methylbutan-2-amine
(iii) N,N-Dimethylmethanamine
Q3. Give one chemical test to distinguish between methylamine and dimethylamine.
Carbylamine test: Methylamine (1° amine) CHCl3 aur alcoholic KOH ke saath foul-smelling isocyanide (methyl carbylamine) banata hai. Dimethylamine (2° amine) ye test nahi deta kyunki iske N pe H hi nahi bacha jo required N-H hai reaction ke mechanism ke liye — koi reaction nahi hota.
CH3NH2 + CHCl3 + 3KOH(alc.) → CH3NC↑(foul smell) + 3KCl + 3H2O
Q4. Give reasons: (i) pKb of aniline is more than that of methylamine. (ii) Ethylamine is soluble in water whereas aniline is not.
(i) Aniline ka pKb methylamine se zyada hai (matlab aniline kam basic hai). Aniline me N ka lone pair benzene ring ke saath resonance me delocalise ho jaata hai, jisse lone pair H+ lene ke liye kam available hota hai. Methylamine me alkyl group ka +I effect N pe electron density badhata hai, jisse lone pair zyada available hota hai — isliye methylamine zyada basic hai (kam pKb).
(ii) Ethylamine (chhoti chain, 2 carbon) paani ke saath strong H-bonding kar sakti hai isliye ghul jaati hai. Aniline me bada hydrophobic benzene ring hai jo H-bonding ko dominate kar deta hai — isliye aniline paani me practically insoluble hai.
Q5. Account for the following: Although amino group is o,p-directing group in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
Nitration strong acidic medium (conc. HNO3 + H2SO4) me hoti hai. Is acidic medium me aniline ka ek bada fraction protonate hokar anilinium ion (-NH3+) ban jaata hai. -NH3+ ek electron-withdrawing group hai jo m-directing hai (o,p position par electron density kam ho jaati hai deactivation ke through), isliye significant m-nitroaniline bhi bin jaata hai — sirf free -NH2 wale molecules hi o,p product dete hain.
Q6. Write chemical equations for the preparation of 1-phenylethanamine using Grignard reagent... (represent as reduction of the corresponding nitrile/general prep)
1-Phenylethanamine C6H5-CH(NH2)-CH3 ko corresponding oxime ya nitrile ke reduction se banaya ja sakta hai — general route: acetophenone se oxime banao (NH2OH ke saath), phir oxime ko H2/Ni ya LiAlH4 se reduce karo.
C6H5COCH3 + NH2OH → C6H5C(=NOH)CH3 --[H2/Ni]--> C6H5CH(NH2)CH3
Q7. Arrange the following in increasing order of their basic strength: aniline, p-toluidine, p-nitroaniline (in aqueous solution).
Increasing basicity: p-nitroaniline < aniline < p-toluidine
-NO2 strong electron-withdrawing group hai (para position pe) jo lone pair ko aur zyada ring ke saath delocalise/withdraw kar deta hai — basicity sabse kam. -CH3 (p-toluidine me) electron-donating (+I) hai jo N pe electron density badhata hai — basicity sabse zyada. Plain aniline in dono ke beech me hai.
Q8. Complete the following reactions: (i) C6H5NH2 + CHCl3 + KOH(alc.) → (ii) C6H5N2Cl + H3PO2 + H2O →
(i) C6H5NC (phenyl isocyanide, foul smell) + 3KCl + 3H2O — carbylamine reaction
C6H5NH2 + CHCl3 + 3KOH(alc.) → C6H5NC + 3KCl + 3H2O
(ii) C6H6 (benzene) + N2↑ + H3PO3 — diazonium group ka reductive removal (deamination)
C6H5N2+Cl- + H3PO2 + H2O → C6H6 + N2↑ + H3PO3 + HCl
Q9. How will you convert ethanamine into methanamine?
Ethanamide (CH3CONH2) se Hofmann bromamide degradation karke methanamine banaya ja sakta hai (ek carbon kam karta hai). Par direct ethanamine se methanamine nahi banta — pehle ethanamine ko ethanoic acid me convert karo, phir amide, phir Hofmann degradation se ek-carbon-kam methanamine milega.
CH3CONH2 + Br2 + 4NaOH → CH3NH2 + 2NaBr + Na2CO3 + 2H2O
Q10. Write structures of different isomeric amines corresponding to the molecular formula C4H11N. Also write their IUPAC names.
Primary (4): CH3CH2CH2CH2NH2 (butan-1-amine), CH3CH2CH(NH2)CH3 (butan-2-amine), (CH3)2CHCH2NH2 (2-methylpropan-1-amine), (CH3)3CNH2 (2-methylpropan-2-amine)
Secondary (3): CH3CH2CH2NHCH3, (CH3)2CHNHCH3, CH3CH2NHCH2CH3 (N-ethylethanamine)
Tertiary (1): CH3CH2N(CH3)2 (N,N-dimethylethanamine)
Q11. Write the IUPAC name of the coupling product formed by reaction of benzenediazonium chloride with N,N-dimethylaniline.
Coupling para position pe hoti hai (N,N-dimethylamino group strongly activating hai): p-(N,N-dimethylamino)azobenzene banta hai, jise 4-(dimethylamino)azobenzene bhi kehte hain — ek important azo dye.
C6H5N2+Cl- + C6H5N(CH3)2 → C6H5-N=N-C6H4-N(CH3)2 (para) + HCl
Q12. Give the structures of A, B and C in the following reactions: (i) CH3CH2Cl --[KCN]--> A --[H2/Ni]--> B (ii) C6H5NO2 --[Fe/HCl]--> C
(i) A = CH3CH2CN (propanenitrile); B = CH3CH2CH2NH2 (propan-1-amine) — nitrile reduction se ek carbon zyada wala 1° amine banta hai.
(ii) C = C6H5NH2 (aniline) — Fe/HCl (ya Sn/HCl) nitro group ko amine me reduce karta hai.
Q13. Complete the following reactions: (i) CH3CH2NH2 + CH3COCl → (ii) C6H5NH2 + Br2(aq) →
(i) N-ethylacetamide banta hai (acylation):
CH3CH2NH2 + CH3COCl → CH3CONHCH2CH3 + HCl
(ii) Directly 2,4,6-tribromoaniline banta hai (bina catalyst ke) kyunki -NH2 strongly activating hai:
C6H5NH2 + 3Br2(aq) → 2,4,6-tribromoaniline↓(white ppt) + 3HBr
Q14. How would you convert aniline into p-bromoaniline (mono-substituted, not tribromo)?
Directly bromination karne pe tribromoaniline ban jaata hai kyunki -NH2 bahut strongly activating hai. Isliye pehle -NH2 ko acetylation se 'protect' karte hain (activating power kam ho jaati hai), phir bromination karte hain, phir hydrolysis se wapas -NH2 free karte hain.
C6H5NH2 --[(CH3CO)2O]--> C6H5NHCOCH3 --[Br2]--> p-BrC6H4NHCOCH3 --[H3O+/hydrolysis]--> p-BrC6H4NH2
Q15. Describe a method for the preparation of pure aliphatic primary amine using Gabriel phthalimide synthesis and explain why it cannot be used to prepare aromatic primary amines.
Phthalimide ke N-H ko ethanolic KOH se deprotonate karke potassium phthalimide banate hain. Ye phir R-X (alkyl halide) ke saath SN2 alkylation karta hai, aur end me alkaline hydrolysis (ya hydrazine treatment) se free 1° amine milta hai, phthalhydrazide byproduct ke saath. Ye pure 1° amine deta hai (2°/3° contamination nahi hota) kyunki N pe already dono H replaced hote hain phthaloyl group se, aur alkylation single step controlled hoti hai.
Aromatic amine (jaise aniline) is method se nahi ban sakti kyunki reaction ke liye R-X ko SN2 nucleophilic substitution dena zaroori hai — aryl halides (Ar-X) SN2 substitution nahi dete (halogen directly sp2 ring carbon se attached hota hai, resonance ke karan C-X bond strong hota hai aur backside attack sterically/electronically blocked hota hai).
Q16. Write short notes on: (i) carbylamine reaction (ii) diazotisation
(i) Carbylamine reaction: 1° amine (aliphatic ya aromatic) ko CHCl3 aur alcoholic KOH ke saath garam karne par foul-smelling isocyanide (carbylamine) banta hai. Ye 1° amine ka specific test hai — 2°/3° amine ye nahi dete.
RNH2 + CHCl3 + 3KOH → RNC + 3KCl + 3H2O
(ii) Diazotisation: Aromatic 1° amine ko NaNO2 + HCl (ya H2SO4) ke saath 273-278 K (0-5°C) pe treat karke diazonium salt banane ka process hai. Low temperature isliye zaroori hai kyunki diazonium salt garam hone pe (5°C se upar) hydrolyse hokar phenol bana deta hai — salt thermally unstable hota hai.
ArNH2 + NaNO2 + 2HCl --[273-278K]--> ArN2+Cl- + NaCl + 2H2O
Q17. Give the reactions of benzenediazonium chloride with the following reagents and name the products: (i) HBF4 (ii) H2O (iii) KI (iv) CuCN
(i) C6H5N2+BF4- banta hai jo heat karne par fluorobenzene deta hai (Balz-Schiemann reaction) — product: C6H5F
C6H5N2Cl + HBF4 → C6H5N2BF4 --[Δ]--> C6H5F + N2 + BF3
(ii) Warm karne par phenol banta hai — product: C6H5OH
C6H5N2Cl + H2O --[Δ]--> C6H5OH + N2 + HCl
(iii) Product: iodobenzene (C6H5I) — direct reaction, Sandmeyer nahi lagti
C6H5N2Cl + KI → C6H5I + N2 + KCl
(iv) Sandmeyer reaction se product: benzonitrile (C6H5CN)
C6H5N2Cl + CuCN → C6H5CN + N2 + CuCl
Q18. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?
Gabriel synthesis me key step hai potassium phthalimide ka R-X ke saath nucleophilic substitution (SN2) — aryl halides (jo aromatic amine banane ke liye required hote) SN2 substitution nahi dete kyunki halogen sp2 ring carbon se strongly bonded hota hai aur backside nucleophilic attack sterically/electronically resist karta hai. Isliye ye method sirf alkyl halides ke saath kaam karta hai, aromatic amines ke liye nahi.
Important Equations — Ek Nazar Me
| Concept | Detail |
|---|---|
| Basicity order (general, aqueous) | Aliphatic amine > NH3 > Aromatic amine (aniline) |
| Substituent effect on aromatic amine basicity | -CH3/-OCH3 (EDG) basicity badhaate hain (para > meta); -NO2/-Cl/-Br (EWG) basicity ghataate hain |
| Hinsberg test — 1° amine | C6H5SO2Cl ke saath sulphonamide banta hai jiska N-H acidic hai → KOH me ghul jaata hai |
| Hinsberg test — 2° amine | Sulphonamide banta hai par N-H nahi bachta → KOH me insoluble rehta hai |
| Hinsberg test — 3° amine | Reaction hi nahi hota (N-H hai hi nahi) → sulphonyl chloride ke saath unreacted rehta hai |
| Diazotisation condition | Aromatic 1° amine + NaNO2 + HCl at 273-278 K (0-5°C) — is se upar temperature pe salt hydrolyse ho jaata hai |
| ArN2+ + H3PO2 | → Ar-H (N2 group remove, reductive deamination) |
| ArN2+ + H2O (warm) | → Ar-OH (phenol) |
| ArN2+ + CuCN (Sandmeyer) | → Ar-CN (nitrile) |
| ArN2+ + CuCl/CuBr (Sandmeyer) | → Ar-Cl / Ar-Br |
| ArN2+ + KI | → Ar-I (direct, no catalyst) |
| ArN2+ + HBF4 then heat (Balz-Schiemann) | → Ar-F |
| Coupling reaction | ArN2+ + phenol/amine (alkaline medium) → azo compound (dye) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Basicity order galat likhna. Kai students yaad karte hain 'aromatic > aliphatic' — ulta hai. Aromatic amine (aniline) ki basicity kam hoti hai kyunki lone pair resonance me ring ke saath delocalise ho jaata hai.
- Hinsberg test me solubility swap kar dena. 1° amine ka product KOH me GHULTA hai (N-H acidic hai), 2° amine ka product NAHI ghulta. Ise ulta likhna common mistake hai — yaad rakho: jitna N-H bacha, utni acidity, utni solubility.
- Diazonium salt ka storage temperature bhool jaana. Diazotisation aur reaction dono 0-5°C (273-278 K) pe honi chahiye — is temperature se upar salt turant hydrolyse hokar phenol ban jaata hai. Room temperature pe diazonium salt store nahi kiya ja sakta.
- Gabriel synthesis se aromatic amine bana sakte hain — ye galat concept. Ye method sirf pure 1° ALIPHATIC amine ke liye hai. Aryl halides SN2 nahi dete, isliye aniline jaisa amine is method se nahi banta.
- Coupling reaction ka product azo compound (N=N) hai, azoxy nahi. Diazonium salt ka phenol/amine se coupling reaction N=N (azo) bond banata hai — students isse condensation ya azoxy (N=N-O) samajh lete hain jo galat hai.
- tert-Butylamine ko tertiary amine samajh lena. (CH3)3C-NH2 me carbon quaternary hai par NITROGEN pe sirf ek substituent hai (do H baaki hain) — ye primary amine hai. Classification carbon ki tarah nahi, N pe attached groups ki basis pe hoti hai.
Board-Style Important Questions
- 1 mark: Arrange the following in decreasing order of basic strength: aniline, methylamine, N-methylaniline (qualitative reasoning only).
- 2 marks: Write the products formed when aniline reacts with (i) acetic anhydride (ii) bromine water. Give balanced equations.
- 2 marks: Why does aniline not undergo Friedel-Crafts reaction?
- 3 marks: Describe the Hinsberg test and explain how it distinguishes between 1°, 2° and 3° amines with reasoning.
- 3 marks: Explain, with mechanism-in-words, why diazonium salt must be prepared and used below 5°C, and give two synthetic reactions showing its importance in preparing aryl halides.
- 5 marks: Starting from benzene, outline a synthesis route to prepare aniline, and then convert aniline into (i) phenol (ii) bromobenzene (iii) an azo dye, using diazonium salt chemistry.
Aksar Poochhe Jaane Wale Sawaal
Amines me 1°, 2°, 3° classification alcohols se kaise alag hai?
Alcohols me classification carbon pe attached alkyl groups ki basis pe hoti hai. Amines me nitrogen pe kitne H replace hue hain uski basis pe hoti hai — isliye tert-butylamine bhi primary amine hai kyunki N pe sirf ek substituent hai.
Aniline paani me kyun nahi ghulta jabki chhoti chain amine ghul jaati hai?
Aniline ka bada hydrophobic benzene ring H-bonding capacity ko dominate kar deta hai. Chhoti aliphatic amine me N-H bonds paani ke saath effectively H-bond kar paate hain isliye wo ghul jaati hain.
Diazonium salt itna reactive kyun hota hai?
N2+ group ek excellent leaving group hai (stable N2 gas banata hai jaate waqt), isliye diazonium salt easily displace hokar bahut saare nucleophiles se react kar leta hai — yahi iski synthetic utility ka base hai.
Gabriel synthesis kyun 'pure' 1° amine deta hai jabki direct ammonolysis nahi deta?
Ammonolysis (R-X + NH3) me product khud bhi nucleophilic hota hai aur aur R-X se react kar sakta hai, isliye mixture ban jaata hai. Gabriel synthesis me phthalimide nitrogen protected rehta hai jab tak final hydrolysis step nahi hota, isliye controlled single product milta hai.
Carbylamine test sirf 1° amine ke liye specific kyun hai?
Reaction mechanism ko dichlorocarbene intermediate chahiye jo N-H wale nitrogen pe insert ho aur isocyanide bana sake — ye sirf tab possible hai jab N pe do H available hon (1° amine). 2°/3° amine me required H availability nahi hoti isliye reaction nahi hota.
Coupling reaction me para position hi kyun prefer hota hai?
Phenol/aromatic amine ka -OH ya -NR2 group strongly o,p-directing hota hai. Ortho position steric hindrance ke karan kam favourable hota hai bade diazonium electrophile ke saath, isliye para position major product deta hai.
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