Amines: Class 12 Chemistry Chapter 9 Hinglish mein

Chemistry Part 2 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

90 min
27 practice questions
39 MCQs

Is chapter mein aap kya seekhenge

Amines ko ammonia ke derivatives ki tarah samjho, jinmein pyramidal sp3 nitrogen par ek lone pair hota hai
Amines ko primary, secondary ya tertiary mein classify karo aur common aur IUPAC rules se unke naam likho
Preparation ka sahi route chuno (nitro compounds, nitriles ya amides ka reduction, ammonolysis, Gabriel synthesis ya Hoffmann degradation) aur product mein carbons ki ginti predict karo
Hydrogen bonding ke zariye amines ke boiling points aur solubility samjhao
Inductive, solvation, steric aur resonance effects ki madad se gas phase aur paani mein amines ko basic strength ke hisaab se rank karo
Carbylamine, nitrous acid aur Hinsberg tests se primary, secondary aur tertiary amines ko alag-alag pehchano, aur unke acylation aur ring-substitution reactions likho
Arenediazonium salts banao aur unke replacement aur coupling reactions se aromatic compounds banao, azo dyes bhi

NCERT Chemistry Class 12, Chapter 9, pp. 259–280 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Garmi ki dopahar mein kisi fish market se guzro, to jo tez smell tum tak pahunchti hai woh zyadatar ek chhote molecule, trimethylamine, $(C H_{3})_{3} N$ se aati hai. Sadta hua meat do aur chhodta hai, jinke naam bhi seedhe-seedhe hain: putrescine aur cadaverine. Yeh teeno amines (ऐमीन) hain: aise compounds jo ammonia se bante hain, jab uske hydrogen atoms ki jagah carbon groups laga diye jaate hain.

Amines biology aur medicine mein har jagah milte hain. Nerve cells dopamine aur serotonin se ek doosre ko signal bhejte hain; tobacco ki lat nicotine ki wajah se lagti hai; lidocaine jaise local anaesthetics aur kai antihistamines mein amino group hota hai. Industry mein hexane-1,6-diamine nylon-6,6 ke do building blocks mein se ek hai, aur long-chain ammonium salts fabric softeners aur kai disinfectants ka active part hote hain.

Chapter ke do hisse hain. Sections 9.1 se 9.6 khud amines ke baare mein hain: unki shape, classes, naam, preparation, properties aur reactions. Sections 9.7 se 9.10 aromatic amines se bane arenediazonium salts ko follow karte hain. Yeh ions sirf barf jaise thande paani mein tikte hain, phir bhi inki madad se chemists benzene ring par F, I, CN, OH aur doosre groups laga paate hain jahan direct methods fail ho jaate hain, aur yeh do rings ko jodkar chamakdaar azo dyes banate hain.

9.1 Structure of Amines

Shuruaat ammonia se karo. Nitrogen ke paas paanch valence electrons hote hain: teen hydrogen atoms ke electrons ke saath σ bonds mein pair bana lete hain, aur do ek saath lone pair ke roop mein rehte hain. Yeh chaar electron pairs chaar $s p^{3}$ hybrid orbitals mein hote hain jo lagbhag ek tetrahedron ke corners ki taraf point karte hain. Isliye teeno bonded atoms ek neechi pyramid ke base par baithte hain, aur nitrogen uske apex par hota hai.

Amine mein bilkul yahi arrangement rehta hai. Ek, do ya teeno hydrogens ko alkyl ya aryl groups se replace karo, phir bhi nitrogen teen σ bonds banata hai aur chauthe $s p^{3}$ orbital mein ek lone pair rakhta hai (Fig. 9.1).

Definition

Amine

Aisa organic compound jismein ammonia, $N H_{3}$ , ke ek, do ya teen hydrogen atoms alkyl ya aryl groups se replace ho jaate hain. Nitrogen ka lone pair bana rehta hai.

ExampleEthanamine, $C H_{3} C H_{2} N H_{2}$ , aur aniline, $C_{6} H_{5} N H_{2}$ .

3/4Do replace hue: secondary amine.

Fig. 9.1Pyramid aur uska lone pair wahi rehte hain; sirf base par lage groups badalte hain.NCERT Fig. 9.1 · p. 260

Ek nitrogen atom pyramid ke apex par hai, uske upar ek shaded lone-pair lobe hai aur neeche teen bonds hain: ek left ki taraf, ek dashed peeche right ki taraf aur ek wedge dekhne wale ki taraf. Left aur right par counters N–H bonds aur carbon groups ki ginti dikhate hain. Step 1 mein teen hydrogens wala ammonia hai: 3 N–H bonds aur 0 carbon groups. Step 2 mein left hydrogen ki jagah methyl group aata hai, jisse methanamine banta hai, ek primary amine. Step 3 mein right hydrogen bhi badal jaata hai, jisse dimethylamine banta hai, ek secondary amine. Step 4 mein teen methyl groups hain, yaani trimethylamine, ek tertiary amine jismein koi N–H bond nahi bacha.

Lone pair ko sirf ek nucleus pakadta hai, isliye woh bonding pair se zyada phailta hai aur bonds ko thoda paas dhakel deta hai. Isliye nitrogen ke aas-paas ke angles tetrahedral 109.5° se thode kam reh jaate hain; ammonia mein H–N–H angle lagbhag 107° hota hai. Bulky groups is dabaav ke khilaaf peeche dhakelte hain, isliye trimethylamine mein C–N–C angle tetrahedral value ke kareeb rehta hai.NCERT trimethylamine ke liye 108° deta hai; gas-phase measurements lagbhag 111° dete hain. Dono hi tarah molecule pyramid hai, flat nahi.

Poore chapter mein lone pair ko dhyaan mein rakho. Isi ki wajah se amines protons le lete hain (yeh bases hain), electron-poor carbon atoms par attack karte hain (yeh nucleophiles hain), aur isi wajah se amino group benzene ring ko itna reactive bana deta hai.

Chalo explore karein

Nitrogen pyramid banao

Tumhe chahiye

ek chhota aloo ya clay ki goli
chaar toothpicks
teen matar ke daane ya chhote beads
teen bade beads ya angoor
ek cotton ball

Pehle guess karo

Toothpicks tetrahedron ke corners ki taraf point karti hain. Agar model apne teen ends par tika ho, to tumhe kaun si shape dikhti hai?

ek neechi pyramid jiske upar aloo haiek flat triangle jiske beech mein aloo haiteen balls ki ek seedhi line

Aloo mein chaar toothpicks aise ghusao ki woh ek doosre se jitna ho sake door point karein, lagbhag 109° ke angle par.
Teen sticks par ek-ek matar lagao: yeh hydrogen atoms hain. Chauthi stick par cotton ball lagao: yeh lone pair hai.
Model ko uske teen matar ke daanon par khada karo aur side se dekho.
Ek-ek karke matar ki jagah bade beads (methyl groups) lagao aur dekho ki beads kaise ek doosre se takraane lagte hain.

Quick check

3 mein se 1

Trimethylamine mein nitrogen ke aas-paas shape kya hai?

trigonal planartetrahedral, N ke aas-paas chaar atoms ke saathpyramidallinear

Answer dekho

Answer (C) pyramidal

Chaaron electron pairs tetrahedron ke corners ki taraf point karte hain, lekin ek lone pair hai, isliye teen carbon atoms aur nitrogen milkar pyramid banate hain. Trigonal planar ke liye

s p^{2}

nitrogen chahiye hota jiska plane mein koi lone pair na ho.

9.2 Classification

Amines ko nitrogen par lage carbon groups gin kar classify kiya jaata hai. Ek group se primary (1°) amine, $R N H_{2}$ banta hai; do se secondary (2°) amine, $R_{2} N H$ ; teen se tertiary (3°) amine, $R_{3} N$ . Groups alkyl (R) ya aryl (Ar) ho sakte hain aur unka ek jaisa hona zaroori nahi. Jis secondary ya tertiary amine ke saare groups ek jaise hon, use simple kehte hain; alag-alag groups wala mixed hota hai.

Compare

Amines ki teen classes

Primary (1°)

general formula $R N H_{2}$
do N–H bonds
jaise $C H_{3} C H_{2} C H_{2} N H_{2}$

Secondary (2°)

$R_{2} N H$ (groups alag ho sakte hain)
ek N–H bond
jaise $C_{6} H_{5} N H C_{2} H_{5}$

Tertiary (3°)

$R_{3} N$
koi N–H bond nahi
jaise $(C H_{3} C H_{2})_{3} N$

Chautha group tabhi jud sakta hai jab nitrogen apne lone pair se chautha bond banaye. Isse quaternary ammonium ion, $R_{4} N^{+}$ , banta hai, jo tetramethylammonium chloride, $(C H_{3})_{4} N^{+} C l^{-}$ , jaise salts mein milta hai. Iske paas koi lone pair nahi bachta, isliye yeh amine nahi hai, lekin amine ka baar-baar alkylation yahin aakar khatam hota hai (Ammonolysis of alkyl halides).

Ek trap se bacho. Alcohols aur haloalkanes mein 1°, 2° aur 3° us carbon atom ko describe karte hain jis par group laga hai. Amines mein yeh nitrogen ko describe karte hain. Isliye 2-methylbutan-2-amine, $C H_{3} C H_{2} C (C H_{3})_{2} N H_{2}$ , ek primary amine hai, bhale hi uska $N H_{2}$ tertiary carbon par baitha ho, jabki matching alcohol, 2-methylbutan-2-ol, ek tertiary alcohol hai.

2/2Amine ke liye, nitrogen par lage carbons gino: ek, isliye 1°.

Fig. 9.2Skeleton same, ginne ka rule alag: 2-methylbutan-2-ol 3° hai, 2-methylbutan-2-amine 1° hai.NCERT p. 260

Ek central carbon par left mein methyl group, right mein ethyl group, neeche methyl group aur upar ek functional group hai. Step 1 mein upar wala group OH hai; central carbon par gola bana hai aur uske teen carbon neighbours par 1, 2 aur 3 likha hai, isliye label 3° alcohol padhta hai, kyunki alcohols mein carbon par lage carbons gine jaate hain. Step 2 mein upar wala group NH2 hai; nitrogen par gola bana hai aur sirf ek carbon, uske neeche wala, numbered hai, isliye label 1° amine padhta hai, kyunki amines mein nitrogen par lage carbons gine jaate hain.

Definition

Primary, secondary aur tertiary amines

Aise amines jinke nitrogen par ek ( $R N H_{2}$ ), do ( $R_{2} N H$ ) ya teen ( $R_{3} N$ ) carbon groups hon. Label N par lage groups ginta hai, uske bagal wale carbon par nahi.

Quick check

3 mein se 1

Inmein se kaun sa secondary amine hai?

C H_{3} C H_{2} C H (N H_{2}) C H_{3}

(C H_{3} C H_{2})_{3} N

C_{6} H_{5} N H C H_{2} C H_{3}

(C H_{3})_{4} N^{+} C l^{-}

Answer dekho

Answer (C) $C_{6} H_{5} N H C H_{2} C H_{3}$

N-Ethylaniline ke N par ek phenyl aur ek ethyl group hai, isliye yeh 2° hai. Option A ka

N H_{2}

secondary carbon par hai, lekin N par sirf ek group hai, isliye yeh 1° amine hai; option B 3° hai aur D quaternary salt hai.

9.3 Nomenclature

Har amine ke naam ko do baatein batani hoti hain: kaun sa carbon framework nitrogen ko carry karta hai, aur us nitrogen se aur kya juda hai. Do systems yeh kaam karte hain. Puraane common names bas nitrogen par lage groups ko alphabetical order mein likhte hain aur aakhir mein amine jod dete hain, sab ek hi word mein: methylamine, ethylmethylamine, triethylamine (repeated groups ke liye di- aur tri- ke saath).

IUPAC names iski jagah alkane se shuru karte hain aur alkanamine dete hain. Primary amine mein alkane ka aakhri e hat jaata hai, uski jagah amine aata hai, aur ek locant batata hai ki $N H_{2}$ kis carbon par hai: butan-1-amine $C H_{3} C H_{2} C H_{2} C H_{2} N H_{2}$ hai aur butan-2-amine $C H_{3} C H (N H_{2}) C H_{2} C H_{3}$ hai. Jab do amino groups hon to e rehta hai, uske baad diamine aata hai aur dono locants diye jaate hain: butane-1,4-diamine $H_{2} N (C H_{2})_{4} N H_{2}$ hai.

Secondary aur tertiary amines mein sabse bada carbon group (sabse lambi chain wala) parent alkanamine banta hai. Nitrogen par baaki har group locant N ke saath prefix ban jaata hai, aur prefixes alphabetical order mein aate hain. Fig. 9.3 aisa ek naam step by step banata hai.

4/4N-ethyl-N-methylbutan-1-amine.

Fig. 9.3Sabse bada group → parent alkanamine; baaki → N-prefixes.NCERT p. 261

Ek nitrogen atom par left mein chaar-carbon chain, right mein ethyl group aur neeche methyl group hai. Step 1 teeno groups ko box karta hai aur unpar butyl 4 C, ethyl 2 C aur methyl 1 C likhta hai. Step 2 butyl chain ko nitrogen se bonded carbon se shuru karke 1 se 4 tak number karta hai aur parent ka naam butan-1-amine rakhta hai. Step 3 baaki do groups par N-ethyl aur N-methyl likhta hai aur batata hai ki prefixes alphabetical order mein aate hain. Step 4 poora naam dikhata hai, N-ethyl-N-methylbutan-1-amine.

Arylamines mein nitrogen seedha benzene ring se bonded hota hai. Sabse simple, $C_{6} H_{5} N H_{2}$ , apna traditional naam aniline rakhta hai, jise IUPAC bhi maanta hai; systematic naam benzenamine hai. Ring substituents ko us carbon se number kiya jaata hai jis par $N H_{2}$ hai (C-1), aur nitrogen par lage groups ko phir se N milta hai: $C_{6} H_{5} N H C H_{2} C H_{3}$ N-ethylaniline (N-ethylbenzenamine) hai. Benzylamine, $C_{6} H_{5} C H_{2} N H_{2}$ , arylamine nahi hai. Iska nitrogen ek $C H_{2}$ group par hai, isliye IUPAC ise phenylmethanamine kehta hai, aur yeh alkylamine ki tarah behave karta hai.

Table 9.1Kuch amines ke IUPAC aur (common) names

Amine	Naam
$C H_{3} (C H_{2})_{3}$ – $N H_{2}$	butan-1-amine (n-butylamine)
$C H_{3} C H (N H_{2})$ – $C H_{2} C H_{3}$	butan-2-amine (sec-butylamine)
$(C H_{3})_{2} N H$	N-methylmethanamine (dimethylamine)
$C H_{3} C H_{2} C H_{2}$ – $N H C H_{3}$	N-methylpropan-1-amine (methylpropylamine)
$(C_{2} H_{5})_{2} N$ – $C_{3} H_{7}$	N,N-diethylpropan-1-amine (diethylpropylamine)
$H_{2} N (C H_{2})_{4}$ – $N H_{2}$	butane-1,4-diamine (putrescine)
$C H_{2} = C H$ – $C H_{2} C H_{2} N H_{2}$	but-3-en-1-amine
$C_{6} H_{11} N H_{2}$	cyclohexanamine (cyclohexylamine)
$C_{6} H_{5} C H_{2} N H_{2}$	phenylmethanamine (benzylamine)
$C_{6} H_{5} N H C_{2} H_{5}$	N-ethylaniline ya N-ethylbenzenamine
m- $C H_{3} C_{6} H_{4} N H_{2}$	3-methylaniline (m-toluidine)
p- $O_{2} N C_{6} H_{4} N H_{2}$	4-nitroaniline (p-nitroaniline)

Example

Amines ko dono tarah se naam dena

Har amine ka IUPAC naam aur class batao: (a) $(C H_{3})_{2} C H N H C H_{2} C H_{3}$ ; (b) $C H_{3} C H_{2} C H (C H_{3})$ – $C H_{2} N H_{2}$ ; (c) $C_{6} H_{5} N (C H_{3}) C H_{2} C H_{3}$ .

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

C H_{3} C H_{2} C H_{2} N (C H_{3})_{2}

ka IUPAC naam kya hai?

dimethylpropylamineN,N-dimethylpropan-1-amine1-(dimethylamino)propaneN-propyl-N-methylmethanamine

Answer dekho

Answer (B) N,N-dimethylpropan-1-amine

Propyl sabse bada group hai, isliye parent propan-1-amine hai aur do methyl groups N-prefixes hain. Option A common name hai, aur option D galti se methyl ko parent chunta hai.

Practice · 2 questions

Practice: classify karna aur naam dena: answers ke saath

1
Quick answerEasyNCERT Intext Question 9.1
Har species ko uski class se match karo.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
$(C H_{3})_{3} C C H_{2} N H_{2}$ : primary amine
$C_{6} H_{5} C H_{2} N H C H_{3}$ : secondary amine
$C_{6} H_{5} N (C_{2} H_{5})_{2}$ : tertiary amine
$C_{6} H_{5} C H_{2} N (C H_{3})_{3}^{+} B r^{-}$ : quaternary salt
2,2-Dimethylpropan-1-amine ke N par ek group hai (bagal wala crowded carbon matter nahi karta); N-methyl-1-phenylmethanamine par do hain; N,N-diethylaniline par teen hain; benzyltrimethylammonium bromide par chaar hain aur ek positive charge hai.
2
Short answerHardNCERT Intext Question 9.2
Formula $C_{5} H_{13} N$ ke 17 amine isomers hain: aath primary, chhah secondary aur teen tertiary.
1. (a)
  Chhah secondary amines ke condensed formulas aur IUPAC names likho.
  Solution dikhao
  $C H_{3} N H$ – $(C H_{2})_{3}$ – $C H_{3}$ : N-methylbutan-1-amine
  $C H_{3} N H$ – $C H (C H_{3})$ – $C_{2} H_{5}$ : N-methylbutan-2-amine
  $C H_{3} N H$ – $C H_{2}$ – $C H (C H_{3})_{2}$ : N,2-dimethylpropan-1-amine
  $C H_{3} N H$ – $C (C H_{3})_{3}$ : N,2-dimethylpropan-2-amine
  $C_{2} H_{5} N H$ – $C H_{2}$ – $C H_{2} C H_{3}$ : N-ethylpropan-1-amine
  $C_{2} H_{5} N H$ – $C H (C H_{3})_{2}$ : N-ethylpropan-2-amine
2. (b)
  Teen tertiary amines ke liye bhi yahi karo.
  Solution dikhao
  $(C H_{3})_{2} N$ – $C H_{2}$ – $C H_{2} C H_{3}$ : N,N-dimethylpropan-1-amine
  $(C H_{3})_{2} N$ – $C H (C H_{3})_{2}$ : N,N-dimethylpropan-2-amine
  $C H_{3}$ – $N (C_{2} H_{5})_{2}$ : N-ethyl-N-methylethanamine
3. Har pair kis tarah ka structural isomerism dikhata hai, naam batao: (i) N-methylbutan-1-amine aur N-ethylpropan-1-amine; (ii) N-methylbutan-1-amine aur N-methylbutan-2-amine; (iii) N-ethylpropan-1-amine aur N,N-dimethylpropan-1-amine.
  Solution dikhao
  (i) Metamerism: functional group same hai, lekin nitrogen ke dono taraf alag groups hain. (ii) Position isomerism: ek mein nitrogen C-1 par hai aur doosre mein C-2 par. (iii) Functional isomerism: ek secondary amine hai, doosra tertiary.

9.4 Preparation of Amines

Amine tak pahunchne ka har route ya to naya C–N bond banata hai ya carbon se pehle se bonded kisi nitrogen group ko reduce karta hai. Route chunne ke liye do sawaal poochho: tumhe kaun si class ka amine chahiye, aur carbon count same rehna chahiye, ek badhna chahiye ya ek ghatna chahiye? Is section ke end mein Table 9.2 chhahon methods ko saath-saath rakhta hai.

Reduction of nitro compounds

Zyadatar arylamines ki shuruaat nitro compounds se hoti hai, kyunki benzene ring par nitrogen lagane ka sabse aasaan tareeka nitration hai. $- N O_{2}$ ko $- N H_{2}$ mein badalna six-electron reduction hai, aur do tarah ke reagents yeh karte hain: acid ke saath metal (hydrochloric acid ke saath tin ya iron), ya Ni, Pd ya Pt jaise baarik powder wale catalyst par hydrogen gas ((9.1)). Yahi reagents nitroalkanes ko alkylamines mein badalte hain; jaise nitroethane se ethanamine milta hai.

4-nitrotoluene se 4-methylaniline

\begin{aligned} C H_{3} C_{6} H_{4} N O_{2} + 3 H_{2} \\ \to_{e t h a n o l}^{P d} C H_{3} C_{6} H_{4} N H_{2} \\ + 2 H_{2} O \end{aligned}

(9.1)

Bade batches ke liye iron filings aur hydrochloric acid sabse sasti jodi hain, aur inhe hairaani ki had tak kam acid chahiye. Reduction chalte hue jo iron(II) chloride banta hai woh paani se react karke HCl dobara bana deta hai, isliye wahi acid baar-baar use hota hai. Jo groups mushkil se reduce hote hain woh bach jaate hain: 3-chloronitrobenzene se 3-chloroaniline milta hai aur uska chlorine waisa hi rehta hai.

iron aur acid ke saath

\begin{aligned} C l C_{6} H_{4} N O_{2} + 6 [H] \\ \overset{F e, H C l}{\to} C l C_{6} H_{4} N H_{2} \\ + 2 H_{2} O \end{aligned}

Ammonolysis of alkyl halides

Ammonia ek nucleophile hai, isliye yeh Unit 6 ke SN2 route se alkyl ya benzyl halide se halide ion ko hata sakta hai. Halide ko ethanol mein ammonia ke saath sealed tube mein garam karne par (lagbhag 373 K, taaki ammonia ud na jaaye) alkylammonium salt banta hai, aur phir NaOH jaisa strong base free amine ko alag kar deta hai ((9.2)). Ammonia se is tarah C–X bond todne ko ammonolysis kehte hain.

ammonolysis, phir amine ko free karna

\begin{aligned} N H_{3} + R - X \overset{}{\to} R - N H_{3}^{+} X^{-} \\ R - N H_{3}^{+} X^{-} + N a O H \\ \overset{}{\to} R - N H_{2} + H_{2} O + N a X \end{aligned}

(9.2)

Dikkat yeh hai ki product khud bhi nucleophile hai, aur ammonia se thoda behtar, kyunki uska alkyl group nitrogen ko electron density deta hai. Yeh bache hue halide ke liye compete karta hai aur Fig. 9.4 ki seedhiyan chadhta hai: pehle primary, phir secondary, phir tertiary amine, aur aakhir mein quaternary ammonium salt. Isliye ammonolysis se mixture milta hai.

Ammonia bahut zyada excess mein ho to har halide molecule ke amine ke bajaye ammonia molecule se milne ke chances kahin zyada ho jaate hain, isliye primary amine main product ban jaata hai. Iodides sabse tez react karte hain, phir bromides, phir chlorides (RI > RBr > RCl), kyunki teeno mein C–I bond sabse kamzor hai.

5/5Halide ko bahut saare ammonia mein rakhne se zyadatar reaction pehli step par hi ruk jaata hai.

Fig. 9.4Bana hua har amine halide par phir se attack kar sakta hai.NCERT p. 263

Ek seedhi neeche left mein ammonia se upar chadhti hai. Har step par plus RX label wala ek arrow judta hai. Step 1 mein ammonia primary amine RNH2 mein badalta hai. Step 2 secondary amine R2NH jodta hai, step 3 tertiary amine R3N aur step 4 upar right mein quaternary salt R4N+ X−, jin par 1°, 2°, 3° aur 4° likha hai. Step 5 poori seedhi rakhta hai aur ek box jodta hai jo kehta hai ki ammonia ke bade excess se zyadatar RNH2 milta hai.

Example 9.1NCERT Example 9.1 · p. 263

Ammonolysis ki equations likhna

Equations likho aur har organic product ka naam batao: (a) 1-bromopropane ko ethanolic ammonia ke saath garam kiya gaya, aur ise step by step quaternary salt tak follow karo; (b) butan-1-amine ko do moles iodomethane ke saath treat kiya gaya.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Reduction of nitriles

Nitrile, $R - C \equiv N$ , apne triple bond par chaar hydrogen atoms jodta hai aur primary amine deta hai jismein nitrogen ek naye $C H_{2}$ group par hota hai. Lithium aluminium hydride, $L i A l H_{4}$ , yeh karta hai, aur nickel par hydrogen aur ethanol mein sodium bhi. Nitrile alkyl halide aur cyanide ion se aasaani se ban jaata hai, isliye halide → nitrile → amine wala sequence chain mein ek carbon jodta hai. Amine series mein ascend karne ka yahi classic tareeka hai.

nitrile se primary amine

\begin{aligned} R - C \equiv N + 4 [H] \\ \overset{L i A l H_{4}}{\to} R - C H_{2} - N H_{2} \end{aligned}

(9.3)

Reduction of amides

Amides bhi isi tarah $L i A l H_{4}$ aur uske baad paani se reduce hote hain: C=O group $C H_{2}$ ban jaata hai, isliye amine mein amide jitne hi same carbons hote hain. Jis amide ke nitrogen par pehle se ek group laga ho, usse secondary ya tertiary amine milta hai; jaise N-methylethanamide, $C H_{3} C O N H C H_{3}$ , se N-methylethanamine, $C H_{3} C H_{2} N H C H_{3}$ , milta hai.

amide se amine

\begin{aligned} R - C O - N H_{2} \to_{(i i) H_{2} O}^{(i) L i A l H_{4}} \\ R - C H_{2} - N H_{2} \end{aligned}

(9.4)

Example 9.2NCERT Example 9.2 · p. 264

Ek carbon jodna

Dikhao ki (a) 1-bromobutane se pentan-1-amine aur (b) 1-chloro-2-methylpropane se 3-methylbutan-1-amine kaise banaoge.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Gabriel phthalimide synthesis

Gabriel synthesis primary amines banata hai, bina un mixtures ke jo ammonolysis ko bigaad dete hain. Iska nitrogen source phthalimide hai, jismein ek N–H group do C=O groups ke beech baitha hai. Dono carbonyl groups nitrogen se electron density kheenchte hain, isliye hydrogen itna acidic ho jaata hai ki ethanolic potassium hydroxide use hata sake. Isse bana phthalimide anion achha nucleophile hai aur primary alkyl halide se halide ko hata deta hai (Fig. 9.5).

Bana hua N-alkylphthalimide dobara react nahi kar sakta: uske nitrogen par koi hydrogen nahi bacha, aur uska lone pair do carbonyl groups ke saath share hota hai. Phir aqueous sodium hydroxide ke saath garam karne par dono C–N bonds hydrolyse ho jaate hain, aur primary amine aur phthalate ion free ho jaate hain. Yeh method arylamines ke liye fail hota hai: benzene ring par lage halogen ko phthalimide anion hata nahi sakta, un reasons se jo Unit 6 mein dekhe the.

3/4Anion primary halide se $X^{-}$ ko hatata hai (SN2).

Fig. 9.5Nitrogen sirf ek baar diya jaata hai, isliye product pure primary amine hota hai.NCERT p. 264

Ek benzene ring ek five-membered ring se fused hai jismein do C=O groups aur ek nitrogen hai. Step 1, phthalimide, do carbonyl groups ke beech nitrogen par lage hydrogen ko highlight karta hai. Step 2 dikhata hai ki KOH ne woh hydrogen hata diya: nitrogen par negative charge aur bagal mein ek K+ ion. Step 3 mein R–X ke saath garam karne ke baad ek R group nitrogen se bonded hai, yaani N-alkylphthalimide. Step 4 mein NaOH ne ring khol di hai: dono carbonyl carbons par ab O− hai, aur right mein primary amine R–NH2 free ho jaata hai.

Hoffmann bromamide degradation reaction

1881 mein August Wilhelm von Hofmann ne paaya ki bromine aur sodium hydroxide primary amide ko ek carbon kam wale primary amine mein badal dete hain.Chemist apna naam Hofmann likhte the; NCERT reaction ko Hoffmann bromamide degradation kehta hai. Yeh Hoffmann bromamide degradation isliye kaam karta hai kyunki carbonyl carbon se juda group R nitrogen par shift ho jaata hai, aur carbonyl carbon carbonate ban kar nikal jaata hai (Fig. 9.6). Chemists iska use homologous series mein ek step neeche jaane ke liye karte hain.

Hoffmann bromamide degradation

\begin{aligned} R - C O - N H_{2} + B r_{2} \\ + 4 N a O H \overset{}{\to} R - N H_{2} \\ + N a_{2} C O_{3} + 2 N a B r \\ + 2 H_{2} O \end{aligned}

(9.5)

Yeh reaction aromatic amides ke liye bhi kaam karta hai, aur shift hone wala group apna structure nahi badalta. Isliye aryl amide se arylamine milta hai: 4-chlorobenzamide se 4-chloroaniline milta hai.

3/3Carbonyl carbon carbonate ban kar nikal jaata hai: ek carbon kam.

Fig. 9.6Hexanamide (6 C) se pentan-1-amine (5 C) milta hai.NCERT p. 264

Ek line mein paanch carbon circles group R banate hain, jo ek chhathe, dashed carbon se jude hain jis par C=O aur right mein ek NH2 group hai. Step 1 ka heading hexanamide, 6 C hai, aur note kehta hai ki C=O wala carbon nikal jaayega. Step 2, jiska heading Br2 + NaOH hai, chain se nitrogen tak ek curved arrow jodta hai jis par R moves to N likha hai. Step 3 mein paanch-carbon chain nitrogen par khisak jaati hai aur carbonyl carbon door jaakar fade ho jaata hai, jis par carbonate likha hai; heading pentan-1-amine, 5 C padhta hai, aur note kehta hai ek carbon kam.

Example 9.3NCERT Example 9.3 · p. 265

Hoffmann product se ulta sochna

(a) Hoffmann degradation mein kaun sa amide cyclohexanamine, $C_{6} H_{11} N H_{2}$ , deta hai? (b) 4-methylbenzamide bromine aur NaOH ke saath kaun sa amine deta hai, aur $L i A l H_{4}$ ke saath kya deta hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Table 9.2Amine banane ka route chunna

Method	Kya milta hai	Carbons
Nitro compound ko reduce karo	1° (arylamines ka main route)	same
$R - X$ ka ammonolysis	1°, 2°, 3° aur $R_{4} N^{+}$ ka mixture	same
$R - X$ → nitrile, phir reduce karo	1°	$R - X$ se ek zyada
Amide ko reduce karo	1° (N-substituted amides se 2° ya 3°)	same
Gabriel synthesis	pure 1° alkylamine	same
Hoffmann degradation	1° (alkyl ya aryl)	amide se ek kam

Quick check

4 mein se 1

Tumhe 1-bromobutane se butan-1-amine chahiye, jismein secondary aur tertiary amines na hon. Kaun sa route sabse achha hai?

Ethanolic ammonia ke saath garam karoNaCN, phir

L i A l H_{4}

Bromide ka Hoffmann degradationGabriel synthesis

Answer dekho

Answer (D) Gabriel synthesis

Gabriel synthesis sirf primary amine deta hai. Ammonolysis mixture deta hai, nitrile route pentan-1-amine deta hai (ek carbon zyada), aur Hoffmann degradation ko halide nahi, amide chahiye.

Practice · 1 question

Practice: amines banana: answers ke saath

1
Short answerMediumNCERT Intext Question 9.3
Har compound tum kaise banaoge? Har step ke reagents batao.
1. Chlorobenzene se 4-chloroaniline
  Solution dikhao
  Concentrated $H N O_{3}$ aur $H_{2} S O_{4}$ se nitration karo. Chlorine ortho/para direct karta hai, aur main product, 1-chloro-4-nitrobenzene, ko alag kar lo. Use Fe aur HCl (ya Sn aur HCl) se reduce karo aur NaOH daal kar 4-chloroaniline ko free karo.
2. Benzene se N,N-diethylaniline
  Solution dikhao
  Benzene ka nitration karo ( $H N O_{3}$ aur $H_{2} S O_{4}$ , lagbhag 330 K), nitrobenzene ko Sn aur HCl se reduce karo aur NaOH se aniline ko free karo. Phir aniline ko do moles bromoethane aur HBr sokhne ke liye $N a_{2} C O_{3}$ jaise base ke saath garam karo. Aur zyada bromoethane hone par reaction quaternary salt tak chala jaayega.
  $\begin{aligned} C_{6} H_{5} N H_{2} + 2 C_{2} H_{5} B r \\ \overset{}{\to} C_{6} H_{5} N (C_{2} H_{5})_{2} \\ + 2 H B r \end{aligned}$
3. 1,3-dibromopropane se pentane-1,5-diamine
  Solution dikhao
  Product mein paanch carbons hain aur dibromide mein teen, isliye nitriles ke zariye har end par ek carbon jodo. Ethanolic KCN ke saath garam karke pentanedinitrile, $N C (C H_{2})_{3} C N$ , banao, phir dono C≡N groups ko $L i A l H_{4}$ (ya Ni par $H_{2}$ ) se reduce karke $H_{2} N (C H_{2})_{5}$ – $N H_{2}$ banao.

9.5 Physical Properties

Smell se tum pehle hi mil chuke ho. Sabse halke amines (methanamine, N-methylmethanamine, trimethylamine aur ethanamine) room temperature se neeche boil hote hain aur gas ban kar nikal jaate hain, jinki smell ammonia aur baasi machhli jaisi hoti hai. Chain lambi karo to boiling point badhta hai, isliye propan-1-amine aur usse bade amines liquids hain aur long-chain amines waxy solids hote hain. Taaza distil kiya hua aniline colourless hota hai; aadhi bhari bottle mein chhod do to hawa use dheere-dheere oxidise karti hai aur woh yellow se brown hota jaata hai.

Paani mein amino group achhe se ghul-mil jaata hai. Iska lone pair paani ke molecule ke O–H se hydrogen bond leta hai, aur primary aur secondary amines mein N–H hydrogens bhi paani ke oxygen se bond karke ehsaan lauta dete hain. Isliye methanamine, ethanamine aur unke chhote rishtedaar aasaani se ghul jaate hain. Har extra $C H_{2}$ aisa hydrocarbon jodta hai jise paani pakad nahi sakta, aur lagbhag chhah carbons ke baad amine mushkil se ghulta hai. Ethanol, diethyl ether aur hexane jaise organic solvents bhi har size ke amines ko ghol lete hain.

Primary aur secondary amines aapas mein bhi hydrogen bond banate hain: ek molecule ka N–H padosi molecule ke lone pair se bond karta hai. Primary amine ke paas dene ke liye do N–H hydrogens hote hain, secondary amine ke paas ek aur tertiary amine ke paas koi nahi (Fig. 9.7). Molecules jitna zyada associate karte hain, unhe alag karne mein utni zyada energy lagti hai, isliye isomeric amines mein boiling points 1° > 2° > 3° order mein girte hain.

1/3Propan-1-amine: har molecule ke paas dene ke liye do N–H hydrogens hain; b.p. 321 K.

Fig. 9.7Kam N–H hydrogens, kam hydrogen bonds, kam boiling point (teen

C_{3} H_{9} N

isomers).NCERT Fig. 9.2 · p. 266

Chhah amine molecules do rows mein bane hain, har ek nitrogen ke upar ek R group ke saath. Dotted lines N–H hydrogen se padosi nitrogen tak hydrogen bonds dikhati hain. Step 1 (primary) mein har molecule ke paas do N–H hydrogens hain, ek apni row ke agle molecule se bond karta hai aur ek neeche wali row ke molecule se, jisse network banta hai; right mein ek bar gauge 321 K padhta hai. Step 2 (secondary) mein har molecule ke paas ek N–H hai, isliye sirf har row mein chains bachti hain, aur gauge 310 K par gir jaata hai. Step 3 (tertiary) mein koi N–H nahi bachta, koi dotted line nahi rehti, aur gauge 276 K par gir jaata hai.

Table 9.3Lagbhag 59 g mol⁻¹ molar mass M wale compounds ke boiling points

Compound	M	b.p./K
$C H_{3} C H_{2} C H_{2}$ – $N H_{2}$ (1°)	59	321
$C H_{3} C H_{2}$ – $N H C H_{3}$ (2°)	59	310
$(C H_{3})_{3} N$ (3°)	59	276
$C H_{3} C H_{2}$ – $C H_{2} C H_{3}$ (alkane)	58	273
$C H_{3} C H_{2} C H_{2}$ – $O H$ (alcohol)	60	370

Table amines ko milte-julte mass wale padosiyon ke beech bhi rakhta hai. Yeh alkane se kaafi upar boil hote hain, jismein sirf weak dispersion forces hoti hain, lekin alcohol se kaafi neeche. Oxygen (electronegativity 3.5) nitrogen (3.0) se zyada electronegative hai, isliye O–H bond zyada polar hota hai aur O–H⋯O hydrogen bonds N–H⋯N bonds se zyada strong hote hain.

Socho zara

Alcohol ya amine: paani mein kaun behtar ghulta hai?

Amine molecules ke beech weak hydrogen bonds dekh kar lag sakta hai ki amines paani se bhi kam achhe se bond karte honge. Lekin data kuch aur kehta hai: butan-1-amine paani mein har proportion mein mix ho jaata hai, jabki 298 K par 100 g paani mein butan-1-ol ka sirf lagbhag 7 g ghulta hai. Aisa kyun?

Ghulna is par depend karta hai ki molecule paani se kitna achha bond karta hai; boiling is par depend karti hai ki woh khud se kitna achha bond karta hai. Nitrogen apne lone pair ko oxygen jitna kas kar nahi pakadta, isliye amine ka lone pair paani se aane wale hydrogen bond ka excellent acceptor hai. Lekin amine molecules ke beech donor weakly polar N–H hota hai, isliye amine–amine hydrogen bonds weak hote hain aur amines alcohols se kam temperature par boil hote hain.

Quick check

3 mein se 1

Boiling point ke order mein lagao, sabse zyada wala pehle.

propan-1-amine
trimethylamine
propan-1-ol
N-methylethanamine

Answer dekho

Answer

propan-1-ol
propan-1-amine
N-methylethanamine
trimethylamine

370 K, 321 K, 310 K aur 276 K: O–H⋯O bonds N–H⋯N bonds se aage hain, aur amines mein jitne zyada N–H hydrogens, utni zyada hydrogen bonding.

9.6 Chemical Reactions

Amino group ki do khaasiyatein lagbhag uske saare reactions samjha deti hain. Nitrogen par lone pair amines ko bases (yeh protons lete hain) aur nucleophiles (yeh electron-poor carbon par attack karte hain) banata hai. N–H hydrogens ki ginti tay karti hai ki aage kya hoga: primary amine ke paas khone ke liye do hote hain, secondary ke paas ek aur tertiary ke paas koi nahi, isliye teeno classes aksar alag products deti hain. Chemists inhi farkon se unhe alag pehchante hain.

Basic character of amines

Amines acids se waise hi react karte hain jaise ammonia: lone pair se proton ke saath bond bana kar. Products substituted ammonium salts hote hain: propan-1-amine aur hydrogen chloride se propylammonium chloride banta hai, aur aniline se anilinium chloride.

salt banna

\begin{aligned} C H_{3} C H_{2} C H_{2} N H_{2} + H C l \\ \overset{}{\to} C H_{3} C H_{2} C H_{2} N H_{3}^{+} C l^{-} \end{aligned}

Yeh salts ionic hote hain, isliye paani mein ghulte hain lekin ether jaise non-polar solvents mein nahi, aur NaOH jaisa zyada strong base inhe wapas free amine bana deta hai. Yeh baatein milkar ek simple separation deti hain. Ether mein organic compounds ke mixture ko dilute hydrochloric acid ke saath hilao, to sirf amine apne salt ke roop mein water layer mein chala jaata hai; us layer mein NaOH daalne par amine phir se free ho jaata hai.

Try karke dekho

Machhli ke baad nimbu

Tumhe chahiye

kitchen se kachchi machhli ka chhota tukda ya kuch prawns
aadha nimbu
nal ke paani ke do bowls
sabun

Pehle guess karo

Machhli chhoone ke baad tum ek haath saade paani mein aur doosra nimbu paani mein dhote ho. Kis haath se machhli ki smell kam aayegi?

dono se same smell aayegisaade paani mein dhoya haathnimbu paani mein dhoya haath

Dono hatheliyan kuch seconds machhli par ragdo.
Left haath saade paani mein dho lo aur right haath aise paani mein jismein aadhe nimbu ka ras ho.
Dono haath sukhao aur dono ki smell compare karo.
Aakhir mein dono haath sabun se dho lo.

Table 9.4Tumne kya notice kiya?

Kisse dhoya	Baad mein smell
Saada paani
Nimbu paani

Tumhari entries apne aap save hoti hain.

Paani mein amine paani ke molecule se ek proton le leta hai aur peeche hydroxide ions chhod deta hai, isliye uska solution alkaline hota hai. Is reaction ka equilibrium constant, yaani base dissociation constant $K_{b}$ , base ki strength naapta hai ((9.6)).

paani mein ek amine

\begin{aligned} R N H_{2} + H_{2} O \\ \overset{}{⇌} R N H_{3}^{+} + O H^{-} \end{aligned}

base dissociation constant

K_{b} = \frac{[R N H_{3}^{+}] [O H^{-}]}{[R N H_{2}]}

(9.6)

jahan

$K_{b}$: base dissociation constant (paani ka concentration isi mein shaamil hai)
$[X]$: $m o l L^{- 1}$ mein equilibrium concentration

Kyunki $K_{b}$ values ten ki kai powers tak phaili hoti hain, chemists aam taur par p $K_{b}$ $= - \log K_{b}$ batate hain. Zyada $K_{b}$ , ya kam p $K_{b}$ , ka matlab hai zyada strong base. Ammonia ka p $K_{b} = 4.75$ hai. Simple alkylamines lagbhag 3.0 aur 4.2 ke beech aate hain, isliye yeh ammonia se zyada strong bases hain, jabki aniline, 9.38 par, lagbhag ten ki paanch powers se weak hai.

Example

pKb se Kb aur pH tak

Methanamine ka p $K_{b} = 3.38$ hai aur aniline ka 9.38. (a) Dono ka $K_{b}$ nikaalo aur compare karo. (b) 298 K par methanamine ke $0.10 m o l L^{- 1}$ solution ka pH nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Structure aur basicity: alkanamines vs ammonia

Sabse saaf experiment se shuru karo: solvent hata do. Gas phase mein base ko uski proton affinity se rank kiya ja sakta hai, yaani woh energy jo akela molecule proton lene par release karta hai. Table 9.5 dikhata hai ki ammonia mein jode gaye har methyl group se yeh kuch das kilojoules per mole badhti hai, aur ethyl groups ise thoda aur badhate hain: gas mein bina kisi exception ke 3° > 2° > 1° > $N H_{3}$ .

Inductive effect. Yeh lagaatar badhna +I effect ka kaam hai. Alkyl groups electron density release karte hain, isliye nitrogen par jitne zyada groups, uska lone pair utni aasaani se proton ke saath share hota hai, aur naye cation ka positive charge utna achhe se phail kar stable hota hai. Base tab strong hota hai jab uska conjugate cation comfortable ho.

Table 9.5Gas-phase proton affinity, PA, kJ mol⁻¹ mein (gaseous base ke proton lene par release hui energy; zyada = zyada basic) aur 298 K par paani mein p

K_{b}

Base	PA	p $K_{b}$
$N H_{3}$	854	4.75
$C H_{3} N H_{2}$	899	3.38
$(C H_{3})_{2} N H$	930	3.27
$(C H_{3})_{3} N$	949	4.22
$C_{2} H_{5} N H_{2}$	912	3.29
$(C_{2} H_{5})_{2} N H$	952	3.00
$(C_{2} H_{5})_{3} N$	982	3.25
$C_{6} H_{5} C H_{2}$ – $N H_{2}$	–	4.70
$C_{6} H_{5} N H_{2}$	–	9.38

Paani picture badal deta hai, kyunki yeh cations ko bhi stable kar sakta hai. Do aur effects kaam mein aate hain.

Solvation. Paani mein cation apne N–H hydrogens se paani ke molecules tak hydrogen bonds ke zariye bhi stable hota hai; ise solvation kehte hain. $R N H_{3}^{+}$ ke paas aise teen hydrogens hain, $R_{2} N H_{2}^{+}$ ke paas do aur $R_{3} N H^{+}$ ke paas sirf ek (Fig. 9.8). Sirf is hisaab se order ulta ho jaata: 1° > 2° > 3°.

Crowding. Nitrogen ke aas-paas bulky groups un paani ke molecules ke raaste mein aate hain jo cation ko solvate karte hain. Groups jitne bade, yeh utna zyada matter karta hai; yeh steric hindrance tertiary amines ko sabse zyada nuksaan pahunchata hai.

2/4 $R N H_{3}^{+}$ paani ke saath teen hydrogen bonds se juda rehta hai.

Fig. 9.8Gas mein sirf +I effect; paani mein +I, solvation aur crowding teeno saath.NCERT p. 268

Step 1, gas phase: ek tertiary ammonium ion, yaani teen R groups aur ek H wala N+, jiske R–N bonds par arrows dikhate hain ki alkyl groups N+ par electron density push kar rahe hain; heading gas-phase order 3° > 2° > 1° > NH3 deta hai. Steps 2 se 4 paani mein akele ions dikhate hain, jinke har N–H se paani ke molecules (OH2) dotted lines se hydrogen-bonded hain. Step 2, RNH3+, se teen waters jude hain. Step 3, R2NH2+, se do. Step 4, R3NH+, se ek water juda hai aur teen bade R groups nitrogen ko gher rahe hain.

Paani mein teeno effects saath kaam karte hain, aur measured orders ek compromise hote hain. Methyl series ke liye order $(C H_{3})_{2} N H$ > $C H_{3} N H_{2}$ > $(C H_{3})_{3} N$ > $N H_{3}$ hai; ethyl series ke liye $(C_{2} H_{5})_{2} N H$ > $(C_{2} H_{5})_{3} N$ > $C_{2} H_{5} N H_{2}$ > $N H_{3}$ hai. Dono mein secondary amine jeetta hai: use do alkyl groups ka fayda milta hai aur solvation ke liye do N–H hydrogens bhi bache rehte hain. Tertiary amine ki position badalti rehti hai, kyunki uske liye teesre alkyl group ka fayda aur solvation ka nuksaan lagbhag cancel ho jaate hain, isliye groups mein chhota sa badlaav balance ko idhar-udhar kar deta hai.

Arylamines vs ammonia

Cyclohexanamine aur aniline dono mein six-carbon ring par $N H_{2}$ group hai, phir bhi cyclohexanamine (p $K_{b}$ 3.36) base ke roop mein aniline (9.38) se lagbhag das lakh guna strong hai, aur aniline to ammonia se bhi weak hai. Farak ring ke π system ka hai. Aniline mein nitrogen ka lone pair ring ke p orbitals ke bagal mein hota hai aur unmein phail jaata hai, jisse ortho aur para carbons par negative charge aa jaata hai (Fig. 9.9). Resonance ke roop mein likho to aniline ke paanch contributing structures hain, jinmein se teen mein pair ring mein dhakela gaya hai.

Protonation us pair ko ek naye N–H bond mein lock kar deta hai. Anilinium ion, $C_{6} H_{5} N H_{3}^{+}$ , ke paas ek normal benzene ring ke sirf do Kekulé structures bachte hain, isliye proton lete waqt aniline apni extra stability kho deta hai, aur equilibrium kaafi left ki taraf rehta hai. $s p^{2}$ ring carbon bhi inductively nitrogen ke electrons ko kheenchta hai.

4/5Asli molecule: C-2, C-4 aur C-6 par δ−.

Fig. 9.9Ring ke saath share hone se lone pair proton ke liye kam available hota hai.NCERT p. 269

Alternating double bonds wali ek benzene ring ke upar NH2 group hai. Step 1 nitrogen par lone pair ko do dots ki tarah dikhata hai. Step 2 mein ek curved arrow pair ko N–C bond mein le jaata hai, jo ab double bond hai, nitrogen par positive charge aur upar-right wale ortho carbon par negative charge ke saath. Step 3 mein yahi hai, bas negative charge neeche wale para carbon par chala gaya hai. Step 4 asli molecule ka summary hai: dono ortho carbons aur para carbon par delta minus, nitrogen par delta plus. Step 5 anilinium ion, NH3+, dikhata hai, jiska pair extra N–H bond mein use ho chuka hai, isliye ring mein koi charge nahi phailta.

Ring par laga group is effect ko tune karta hai. Para position par methyl group (4-methylaniline, p $K_{b}$ 8.92) ya methoxy group (4-methoxyaniline, 8.66) nitrogen ki taraf electron density wapas deta hai aur basicity thodi badha deta hai. Chlorine atom (4-chloroaniline, lagbhag 10.0) ya nitro group (4-nitroaniline, lagbhag 13.0) density kheenchta hai aur basicity ghatata hai, aur $- C O O H$ aur $- S O_{3} H$ bhi yahi karte hain. Benzylamine (4.70) dikhata hai ki is effect ke liye nitrogen ka khud ring par hona zaroori hai: beech mein ek $C H_{2}$ ho to amine normal alkylamine ki tarah behave karta hai.

InteractiveBasicity explorer: gas ya paani?

Buttons gas phase aur paani ke beech, aur teen series ke beech switch karte hain: methylamines ke saath ammonia, ethylamines ke saath ammonia, aur benzylamine aur cyclohexanamine ke saath ring-substituted anilines (sirf paani). Learner pehle apni samajh ke order mein bases par tap karta hai, sabse strong pehle, phir Reveal dabata hai. Phir horizontal bars measured values dikhate hain, gas ke liye kilojoules per mole mein proton affinity ya paani ke liye pKb, sabse strong pehle, aur learner ke guesses sahi ya galat mark hote hain. Kisi bhi base par tap karne se ek factor panel bhar jaata hai jo uske alkyl groups, uska cation paani ko kitne N–H hydrogens de sakta hai, nitrogen kitna crowded hai aur lone pair ring ke saath share hota hai ya nahi, yeh sab batata hai, saath mein ek sentence jo order mein uski jagah samjhata hai.

Fig. 9.10Ek phase aur ek series chuno, khud bases ko rank karo, phir measured order aur reasons reveal karo.

Example 9.4NCERT Example 9.4 · p. 270

Paani mein bases ko rank karna

Paani mein in chaar bases ko sabse strong se sabse weak tak rank karo: 4-nitroaniline, ammonia, N-methylmethanamine aur 4-methylaniline.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Alkylation

Kyunki amines nucleophiles hain, yeh alkyl halides se bilkul ammonia ki tarah react karte hain (Unit 6), aur har step mein ek N–H hydrogen ki jagah alkyl group aa jaata hai. Kaafi halide ho to reaction quaternary ammonium salt tak chalta jaata hai. Jaise, propan-1-amine ko excess iodomethane aur HI hataane wale base ke saath garam karne par N,N,N-trimethylpropan-1-aminium iodide, $C_{3} H_{7} N (C H_{3})_{3}^{+} I^{-}$ , milta hai.

Acylation

Primary ya secondary amine ko acid chloride ke saath mix karo to nitrogen ka lone pair carbonyl carbon par attack karta hai; chloride nikal jaata hai, aur ek N–H hydrogen HCl ke roop mein chala jaata hai. Product ek amide hota hai, aur N–H hydrogen ki jagah acyl group, $R C O^{-}$ , lagne ke is poore badlaav ko acylation kehte hain. Acid anhydrides yahi kaam zyada narmi se karte hain (chloride ki jagah carboxylate nikalta hai), aur esters sirf dheere react karte hain.

N-propylethanamide

\begin{aligned} C_{3} H_{7} N H_{2} + C H_{3} C O C l \\ \overset{p y r i d i n e}{\to} C_{3} H_{7} N H C O C H_{3} \\ + H C l \end{aligned}

(9.7)

Acid chloride ka har mole ek mole HCl release karta hai, aur woh acid bache hue amine ko protonate kar dega, use aisa ammonium ion bana dega jiske paas attack karne ke liye koi lone pair nahi bachta. Isliye reaction pyridine mein, ya amine se zyada strong kisi aur base ke saath chalaya jaata hai, jo HCl ko sokh leta hai aur reaction ko amide ki taraf badhata rehta hai. Tertiary amines ke nitrogen par koi hydrogen nahi hota, isliye woh amides nahi ban sakte.

Do acylations aage kaam aayenge. Ethanoic anhydride aniline ko N-phenylethanamide (acetanilide) mein badalta hai, jo Electrophilic substitution mein use hone wala protected form hai. Benzoyl chloride, $C_{6} H_{5} C O C l$ , se benzamides bante hain, aur is reaction ko benzoylation kehte hain; jaise N-methylmethanamine N,N-dimethylbenzamide ban jaata hai. Room temperature par carboxylic acid khud bahut weak acylating agent hai: amine ke saath woh bas salt banata hai, aur amide sirf tez garam karne par banta hai.

acetanilide

\begin{aligned} C_{6} H_{5} N H_{2} + (C H_{3} C O)_{2} O \\ \overset{}{\to} C_{6} H_{5} N H C O C H_{3} \\ + C H_{3} C O O H \end{aligned}

Carbylamine reaction

Organic chemistry ka ek sabse yaadgaar test smell par tika hai. Thode se amine ko chloroform aur ethanolic KOH ke saath garam karo: agar bahut tez, ulti laane wali badboo aaye, to amine primary hai. Yeh smell ek isocyanide (ya carbylamine), $R - N \equiv C$ , se aati hai, jismein dono N–H hydrogens hatne ke baad chloroform ka carbon nitrogen par aa jaata hai. Secondary aur tertiary amines ke paas yeh do hydrogens nahi hote aur koi smell nahi aati. Aliphatic aur aromatic dono primary amines yeh test dete hain, aur test ke do naam hain: carbylamine reaction aur isocyanide test.

carbylamine reaction

\begin{aligned} R - N H_{2} + C H C l_{3} \\ + 3 K O H \overset{h e a t}{\to} R - N C \\ + 3 K C l + 3 H_{2} O \end{aligned}

(9.8)

Reaction with nitrous acid

Nitrous acid, $H N O_{2}$ , unstable hota hai, isliye ise zaroorat ke waqt flask mein hi sodium nitrite aur HCl jaise mineral acid se banaya jaata hai. Amines ki teeno classes isse alag-alag react karti hain.

Primary aliphatic amines alkyldiazonium salts banate hain jo turant toot jaate hain. Nitrogen gas bubbles ban kar nikalti hai aur alkyl hissa zyadatar alcohol ban jaata hai, saath mein alkenes aur doosre by-products bhi. Har $N H_{2}$ group ek $N_{2}$ molecule deta hai, isliye gas naapne se amino groups naape ja sakte hain; ek samay amino acids aur proteins ka estimate isi method se kiya jaata tha.
Primary aromatic amines 273–278 K par arenediazonium salts dete hain jo tab tak solution mein tike rehte hain jab tak unhe thanda rakha jaaye (Section 9.7).
Secondary amines N-nitrosamines, $R_{2} N - N = O$ , dete hain, jo aam taur par yellow oils hote hain, jabki tertiary aliphatic amines bas apne salts ke roop mein acid mein ghul jaate hain.

primary aliphatic amine aur nitrous acid

\begin{aligned} R - N H_{2} \overset{N a N O_{2} + H C l}{\to} [R - N_{2}^{+} C l^{-}] \\ \overset{H_{2} O}{\to} R - O H + N_{2} ↑ + H C l \end{aligned}

(9.9)

Reaction with arylsulphonyl chloride

Primary aur secondary amines ko sulphonylate bhi kiya ja sakta hai. Reagent benzenesulphonyl chloride, $C_{6} H_{5} S O_{2} C l$ , hai, jise Hinsberg's reagent kehte hain; yeh acid chloride ki tarah behave karta hai aur in amines ko sulphonamides mein badal deta hai. Test ki asli value is baat mein hai ki yeh sulphonamides phir alkali mein kya karte hain (Fig. 9.11).

Primary amine: product, jaise butan-1-amine se $C_{6} H_{5} S O_{2} N H C_{4} H_{9}$ , ke nitrogen par ab bhi ek H hai. Strongly electron-withdrawing $S O_{2}$ group us H ko acidic bana deta hai, isliye sulphonamide aqueous KOH mein apne potassium salt ke roop mein ghul jaata hai. Acid daalne par yeh wapas solid ban kar nikal aata hai.
Secondary amine: product, jaise N-methylbutan-1-amine se $C_{6} H_{5} S O_{2} N (C H_{3}) C_{4} H_{9}$ , ke nitrogen par koi H nahi hai. Alkali ko dene ke liye kuch nahi hone se yeh KOH mein bina ghule rehta hai.
Tertiary amine: koi N–H nahi, koi reaction nahi. Amine waisa hi rehta hai aur sirf acid daalne par apne salt ke roop mein ghulta hai.

primary amine aur Hinsberg's reagent

\begin{aligned} C_{6} H_{5} S O_{2} C l + H_{2} N C_{4} H_{9} \\ \overset{}{\to} C_{6} H_{5} S O_{2} N H C_{4} H_{9} \\ + H C l \end{aligned}

2/31°: sulphonamide ghul jaata hai. 2°: solid. 3°: koi reaction nahi.

Fig. 9.11Acid daalne se pehle aur baad KOH mein solubility teeno classes ko alag kar deti hai.NCERT p. 272

Teen test tubes par 1° amine, 2° amine aur 3° amine likha hai. Step 1: har ek mein benzenesulphonyl chloride daal kar tubes hilayi jaati hain; har liquid amine oily layer ki tarah upar tairta hai. Step 2: aqueous KOH ke saath hilane ke baad 1° tube clear hai, 2° tube mein neeche solid hai aur 3° tube mein ab bhi oily layer hai. Step 3: dilute hydrochloric acid daalne ke baad 1° tube mein solid banta hai, 2° tube mein solid bana rehta hai aur 3° tube clear ho jaati hai kyunki amine apne salt ke roop mein ghul jaata hai.

Mixture par yahi reactions chalane se teeno amines alag bhi ho jaate hain: bina react hue tertiary amine ko distil ya extract karke nikala ja sakta hai, aur dono sulphonamides ko alkali mein unki solubility se alag karke phir hydrolyse karke wapas amines banaya jaata hai. Aajkal ki laboratories p-toluenesulphonyl chloride, $C H_{3} C_{6} H_{4} S O_{2} C l$ , prefer karti hain; yeh same tarah react karta hai aur iske sulphonamides behtar crystallise hote hain.

Example

Kaun sa amine kaun sa hai?

Bottles X, Y aur Z mein kisi order mein pentan-1-amine, N-ethylpropan-1-amine aur N,N-dimethylpropan-1-amine (sab $C_{5} H_{13} N$ ) hain. Chloroform aur ethanolic KOH ke saath garam karne par X badboo deta hai; Y aur Z nahi dete. Benzenesulphonyl chloride aur KOH ke saath Y aisa solid deta hai jo ghulta nahi, aur Z mein koi badlaav nahi dikhta. X, Y aur Z ko pehchano.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Electrophilic substitution

Fig. 9.9 ko phir se dekho: $- N H_{2}$ ka lone pair ring mein electron density deta hai, zyadatar ortho aur para carbons par. Isliye amino group ek powerful activating group aur ortho/para director hai. Aniline electrophiles se benzene se itna zyada tez react karta hai ki asli problem ek substitution ke baad rokna hai.

Bromination. Room temperature par aniline mein bromine water daalo to turant ek white solid dikhta hai: 2,4,6-tribromoaniline. Teeno activated positions par bromine lag jaata hai, aur kisi catalyst ki zaroorat nahi padti.

2,4,6-tribromoaniline

\begin{aligned} C_{6} H_{5} N H_{2} + 3 B r_{2} \\ \overset{}{\to} C_{6} H_{2} B r_{3} N H_{2} + 3 H B r \end{aligned}

Ek bromine par rokne ke liye chemists amino group ko ethanoic anhydride se acetylate karke protect karte hain (acetanilide). Acetanilide mein nitrogen ka lone pair bagal wale C=O group ke saath bhi share hota hai (pair ko N=C double bond aur oxygen par negative charge ki tarah draw kiya ja sakta hai), isliye yeh ring ko kam deta hai. Ring ab bhi activated hai, lekin halke se. Ab ethanoic acid mein bromine sirf ek bromine jodta hai, zyadatar para par, kyunki bulky $- N H C O C H_{3}$ group ortho positions ko dhak leta hai. Phir acid ya alkali se hydrolysis acetyl group ko hata deta hai aur 4-bromoaniline milta hai (Fig. 9.12).

3/4Ek Br, bulky amide group ke para par.

Fig. 9.12Cap lagao, ek baar substitute karo, cap hatao: amino group ko kaabu karne ka standard tareeka.NCERT p. 272

Ek benzene ring poore samay beech mein rehti hai. Step 1: bromine water ke saath aniline teen bromine atoms le leta hai, dono ortho positions aur para position par, jis par too reactive likha hai. Step 2: ethanoic anhydride NH2 group par NHCOCH3 ka cap laga deta hai, jis par tamer ring likha hai. Step 3: ethanoic acid mein bromine ke saath sirf ek bromine, para position par, judta hai. Step 4: hydrolysis cap hata deta hai aur 4-bromoaniline milta hai.

Nitration. Mixed acid aniline par do unwanted tareekon se attack karta hai. Nitric acid uske kaafi hisse ko oxidise karke tar jaise by-products bana deta hai, aur strongly acidic mixture mein kaafi aniline protonate hokar anilinium ion ban jaata hai. $- N H_{3}^{+}$ group ke paas koi lone pair nahi hota aur yeh electrons kheenchta hai, isliye yeh deactivating aur meta-directing hai. Isliye nitro products mein lagbhag utna hi meta isomer hota hai jitna para, aur ortho bahut kam. Amino group ko pehle acetanilide ke roop mein protect karne se dono problems hal ho jaati hain: phir lagbhag 288 K par $H N O_{3}$ aur $H_{2} S O_{4}$ se nitration zyadatar 4-nitroacetanilide deta hai, jise hydrolyse karke 4-nitroaniline banaya jaata hai.

Sulphonation. Concentrated sulphuric acid pehle anilinium hydrogensulphate salt banata hai. Ise 453–473 K tak garam karne par 4-aminobenzenesulphonic acid milta hai, jise sulphanilic acid kehte hain. Is molecule par ek acidic $- S O_{3} H$ group aur ek basic $- N H_{2}$ group dono hote hain, isliye yeh zyadatar zwitterion, $H_{3} N^{+} C_{6} H_{4} S O_{3}^{-}$ , ke roop mein rehta hai.

sulphanilic acid

\begin{aligned} C_{6} H_{5} N H_{2} + H_{2} S O_{4} \\ \overset{}{\to} C_{6} H_{5} N H_{3}^{+} H S O_{4}^{-} \\ \overset{453-473 K}{\to} H_{2} N C_{6} H_{4} S O_{3} H \\ + H_{2} O \end{aligned}

Friedel–Crafts reactions nahi hote. Aluminium chloride ke saath ring alkylation ya acylation aniline ke liye fail ho jaata hai. Lewis acid $A l C l_{3}$ nitrogen ka lone pair pakad leta hai, jisse nitrogen par positive charge aa jaata hai, aur strongly activating group strongly deactivating ban jaata hai.

Practice · 5 questions

Practice: amines ke reactions: answers ke saath

1
Quick answerMediumNCERT Intext Question 9.4
Har set ko jaisa kaha gaya hai waise lagao.
1. (a)
  Paani mein badhti basic strength
  aniline
  4-methylaniline
  4-nitroaniline
  cyclohexanamine
  Answer dekho
  Answer
  4-nitroaniline
  aniline
  4-methylaniline
  cyclohexanamine
  p $K_{b}$ lagbhag 13.0, 9.38, 8.92 aur 3.36. Ring delocalisation anilines ko weak banata hai, nitro group aur weak kar deta hai, methyl group thoda strong karta hai, aur cyclohexanamine ke nitrogen par koi ring hai hi nahi.
2. (b)
  Paani mein badhti basic strength
  $(C H_{3})_{3} N$
  $N H_{3}$
  $C H_{3} N H_{2}$
  $(C H_{3})_{2} N H$
  Answer dekho
  Answer
  $N H_{3}$
  $(C H_{3})_{3} N$
  $C H_{3} N H_{2}$
  $(C H_{3})_{2} N H$
  p $K_{b}$ 4.75, 4.22, 3.38 aur 3.27: paani mein secondary amine sabse strong hai aur tertiary amine primary se neeche aa jaata hai kyunki uska cation theek se solvate nahi hota.
2
Short answerEasyNCERT Intext Question 9.5
Har acid–base reaction poora karo aur product ka naam batao.
1. $H B r$ ke saath $(C H_{3})_{2} N H$
  Solution dikhao
  $(C H_{3})_{2} N H + H B r \overset{}{\to} (C H_{3})_{2} N H_{2}^{+} B r^{-}$
  Dimethylammonium bromide (N-methylmethanaminium bromide).
2. $H_{2} S O_{4}$ ke saath $C_{6} H_{5} N H_{2}$
  Solution dikhao
  $\begin{aligned} C_{6} H_{5} N H_{2} + H_{2} S O_{4} \\ \overset{}{\to} C_{6} H_{5} N H_{3}^{+} H S O_{4}^{-} \end{aligned}$
  Anilinium hydrogensulphate.
Short answerMediumNCERT Intext Question 9.6
Benzylamine ko excess iodoethane ke saath garam kiya jaata hai, aur acid ko neutralise karne ke liye sodium carbonate maujood hai. Overall equation likho aur final product ka naam batao.
Solution dikhao
$\begin{aligned} C_{6} H_{5} C H_{2} N H_{2} + 3 C_{2} H_{5} I \\ + N a_{2} C O_{3} \\ \overset{}{\to} C_{6} H_{5} C H_{2} N (C_{2} H_{5})_{3}^{+} I^{-} \\ + 2 N a I + C O_{2} + H_{2} O \end{aligned}$
Do ethyl groups dono N–H hydrogens ko replace karte hain, aur phir tertiary amine teesra ethyl group le leta hai. Product N-benzyl-N,N-diethylethanaminium iodide (benzyltriethylammonium iodide) hai. Sodium carbonate pehle do steps mein bane do moles HI ko neutralise karta hai.
Short answerEasyNCERT Intext Question 9.7
Pyridine mein 4-methylaniline ke ethanoyl chloride ke saath reaction ki equation likho, aur product ka naam batao.
Solution dikhao
$\begin{aligned} C H_{3} C_{6} H_{4} N H_{2} + C H_{3} C O C l \\ \overset{}{\to} C H_{3} C_{6} H_{4} N H C O C H_{3} \\ + H C l \end{aligned}$
Product N-(4-methylphenyl)ethanamide (4-methylacetanilide) hai; pyridine HCl ko le leta hai.
5
Short answerMediumNCERT Intext Question 9.8
Paanch aromatic amines ka formula $C_{7} H_{9} N$ hai.
1. (a)
  Unke structures aur names likho.
  Solution dikhao
  benzylamine (phenylmethanamine), $C_{6} H_{5}$ – $C H_{2} N H_{2}$
  N-methylaniline, $C_{6} H_{5}$ – $N H C H_{3}$
  2-, 3- aur 4-methylaniline (o-, m- aur p-toluidine), $C H_{3} C_{6} H_{4}$ – $N H_{2}$
2. 273 K par $N a N O_{2}$ aur HCl ke saath kaun sa turant nitrogen gas chhodta hai, aur kyun?
  Solution dikhao
  Benzylamine. Iska $N H_{2}$ ring par nahi, $C H_{2}$ group par hai, isliye yeh primary aliphatic amine ki tarah behave karta hai: iske diazonium ion ko stable karne ke liye koi ring nahi hai aur yeh turant $N_{2}$ kho deta hai. Teeno methylanilines aise diazonium salts banate hain jo thande solution mein tike rehte hain, aur N-methylaniline, jo secondary amine hai, N-nitrosamine deta hai.

Diazonium Salts

Diazonium salt mein $- N_{2}^{+}$ wala aryl group ek anion ke saath hota hai: $A r N_{2}^{+} X^{-}$ , jahan $X^{-}$ $C l^{-}$ , $B r^{-}$ , $H S O_{4}^{-}$ , $B F_{4}^{-}$ waghairah ho sakta hai. $- N_{2}^{+}$ group, yaani $- \overset{+}{N} \equiv N$ , diazonium group hai. Naam arene, word diazonium aur anion se bante hain: $C_{6} H_{5} N_{2}^{+} C l^{-}$ benzenediazonium chloride hai, aur 4-nitroaniline aur fluoroboric acid se bana salt 4-nitrobenzenediazonium tetrafluoroborate hai.

Primary aliphatic amines bhi diazonium ions banate hain, lekin yeh bante hi nitrogen kho dete hain (Reaction with nitrous acid). Arenediazonium ions kaafi der tikte hain, bas unhe thanda rakhna hota hai. Agle chaar sections dikhate hain ki yeh kaise bante hain, kaise hote hain, kaise react karte hain aur itne useful kyun hain.

9.7 Method of Preparation of Diazonium Salts

Diazonium salt banana jaldi ho jaata hai, lekin ise thande mein hi karna padta hai. Amine ko excess dilute hydrochloric acid mein ghol kar barf mein 273–278 K (0–5 °C) tak thanda kiya jaata hai; phir sodium nitrite solution daala jaata hai, aur acid mein isse bana nitrous acid $- N H_{2}$ ko $- N_{2}^{+}$ mein badal deta hai. Is step ko diazotisation kehte hain. Product ko koi bottle mein nahi rakhta: thanda solution seedha agle reaction mein jaata hai (neeche lab steps).

aniline ka diazotisation

\begin{aligned} C_{6} H_{5} N H_{2} + N a N O_{2} \\ + 2 H C l \overset{273-278 K}{\to} \\ C_{6} H_{5} N_{2}^{+} C l^{-} + N a C l \\ + 2 H_{2} O \end{aligned}

(9.10)

Itna thanda kyun? Arenediazonium ion isliye stable hai kyunki uska positive charge resonance se ring ke saath share hota hai, lekin bas thoda sa (Fig. 9.13). Lagbhag 278 K se upar paani is par attack karta hai, nitrogen gas nikal jaati hai aur phenol banta hai. Alkyldiazonium ion ke paas charge share karne ke liye koi ring nahi hoti, isliye woh thande mein bhi turant $N_{2}$ kho deta hai.

2/3+ charge ring par phail jaata hai; ion 273–278 K par tikta hai.

Fig. 9.13Share hua charge arenediazonium ion ko thande mein waqt deta hai, lekin room temperature par nahi.NCERT p. 274

Step 1 mein ek alkyldiazonium ion, R–N2+, 273 K par bhi turant R+ aur nitrogen gas mein toot jaata hai, bubbles ke saath. Step 2 mein ek benzene ring par N+≡N hai, aur dono ortho carbons aur para carbon par chhote plus signs dikhate hain ki charge ring ke saath share hai, saath mein 273–278 K wala cold label hai. Step 3 mein wahi ring 278 K se upar garam karne ke baad hai: group ab OH hai, nitrogen ke bubbles upar uth rahe hain, aur label phenol plus N2 plus HCl padhta hai.

Step by step

Lab mein aniline ka diazotisation

Step 1 / 5

Gholo

Aniline ko lagbhag teen equivalents dilute hydrochloric acid mein chala kar milao; yeh anilinium chloride ban kar ghul jaata hai.

1 / 5

Quick check

3 mein se 1

Diazotisation 273–278 K par kyun kiya jaata hai?

taaki sodium nitrite ghul jaayereaction ko tez karne ke liyeaniline ko solid rakhne ke liyetaaki diazonium salt

N_{2}

kho kar phenol na bana de

Answer dekho

Answer (D) taaki diazonium salt $N_{2}$ kho kar phenol na bana de

Arenediazonium ion bas thoda sa stable hota hai; garam paani is par attack karta hai, nitrogen nikal jaati hai aur phenol banta hai. Thanda karne se reaction tez nahi, dheema hota hai.

9.8 Physical Properties

Solid benzenediazonium chloride tumhe shayad hi kabhi dikhe. Yeh colourless crystals banata hai, lekin chemists inhe lagbhag kabhi isolate nahi karte, kyunki sookha salt zor se decompose ho sakta hai. Paani mein yeh aasaani se ghul jaata hai aur jab tak solution thanda rahe tab tak tika rehta hai; garam karo to salt paani se react karne lagta hai. Anion badalne se sab badal jaata hai: benzenediazonium tetrafluoroborate, $C_{6} H_{5} N_{2}^{+} B F_{4}^{-}$ , paani mein nahi ghulta aur itna stable hai ki ise filter karke room temperature par rakha ja sakta hai.

9.9 Chemical Reactions

Diazonium salts mote taur par do tarah se react karte hain. Zyadatar reactions mein $- N_{2}^{+}$ group nitrogen gas ban kar nikal jaata hai aur ring par uski jagah koi aur group aa jaata hai. Coupling reactions mein dono nitrogen atoms rehte hain, do aromatic rings ke beech pul ki tarah.

Reactions jinmein nitrogen replace hota hai

Diazonium group ek behtareen leaving group hai. Yeh $N_{2}$ ban kar nikalta hai, jo sabse stable molecules mein se ek hai, aur gas solution se bubbles ban kar nikal jaati hai isliye reaction ulta nahi chal sakta. Isi wajah se itne alag-alag groups iski jagah le sakte hain (Table 9.6).

Table 9.6

A r N_{2}^{+}

ke diazonium group ko replace karna

Reagent	Product (note)
$C u C l$ / $H C l$	$A r C l$ (Sandmeyer)
$C u B r$ / $H B r$	$A r B r$ (Sandmeyer)
$C u C N$ / $K C N$	$A r C N$ (Sandmeyer)
HCl ya HBr ke saath Cu powder	$A r C l$ ya $A r B r$ (Gattermann, kam yield)
$K I$ solution	$A r I$ (catalyst ki zaroorat nahi)
$H B F_{4}$ , phir garam karo	$A r F$ ( $A r N_{2}^{+} B F_{4}^{-}$ ke zariye)
$H_{3} P O_{2}$ aur paani, ya ethanol	$A r H$ (group hat jaata hai)
paani, garam	$A r O H$ (ek phenol)
$H B F_{4}$ , phir $N a N O_{2}$ aur Cu, garam karo	$A r N O_{2}$ (ek nitro compound)

Chloride, bromide aur cyanide. Yeh teeno matching copper(I) salt ke saath aate hain: CuCl, CuBr ya CuCN. Copper(I) hi in Sandmeyer reactions ko kaam karwata hai; akele chloride ya cyanide ions kuch khaas nahi karte. Ek puraana variant, Gattermann reaction, iski jagah HCl ya HBr ke saath copper powder use karta hai, lekin uski yields aam taur par kam hoti hain.

cyanide ke saath Sandmeyer reaction

\begin{aligned} A r N_{2}^{+} C l^{-} + C u C N \\ \overset{}{\to} A r C N + N_{2} ↑ + C u C l \end{aligned}

Iodide. Iodide ko copper ki bilkul zaroorat nahi. Diazonium salt mein potassium iodide solution mila kar chalana hi aryl iodide dene ke liye kaafi hai, jo aisa compound hai jise direct iodination se banana mushkil hai.

aryl iodide

\begin{aligned} A r N_{2}^{+} C l^{-} + K I \\ \overset{}{\to} A r I + K C l + N_{2} ↑ \end{aligned}

(9.11)

Fluoride. Fluorine do stages mein lagta hai. Pehle, fluoroboric acid anion ko badal deta hai, aur insoluble arenediazonium tetrafluoroborate solid ban kar nikal aata hai. Phir sookhe solid ko garam kiya jaata hai, aur yeh aryl fluoride, boron trifluoride aur nitrogen mein toot jaata hai.

aryl fluoride

\begin{aligned} A r N_{2}^{+} C l^{-} + H B F_{4} \\ \overset{}{\to} A r N_{2}^{+} B F_{4}^{-} + H C l \\ A r N_{2}^{+} B F_{4}^{-} \overset{h e a t}{\to} \\ A r F + B F_{3} + N_{2} ↑ \end{aligned}

Hydrogen. Group ko poori tarah hataane ke liye mild reducing agent use karo. Phosphinic (hypophosphorous) acid, $H_{3} P O_{2}$ , phosphorous acid ban jaata hai aur ethanol ethanal ban jaata hai; dono hi tarah ring ko wahan ek hydrogen mil jaata hai jahan nitrogen tha. Ring se amino group hataana bekaar lagta hai, lekin synthesis ki yeh sabse useful tricks mein se ek hai (Section 9.10).

group ko hataana

\begin{aligned} A r N_{2}^{+} C l^{-} + H_{3} P O_{2} \\ + H_{2} O \overset{}{\to} A r H + N_{2} \\ + H_{3} P O_{3} + H C l \end{aligned}

Hydroxyl. Phenol ke liye kisi extra reagent ki zaroorat nahi: solution lagbhag 283 K se zyada garam hote hi paani khud diazonium group ko replace kar deta hai.

phenol

\begin{aligned} A r N_{2}^{+} C l^{-} + H_{2} O \\ \overset{}{\to} A r O H + N_{2} ↑ + H C l \end{aligned}

Nitro. Nitro group ko wapas wahan bhi lagaya ja sakta hai jahan se amine aaya tha: tetrafluoroborate salt ko aqueous sodium nitrite aur copper powder ke saath garam karne par nitroarene milta hai.

Coupling reactions

Coupling reaction mein diazonium ion apna nitrogen rakhta hai aur electrophile ki tarah kaam karta hai. Yeh weak electrophile hai, isliye sirf strongly activated rings, yaani phenols aur arylamines ki rings, par attack karta hai, lagbhag hamesha para position par (ya para bhari ho to ortho par). Dono rings aakhir mein ek azo group, –N=N–, se jud jaati hain. Isse bana lamba conjugated system visible light absorb karta hai, isliye azo compounds chamakdaar rang ke hote hain (Fig. 9.14).

3/3 $H^{+}$ nikalne se ring wapas ban jaati hai: orange 4-hydroxyazobenzene.

Fig. 9.14Diazonium ion electrophile hai; activated ring electrons deti hai.NCERT p. 276

Step 1: right mein N+≡N wali ek benzene ring thodi door rakhi phenoxide ring ki taraf mooh kiye hai, jiske bilkul right mein O− hai; phenoxide ke para carbon par ek H hai, aur us ring se end wale nitrogen tak ek curved arrow jaata hai. Step 2: phenoxide ring paas aa gayi hai aur ek naya bond end wale nitrogen ko uske para carbon se jodta hai, jis par ab nitrogen aur H dono hain. Step 3: H, H+ ban kar nikal gaya hai, nitrogen–nitrogen bond ab double bond hai, O− ab OH ban gaya hai, aur product par 4-hydroxyazobenzene likha hai, ek orange azo dye jiska –N=N– group dono rings ko jodta hai.

Mildly alkaline solution mein phenol ke saath, jahan woh bahut reactive phenoxide ion ke roop mein hota hai, benzenediazonium chloride orange 4-hydroxyazobenzene deta hai. Mildly acidic solution mein aniline ke saath yeh yellow 4-aminoazobenzene deta hai. Dono phenol ya aniline ring par electrophilic substitutions hain. Primary aromatic amine ka classic laboratory test 2-naphthol use karta hai: amine ko 273–278 K par diazotise karo, thande solution ko NaOH mein 2-naphthol mein daalo, aur turant orange-red dye precipitate ho jaata hai ((9.12)).

2-naphthol dye test

\begin{aligned} C_{6} H_{5} N_{2}^{+} + C_{10} H_{7} O H \\ + O H^{-} \overset{}{\to} \\ C_{6} H_{5} N = N C_{10} H_{6} O H + H_{2} O \end{aligned}

(9.12)

Chalo explore karein

Tumhari chemistry lab ka ek azo dye

Tumhe chahiye

methyl orange indicator solution
dilute hydrochloric acid
dilute sodium hydroxide solution
distilled water
teen test tubes
safety goggles

Pehle guess karo

Methyl orange ek azo dye hai. Jab iski kuch boondein acid, paani aur alkali mein jaati hain to tum kya expect karte ho?

teeno mein same orange colouracid mein red, paani mein yellow-orange aur alkali mein yellowacid mein colourless aur alkali mein orange

Teen test tubes ko aadha dilute HCl, distilled water aur dilute NaOH se bharo.
Har ek mein methyl orange ki do boondein daalo aur hilao.
Har colour note karo, phir alkali wali tube mein boond-boond acid daalo aur dekho.

Table 9.7Methyl orange ke colours

Tube	Colour
Acid
Paani
Alkali

Tumhari entries apne aap save hoti hain.

Methyl orange acid mein red aur alkali mein yellow hota hai; yeh lagbhag pH 3.1 aur 4.4 ke beech badalta hai, isliye distilled water (pH lagbhag 5 se 7) yellow-orange dikhata hai. Yeh dye diazotised sulphanilic acid ko N,N-dimethylaniline ke saath couple karke banta hai, yaani is chapter ke do reactions ek hi molecule mein mil jaate hain. Azo group ko protonate karne se conjugated system badalta hai, jisse absorb hone wali light aur tumhe dikhne wala colour badal jaata hai.

InteractiveDiazonium reaction map

Upar ek card 273 se 278 K par rakha benzenediazonium chloride dikhata hai. Uske neeche gyarah reagent buttons hain: copper(I) chloride, bromide aur cyanide ke saath Sandmeyer reactions, Gattermann reaction, potassium iodide, fluoroboric acid phir garam karna, phosphinic acid, garam paani, fluoroborate phir copper ke saath sodium nitrite, aur alkali mein phenol ya acid mein aniline ke saath coupling. Reagent chunne par benzene ring par product ban jaata hai aur uska naam, balanced equation, reaction ka naam, nitrogen gas nikalti hai ya nahi, aur route kyun useful hai, yeh sab dikhta hai. Play dabane par nitrogen ke bubbles nikalte ya azo link banta dikhta hai. Target practice mode ek product ka naam deta hai, jaise fluorobenzene ya 4-hydroxyazobenzene, aur learner se sahi reagent chunne ko kehta hai, har choice ke baad feedback ke saath.

Fig. 9.15Ek reagent chuno aur dekho ki diazonium group ki jagah kya aata hai ya woh bana rehta hai, ya target practice par switch karo.

Timeline

Amines, dyes aur named reactions

1856
Perkin's mauve
Atharah saal ke William Perkin ne quinine banane ki koshish mein impure aniline ko oxidise kiya aur mauveine paaya, jo pehla synthetic dye tha.
1858
Diazonium salts
Peter Griess ne aromatic amines ko nitrous acid se treat kiya aur azo dyes ka raasta khol diya.
1881
Hofmann's degradation
Pata chala ki amides ek carbon atom kho kar amines dete hain.
1884
Sandmeyer reaction
Traugott Sandmeyer phenylacetylene banane ki koshish kar rahe the, aur unhone paaya ki copper(I) chloride diazonium salt se chlorobenzene deta hai.
1887
Gabriel synthesis
Siegmund Gabriel ne pure primary amines banane ke liye phthalimide use kiya.
1890
Hinsberg test
Oscar Hinsberg ne amines ki classes ko alag pehchanne ke liye benzenesulphonyl chloride use kiya.
1932
Prontosil
Ek azo dye pehli sulpha drug bani.

9.10 Importance of Diazonium Salts in Synthesis of Aromatic Compounds

Saare replacement reactions ko saath rakho to diazonium salt ek hub ban jaata hai. Ek aromatic amine se tum aryl fluoride, chloride, bromide, iodide, nitrile, phenol, nitro compound ya khaali arene tak pahunch sakte ho. Inmein se kai ko direct banana mushkil ya namumkin hai. Fluorine benzene ke saath explosively react karta hai, isliye aryl fluorides direct fluorination se nahi ban sakte. Sirf iodine se iodination dheema aur reversible hai aur ise oxidising agent chahiye. Aur chlorobenzene cyanide ion ko bhaav hi nahi deta, isliye benzonitrile tak ka akela aasaan raasta diazonium salt se hokar jaata hai.

Doosra bada use hai steering. Amino group, ya uska tamer acetyl form, ek strong ortho/para director hai. Tum ise ring par laga sakte ho, isse naye substituent ko sahi jagah guide kar sakte ho, aur phir diazonium salt aur $H_{3} P O_{2}$ se ise hata sakte ho. Isse aisa substitution pattern milta hai jo koi direct reaction nahi deta, jaisa Example 9.5 dikhata hai.

Example 9.5NCERT Example 9.5 · p. 277

3-bromotoluene banana

Toluene ka bromination 2- aur 4-bromotoluene deta hai, 3-isomer kabhi nahi, kyunki $- C H_{3}$ ortho/para direct karta hai. Dikhao ki 4-methylaniline se 3-bromotoluene kaise banaoge.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

6/6 $H_{3} P O_{2}$ $- N_{2}^{+}$ ki jagah H laga deta hai: 3-bromotoluene.

Fig. 9.16Amino group se bromine ko sahi jagah lagao, phir use hata do.NCERT Example 9.5 · p. 277

Ek benzene ring par poore samay neeche methyl group rehta hai. Step 1: upar NH2, yaani 4-methylaniline. Step 2: ethanoic anhydride NH2 par NHCOCH3 ka cap lagata hai. Step 3: bromine upar-right wale carbon par, cap ke bagal mein, judta hai. Step 4: hydrolysis cap ko wapas NH2 bana deta hai. Step 5: 273 se 278 K par sodium nitrite aur HCl se NH2, N2+ ban jaata hai. Step 6: phosphinic acid N2+ ko H se replace karta hai, aur 3-bromotoluene bachta hai jismein Br, CH3 ke meta par hai.

Quick check

3 mein se 1

Kaun sa compound benzene par direct substitution se theek se nahi banta, lekin benzenediazonium chloride se aasaani se ban jaata hai?

chlorobenzenenitrobenzenefluorobenzenebromobenzene

Answer dekho

Answer (C) fluorobenzene

Direct fluorination explosive aur uncontrollable hai, lekin benzenediazonium tetrafluoroborate ko garam karne par saaf-saaf fluorobenzene milta hai. Chlorobenzene, nitrobenzene aur bromobenzene sab direct bante hain.

Practice · 1 question

Practice: diazonium salts ka use: answers ke saath

1
Short answerHardNCERT Intext Question 9.9
Dikhao ki har conversion kaise karoge.
1. 4-Chloroaniline se 1-chloro-4-fluorobenzene
  Solution dikhao
  Diazotise karo ( $N a N O_{2}$ aur HCl, 273–278 K), phir $H B F_{4}$ daal kar 4-chlorobenzenediazonium tetrafluoroborate ko precipitate karo. Filter karo, dhyaan se sukhao aur garam karo: $N_{2}$ aur $B F_{3}$ nikal jaate hain aur diazonium group ki jagah fluorine aa jaata hai.
2. 4-Nitroaniline se 1,3-dibromo-5-nitrobenzene
  Solution dikhao
  Bromine water ke saath hilao. Strongly activating $- N H_{2}$ dono khaali ortho positions par Br laga deta hai (iski para position par pehle se $N O_{2}$ hai), aur 2,6-dibromo-4-nitroaniline milta hai. Strong acid mein 273–278 K par diazotise karo aur $H_{3} P O_{2}$ se treat karke $- N_{2}^{+}$ ko H se replace karo. Ab dono bromine atoms aur nitro group sab ek doosre ke meta par hain.

Recap

Summary aur key points

Amines mein ammonia ke ek, do ya teen hydrogens alkyl ya aryl groups se replace hote hain; nitrogen $s p^{3}$ aur pyramidal hota hai aur lone pair rakhta hai (Section 9.1).
Primary, secondary aur tertiary amines ke N par ek, do ya teen carbon groups hote hain; chaar groups se quaternary ammonium ion banta hai (Section 9.2).
IUPAC names alkanamines hote hain, jinmein nitrogen par chhote groups ke liye N-prefixes hote hain; aniline (benzenamine) parent arylamine hai (Section 9.3).
Amines nitro compounds, nitriles (ek carbon zyada) ya amides ko reduce karke, ammonolysis (mixture) se, Gabriel synthesis (pure 1°) se aur Hoffmann degradation (ek carbon kam) se bante hain (Section 9.4).
N–H ke zariye hydrogen bonding se boiling points 1° > 2° > 3° order mein girte hain aur chhote amines paani mein ghul paate hain; alcohols amines se zyada temperature par boil hote hain (Section 9.5).
Amines bases hain jo acids ke saath salts banate hain; kam p $K_{b}$ ka matlab zyada strong base (Basic character of amines).
Gas phase mein basicity +I effect follow karti hai, 3° > 2° > 1° > $N H_{3}$ ; paani mein solvation aur crowding order badal dete hain, aur methyl aur ethyl dono series mein secondary amine sabse strong hai (Structure aur basicity: alkanamines vs ammonia).
Arylamines kaafi weak bases hain kyunki lone pair ring mein delocalised hota hai; donor groups inki basicity badhate hain aur acceptor groups ghatate hain (Arylamines vs ammonia).
Amines quaternary salts tak alkylate hote hain, aur primary aur secondary amines acylate hote hain; carbylamine test 1° amines ko pakadta hai aur Hinsberg test 1°, 2° aur 3° amines ko alag karta hai (Section 9.6).
$- N H_{2}$ ring ko ortho/para attack ke liye strongly activate karta hai; acetylation ise kaabu mein laata hai, aur strong acid mein anilinium ion meta direct karta hai (Electrophilic substitution).
273–278 K par diazotisation $A r N H_{2}$ ko $A r N_{2}^{+} X^{-}$ mein badalta hai, jo resonance se stable hota hai lekin garam ya sookha hone par unstable (Section 9.7).
$- N_{2}^{+}$ ko replace karne se aryl chlorides, bromides aur nitriles (Sandmeyer), iodides, fluorides, arenes, phenols aur nitro compounds milte hain; phenols ya arylamines ke saath coupling se azo dyes milte hain (Section 9.9, Section 9.10).

Practice · 18 questions

Practice questions answers ke saath (NCERT “Exercises”)

1
Short answerEasyNCERT Exercises Q9.1
Har amine ka IUPAC naam batao aur bolo ki woh primary, secondary ya tertiary hai.
1. $(C H_{3})_{2} C H$ – $C H_{2} C H_{2} N H_{2}$
  Solution dikhao
  3-Methylbutan-1-amine; primary.
2. $C H_{3} C H_{2} C H_{2}$ – $N H C H_{2} C H_{3}$
  Solution dikhao
  N-Ethylpropan-1-amine; secondary.
3. $(C H_{3})_{2} N$ – $(C H_{2})_{3} C H_{3}$
  Solution dikhao
  N,N-Dimethylbutan-1-amine; tertiary.
4. $C_{6} H_{5}$ – $C H_{2} C H_{2} N H_{2}$
  Solution dikhao
  2-Phenylethan-1-amine (2-phenylethanamine); primary.
5. p- $C H_{3} O C_{6} H_{4} N H_{2}$
  Solution dikhao
  4-Methoxyaniline (4-methoxybenzenamine); primary.
6. $C_{6} H_{5} N H$ – $C H (C H_{3})_{2}$
  Solution dikhao
  N-(Propan-2-yl)aniline, jise N-isopropylaniline bhi kehte hain; secondary.
7. o- $C l C_{6} H_{4} N H_{2}$
  Solution dikhao
  2-Chloroaniline (2-chlorobenzenamine); primary.
2
Short answerMediumNCERT Exercises Q9.2
Har pair ke liye ek aisa chemical test batao jiska result dono compounds ke liye alag ho, aur batao ki tum kya observe karoge.
1. Propan-1-amine aur N-methylethanamine
  Solution dikhao
  Carbylamine test: har ek ko $C H C l_{3}$ aur ethanolic KOH ke saath garam karo. Propan-1-amine (1°) propyl isocyanide ki badboo deta hai; N-methylethanamine (2°) nahi deta.
2. N-Methylaniline aur N,N-dimethylaniline
  Solution dikhao
  Hinsberg test: har ek ko $C_{6} H_{5} S O_{2} C l$ aur KOH ke saath hilao. N-Methylaniline (2°) aisa sulphonamide deta hai jo solid ke roop mein bina ghule rehta hai; N,N-dimethylaniline (3°) react nahi karta.
3. Cyclohexanamine aur aniline
  Solution dikhao
  Azo-dye test: har ek ko 273–278 K par $N a N O_{2}$ aur HCl se treat karo, phir solution ko NaOH mein 2-naphthol mein daalo. Aniline orange-red precipitate deta hai; cyclohexanamine $N_{2}$ chhodta hai aur koi dye nahi deta.
4. 4-Methylaniline aur benzylamine
  Solution dikhao
  Wahi azo-dye test: sirf 4-methylaniline, jo arylamine hai, coloured dye deta hai. Bromine water bhi kaam karta hai: yeh sirf 4-methylaniline ke saath 2,6-dibromo-4-methylaniline ka white precipitate deta hai.
5. Aniline aur N-ethylaniline
  Solution dikhao
  Carbylamine test: sirf aniline (1°) isocyanide ki smell deta hai; N-ethylaniline (2°) nahi deta.
3
Long answerMediumNCERT Exercises Q9.3
Har observation ki wajah batao.
1. 4-Methylaniline aniline se thoda zyada strong base hai, lekin 4-nitroaniline kaafi zyada weak hai.
  Solution dikhao
  Teeno mein lone pair kuch had tak ring mein delocalised hai. Para methyl group electrons release karta hai, density ko nitrogen ki taraf wapas dhakelta hai aur cation ko stable karta hai, isliye 4-methylaniline thoda zyada strong hai (p $K_{b}$ 8.92, 9.38 ke muqable). Para nitro group resonance aur induction se electrons zor se kheenchta hai, lone pair ko ring mein aur andar kheench leta hai, isliye 4-nitroaniline lagbhag 3.6 p $K_{b}$ units weak hai (p $K_{b}$ lagbhag 13.0).
2. Propan-1-amine paani mein har proportion mein mix ho jaata hai, lekin N,N-dimethylaniline mushkil se ghulta hai.
  Solution dikhao
  Propan-1-amine chhota hai: iska lone pair paani se hydrogen bonds leta hai, iske do N–H hydrogens hydrogen bonds dete hain, aur iska hydrocarbon hissa chhota hai. N,N-Dimethylaniline mein koi N–H hydrogen nahi hai, iska lone pair kuch had tak ring se bandha hai, aur iska eight-carbon hydrocarbon hissa paani ko repel karta hai.
3. Iron(II) sulphate solution mein ethanamine solution daalne par dirty green precipitate banta hai.
  Solution dikhao
  Ethanamine ek base hai aur paani mein hydroxide ions banata hai:
  $\begin{aligned} C_{2} H_{5} N H_{2} + H_{2} O \\ \overset{}{⇌} C_{2} H_{5} N H_{3}^{+} + O H^{-} \end{aligned}$
  Hydroxide ions dirty green iron(II) hydroxide ko precipitate karte hain:
  $F e^{2 +} + 2 O H^{-} \overset{}{\to} F e (O H)_{2} ↓$
4. Concentrated acid mein N,N-dimethylaniline ka nitration zyadatar meta nitro compound deta hai, jabki $- N (C H_{3})_{2}$ ortho/para direct karta hai.
  Solution dikhao
  Concentrated $H_{2} S O_{4}$ mein zyadatar amine protonate hokar $C_{6} H_{5}$ – $N H (C H_{3})_{2}^{+}$ ban jaata hai. Positively charged group ke paas koi free lone pair nahi hota aur yeh electrons kheenchta hai, isliye ring ko deactivate karta hai aur meta direct karta hai; nitration isi cation par hota hai.
5. Anisole, $C_{6} H_{5} O C H_{3}$ , Friedel–Crafts methylation karta hai, lekin aniline nahi karta.
  Solution dikhao
  Aniline ka nitrogen anisole ke oxygen se kaafi zyada strong Lewis base hai. Yeh catalyst $A l C l_{3}$ se bond bana leta hai, jisse nitrogen par positive charge aata hai jo ring ko strongly deactivate karta hai, aur yeh catalyst ko khatam kar deta hai.
6. Benzenediazonium chloride barf jaise thande paani mein kuch der tikta hai, lekin ethanediazonium chloride 273 K par bhi rakha nahi ja sakta.
  Solution dikhao
  Arenediazonium ion mein positive charge ring par delocalised hota hai, aur $N_{2}$ khone par bahut unstable aryl cation bachega. Ethanediazonium ion mein aisa koi stabilisation nahi hota aur yeh turant $N_{2}$ kho deta hai.
7. Gabriel synthesis ammonolysis se zyada pure primary amine deta hai.
  Solution dikhao
  Ammonolysis mein bana primary amine khud nucleophile hai aur aage secondary, tertiary aur quaternary products de deta hai. Gabriel synthesis mein N-alkylphthalimide ke nitrogen par koi H nahi bacha hota aur uska lone pair do C=O groups ke saath share hota hai, isliye use dobara alkylate nahi kiya ja sakta; phir hydrolysis sirf primary amine ko free karta hai.
4
Quick answerMediumNCERT Exercises Q9.4
Har set ko jaisa kaha gaya hai waise lagao.
1. (a)
  Ghatta p $K_{b}$ (sabse weak base pehle)
  $(C H_{3})_{2} N H$
  $C H_{3} N H_{2}$
  $C_{6} H_{5} N H_{2}$
  $C_{6} H_{5} N (C H_{3})_{2}$
  Answer dekho
  Answer
  $C_{6} H_{5} N H_{2}$
  $C_{6} H_{5} N (C H_{3})_{2}$
  $C H_{3} N H_{2}$
  $(C H_{3})_{2} N H$
  p $K_{b}$ 9.38 > 8.92 > 3.38 > 3.27. Dono anilines delocalisation ki wajah se weak hain; N-methyl groups thodi madad karte hain.
2. (b)
  Paani mein badhti basic strength
  4-nitroaniline
  4-methoxyaniline
  aniline
  Answer dekho
  Answer
  4-nitroaniline
  aniline
  4-methoxyaniline
  p $K_{b}$ lagbhag 13.0, 9.38 aur 8.66: electron kheenchne wala nitro group basicity ghatata hai aur electron dene wala methoxy group ise badhata hai.
3. (c)
  Paani mein badhti basic strength
  $C_{6} H_{11} N H_{2}$
  $C_{6} H_{5} C H_{2} N H_{2}$
  $C_{6} H_{5} N H_{2}$
  Answer dekho
  Answer
  $C_{6} H_{5} N H_{2}$
  $C_{6} H_{5} C H_{2} N H_{2}$
  $C_{6} H_{11} N H_{2}$
  Aniline 9.38 (lone pair delocalised), benzylamine 4.70 ( $C H_{2}$ nitrogen ko insulate karta hai, lekin phenyl ab bhi thoda kheenchta hai), cyclohexanamine 3.36 (alkyl group electrons release karta hai).
4. (d)
  Gas phase mein ghatti basic strength
  $N H_{3}$
  $C_{2} H_{5} N H_{2}$
  $(C H_{3})_{2} N H$
  $(C_{2} H_{5})_{3} N$
  Answer dekho
  Answer
  $(C_{2} H_{5})_{3} N$
  $(C H_{3})_{2} N H$
  $C_{2} H_{5} N H_{2}$
  $N H_{3}$
  Proton affinities 982, 930, 912 aur 854 kJ mol⁻¹. Gas mein sirf alkyl groups ka electron release matter karta hai: teen groups do se aage hain, aur do ek se.
5. (e)
  Badhta boiling point
  $(C H_{3})_{2} N (C H_{2})_{2} C H_{3}$
  $C H_{3} (C H_{2})_{4} N H_{2}$
  $C H_{3} (C H_{2})_{4} O H$
  $C H_{3} N H (C H_{2})_{3} C H_{3}$
  Answer dekho
  Answer
  $(C H_{3})_{2} N (C H_{2})_{2} C H_{3}$
  $C H_{3} N H (C H_{2})_{3} C H_{3}$
  $C H_{3} (C H_{2})_{4} N H_{2}$
  $C H_{3} (C H_{2})_{4} O H$
  N,N-Dimethylpropan-1-amine 338 K (koi N–H nahi), N-methylbutan-1-amine 364 K (ek N–H), pentan-1-amine 377 K (do N–H), pentan-1-ol 411 K (zyada strong O–H⋯O bonds).
6. (f)
  Paani mein badhti solubility
  $C H_{3} C H_{2} C H_{2} N H_{2}$
  $C_{6} H_{5} N H_{2}$
  $C_{6} H_{5} N (C H_{3})_{2}$
  Answer dekho
  Answer
  $C_{6} H_{5} N (C H_{3})_{2}$
  $C_{6} H_{5} N H_{2}$
  $C H_{3} C H_{2} C H_{2} N H_{2}$
  N,N-Dimethylaniline (koi N–H nahi aur bada hydrocarbon hissa) < aniline (lagbhag 3.6 g per 100 g paani) < propan-1-amine (har proportion mein mix hota hai).
5
Short answerHardNCERT Exercises Q9.5
Dikhao ki har conversion kaise karoge, reagents aur conditions ke saath.
1. Propanoic acid se ethanamine
  Solution dikhao
  Acid ko propanamide mein badlo ( $S O C l_{2}$ , phir $N H_{3}$ ), phir $B r_{2}$ aur NaOH use karo: Hoffmann degradation carbonyl carbon ko hata deta hai aur $C H_{3} C H_{2} N H_{2}$ deta hai.
2. Pentanenitrile se butan-1-amine
  Solution dikhao
  Nitrile ko alkaline $H_{2} O_{2}$ se partly hydrolyse karke pentanamide, $C_{4} H_{9} C O N H_{2}$ , banao; phir $B r_{2}$ aur NaOH se $C_{4} H_{9} N H_{2}$ milta hai.
3. Ethanol se propanoic acid
  Solution dikhao
  $P B r_{3}$ ethanol ko bromoethane mein badalta hai; ethanol mein KCN se propanenitrile, $C_{2} H_{5} C N$ , milta hai; garam dilute acid se hydrolysis $C_{2} H_{5} C O O H$ deta hai.
4. Propan-1-amine se ethanamine
  Solution dikhao
  Nitrous acid ( $N a N O_{2}$ aur HCl) propan-1-ol aur $N_{2}$ deta hai; acidified $K M n O_{4}$ ise propanoic acid mein oxidise karta hai; $S O C l_{2}$ phir $N H_{3}$ se propanamide milta hai; $B r_{2}$ aur NaOH se ethanamine milta hai.
5. Ethanamine se propan-1-amine
  Solution dikhao
  Nitrous acid se ethanol milta hai; $P B r_{3}$ se bromoethane; KCN se propanenitrile; $L i A l H_{4}$ ise propan-1-amine mein reduce karta hai.
6. Nitroethane se N-methylethanamine
  Solution dikhao
  Sn aur HCl (phir NaOH) se reduce karke ethanamine banao; $C H C l_{3}$ aur ethanolic KOH ke saath garam karke ethyl isocyanide, $C_{2} H_{5} N C$ , banao; ise Pt par $H_{2}$ (ya $L i A l H_{4}$ ) se reduce karke $C_{2} H_{5} N H C H_{3}$ banao.
7. Butanoic acid se propanoic acid
  Solution dikhao
  $S O C l_{2}$ phir $N H_{3}$ se butanamide milta hai; $B r_{2}$ aur NaOH se propan-1-amine; nitrous acid se propan-1-ol; acidified $K M n O_{4}$ se propanoic acid.
8. Benzoic acid se phenylmethanamine
  Solution dikhao
  $S O C l_{2}$ se benzoyl chloride milta hai; $N H_{3}$ se benzamide; $L i A l H_{4}$ aur phir paani se $C_{6} H_{5} C H_{2} N H_{2}$ . (Hoffmann degradation iski jagah aniline dega.)
Long answerMediumNCERT Exercises Q9.6
Bina label ki bottles mein butan-1-amine, N-methylpropan-1-amine aur triethylamine hain. Batao ki p-toluenesulphonyl chloride (TsCl, $C H_{3} C_{6} H_{4} S O_{2} C l$ ) aur aqueous KOH se tum har ek ko kaise pehchanoge, equations ke saath.
Solution dikhao
Har amine ka thoda sa hissa TsCl aur excess aqueous KOH ke saath hilao, phir koi bhi clear solution ho to use dilute HCl se acidify karo.
Butan-1-amine (1°) aisa sulphonamide banata hai jismein ab bhi ek N–H hai. $S O_{2}$ group us H ko acidic bana deta hai, isliye product KOH mein salt ban kar ghul jaata hai aur clear solution deta hai; acid daalne par yeh phir solid ban kar nikal aata hai.
$\begin{aligned} T s C l + H_{2} N C_{4} H_{9} \\ \overset{}{\to} T s N H C_{4} H_{9} + H C l \\ T s N H C_{4} H_{9} + K O H \\ \overset{}{\to} [T s N C_{4} H_{9}]^{-} K^{+} + H_{2} O \end{aligned}$
N-Methylpropan-1-amine (2°) aisa sulphonamide banata hai jismein koi N–H nahi, jo KOH mein nahi ghulta aur solid ya oil ban kar alag ho jaata hai.
$\begin{aligned} T s C l + H N (C H_{3}) C_{3} H_{7} \\ \overset{}{\to} T s N (C H_{3}) C_{3} H_{7} + H C l \end{aligned}$
Triethylamine (3°) mein koi N–H nahi hai aur yeh react nahi karta. Yeh alkali mein alag layer ki tarah rehta hai aur HCl daalne par ghul jaata hai:
$\begin{aligned} (C_{2} H_{5})_{3} N + H C l \\ \overset{}{\to} (C_{2} H_{5})_{3} N H^{+} C l^{-} \end{aligned}$
7
Short answerMediumNCERT Exercises Q9.7
Har named reaction ke liye ek sentence mein samjhao ki woh kya karta hai aur diye gaye compound ke liye ek balanced equation do.
1. Carbylamine reaction, 4-methylaniline ke saath
  Solution dikhao
  Primary amines chloroform aur ethanolic KOH ke saath badboodaar isocyanides dete hain; yeh 1° amines ka test hai.
  $\begin{aligned} C H_{3} C_{6} H_{4} N H_{2} + C H C l_{3} \\ + 3 K O H \overset{}{\to} C H_{3} C_{6} H_{4} N C \\ + 3 K C l + 3 H_{2} O \end{aligned}$
2. Diazotisation, 4-chloroaniline ke saath
  Solution dikhao
  Primary arylamine aur nitrous acid 273–278 K par diazonium salt dete hain, jise turant use kiya jaata hai.
  $\begin{aligned} C l C_{6} H_{4} N H_{2} + N a N O_{2} \\ + 2 H C l \overset{}{\to} C l C_{6} H_{4} N_{2}^{+} C l^{-} \\ + N a C l + 2 H_{2} O \end{aligned}$
3. Hoffmann bromamide reaction, propanamide ke saath
  Solution dikhao
  Primary amide apna carbonyl carbon kho deta hai aur ek carbon kam wala amine deta hai.
  $\begin{aligned} C H_{3} C H_{2} C O N H_{2} + B r_{2} \\ + 4 N a O H \overset{}{\to} C H_{3} C H_{2} N H_{2} \\ + N a_{2} C O_{3} + 2 N a B r \\ + 2 H_{2} O \end{aligned}$
4. Coupling reaction, 4-methylbenzenediazonium chloride ka phenol ke saath
  Solution dikhao
  Diazonium ion phenoxide ki para position par attack karta hai aur orange azo dye deta hai.
  $\begin{aligned} C H_{3} C_{6} H_{4} N_{2}^{+} C l^{-} \\ + C_{6} H_{5} O H + N a O H \\ \overset{}{\to} C H_{3} C_{6} H_{4} N = N C_{6} H_{4} O H \\ + N a C l + H_{2} O \end{aligned}$
5. Ammonolysis, bromoethane ka
  Solution dikhao
  Ammonia bromide ko hatata hai (SN2); NaOH amine ko free karta hai, aur excess ammonia aage ke alkylation ko kam karta hai.
  $\begin{aligned} C_{2} H_{5} B r + N H_{3} \\ \overset{}{\to} C_{2} H_{5} N H_{3}^{+} B r^{-} \end{aligned}$
6. Acetylation, N-methylaniline ka ethanoic anhydride ke saath
  Solution dikhao
  Acetyl group N–H hydrogen ko replace karta hai aur amide banta hai.
  $\begin{aligned} C_{6} H_{5} N H C H_{3} + (C H_{3} C O)_{2} O \\ \overset{}{\to} C_{6} H_{5} N (C H_{3}) C O C H_{3} \\ + C H_{3} C O O H \end{aligned}$
7. Gabriel phthalimide synthesis, 2-methylpropan-1-amine banana
  Solution dikhao
  Potassium phthalimide ko 1-bromo-2-methylpropane ke saath garam karke N-(2-methylpropyl)phthalimide banao, phir garam NaOH(aq) se hydrolyse karo: $(C H_{3})_{2} C H$ – $C H_{2} N H_{2}$ aur sodium phthalate bante hain. Sirf primary amine banta hai.
8
Short answerHardNCERT Exercises Q9.8
Har synthesis plan karo: reagents aur intermediate compounds batao.
1. 4-Nitrotoluene se 4-methylbenzoic acid
  Solution dikhao
  Fe aur HCl se 4-methylaniline milta hai; 273–278 K par $N a N O_{2}$ aur HCl se diazonium salt; CuCN aur KCN se 4-methylbenzonitrile; garam dilute acid ise hydrolyse karke 4-methylbenzoic acid banata hai.
2. Nitrobenzene se 3-chlorophenol
  Solution dikhao
  $F e C l_{3}$ ke saath $C l_{2}$ nitro group ke meta par chlorinate karta hai, aur 1-chloro-3-nitrobenzene milta hai; Sn aur HCl ise 3-chloroaniline mein reduce karte hain; 273–278 K par diazotise karo; paani ke saath garam karke 3-chlorophenol lo.
3. Phenylethanoic acid se phenylmethanamine
  Solution dikhao
  $S O C l_{2}$ phir $N H_{3}$ se 2-phenylethanamide, $C_{6} H_{5} C H_{2}$ – $C O N H_{2}$ , milta hai; $B r_{2}$ aur NaOH se $C_{6} H_{5} C H_{2} N H_{2}$ , ek carbon kam.
4. Aniline se 2,4,6-tribromobenzonitrile
  Solution dikhao
  Bromine water se 2,4,6-tribromoaniline milta hai; ise strong acid mein 273–278 K par diazotise karo; CuCN aur KCN $- N_{2}^{+}$ ko CN se replace karte hain.
5. 1-Bromo-3-phenylpropane se 4-phenylbutan-1-amine
  Solution dikhao
  Ethanol mein KCN se 4-phenylbutanenitrile, $C_{6} H_{5} (C H_{2})_{3}$ – $C N$ , milta hai; $L i A l H_{4}$ ise $C_{6} H_{5} (C H_{2})_{4}$ – $N H_{2}$ mein reduce karta hai.
6. Toluene se 4-methylaniline
  Solution dikhao
  Concentrated $H N O_{3}$ aur $H_{2} S O_{4}$ se nitration karo ( $C H_{3}$ ortho/para direct karta hai), 4-nitrotoluene ko alag karo, aur use Sn aur HCl ya Pd par $H_{2}$ se reduce karo.
7. Aniline se 4-nitroaniline
  Solution dikhao
  Acetylate karke ( $(C H_{3} C O)_{2} O$ , pyridine) acetanilide banao; nitrate karke ( $H N O_{3}$ aur $H_{2} S O_{4}$ , lagbhag 288 K) 4-nitroacetanilide banao; acid ya alkali se hydrolyse karo. Direct nitration ring ko oxidise kar dega aur kaafi meta product dega.
8. 4-Methylbenzamide se toluene
  Solution dikhao
  $B r_{2}$ aur NaOH se 4-methylaniline milta hai; 273–278 K par $N a N O_{2}$ aur HCl se diazonium salt; $H_{3} P O_{2}$ (ya ethanol) $- N_{2}^{+}$ ko H se replace karke toluene deta hai.
9
Short answerMediumNCERT Exercises Q9.9
Har sequence mein A, B aur C ko pehchano.
1. 1-Bromobutane ko KCN ke saath garam karne par A milta hai; alkaline $H_{2} O_{2}$ A ko partly hydrolyse karke B banata hai; $B r_{2}$ aur NaOH B ko C mein badalte hain.
  Solution dikhao
  A pentanenitrile, $C_{4} H_{9} C N$ , hai; B pentanamide, $C_{4} H_{9} C O N H_{2}$ , hai; C butan-1-amine, $C_{4} H_{9} N H_{2}$ , hai.
2. 273 K par $N a N O_{2}$ aur HCl ke saath aniline A deta hai; CuBr aur HBr ke saath A, B deta hai; dry ether mein Mg ke saath B, C deta hai.
  Solution dikhao
  A benzenediazonium chloride hai; B bromobenzene hai; C phenylmagnesium bromide, $C_{6} H_{5} M g B r$ , hai, ek Grignard reagent (Unit 6).
3. KCN ke saath benzyl chloride A deta hai; $L i A l H_{4}$ A ko B mein reduce karta hai; $C H C l_{3}$ aur ethanolic KOH ke saath garam karne par B, C deta hai.
  Solution dikhao
  A phenylethanenitrile, $C_{6} H_{5} C H_{2} C N$ , hai; B 2-phenylethanamine, $C_{6} H_{5}$ – $C H_{2} C H_{2} N H_{2}$ , hai; C isocyanide $C_{6} H_{5}$ – $C H_{2} C H_{2} N C$ hai.
4. Sn aur HCl ke saath 3-nitrotoluene A deta hai; 273 K par $N a N O_{2}$ aur HCl ke saath A, B deta hai; ethanol ke saath garam karne par B, C deta hai.
  Solution dikhao
  A 3-methylaniline hai; B 3-methylbenzenediazonium chloride hai; C toluene hai (ethanol oxidise hokar ethanal ban jaata hai).
5. $N H_{3}$ ke saath benzoyl chloride A deta hai; $B r_{2}$ aur NaOH ke saath A, B deta hai; ethanoic anhydride ke saath B, C deta hai.
  Solution dikhao
  A benzamide hai; B aniline hai; C acetanilide (N-phenylethanamide) hai.
6. Toluene ka nitration A (para isomer) deta hai; Fe aur HCl A ko B mein reduce karte hain; B ko 273 K par diazotise karke NaOH mein phenol ke saath couple karne par C milta hai.
  Solution dikhao
  A 4-nitrotoluene hai; B 4-methylaniline hai; C orange azo dye $C H_{3} C_{6} H_{4}$ – $N = N$ – $C_{6} H_{4} O H$ , yaani 4-[(4-methylphenyl)diazenyl]phenol hai.
Short answerHardNCERT Exercises Q9.10
Ek aromatic compound P, $C_{8} H_{8} O_{2}$ , ko aqueous ammonia ke saath garam karne par Q milta hai. Q bromine aur NaOH se react karke R, $C_{7} H_{9} N$ , deta hai. R ko 273 K par diazotise karke NaOH mein phenol mein daalne par orange dye milta hai. Alkaline $K M n O_{4}$ P ko benzene-1,4-dicarboxylic acid mein oxidise karta hai. P, Q aur R ko pehchano aur unke naam batao.
Solution dikhao
R mein Q se ek carbon kam hai aur yeh primary amine hai, isliye Q ek amide hai aur yeh step Hoffmann degradation hai. Kyunki R azo dye banata hai, iska NH₂ ring par hai.
P oxidise hokar 1,4-dicarboxylic acid banta hai, isliye P par do carbon groups ek doosre ke para par hain: ek $C H_{3}$ aur ek $C O O H$ .
P 4-methylbenzoic acid, $C H_{3} C_{6} H_{4} C O O H$ ( $C_{8} H_{8} O_{2}$ ), hai; Q 4-methylbenzamide, $C H_{3} C_{6} H_{4} C O N H_{2}$ , hai (garam karne par ammonium salt paani kho deta hai); R 4-methylaniline, $C H_{3} C_{6} H_{4} N H_{2}$ ( $C_{7} H_{9} N$ ), hai.
11
Short answerMediumNCERT Exercises Q9.11
Har reaction poora karo aur organic product ka naam batao.
1. $C H C l_{3}$ aur ethanolic KOH ke saath garam kiya gaya propan-1-amine
  Solution dikhao
  $\begin{aligned} C_{3} H_{7} N H_{2} + C H C l_{3} \\ + 3 K O H \overset{}{\to} C_{3} H_{7} N C \\ + 3 K C l + 3 H_{2} O \end{aligned}$
  Propyl isocyanide.
2. 4-Chlorobenzenediazonium chloride + KI
  Solution dikhao
  $\begin{aligned} C l C_{6} H_{4} N_{2}^{+} C l^{-} + K I \\ \overset{}{\to} C l C_{6} H_{4} I + K C l + N_{2} \end{aligned}$
  1-Chloro-4-iodobenzene.
3. 4-Methylaniline + HBr
  Solution dikhao
  $\begin{aligned} C H_{3} C_{6} H_{4} N H_{2} + H B r \\ \overset{}{\to} C H_{3} C_{6} H_{4} N H_{3}^{+} B r^{-} \end{aligned}$
  4-Methylanilinium bromide.
4. 3-Nitrobenzenediazonium chloride + paani, garam
  Solution dikhao
  $\begin{aligned} O_{2} N C_{6} H_{4} N_{2}^{+} C l^{-} \\ + H_{2} O \overset{}{\to} O_{2} N C_{6} H_{4} O H \\ + N_{2} + H C l \end{aligned}$
  3-Nitrophenol.
5. 4-Methylaniline + excess bromine water
  Solution dikhao
  $\begin{aligned} C H_{3} C_{6} H_{4} N H_{2} + 2 B r_{2} \\ \overset{}{\to} C H_{3} C_{6} H_{2} B r_{2} N H_{2} \\ + 2 H B r \end{aligned}$
  2,6-Dibromo-4-methylaniline; para position par pehle se methyl group hai.
6. N-Methylaniline + benzoyl chloride
  Solution dikhao
  $\begin{aligned} C_{6} H_{5} N H C H_{3} + C_{6} H_{5} C O C l \\ \overset{}{\to} C_{6} H_{5} N (C H_{3}) C O C_{6} H_{5} \\ + H C l \end{aligned}$
  N-Methyl-N-phenylbenzamide.
7. 4-Methylbenzenediazonium chloride: (i) $H B F_{4}$ , (ii) garam karo
  Solution dikhao
  $\begin{aligned} C H_{3} C_{6} H_{4} N_{2}^{+} C l^{-} + H B F_{4} \\ \overset{}{\to} C H_{3} C_{6} H_{4} N_{2}^{+} B F_{4}^{-} + H C l \\ \overset{h e a t}{\to} C H_{3} C_{6} H_{4} F \\ + B F_{3} + N_{2} \end{aligned}$
  1-Fluoro-4-methylbenzene (4-fluorotoluene).
Short answerMediumNCERT Exercises Q9.12
Ek student 4-bromotoluene ko potassium phthalimide ke saath garam karke aur phir hydrolyse karke 4-methylaniline banana chahta hai. Samjhao ki yeh kyun fail hota hai, aur ek kaam karne wala route suggest karo.
Solution dikhao
Gabriel synthesis mein phthalimide anion ko SN2 reaction se bromide ko hataana hota hai. 4-bromotoluene mein C–Br carbon ek $s p^{2}$ ring carbon hai, bond mein partial double-bond character hai, aur ring peeche se attack ko block karti hai, isliye koi substitution nahi hota.
Ek kaam karne wala route: toluene ka nitration karo, 4-nitrotoluene ko alag karo, aur use Fe aur HCl (ya Sn aur HCl) se reduce karke 4-methylaniline banao.
Short answerMediumNCERT Exercises Q9.13
Butan-1-amine aur 4-chloroaniline, dono ko 273–278 K par $N a N O_{2}$ aur dilute HCl se treat kiya jaata hai, aur baad mein dono solutions ko garam hone diya jaata hai. Equations likho aur compare karo ki tumhe kya dikhega.
Solution dikhao
Butan-1-amine: thande mein bhi turant nitrogen ke bubbles nikalte hain, aur butan-1-ol banta hai (doosre products ke saath).
$\begin{aligned} C_{4} H_{9} N H_{2} + H N O_{2} \\ \overset{}{\to} C_{4} H_{9} O H + N_{2} ↑ + H_{2} O \end{aligned}$
4-Chloroaniline: thande mein koi gas nahi; 4-chlorobenzenediazonium chloride ka clear solution banta hai.
$\begin{aligned} C l C_{6} H_{4} N H_{2} + N a N O_{2} \\ + 2 H C l \overset{}{\to} C l C_{6} H_{4} N_{2}^{+} C l^{-} \\ + N a C l + 2 H_{2} O \end{aligned}$
Garam karne par ab nitrogen nikalti hai aur 4-chlorophenol banta hai.
$\begin{aligned} C l C_{6} H_{4} N_{2}^{+} C l^{-} + H_{2} O \\ \overset{}{\to} C l C_{6} H_{4} O H + N_{2} ↑ + H C l \end{aligned}$
14
Short answerMediumNCERT Exercises Q9.14
Har ek ki wajah batao.
1. Ethanol aur ethanamine ka molar mass lagbhag same hai, phir bhi ethanol O–H se proton kahin zyada aasaani se chhodta hai jitna ethanamine N–H se chhodta hai.
  Solution dikhao
  Oxygen nitrogen se zyada electronegative hai, isliye O–H bond zyada polar hai aur ethoxide ion apna negative charge $C_{2} H_{5} N H^{-}$ ion se behtar sambhalta hai. Zyada stable conjugate base ka matlab zyada strong acid, isliye amines alcohols se kaafi weak acids hain.
2. Propan-1-amine apne isomer trimethylamine se lagbhag 45 K zyada par boil hota hai.
  Solution dikhao
  Propan-1-amine mein do N–H hydrogens hain aur uske molecules aapas mein hydrogen bond banate hain. Trimethylamine mein koi nahi hai, isliye uske molecules ko sirf weak dipole aur dispersion forces jode rakhti hain (321 K vs 276 K).
3. Base ke roop mein cyclohexanamine aniline se lagbhag das lakh guna strong hai.
  Solution dikhao
  Inki p $K_{b}$ values 3.36 aur 9.38 hain, chhah units ka farak. Aniline mein lone pair ring mein delocalised hai, aur protonation us resonance ko khatam kar dega; cyclohexanamine mein lone pair localised hai aur alkyl group electrons release karta hai.
Quick answerEasy
Kaun sa reagent benzenediazonium chloride ko iodobenzene mein badalta hai?
$K I$ solution $F e C l_{3}$ ke saath $I_{2}$ copper powder ke saath $H I$ acetone mein $N a I$
Answer dekho
Answer (A) $K I$ solution
Potassium iodide solution daalne par iodide ion seedha diazonium group ko replace kar deta hai; nitrogen gas release hoti hai.
Quick answerEasy
Primary aromatic amine ko diazonium salt mein badalne ko kehte hain.
Answer dekho
Answer diazotisation
Diazotisation mein $A r N H_{2}$ 273–278 K par $N a N O_{2}$ aur do moles HCl se react karke $A r N_{2}^{+} C l^{-}$ , NaCl aur paani deta hai.
Quick answerMedium
3-phenylpropanamide, $C_{6} H_{5} C H_{2} C H_{2} C O N H_{2}$ , ka Hoffmann degradation ek amine deta hai. Amine mein kitne carbon atoms hain?
Answer dekho
Answer 8
Amide mein 9 carbons hain; carbonyl carbon nikal jaata hai, aur 2-phenylethanamine, $C_{6} H_{5} C H_{2} C H_{2} N H_{2}$ , bachta hai jismein 8 hain.
18
Quick answerMedium
Inmein se kaun carbylamine test positive dete hain? Jitne sahi hon sab chuno.
anilineN-methylanilinebenzylaminetriethylamine
Saare sahi options chuno
Answer dekho
Answer
(A) aniline
(C) benzylamine
Aniline aur benzylamine dono primary amines hain (aromatic aur aliphatic), isliye dono isocyanides dete hain. N-Methylaniline secondary hai aur triethylamine tertiary.

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Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Intext Question 9.1	Practice: classify karna aur naam dena · Q1
NCERT Intext Question 9.2	Practice: classify karna aur naam dena · Q2
NCERT Intext Question 9.3	Practice: amines banana · Q1
NCERT Intext Question 9.4	Practice: amines ke reactions · Q1
NCERT Intext Question 9.5	Practice: amines ke reactions · Q2
NCERT Intext Question 9.6	Practice: amines ke reactions · Q3
NCERT Intext Question 9.7	Practice: amines ke reactions · Q4
NCERT Intext Question 9.8	Practice: amines ke reactions · Q5
NCERT Intext Question 9.9	Practice: diazonium salts ka use · Q1
NCERT Exercises Q9.1	Exercises · Q1
NCERT Exercises Q9.2	Exercises · Q2
NCERT Exercises Q9.3	Exercises · Q3
NCERT Exercises Q9.4	Exercises · Q4
NCERT Exercises Q9.5	Exercises · Q5
NCERT Exercises Q9.6	Exercises · Q6
NCERT Exercises Q9.7	Exercises · Q7
NCERT Exercises Q9.8	Exercises · Q8
NCERT Exercises Q9.9	Exercises · Q9
NCERT Exercises Q9.10	Exercises · Q10
NCERT Exercises Q9.11	Exercises · Q11
NCERT Exercises Q9.12	Exercises · Q12
NCERT Exercises Q9.13	Exercises · Q13
NCERT Exercises Q9.14	Exercises · Q14

Quick check: MCQs

9.1 Structure of Amines3 questions
9.2 Classification3 questions
9.3 Nomenclature3 questions
Hoffmann bromamide degradation reaction4 questions
9.5 Physical Properties3 questions
Arylamines vs ammonia4 questions
Reaction with arylsulphonyl chloride4 questions
Electrophilic substitution4 questions
9.7 Method of Preparation of Diazonium Salts3 questions
Coupling reactions5 questions
9.10 Importance of Diazonium Salts in Synthesis of Aromatic Compounds3 questions

Glossary · 15

Key terms (English + Hindi)

acylationऐसिलनaisilan: Acid chloride ya anhydride se nitrogen (ya oxygen) par lage hydrogen ko acyl group, $R C O^{-}$ , se replace karna.Chapter mein pehli baar yahan aaya →
amineऐमीनaimeen: Ammonia ke ek, do ya teen hydrogens ko alkyl ya aryl groups se replace karke bana compound; nitrogen par lone pair bana rehta hai.Chapter mein pehli baar yahan aaya →
ammonolysisअमोनी-अपघटनamoni-apaghatan: Ammonia se C–X bond todna, jismein halogen ki jagah amino group aa jaata hai.Chapter mein pehli baar yahan aaya →

carbylamine reaction: Jab primary amine ko $C H C l_{3}$ aur ethanolic KOH ke saath garam kiya jaata hai to badboodaar isocyanide banna; 1° amines ka test.Chapter mein pehli baar yahan aaya →
coupling reactionयुग्मन अभिक्रियाyugman abhikriya: Phenol ya arylamine ring par diazonium ion ka attack, jo rings ko –N=N– se jod kar coloured azo compound deta hai.Chapter mein pehli baar yahan aaya →

diazonium saltडाइऐज़ोनियम लवणdaaiazoniyam lavan: Primary aromatic amine se bana salt $A r N_{2}^{+} X^{-}$ ; sirf thande solution mein stable.Chapter mein pehli baar yahan aaya →
diazotisationडाइऐज़ोटीकरणdaaiazoteekaran: 273–278 K par $N a N O_{2}$ aur acid se primary aromatic amine ko diazonium salt mein badalna.Chapter mein pehli baar yahan aaya →

Gabriel phthalimide synthesis: Pure primary amines banane ka route: potassium phthalimide ko $R - X$ se alkylate kiya jaata hai aur product ko hydrolyse kiya jaata hai.Chapter mein pehli baar yahan aaya →

Hinsberg's reagent: Benzenesulphonyl chloride, $C_{6} H_{5} S O_{2} C l$ , jo 1°, 2° aur 3° amines ko pehchanne aur alag karne mein use hota hai.Chapter mein pehli baar yahan aaya →
Hoffmann bromamide degradation: $B r_{2}$ aur NaOH se primary amide ko ek carbon kam wale primary amine mein badalna.Chapter mein pehli baar yahan aaya →

pKb: $- \log K_{b}$ , base strength ka ek naap: pKb jitna chhota, base utna strong.Chapter mein pehli baar yahan aaya →
primary, secondary aur tertiary aminesप्राथमिक, द्वितीयक और तृतीयक ऐमीनprimary, secondary and tertiary aminespraathmik, dviteeyak aur triteeyak aimeen: Aise amines jinke nitrogen par ek, do ya teen carbon groups hon: $R N H_{2}$ , $R_{2} N H$ aur $R_{3} N$ .Chapter mein pehli baar yahan aaya →

quaternary ammonium saltचतुष्क अमोनियम लवणchatushk amoniyam lavan: Ion $R_{4} N^{+}$ ka salt, jismein nitrogen par chaar carbon groups hote hain aur koi lone pair nahi.Chapter mein pehli baar yahan aaya →

Sandmeyer reaction: Matching copper(I) salt se $- N_{2}^{+}$ ko Cl, Br ya CN se replace karna.Chapter mein pehli baar yahan aaya →
solvationविलायकनvilaayakan: Ghule hue ion ke chaaron taraf solvent molecules ke jamaa hone se uska stable hona; paani mein zyadatar hydrogen bonds ke zariye.Chapter mein pehli baar yahan aaya →

9.1 Structure of Amines

9.2 Classification

9.3 Nomenclature

9.4 Preparation of Amines

Reduction of nitro compounds

Ammonolysis of alkyl halides

Reduction of nitriles

Reduction of amides

Gabriel phthalimide synthesis

Hoffmann bromamide degradation reaction

9.5 Physical Properties

9.6 Chemical Reactions

Basic character of amines

Structure aur basicity: alkanamines vs ammonia

Arylamines vs ammonia

Alkylation

Acylation

Carbylamine reaction

Reaction with nitrous acid

Reaction with arylsulphonyl chloride

Electrophilic substitution

Diazonium Salts

9.7 Method of Preparation of Diazonium Salts

9.8 Physical Properties

9.9 Chemical Reactions

Reactions jinmein nitrogen replace hota hai

Coupling reactions

9.10 Importance of Diazonium Salts in Synthesis of Aromatic Compounds

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