Organic Chemistry – Some Basic Principles and Techniques (Organic Chemistry – Kuch Basic Principles aur Techniques): Class 11 Chemistry Chapter 8 Hinglish mein

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

Published 22 Oct 2026 · Updated 22 Oct 2026

115 min
46 practice questions
41 MCQs

Is chapter mein aap kya seekhenge

Samjhao ki sp³, sp² aur sp hybridisation carbon compounds ke shapes kaise tay karte hain, aur π bonds ki khaas features kya hain
Organic molecules ke complete, condensed, bond-line aur wedge–dash representations banao aur ek ko doosre mein badlo
Organic compounds ko skeleton aur functional group ke hisaab se classify karo, aur homologous series pehchano
IUPAC rules se compounds ke naam rakho, functional groups aur substituted benzenes samet, aur naam se structures banao
Structural isomerism aur stereoisomerism ke types pehchano
Bond fission, reactive intermediates, nucleophiles aur electrophiles ko describe karo, aur curved arrows use karo
Stability aur reactivity samjhane ke liye inductive, resonance, electromeric aur hyperconjugation effects use karo
Purification techniques chuno, elements ke tests describe karo aur analytical data se percentage composition calculate karo

NCERT Chemistry Class 11, Chapter 8, pp. 256–294 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Paracetamol ki tablet ko uski strip se dabaakar nikalo, aur tum ek saath kai carbon compounds ko haath laga rahe ho: drug, PVC blister aur foil par laga lacquer. Koi aur element apne hi atoms se itni lambi, stable chains aur rings nahi banata (catenation), aur carbon in skeletons ko hydrogen, oxygen, nitrogen aur gine-chune doosre elements se sajata hai. Isse laakhon compounds bante hain, aur unki study organic chemistry hai.

Yeh chapter baaki organic chemistry ka toolkit hai: bonds molecule ka shape kaise tay karte hain, molecules ko kaise draw karein aur unke naam kaise rakhein, reactions mein electrons kaise move karte hain, aur chemists kisi compound ko kaise purify karte hain aur kaise pata lagate hain ki usmein kaun se elements hain, aur har ek kitna hai.

Ek living cell ke andar dekho to lagbhag har kaam karne wala part ek organic molecule hai: reactions ko tez karne wale enzymes, membranes banane wale fats, instructions le jaane wala DNA. Body ke bahar, organic chemistry humein petrol aur LPG, nylon aur polyester, paints, detergents aur zyaadatar medicines deti hai.

8.1 General Introduction

Yeh shabd is science se bhi purana hai. Pehle chemists compounds ko unke source ke hisaab se baantte the, living things se ya minerals se, aur Berzelius maante the ki organic compounds ke liye ek vital force chahiye jo sirf living cells mein milti hai. 1828 mein Wöhler ne urea, jo urine ka ek compound hai, sirf ammonium cyanate ko garam karke bana diya:

Wöhler ka urea synthesis

\begin{aligned} N H_{4} C N O \\ \overset{heat}{\to} N H_{2} C O N H_{2} \end{aligned}

(8.1)

Jaldi hi doosre log aur aage gaye: Hermann Kolbe ne 1845 mein carbon disulphide se acetic acid banaya, aur Marcellin Berthelot ne 1856 mein laboratory mein methane banaya. Aaj 'organic' ka matlab bas carbon ka compound hai, kuch simple compounds jaise $C O_{2}$ aur carbonates ko chhodkar. In compounds ka behaviour predict karne mein jo cheez madad karti hai woh hai electron-pair bond, jo is poore chapter mein ek dhaage ki tarah chalta hai.

8.2 Carbon ki Tetravalence: Organic Compounds ke Shapes

8.2.1 Carbon Compounds ke Shapes

Unit 4 mein tumne carbon ke chaar valence electrons aur uske banaye hybrid orbitals dekhe the. Yahi uska shape tay karte hain. Chaar single bonds wala carbon sp³ aur tetrahedral hota hai, jaise methane mein. Ek double bond wala carbon sp² aur trigonal planar hota hai, jaise ethene mein. Triple bond (ya do double bonds) wala carbon sp aur linear hota hai, jaise ethyne mein. Fig. 8.1 teeno ko compare karta hai.

2/3Ek plane mein 120° par teen σ bonds; π bond ke liye ek p orbital bachta hai.

Fig. 8.1Zyada s character bonding electrons ko carbon ke kareeb kheenchta hai.NCERT p. 257

Ek data card ke paas teen states mein ek carbon atom. Step 1, sp³: chaar bonds (plain, wedge, hashed), 109.5°, C–H 109.4 pm, s-character bar 25 per cent par. Step 2, sp²: dashed plane mein 120° par teen bonds, upar aur neeche ek p orbital, C–H 108.7 pm, bar 33 per cent par. Step 3, sp: 180° par ek line mein do bonds, do p orbitals, C–H 106.3 pm, bar 50 per cent par.

Table 8.1 ki C–H row ko upar se neeche dekho: carbon jaise-jaise sp³ se sp ki taraf jaata hai, bond 109.4 pm se 106.3 pm tak chhota hota jaata hai. Iska reason hai s character (25 %, 33 %, 50 %). s orbital ke electrons p orbital ke electrons se zyada time nucleus ke paas bitaate hain, isliye s-rich hybrid apne bonding pair ko kas ke pakadta hai: bond chhota aur mazboot hota hai, aur carbon aise behave karta hai jaise woh zyada electronegative ho. Isi wajah se ek strong base ethyne se proton kheench sakta hai, lekin ethane se nahi (Unit 9).

Table 8.1Hybridisation carbon ke bonds ko kaise badalta hai

	sp³	sp²	sp
s character	25 %	33 %	50 %
Carbon par shape	tetrahedral	trigonal planar	linear
Bond angle	109.5°	120°	180°
C–H length (pm)	109.4 (ethane)	108.7 (ethene)	106.3 (ethyne)
C–C length (pm)	154 (C–C)	134 (C=C)	120 (C≡C)

8.2.2 π Bonds ki Kuch Khaas Features

Do pencils ko ek doosre ke bagal mein rakho to woh apni poori length par chhooti hain; unhe cross karo to woh sirf ek point par milti hain. π bond do p orbitals ka side-by-side overlap hai, isliye yeh tabhi banta hai jab padosi carbons ke p orbitals parallel hon. Ethene mein har carbon ka bacha hua p orbital molecular plane se seedha bahar ki taraf point karta hai, aur π electrons do lobes mein rehte hain, ek us plane ke upar aur ek neeche (Fig. 8.2).

2/3Side-by-side overlap se π electron density plane ke upar aur neeche aati hai.

Fig. 8.2Parallel p orbitals overlap karte hain; 90° ka twist π bond tod deta hai.NCERT p. 257

σ bond se jude do carbons, har ek ke saath dashed plane mein do hydrogens. Step 1: har carbon par ek seedha khada p orbital, doosre ke parallel. Step 2: lobes milkar C–C axis ke upar aur neeche ek π cloud banate hain. Step 3: right wala CH₂ 90 degrees ghuma diya gaya hai, uska p orbital leta hua hai, aur ek cross dikhata hai ki koi overlap nahi hai.

Ab ek $C H_{2}$ group ko C=C axis ke around twist karne ki koshish karo. p orbitals ek doosre se door mud jaate hain, 90° par overlap zero ho jaata hai, aur π bond khatam ho jaata hai. Isliye twist karne mein lagbhag ek π bond jitni energy lagti hai, to double bond ke ends locked rehte hain (restricted rotation) aur ethene ke saare chhe atoms ek hi plane mein hote hain; yahi locking Section 8.6.2 ke geometrical isomers deti hai. Aur kyunki π electrons C–C axis se door rehte hain, aur σ electrons se kam kas ke pakde hote hain, electron dhoondhne wale reagents unhe aasani se pakad lete hain: multiple bonds hi aksar woh jagah hote hain jahan reactions hote hain.

Example 8.1NCERT Problem 8.1 · p. 257

σ aur π bonds ginna

Har molecule mein kitne σ aur π bonds hain?

$C H_{2} = C H - C \equiv N$ (acrylonitrile, acrylic fibres banane mein use hota hai)
$H C \equiv C - C \equiv C H$
$C H_{3} - C H = C H - C H O$

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

Example 8.2NCERT Problems 8.2, 8.3 · p. 257

Hybridisation aur shape

(i) $C H_{2} = C H - C H_{2} O H$ , (ii) $H C \equiv C - C H = C H_{2}$ aur (iii) $C H_{3} C O O H$ mein har carbon ki hybridisation left se right batao. Phir (iv) $C O C l_{2}$ aur (v) $C S_{2}$ mein carbon par hybridisation aur shape batao.

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

8.3 Organic Compounds ke Structural Representations

8.3.1 Complete, Condensed aur Bond-line Structural Formulas

$C_{4} H_{10} O$ jaisa molecular formula yeh nahi batata ki atoms kaise jude hain, aur kai compounds ka yahi formula hota hai, isliye chemists structures draw karte hain. Lewis structure ke har shared pair ki jagah ek dash laga do aur tumhe complete structural formula mil jaata hai; O, N, S ya halogens par lone pairs dikha bhi sakte ho aur chhod bhi sakte ho.

Condensed formula. Har carbon ko uske hydrogens ke saath ek group ki tarah likho aur zyaadatar dashes hata do: butan-2-ol ban jaata hai $C H_{3} C H (O H) C H_{2} C H_{3}$ . Brackets branch dikhate hain, aur baar-baar aane wali unit ko subscript milta hai, isliye hexane $C H_{3} (C H_{2})_{4} C H_{3}$ hai.

Bond-line formula. Sirf carbon skeleton ko zig-zag ki tarah draw karo. Har corner aur line ka end ek carbon hai, jiske paas chaar bonds poore karne jitne hydrogens hain; C aur H ke alawa baaki atoms likhe jaate hain, unpar lage H ke saath. Khaali line end ek $C H_{3}$ group hai, aur rings polygons ban jaati hain: cyclohexanol ek hexagon hai jiske ek corner par OH hai. Fig. 8.3 butan-2-ol ke liye teeno styles dikhata hai.

3/3Sirf skeleton: corners aur ends carbons hain; OH likha jaata hai.

Fig. 8.3Ek molecule, detail ke teen levels.NCERT p. 258

Butan-2-ol teen tarah se. Step 1: complete formula, ek row mein chaar carbons, har bond dash ki tarah aur doosre carbon ke upar OH. Step 2: condensed formula CH₃CH(OH)CH₂CH₃. Step 3: bond-line formula, teen lines ka zig-zag jiske doosre corner par OH hai, aur har corner aur end par group ka naam batane wale chhote labels hain.

Example 8.4NCERT Problems 8.4, 8.5 · p. 259

Complete, condensed aur bond-line formulas

(i) Propanal, $C H_{3} C H_{2} C H O$ , ka complete structural formula likho. (ii) $C H_{3} C H_{2} C H_{2} C H (C H_{3}) C H_{2} O H$ ka ek compact condensed formula likho aur uske bond-line formula ko describe karo.

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

Example 8.6NCERT Problem 8.6 · p. 260

Bond-line formulas padhna

Fig. 8.4 mein har compound ka condensed aur molecular formula likho.

Fig. 8.4Example 8.6 ke bond-line formulas.NCERT p. 260

Chaar skeletal drawings. (i) Ek pentagon jiske ek corner se ek chhoti line nikalti hai. (ii) Paanch lines ka zig-zag jiske left end par double line hai. (iii) Teen lines ka zig-zag jiske doosre corner par O tak double bond hai. (iv) Ek zig-zag jiske right end par CN likha hai aur left se doosre corner par ek chhoti branch line hai.

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

8.3.2 Organic Molecules ka Three-Dimensional Representation

Paper flat hota hai; molecules nahi. Wedge–dash convention teesri dimension ko waapas le aata hai. Plain line page ke plane mein hoti hai, solid wedge page se bahar tumhari taraf aata hai (chauda end tumhare sabse paas), aur hashed wedge page ke peeche, tumse door jaata hai. Fig. 8.5 ka step 1 bromochlorofluoromethane, $C H B r C l F$ , ko isi style mein draw karta hai.

3/4Atoms balls ki tarah, bonds sticks ki tarah: angles aasani se dikhte hain.

Fig. 8.5Har model ek hi molecule ki alag feature dikhata hai.NCERT Figs. 8.1, 8.2 · p. 260

CHBrClF chaar tarah se. Step 1: wedge–dash drawing, H aur Br page mein, Cl solid wedge par tumhari taraf, F hashed wedge par door. Step 2: framework model, sirf chaar rods. Step 3: ball-and-stick model, alag-alag size ki labelled balls ke saath. Step 4: space-filling model, van der Waals radii ke scale par overlap karte spheres, H sabse chhota aur Br sabse bada.

Kaun sa model chunna chahiye? Kaun sa atom kis se juda hai, yeh dekhne ke liye sirf sticks kaafi hain: ek framework model. Bond angles padhne ke liye atoms ke liye balls jodo: ek ball-and-stick model (double bonds ki jagah springs lagte hain). Do molecules ek doosre mein fit ho sakte hain ya nahi, yeh judge karne ke liye, jaise ek drug ko apne target mein fit hona hota hai, tumhe atoms unke asli relative sizes mein chahiye, asli molecule ki tarah overlap karte hue: van der Waals radii se bana ek space-filling model. Free molecule-viewing software teeno draw karta hai aur unhe ghumane deta hai.

Try karke dekho

Carbon ke clay-ball models

Tumhe chahiye

do colours ki modelling clay
toothpicks
ek protractor

Pehle guess karo

Do carbon balls do toothpicks se jude hain. Kya tum ek end ko pakad kar doosre end ko twist kar sakte ho?

Haan, aaraam seSirf toothpick ko mod kar ya tod karHaan, lekin sirf aadha turn

Ek badi clay ball (carbon) mein chaar toothpicks aise ghusao ki woh ek doosre se jitna ho sake utna door point karein; chhoti balls (H) lagao aur ek angle naapo.
Aise do carbons ko ek toothpick se jodo (ethane) aur ek end ko twist karo.
Teen-direction wale do carbons ko do toothpicks se jodo (ethene) aur wahi twist try karo.

Quick check

4 mein se 1

Bond-line formula mein jis corner par teen lines milti hain, woh dikhata hai

ek CH groupek

C H_{2}

groupbina H wala carbonek

C H_{3}

group

Answer dekho

Answer (A) ek CH group

Teen bonds draw kiye gaye hain, to ek H chaar poore karta hai. Do lines ka matlab

C H_{2}

; line end

C H_{3}

hai.

8.4 Organic Compounds ka Classification

Laakhon known compounds hone ki wajah se chemists unhe do baar sort karte hain: ek baar carbon skeleton ke shape se aur ek baar functional group se. Fig. 8.6 skeleton wala tree banata hai.

3/3Aromatic: benzenoid, non-benzenoid, ya ring mein heteroatom ke saath.

Fig. 8.6Skeleton se sorting: chains, simple rings aur aromatic rings.NCERT p. 261

Teen steps mein ek classification tree. Step 1: organic compounds acyclic (example propanone) aur cyclic (example cyclobutane) mein bat jaate hain. Step 2: alicyclic rings homocyclic (cyclopentene) aur heterocyclic (oxane, ring mein ek oxygen) mein bat jaati hain. Step 3: aromatic rings benzenoid (toluene), non-benzenoid (azulene, judi hui five- aur seven-membered rings) aur heterocyclic (pyrrole) mein bat jaati hain.

Acyclic (open-chain) compounds, jinhe aliphatic compounds bhi kehte hain, mein seedhi ya branched chains hoti hain, jaise gas lighter ka butane, sanitiser ka ethanol aur lemon juice ka citric acid.

Cyclic (closed-chain) compounds mein kam se kam ek ring hoti hai. Alicyclic rings kaafi had tak aliphatic chains jaisa behave karti hain. Zyaadatar homocyclic (carbocyclic) hoti hain, jinki ring mein sirf carbon hota hai, jaise cyclobutane, methylcyclopentane aur cyclooctane. Kuch heterocyclic hoti hain, jinki ring mein O, N ya S hota hai: oxane (six-membered ring mein ek O) aur piperidine (ek N) non-aromatic heterocycles hain.

Aromatic compounds ki ek khaas stability hoti hai (Unit 9). Benzenoid compounds mein benzene rings hoti hain, jaise toluene, phenol aur anthracene (teen judi hui rings). Azulene, five- aur seven-membered judi hui rings wala ek gehra neela hydrocarbon, bina benzene ring ke aromatic hai (non-benzenoid). Agar aromatic ring mein heteroatom ho to compound heterocyclic aromatic hota hai: pyrrole mein five-membered ring mein ek NH hai, aur pyrimidine, jo DNA bases cytosine aur thymine ka parent hai, mein six-membered ring mein do N hain.

Doosri sorting functional groups se hoti hai, jo compounds ko homologous series naam ki families mein baantti hai.

8.4.1 Functional Group

Definition

Functional group

Carbon skeleton se juda ek atom ya atoms ka group, jo compound ko uske characteristic chemical reactions deta hai.

Example–OH (hydroxyl), >C=O (carbonyl), –COOH (carboxyl), $- N H_{2}$ (amino).

Kyunki zyaadatar chemistry functional group karta hai, baaki skeleton nahi, isliye ek hi functional group wale compounds ek jaisa react karte hain. Methanoic, ethanoic aur propanoic acids sab mein –COOH hai, aur teeno baking soda (sodium hydrogencarbonate) ke saath fizz karte hain aur carbon dioxide chhodte hain.

8.4.2 Homologous Series

Definition

Homologous series

Same functional group aur ek general formula wale compounds ki family, jismein har member agle se ek $C H_{2}$ unit (14 u) se alag hota hai.

ExampleAlkanols, $C_{n} H_{2 n + 1} O H$ : $C H_{3} O H$ , $C_{2} H_{5} O H$ , $C_{3} H_{7} O H$ , …

Members (homologues) ki chemical properties ek jaisi hoti hain, jabki physical properties size ke saath dheere-dheere badalti hain: methanol, ethanol, propan-1-ol aur butan-1-ol 65, 78, 97 aur 118 °C par boil hote hain. Alkanols ke alawa tum amines, alkanoic acids, carbonyl series (alkanals aur alkanones), haloalkanes, aur hydrocarbon series: alkanes, alkenes aur alkynes se bhi miloge.

Ek molecule par do ya zyada functional groups ho sakte hain, ek jaise ya alag; tab woh polyfunctional hota hai. Dahi ka lactic acid, $C H_{3} C H (O H) C O O H$ , mein ek hydroxyl aur ek carboxyl group hai; lemons ke citric acid mein ek hydroxyl aur teen carboxyl groups hain.

8.5 Organic Compounds ki Nomenclature

Naam tabhi kaam ka hai jab woh theek ek structure ki taraf point kare aur us naam se structure dobara banaya ja sake. IUPAC (International Union of Pure and Applied Chemistry) ke rules yahi karte hain, isliye Chennai ka chemist aur Chile ka chemist ek hi naam ke liye ek hi structure draw karte hain.

Purane common (trivial) naam aksar yeh batate hain ki compound kahan mila tha: khatte doodh mein lactic acid (Latin lac), apples mein malic acid (malum), kharab butter mein butyric acid (butyrum). Cubane, $C_{8} H_{8}$ , ka naam uske cube-shaped skeleton par rakha gaya hai. Common names isliye chalte rehte hain kyunki woh chhote hote hain.

Table 8.2Kuch compounds ke common aur IUPAC namesNCERT Table 8.1 · p. 262

Formula	Common name	IUPAC name
$H C \equiv C H$	acetylene	ethyne
$(C H_{3})_{2} C H O H$	isopropyl alcohol	propan-2-ol
$C H_{3} C H O$	acetaldehyde	ethanal
$C H_{3} C O C H_{2} C H_{3}$	methyl ethyl ketone	butan-2-one
$C H_{3} C H_{2} O C H_{2} C H_{3}$	diethyl ether	ethoxyethane
$H O C H_{2} C H (O H) C H_{2} O H$	glycerol	propane-1,2,3-triol
$C H_{3} C O O C H_{2} C H_{3}$	ethyl acetate	ethyl ethanoate
$C_{6} H_{5} C H_{3}$	toluene	methylbenzene

8.5.1 Nomenclature ka IUPAC System

IUPAC naam in hisson se banta hai: parent chain mein carbons ki ginti ke liye ek word root, saturation ke liye ek primary suffix (-ane, -ene, -yne), main functional group ke liye ek secondary suffix, aur baaki sab ke liye prefixes, har ek apne locant ke saath. 5-chloro-4-methylhexan-2-one mein 'hex' six-carbon chain hai, '-an-' batata hai ki har C–C bond single hai, '-2-one' C2 par ketone C=O lagata hai, aur prefixes C5 par Cl aur C4 par $C H_{3}$ lagate hain: $C H_{3} C O C H_{2} C H (C H_{3}) C H C l C H_{3}$ .

Sirf carbon aur hydrogen wale compounds hydrocarbons hain: agar har C–C bond single ho to saturated (yeh alkanes hain; inka purana naam paraffins Latin parum affinis, yaani 'affinity ki kami', se aata hai, kyunki yeh bahut kam react karte hain) aur agar inmein C=C ya C≡C ho to unsaturated.

8.5.2 Alkanes ki IUPAC Nomenclature

Unbranched alkanes ka naam chain length ke root aur -ane se banta hai. Pehle chaar roots historical hain; paanch carbons se aage woh Greek numbers se aate hain. Alkane se ek hydrogen hatane par ek alkyl group milta hai, jiska naam -ane ko -yl mein badal kar banta hai aur general formulas mein ise R likhte hain. Alkanes ek homologous series banate hain jiska general formula $C_{n} H_{2 n + 2}$ hai.

Table 8.3Roots, unbranched alkanes aur alkyl groupsNCERT Tables 8.2, 8.3 · p. 263

C atoms	Root	Alkane	Formula	Alkyl group
1	meth	methane	$C H_{4}$	methyl (Me)
2	eth	ethane	$C_{2} H_{6}$	ethyl (Et)
3	prop	propane	$C_{3} H_{8}$	propyl (Pr)
4	but	butane	$C_{4} H_{10}$	butyl (Bu)
5	pent	pentane	$C_{5} H_{12}$	pentyl
6	hex	hexane	$C_{6} H_{14}$	hexyl
7	hept	heptane	$C_{7} H_{16}$	heptyl
8	oct	octane	$C_{8} H_{18}$	octyl
9	non	nonane	$C_{9} H_{20}$	nonyl
10	dec	decane	$C_{10} H_{22}$	decyl
20	icos	icosane	$C_{20} H_{42}$	icosyl

Branched alkyl groups. Teen aur chaar carbon wale groups alag-alag carbons ke through jude ho sakte hain. IUPAC inke common names ki permission deta hai; systematic names mein group ki numbering us carbon se hoti hai jo main chain se juda hai.

Table 8.4Branched alkyl groupsNCERT p. 263

Group	Structure	Systematic name	Juda hua C
isopropyl	$(C H_{3})_{2} C H^{-}$	1-methylethyl	2°
sec-butyl	$C H_{3} C H_{2} C H (C H_{3})^{-}$	1-methylpropyl	2°
isobutyl	$(C H_{3})_{2} C H C H_{2}^{-}$	2-methylpropyl	1°
tert-butyl	$(C H_{3})_{3} C^{-}$	1,1-dimethylethyl	3°
neopentyl	$(C H_{3})_{3} C C H_{2}^{-}$	2,2-dimethylpropyl	1°

Branched alkane ka naam rakhna. Yeh rules isi order mein lagao:

Carbons ki sabse lambi continuous chain dhoondho, drawing chahe jaise bhi mudi ho; wahi parent hai. Agar do chains barabar lambi hon, to woh lo jismein zyada substituents hon.
Chain ki numbering us end se karo jisse substituents ko lowest locants milein. Locant sets ko term by term compare karo; pehle fark par jo set chhota hai woh jeetta hai, isliye {2, 3, 5} {2, 4, 4} se behtar hai.
Har substituent ko uske locant ke saath, alphabetical order mein, parent naam se pehle likho. Hyphens numbers ko words se alag karte hain, commas numbers ko alag karte hain, aur aakhri prefix bina space ke parent se judta hai.
Ek jaise substituents ko di-, tri-, tetra- se milao aur har locant likho (2,2-dimethyl). Alphabetical order lagaate waqt in multiplying prefixes ko ignore kiya jaata hai.
Agar dono ends se same locant set mile, to chhota number us substituent ko milta hai jo alphabetically pehle aata hai.
Branched substituent ki numbering uske attachment point (C1) se karo aur uska naam brackets mein likho: (2,2-dimethylpropyl). Iska poora naam, di- ya tri- samet, uske pehle letter ke hisaab se alphabetise hota hai, yahan d.
Common prefixes mein iso- aur neo- alphabetising mein gine jaate hain (isopropyl i ke under), lekin sec- aur tert- nahi (tert-butyl b ke under).

Fig. 8.7 yeh rules ek aise molecule par lagata hai jiski drawing mein do alag seven-carbon chains chhupi hain.

3/4Left se numbering {2, 3, 5} deti hai, jo {3, 5, 6} se chhota hai.

Fig. 8.7Sabse lambi chain, sabse zyada substituents, lowest locants, alphabetical order.NCERT p. 264

Ek condensed structure: seven-carbon horizontal chain, C2 par methyl, C3 par ethyl aur C5 par methyl. Step 1 do seven-carbon chains highlight karta hai, jinmein se ek ethyl group se hokar upar jaati hai. Step 2 horizontal chain rakhta hai, jismein teen branches hain. Step 3 left se numbering karta hai, locants 2, 3, 5, aur right wala set 3, 5, 6 kata hua hai. Step 4 naam 3-ethyl-2,5-dimethylheptane dikhata hai.

Cycloalkanes ka naam usi carbon count wale alkane se pehle cyclo lagakar banta hai: cyclopentane. Ek substituent ke liye locant ki zaroorat nahi (methylcyclopentane). Zyada substituents hon to lowest locant set use karo, phir C1 us group ko do jo alphabetical order mein pehle aata hai: 1-ethyl-2-methylcyclohexane; 1,1,3-trimethylcyclohexane, 1,3,3- nahi.

Example 8.7NCERT Problem 8.7 · p. 265

In names mein kya galat hai?

Har compound ko brackets mein diya gaya naam diya gaya hai. Galti samjhao aur sahi naam batao.

$C_{2} H_{5} C (C H_{3})_{2} C H (C H_{3}) C_{2} H_{5}$ (3,4,4-trimethylhexane)
$(C_{2} H_{5})_{2} C H C H_{2} C (C H_{3})_{3}$ (2,2-dimethyl-4-ethylhexane)
$C H_{3} C H (C_{2} H_{5}) C H_{2} C H_{2} C H_{3}$ (2-ethylpentane)

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

8.5.3 Functional Group(s) wale Organic Compounds ki Nomenclature

Jab functional group maujood ho, to naam usi ke control mein hota hai. Pehle use pehchano, kyunki wahi suffix tay karta hai. Phir woh sabse lambi chain chuno jismein woh shaamil ho, aur us chain ki numbering aise karo ki functional group ko sabse chhota possible locant mile, chahe phir kisi branch ko bada locant mile.

Table 8.5Functional groups, prefixes aur suffixesNCERT Table 8.4 · p. 267

Class	Group	Prefix	Suffix	Example
Alkene	>C=C<	–	-ene	propene, $C H_{2} = C H C H_{3}$
Alkyne	–C≡C–	–	-yne	propyne, $C H \equiv C C H_{3}$
Halide	–X	halo-	–	2-chloropropane, $C H_{3} C H C l C H_{3}$
Alcohol	–OH	hydroxy-	-ol	propan-1-ol, $C H_{3} C H_{2} C H_{2} O H$
Aldehyde	–CHO	formyl-, oxo-	-al	propanal, $C H_{3} C H_{2} C H O$
Ketone	>C=O	oxo-	-one	pentan-3-one, $(C_{2} H_{5})_{2} C O$
Carboxylic acid	–COOH	carboxy-	-oic acid	propanoic acid, $C_{2} H_{5} C O O H$
Carboxylate ion	$- C O O^{-}$	–	-oate	sodium ethanoate, $C H_{3} C O O^{-} N a^{+}$
Ester	–COOR	alkoxycarbonyl-	-oate	methyl ethanoate, $C H_{3} C O O C H_{3}$
Acyl halide	–COX	halocarbonyl-	-oyl halide	ethanoyl chloride, $C H_{3} C O C l$
Amide	$- C O N H_{2}$	carbamoyl-	-amide	propanamide, $C_{2} H_{5} C O N H_{2}$
Nitrile	–C≡N	cyano-	-nitrile	propanenitrile, $C_{2} H_{5} C N$
Amine	$- N H_{2}$	amino-	-amine	propan-2-amine, $(C H_{3})_{2} C H N H_{2}$
Ether	–OR	alkoxy-	–	methoxyethane, $C H_{3} O C_{2} H_{5}$
Nitro compound	$- N O_{2}$	nitro-	–	2-nitropropane, $(C H_{3})_{2} C H N O_{2}$
Sulphonic acid	$- S O_{3} H$	sulpho-	-sulphonic acid	benzenesulphonic acid, $C_{6} H_{5} S O_{3} H$

Principal group chunna. Jab kai functional groups hon, to is order mein jo sabse upar ho woh suffix banta hai aur baaki prefixes bante hain: –COOH, $- S O_{3} H$ , –COOR, –COCl, $- C O N H_{2}$ , –CN, –CHO, >C=O, –OH, $- N H_{2}$ , phir C=C aur C≡C. Kuch groups hamesha prefixes hote hain: alkyl (–R), phenyl ( $- C_{6} H_{5}$ ), halo (–F, –Cl, –Br, –I), nitro ( $- N O_{2}$ ) aur alkoxy (–OR).

Isliye $H_{2} N C H_{2} C H_{2} O H$ (ethanolamine) 2-aminoethan-1-ol hai: alcohol ki rank amine se upar hai aur woh suffix leta hai. $C H_{3} C H (N O_{2}) C H = C H_{2}$ mein nitro group sirf prefix ho sakta hai, isliye double bond numbering tay karta hai: 3-nitrobut-1-ene.

Kai ek jaise groups di-, tri- wagairah lete hain; alkane ka aakhri 'e' consonant se pehle rehta hai (propane-1,2,3-triol, butane-2,3-dione, popcorn ki buttery smell) aur vowel se pehle hat jaata hai (propan-1-ol). Do ya zyada multiple bonds hon to ek 'a' joda jaata hai: $C H_{2} = C H C H_{2} C H = C H_{2}$ penta-1,4-diene hai.

Example 8.8NCERT Problem 8.8 · p. 266

Functional groups wale compounds ka naam rakhna

Har compound ka IUPAC naam batao.

$(C H_{3})_{2} C H C H_{2} C H (O H) C H_{3}$
$O H C - C H_{2} C H_{2} - C O C H_{3}$
$C H_{3} C H C l C H_{2} C O O H$
$C H_{2} = C H C H_{2} C H_{2} C \equiv C H$

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

Example 8.9NCERT Problem 8.9 · p. 268

Names se structures

(i) 3-bromo-2-methylpentane, (ii) 4-methylpent-3-en-2-one, (iii) 2-aminopropanoic acid aur (iv) 2-chlorocyclopentan-1-ol ka structure likho.

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

8.5.4 Substituted Benzene Compounds ki Nomenclature

Ek substituent ka naam benzene se pehle prefix ki tarah aata hai: chlorobenzene $C_{6} H_{5} C l$ , ethylbenzene $C_{6} H_{5} C_{2} H_{5}$ , iodobenzene $C_{6} H_{5} I$ . Kai common names IUPAC maanta hai aur unhe parents ki tarah use karta hai: phenol ( $C_{6} H_{5} O H$ ), toluene ( $C_{6} H_{5} C H_{3}$ ), benzaldehyde ( $C_{6} H_{5} C H O$ ) aur benzoic acid ( $C_{6} H_{5} C O O H$ ).

Do substituents. Ring ki numbering aise karo ki unhe lowest locants milein: 1,2, 1,3 ya 1,4. Common names in teen patterns ke liye prefixes ortho (o-), meta (m-) aur para (p-) use karte hain, jaise Fig. 8.8 dimethylbenzenes (xylenes) ke liye dikhata hai.

Fig. 8.8Locants 1,2, 1,3 aur 1,4 ortho, meta aur para ke barabar hain.NCERT p. 269

Teen benzene rings ek ke bagal mein ek, har ek ke top carbon par ek methyl group, jise 1 number diya gaya hai. Pehli ring mein doosra methyl agle carbon, 2, par hai, label 1,2-dimethylbenzene, ortho-xylene. Doosri mein woh carbon 3 par hai, label 1,3, meta. Teesri mein woh bilkul saamne carbon 4 par hai, label 1,4, para.

Teen ya zyada substituents. Ab o, m aur p kaam nahi karte. Locants ka lowest set use karo aur prefixes alphabetically likho; agar parent ka common name hai, to uska apna group C1 hota hai. Examples: 1,2,4-trichlorobenzene (1,3,4- ya 1,2,5- nahi), 4-chloro-2-nitrophenol (C1 par OH) aur 2,4,6-trinitrotoluene, yaani explosive TNT (C1 par $C H_{3}$ ).

Jab benzene ring kisi aisi chain se judi ho jis par functional group hai, to ring ka naam substituent phenyl, $C_{6} H_{5}^{-}$ (Ph), ki tarah rakha jaata hai. 2-Phenylethanol, $C_{6} H_{5} C H_{2} C H_{2} O H$ , gulaab ki khushboo ka bada hissa deta hai.

Example 8.10NCERT Problem 8.10 · p. 270

Benzene derivatives ke structures

(i) m-chlorotoluene, (ii) p-aminophenol, (iii) 1-phenylpropan-2-one aur (iv) 1,3-dichloro-5-nitrobenzene ke structures likho.

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

Quick check

4 mein se 1

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

ka IUPAC naam hai

4-oxobutan-2-ol3-hydroxybutanal2-hydroxybutan-4-al1-oxobutan-3-ol

Answer dekho

Answer (B) 3-hydroxybutanal

–CHO ki rank –OH se upar hai, isliye woh suffix (-al) hai aur uska carbon C1 hai; OH 3-hydroxy ban jaata hai. Ise alcohol ki tarah naam dena priority order ko ignore karna hai.

8.6 Isomerism

Ethanol, jo hand sanitisers ka alcohol hai, aur methoxymethane, jo aerosol propellant ki tarah use hone wali gas hai, dono ka molecular formula $C_{2} H_{6} O$ hai. Phir bhi ethanol ek liquid hai jo 78 °C par boil hota hai, jabki methoxymethane −24 °C par boil hoti hai.

Definition

Isomerism

Aise do ya zyada compounds ka hona jinka molecular formula same ho lekin structures alag hon, aur isliye properties bhi alag hon. Aise compounds isomers kehlate hain.

Isomers do bade types ke hote hain. Structural isomers mein fark hota hai ki kaun sa atom kis se bonded hai; stereoisomers mein bonds same hote hain lekin space mein atoms ka arrangement alag hota hai.

Structural isomerism
- chain
- position
- functional group
- metamerism
Stereoisomerism
- geometrical
- optical

8.6.1 Structural Isomerism

Fig. 8.9 ek hi formula $C_{4} H_{10} O$ ke andar structural isomerism ke chaaron types dhoondhta hai.

3/4Same formula, alag functional group: alcohol vs ether.

Fig. 8.9Ek formula, C₄H₁₀O, structural isomer ke chaar types.NCERT p. 270

Left mein butan-1-ol ko right mein ek doosre C₄H₁₀O compound se compare kiya gaya hai. Step 1, chain isomer: 2-methylpropan-1-ol. Step 2, position isomer: butan-2-ol. Step 3, functional group isomer: ethoxyethane, ek ether. Step 4, metamers: left mein ethoxyethane aur right mein 2-methoxypropane, aise ethers jinke oxygen ke dono taraf alag groups hain.

(i) Chain isomers ke skeletons alag hote hain: $C_{6} H_{14}$ ke paanch hain (hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, 2,3-dimethylbutane). (ii) Position isomers mein fark hota hai ki group kahan hai: 1-bromopropane aur 2-bromopropane. (iii) Functional group isomers ke groups alag hote hain: ethanol aur methoxymethane, ya propanoic acid aur methyl ethanoate. (iv) Metamers same class mein hote hain lekin functional group ke dono taraf alag alkyl groups hote hain: 1-methoxybutane aur ethoxypropane, dono $C_{5} H_{12} O$ .

8.6.2 Stereoisomerism

Stereoisomers mein same atoms same order mein jude hote hain, lekin space mein alag tarah se arranged hote hain. C=C ke around restricted rotation (Section 8.2.2) se geometrical isomers milte hain: cis-but-2-ene mein dono $C H_{3}$ groups double bond ke ek hi taraf hain, trans-but-2-ene mein ulti taraf. Optical isomers aise mirror images hain jo ek doosre par superimpose nahi hote, jaise left aur right haath (Unit 9 aur Class 12).

8.7 Organic Reaction Mechanism ke Fundamental Concepts

Zyaadatar organic reactions mein ek organic molecule, yaani substrate, ek attack karne wale reagent se milta hai. Bonds tootte aur bante hain, aksar thodi der tikne wale intermediates ke through, aur products bante hain, kabhi-kabhi by-products ke saath.

Overall equation batati hai ki reaction kahan se shuru hota hai aur kahan khatam. Reaction mechanism beech ki kahani batata hai: bonds kis order mein tootte aur bante hain, har electron pair ka raasta, aur har step ki energy aur speed. Kuch mechanisms samajh lo, to hazaaron reactions alag-alag ratne wale facts nahi rehte.

8.7.1 Covalent Bond ka Fission

Covalent bond ek shared electron pair hai, aur yeh do tarah se toot sakta hai (Fig. 8.10). Heterolytic cleavage (heterolysis) mein ek fragment dono electrons rakh leta hai, to ek cation ban jaata hai aur doosra anion. Homolytic cleavage (homolysis) mein har fragment ek electron rakhta hai.

2/4X pair le leta hai: empty p orbital wala ek planar, sp² carbocation.

Fig. 8.10Electrons kaun rakhta hai, isi se intermediate tay hota hai.NCERT Fig. 8.3 · p. 271

Teen R groups wala ek carbon, X se bonded. Step 1: bond aur uska shared pair do dots ki tarah. Step 2: ek arrow pair ko X tak le jaata hai, peeche empty p orbital wala planar carbocation aur X minus bachte hain. Step 3: carbon pair ko lone pair ki tarah rakh leta hai, ek pyramidal carbanion, Z plus ke bagal mein. Step 4: do fish-hook arrows har side ko ek electron dete hain: do free radicals.

Carbocations. Agar jaane wala group pair le le, to carbon ke paas sirf chhe valence electrons aur ek positive charge bachta hai: ek carbocation (purana naam: carbonium ion). Jaise, 2-bromo-2-methylpropane ka heterolysis tert-butyl cation deta hai:

\begin{aligned} (C H_{3})_{3} C - B r \\ \overset{}{\to} (C H_{3})_{3} C^{+} \\ + B r^{-} \end{aligned}

(8.2)

Chhe valence electrons ke saath positive carbon sp² hybridised hota hai: uske teen bonds 120° par flat phail jaate hain, aur bacha hua p orbital, us plane se right angle par, khaali hota hai. Carbocation kitni der tikta hai, yeh is par depend karta hai ki charge ke around kya hai. C⁺ se bonded carbons gino: ek ho to primary ( $C H_{3} C H_{2} C H_{2}^{+}$ ), do ho to secondary ( $C H_{3} C H^{+} C H_{2} C H_{3}$ ), teen ho to tertiary ( $(C H_{3})_{2} C^{+} C H_{2} C H_{3}$ ). Har alkyl group charge ki taraf electron density bhejta hai, isliye stability mein 3° > 2° > 1° > $C H_{3}^{+}$ ; Section 8.7.5 aur Section 8.7.9 samjhate hain ki kaise.

Carbanions. Agar carbon pair rakh le, to us par negative charge aur ek lone pair hota hai: ek carbanion. $C H_{3}^{-}$ jaisa simple carbanion sp³ aur trigonal pyramidal hota hai. Alkyl groups charge par electrons dhakelte hain aur use destabilise karte hain, isliye order ulta ho jaata hai: 3° < 2° < 1° < methyl. Heterolysis se hone wale reactions ionic, heteropolar ya polar reactions kehlate hain.

Free radicals. Chlorine par padne wali dhoop kuch $C l_{2}$ molecules ko barabar todti hai, har atom ko ek electron ((8.3)). Is tarah ka bond-breaking, jo heat ya light se hota hai aur half-headed 'fish-hook' arrows se draw kiya jaata hai, unpaired electron wale neutral fragments deta hai: free radicals. $C H_{3}^{∙}$ jaise carbon radicals thodi der hi tikte hain; carbocations ki tarah, inpar jitne zyada alkyl groups hon, yeh utne zyada stable hote hain (3° > 2° > 1° > methyl). Radicals ke through chalne wale reactions free-radical, homopolar ya nonpolar reactions kehlate hain.

C l_{2} \overset{light}{\to} 2 C l^{∙}

(8.3)

8.7.2 Substrate aur Reagent

Substrate kaun hai aur reagent kaun? Aam rule: substrate woh molecule hai jiska carbon naye bond mein jaata hai, aur doosra partner reagent hai. (8.4) aur (8.5) mein propene aur bromoethane substrates hain. Jab naya C–C bond banta hai, to dono partners carbon dete hain, aur jis molecule ko hum follow kar rahe hain use substrate kehte hain. Partners aksar neutral molecules hote hain, free ions nahi.

\begin{aligned} C H_{2} = C H C H_{3} \\ + H B r \\ \overset{}{\to} C H_{3} C H B r C H_{3} \end{aligned}

(8.4)

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

(8.5)

Nucleophiles aur electrophiles. Yahan bhi opposites attract karte hain. Jis reagent ke paas extra electrons hain (ek lone pair ya negative charge), woh substrate ke electron-poor hisse ki taraf jaata hai: polar bond ka positive end, ya woh atom jiska octet adhoora hai. Jis reagent ke paas electrons kam hain, woh ulti taraf jaata hai, lone pairs ya multiple bond ke khule π electrons ki taraf.

Definition

Nucleophile (Nu:)

Aisa reagent jo reaction site par ek electron pair laata hai ('nucleus-loving'). Tab reaction nucleophilic kehlata hai.

Example $H O^{-}$ , $C N^{-}$ , $C H_{3} O^{-}$ , carbanions; neutral $H_{2} O$ , $N H_{3}$ , $R N H_{2}$ apne lone pairs ke through.

Definition

Electrophile (E⁺)

Aisa reagent jo reaction site se ek electron pair leta hai ('electron-loving'). Tab reaction electrophilic kehlata hai.

Example $H^{+}$ , $N O_{2}^{+}$ , carbocations, $B F_{3}$ aur $A l C l_{3}$ (adhoore octets); C=O ya C–X ka carbon.

Electrophiles par charge hona zaroori nahi. $B F_{3}$ mein boron ke octet mein do electrons kam hain, aur C=O ya C–Cl bond ka carbon apne electronegative partner ki wajah se thoda positive reh jaata hai; dono electron pairs accept karte hain. Nucleophile se inka milna nucleophile ke pair se electrophilic centre tak ek curved arrow ki tarah draw kiya jaata hai.

Example 8.11NCERT Problem 8.11 · p. 273

Curved arrows ke saath heterolysis

Mark kiye gaye bond ka heterolytic cleavage dikhao aur bane intermediate ka naam batao: (i) $(C H_{3})_{3} C - C l$ , (ii) $C H_{3} - O C H_{3}$ , (iii) $C H_{3} C H_{2} - L i$ .

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

Example 8.12NCERT Problems 8.12, 8.13 · p. 273

Nucleophiles, electrophiles aur electrophilic centres

(i) $N H_{3}$ , $A l C l_{3}$ , $C N^{-}$ , $N O_{2}^{+}$ , $C H_{3} O^{-}$ , $(C H_{3})_{3} C^{+}$ aur $H_{2} O$ ko nucleophiles ya electrophiles mein classify karo. (ii) $C H_{3} C O C H_{3}$ , $C H_{3} C H_{2} C l$ aur $C H_{3} C O C l$ mein electrophilic centre pehchano.

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

8.7.3 Organic Reactions mein Electron Movement

Curved arrows mechanisms ki shorthand hain. Full arrow ek electron pair ko move karta hai: uski tail wahan se shuru hoti hai jahan pair abhi hai (lone pair ya bond) aur uska head wahan point karta hai jahan pair pahunchta hai (ek atom, ya do atoms ke beech jahan naya bond banta hai). Fig. 8.11 woh teen moves dikhata hai jo tum sabse zyada draw karoge.

Fig. 8.11Arrows electrons se shuru hote hain aur wahan khatam hote hain jahan electrons jaate hain.NCERT p. 274

Teen rows. Row 1: CH₂=CH–CH₂⁺ mein ek arrow π pair ko C1=C2 se C2–C3 position mein le jaata hai, aur ⁺CH₂–CH=CH₂ banta hai. Row 2: H₂C=O mein ek arrow π pair ko oxygen par le jaata hai, aur H₂C⁺–O⁻ banta hai. Row 3: H₂N–CH=CH₂ mein ek arrow nitrogen ke lone pair ko N–C bond position mein le jaata hai jabki doosra arrow C=C π pair ko end carbon par le jaata hai, aur H₂N⁺=CH–CH₂⁻ banta hai.

Half-headed fish-hook arrow ek single electron ko move karta hai, jaise homolysis mein. Substitution $H O^{-}$ + $C H_{3} C H_{2} I$ → $C H_{3} C H_{2} O H$ + $I^{-}$ mein ek arrow oxygen ke lone pair se carbon tak jaata hai aur doosra C–I bond se iodine tak. $C l_{2}$ → $2 C l^{∙}$ mein do fish-hooks Cl–Cl bond ke beech se shuru hote hain, ek-ek har atom ki taraf point karta hai.

8.7.4 Covalent Bonds mein Electron Displacement Effects

Bonding electrons ka permanent displacement molecule mein hamesha rehta hai: inductive aur resonance effects (aur hyperconjugation). Temporary displacement sirf tab aata hai jab reagent attack karta hai, aur uske jaate hi khatam ho jaata hai: electromeric (polarisability) effect.

Compare

Permanent aur temporary displacements

Permanent

Molecule mein hamesha maujood
Inductive effect: σ bonds ke along
Resonance effect: π systems ke through
Dipole moments aur acid strength mein dikhte hain

Temporary

Sirf jab reagent attack kare
Electromeric effect: π pair ka poora shift
Reagent hatane par khatam ho jaata hai

8.7.5 Inductive Effect

Ek line mein aage badhti rassa-kashi socho. 1-fluoropropane mein fluorine C–F electron pair ko apni taraf kheenchta hai (F par δ−, C1 par δ+). C1, jo ab thoda electron-poor hai, C1–C2 bond ko kheenchta hai, jisse C2 par chhota positive charge (δδ+) aata hai, aur C3 par usse bhi kam (δδδ+). σ bonds ka yeh step-by-step polarisation inductive effect hai. Har relay kheenchav ka sirf kuch hissa aage bhejta hai, isliye lagbhag teen bonds ke baad yeh na ke barabar hota hai.

\overset{δ δ δ +}{C H_{3}} \to \overset{δ δ +}{C H_{2}} \to \overset{δ +}{C H_{2}} \to \overset{δ -}{F}

(8.6)

Groups ko hydrogen se compare kiya jaata hai. Jo groups carbon se electron density door kheenchte hain woh −I effect dikhate hain: $- N O_{2}$ , –CN, –COOH, –COOR, halogens (F > Cl > Br > I), –OR aur $- O C_{6} H_{5}$ . Alkyl groups jis carbon se jude hain uski taraf electron density dhakelte hain aur +I effect dikhate hain, jo branching ke saath badhta hai: $- C H_{3}$ < $- C_{2} H_{5}$ < $- C H (C H_{3})_{2}$ < $- C (C H_{3})_{3}$ .

Yeh effect acid strength mein saaf dikhta hai. –COOH ke paas ek −I group carboxylate ion ke negative charge ko phaila deta hai aur use stabilise karta hai, isliye acid zyada ionise hota hai aur uska pKa kam hota hai. Chlorine ko –COOH se door le jaane par effect lagbhag khatam ho jaata hai (Table 8.6).

Table 8.6Chlorine ka −I effect doori ke saath kamzor hota hai (25 °C par pKa)

Acid	Formula	pKa
butanoic acid	$C H_{3} C H_{2} C H_{2} C O O H$	4.8
4-chlorobutanoic acid	$C l C H_{2} C H_{2} C H_{2} C O O H$	4.5
3-chlorobutanoic acid	$C H_{3} C H C l C H_{2} C O O H$	4.1
2-chlorobutanoic acid	$C H_{3} C H_{2} C H C l C O O H$	2.9

Example 8.14NCERT Problems 8.14, 8.15 · p. 275

Bond polarity aur induction ki pahunch

(i) Kaun sa bond zyada polar hai: $H_{3} C - C l$ ya $H_{3} C - I$ ; $H_{3} C - F$ ya $H_{3} C - O H$ ; $H_{3} C - B r$ ya $H_{3} C - L i$ ? (ii) 1-nitropropane, $C H_{3} C H_{2} C H_{2} N O_{2}$ , mein kis carbon par sabse chhota induced positive charge hai?

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

8.7.6 Resonance Structure

Kuch molecules ko kisi ek Lewis structure se describe nahi kiya ja sakta. Formate ion, $H C O O^{-}$ , mein Lewis structure ek C=O aur ek C–O⁻ bond dikhata hai, jinki length alag honi chahiye (lagbhag 120 pm aur 134 pm, jaise formic acid mein). Experiment mein dono C–O bonds ek jaise milte hain, lagbhag 125 pm. Do barabar achhe Lewis structures draw kiye ja sakte hain, aur asli ion dono ka mixture hai (Fig. 8.12).

3/3Asli ion: dono C–O bonds 125 pm, har O par −½.

Fig. 8.12Do equivalent drawings; barabar bonds wala ek asli ion.NCERT p. 275

Formate ion. Step 1, structure I: upar C=O aur neeche C–O minus, arrows neeche wale oxygen ke lone pair ko bond mein aur C=O π pair ko upar wale oxygen par le jaate hain. Step 2, structure II: double bond aur charge ki jagah badal gayi hai. Step 3, hybrid: har C–O bond solid plus dashed draw kiya gaya, har oxygen par minus aadha, dono bonds 125 pm.

Yeh drawings resonance structures (canonical ya contributing structures) hain. Yeh imaginary hain: ion inke beech palatta nahi rehta. Asli species, yaani resonance hybrid, ka ek hi structure hai jise koi ek drawing poori tarah nahi dikha sakti, aur contributors ko double-headed arrow (↔) se joda jaata hai.

Benzene iska classic example hai: iske chhe C–C bonds barabar hain, aur yeh Kekulé structure ke andaaze se kahin zyada stable hai. Cyclohexene ko hydrogenate karne par lagbhag 120 kJ/mol nikalta hai, to teen aise double bonds se lagbhag 360 kJ/mol nikalna chahiye, lekin benzene sirf lagbhag 208 kJ/mol deta hai. Yeh fark, lagbhag 150 kJ/mol, iski resonance energy hai: woh energy jitna hybrid sabse stable contributor se neeche hota hai. Yeh tab sabse zyada hoti hai jab milti-julti energy wale kai important contributors hon.

Resonance structures likhna. Sirf electrons (lone pairs aur π electrons) move karo, nuclei kabhi nahi, aur unpaired electrons ki ginti same rakho. Valid structures barabar important nahi hote. Ek structure zyada count karta hai jab:

har atom ka octet poora ho (H ka duplet);
usmein sabse kam separated charges hon;
negative charge zyada electronegative atom par ho, aur positive charge zyada electropositive atom par;
usmein covalent bonds ki ginti zyada ho;
charge ek atom par jama hone ke bajaye kai atoms par phaila ho.

Example 8.16NCERT Problems 8.16, 8.17 · p. 276

Resonance structures draw karna aur rank karna

(i) Allyl anion, $C H_{2} = C H - C H_{2}^{-}$ , ke resonance structures curved arrows ke saath likho. (ii) Propanone ke enolate ion, $C H_{2} = C (O^{-}) C H_{3}$ , ke do structures likho aur batao kaun zyada contribute karta hai.

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

Example 8.18NCERT Problem 8.18 · p. 276

Koi structure sirf minor contributor kyun hota hai

Methoxyethene ke liye samjhao ki $_{}^{}_{}^{-} C H_{2} - C H = O^{+} - C H_{3}$ (II), $C H_{2} = C H - O - C H_{3}$ (I) ke comparison mein sirf minor contributor kyun hai.

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

8.7.7 Resonance Effect

Delocalisation sirf stabilise hi nahi karta: yeh conjugated system ke ek end se doosre end tak electron density shift karta hai. Yeh shift, yaani resonance (mesomeric) effect, R ya M, tab dikhta hai jab ek π bond doosre π bond ke bagal mein ho ya lone pair wale atom ke bagal mein ho.

+R effect. Ek group conjugated system mein electron density donate karta hai, aksar lone pair se. Methoxybenzene mein oxygen ka lone pair ring mein dhakela jaata hai, aur resonance structures do ortho carbons aur para carbon par negative charge rakhte hain (Fig. 8.13). +R groups: halogens, –OH, –OR, –OCOR, $- N H_{2}$ , –NHR, $- N R_{2}$ aur –NHCOR.

−R effect. Ek group conjugated system se π electrons kheenchta hai. Benzaldehyde mein ring ke π electrons carbonyl oxygen ki taraf kheenche jaate hain, aur ortho aur para carbons par positive charge reh jaata hai. −R groups: –COOH, –CHO, >C=O, –CN aur $- N O_{2}$ .

Fig. 8.13+R ortho aur para carbons ko electron-rich banata hai; −R unhe electron-poor banata hai.NCERT p. 276

Do benzene rings ek ke bagal mein ek. Left, methoxybenzene: OCH₃ oxygen ke lone pair se ring mein ek curved arrow, aur do ortho carbons aur para carbon par δ− labels. Right, benzaldehyde: ring se CHO oxygen ki taraf ek curved arrow, aur ortho aur para carbons par δ+ labels.

Ek chhod kar ek single aur double bonds, jaise β-carotene ki lambi chain mein jo gaajar ko orange rang deti hai, ya lone pair ke bagal mein ek π bond, ek conjugated system banate hain. Iske delocalised π electrons ise aisi polarity aur stability dete hain jo koi ek Lewis structure predict nahi karta.

8.7.8 Electromeric Effect (E effect)

Double bond kisi aane wale par bhi react kar sakta hai. Jaise hi koi reagent C=C ya C=O bond ke paas aata hai, poora π pair do atoms mein se ek par shift ho sakta hai, aur agar reagent bina react kiye chala jaaye to waapas aa jaata hai. Yeh temporary, poora-ya-kuch-nahi wala shift, jo π bond se ek curved arrow ki tarah draw hota hai, electromeric effect (E) hai. Multiple bond aur attack karne wale reagent ke bina koi E effect nahi hota.

+E effect mein pair us atom par jaata hai jisse reagent judta hai. Jab $H^{+}$ propene par attack karta hai, to π pair C1 ko proton se jodta hai aur C2 positive reh jaata hai:

\begin{aligned} C H_{2} = C H C H_{3} \\ + H^{+} \\ \overset{}{\to} C H_{3} C H^{+} C H_{3} \end{aligned}

(8.7)

−E effect mein pair us atom se door jaata hai jisse reagent judta hai. Jab cyanide propanone ke carbonyl carbon par attack karta hai, to π pair oxygen par chala jaata hai:

\begin{aligned} (C H_{3})_{2} C = O \\ + C N^{-} \\ \overset{}{\to} (C H_{3})_{2} C \\ (C N) O^{-} \end{aligned}

(8.8)

Jab inductive aur electromeric effects ulti directions mein kaam karte hain, to electromeric effect jeetta hai: attack ke pal mein wahi bada shift hota hai.

8.7.9 Hyperconjugation

Alkyl groups ke paas ek aur trick hai. Empty p orbital ya π bond ke bagal wale carbon par ek C–H bond apne σ electrons ko side mein us orbital mein jhuka sakta hai, bas shart yeh hai ki woh us orbital ke parallel line up ho. σ density ka yeh rasaav, hyperconjugation, molecule ki ek permanent, stabilising feature hai. Fig. 8.14 ise isopropyl cation, $(C H_{3})_{2} C H^{+}$ , mein dikhata hai.

3/3Partial overlap positive charge ko C–H hydrogen par phaila deta hai.

Fig. 8.14Line up hua C–H bond empty p orbital mein σ electron density bhejta hai.NCERT Fig. 8.4 · p. 277

Isopropyl cation: empty p orbital wala ek central carbon, H aur do methyl carbons se bonded. Step 1: methyl ka ek C–H bond side mein point karta hai. Step 2: woh C–H bond empty p orbital ke parallel ghoom gaya hai. Step 3: ek shaded region C–H bond aur p orbital ko jodta hai, hydrogen aur central carbon par δ+ ke saath.

Ise aksar 'no-bond resonance' ki tarah likha jaata hai, aise structures ke saath jinmein ek α hydrogen carbon se bina bond ke $H^{+}$ ki tarah dikhta hai. Ethyl cation ke liye:

\begin{aligned} H - C H_{2} - C H_{2}^{+} \\ ⟷ H^{+} C H_{2} = C H_{2} \end{aligned}

(8.9)

Jitne zyada α hydrogens, utna zyada stabilisation: $(C H_{3})_{3} C^{+}$ (9 α-H) > $(C H_{3})_{2} C H^{+}$ (6) > $C H_{3} C H_{2}^{+}$ (3) > $C H_{3}^{+}$ (0). Alkenes ko bhi fayda hota hai: but-2-ene (6 α-H) but-1-ene (2) se zyada stable hai. Isi wajah se π system se juda alkyl group electron donor ki tarah kaam karta hai.

Example 8.19NCERT Problem 8.19 · p. 278

Hyperconjugation se carbocations ka order

$C H_{3} C H_{2}^{+}$ , $(C H_{3})_{2} C^{+} C H_{2} C H_{3}$ aur $(C H_{3})_{2} C H^{+}$ ko stability ke order mein lagao, aur samjhao.

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

8.7.10 Organic Reactions aur Mechanisms ke Types

Zyaadatar organic reactions chaar types mein aate hain, jinhe tum Unit 9 aur Class 12 mein detail mein padhoge.

Table 8.7Organic reaction ke chaar typesNCERT p. 278

Type	Kya hota hai	Example
Substitution	ek atom ya group doosre ki jagah le leta hai	$(C H_{3})_{3} C B r$ + $H_{2} O$ → $(C H_{3})_{3} C O H$ + $H B r$
Addition	do parts multiple bond ke across judte hain	$C H_{3} C H = C H_{2}$ + $B r_{2}$ → $C H_{3} C H B r C H_{2} B r$
Elimination	padosi carbons se atoms ya groups nikalte hain, aur multiple bond banta hai	$C H_{3} C H_{2} C H_{2} B r$ + $K O H$ → $C H_{3} C H = C H_{2}$ + $K B r$ + $H_{2} O$ (ethanol mein)
Rearrangement	ek molecule ke atoms rearrange hokar ek isomer banate hain	butane → 2-methylpropane ( $A l C l_{3}$ , HCl, heat)

8.8 Organic Compounds ki Purification ke Methods

Chahe chemist neem ki pattiyon se compound nikale ya flask mein banaye, pehla product hamesha ek mixture hota hai. Use saaf karne ka tareeka chunne ke liye poochho: compound solid hai ya liquid? Kya woh aasani se vapour banta hai? Kya woh paani mein ghulta hai? Jawab paanch methods mein se ek ki taraf le jaate hain: sublimation, crystallisation, distillation, differential extraction ya chromatography.

Humein kaise pata chale ki compound pure hai? Pure solid ek fixed temperature par ekdum se pighalta hai aur pure liquid ek fixed temperature par boil hota hai, jabki impurities melting range ko neeche le aati hain aur chauda kar deti hain. Aaj chemists purity chromatography aur spectroscopy se bhi check karte hain.

8.8.1 Sublimation

Kuch solids garam karne par bina pighle seedhe vapour ban jaate hain, aur vapour thandi surface par seedha waapas solid ban jaata hai. Isliye ulte funnel ya thandi dish ke neeche halka garam karne se sublime hone wala solid, jaise camphor (kapoor) ya naphthalene, un impurities se alag ho jaata hai jo sublime nahi hoti.

8.8.2 Crystallisation

Crystallisation mein aisa solvent use hota hai jismein compound thanda hone par kam ghulta hai lekin garam hone par achhe se ghulta hai. Impure solid ko kam se kam garam solvent mein ghulao, garam hi filter karo, aur thanda karo: pure compound crystallise hota hai aur filter kar liya jaata hai, jabki mother liquor mein soluble impurities reh jaati hain.

Benzoic acid ke liye paani theek hai: 100 °C par 100 mL mein lagbhag 5.6 g ghulta hai lekin 25 °C par sirf lagbhag 0.34 g. Coloured impurities ko filter karne se pehle thode activated charcoal ke saath boil karke hataya jaata hai. Agar koi ek solvent kaam na kare, to ek pair use hota hai, ek jo compound ko aasani se ghulaye aur ek jo mushkil se. Jin impurities ki solubility compound ke kareeb ho, unke liye baar-baar crystallisation karni padti hai.

8.8.3 Distillation

Simple distillation (i) volatile liquid ko non-volatile impurities se aur (ii) aise liquids ko jinke boiling points kaafi door hon, alag karta hai. Kam boiling point wala liquid pehle vapour banta hai; uska vapour water condenser mein thanda hota hai aur liquid collect kiya jaata hai. Jab thermometer reading achanak badhti hai, to receiver badal diya jaata hai. Hexane (342 K) aur nitrobenzene (484 K) aise hi alag hote hain (Fig. 8.15).

3/3Fractionating column do saaf plateaus waapas le aata hai.

Fig. 8.15Flat steps ka matlab hai ki pure liquid distil ho raha hai.NCERT Figs. 8.5, 8.6 · p. 279

Still head par temperature, collect kiye gaye volume ke against. Step 1: do door-door boiling points par do flat steps, beech mein tez badhat, har fraction labelled. Step 2: paas-paas boil hone wale do liquids ke liye ek dashed curve dheere-dheere badhta hai, koi flat step nahi: mixed fractions. Step 3: fractionating column ke saath paas-paas boil hone wala pair phir se do saaf flat steps deta hai.

Fractional distillation. Paas-paas boiling points hon to dono liquids saath mein vapour bante hain. Flask aur condenser ke beech ek fractionating column (glass beads se bhara ya plates wala) upar jaate vapour ko neeche girte condensate se ek badi surface par milwata hai, isliye condensation aur re-vaporisation kai baar repeat hote hain; har stage ek theoretical plate hai. Top se nikalne wala vapour lagbhag pure kam boiling point wala liquid hota hai. Gujarat ke Jamnagar jaisi refineries saikdon plates wale columns se crude oil ko LPG, petrol, kerosene, diesel aur bhaari fractions mein alag karti hain.

Reduced pressure par distillation. Liquid tab boil hota hai jab uska vapour pressure uske upar ke pressure ke barabar ho, isliye water ya vacuum pump use uske normal boiling point se kaafi neeche boil karwa deta hai. Yeh un liquids ke liye theek hai jo bahut zyada temperature par boil hote hain ya boil hone se pehle decompose ho jaate hain; jaise refineries crude oil ke bhaari residue ko vacuum mein distil karti hain taaki lubricating oils bina crack hue mil jaayein.

Steam distillation. 405 K par boil hone wale liquid ko, ya aise liquid ko jo apne boiling point par jal jaaye, utna garam kiye bina kaise distil karein? Agar woh paani mein nahi milta aur steam mein volatile hai, to usmein se steam guzaaro. Do immiscible liquids apna-apna vapour pressure rakhte hain, isliye mixture tabhi boil ho jaata hai jab organic liquid ka vapour pressure $p_{1}$ aur paani ka $p_{2}$ milkar uske upar ke pressure $p$ ke barabar ho jaayein:

p = p_{1} + p_{2}

(8.10)

Kyunki $p_{1}$ paani ke vapour pressure mein judta hai, mixture 373 K se neeche boil hota hai, compound ke apne boiling point se kaafi neeche (Fig. 8.16). Condensate separating funnel mein do layers mein alag ho jaata hai. Eucalyptus aur lemongrass oil jaise essential oils aise hi banaye jaate hain, aur laboratory mein o-nitrophenol ko p-nitrophenol se alag kiya jaata hai, jo steam volatile nahi hai.

Fig. 8.16Chlorobenzene (b.p. 405 K) lagbhag 364 K par steam-distil hota hai.NCERT p. 281

323 K se 378 K tak temperature ke against mm Hg mein vapour pressure ka graph. Paani ka curve 373 K par 760 mm line tak pahunchta hai. Chlorobenzene ka neeche wala curve dheere badhta hai. Dono curves ka sum lagbhag 364 K par 760 mm ko cross karta hai, jahan ek dot laga hai: yeh mixture ka boiling point hai, chlorobenzene ke apne boiling point 405 K se kaafi neeche.

8.8.4 Differential Extraction

Chai ki pattiyon mein caffeine hota hai, jo paani se zyada dichloromethane mein ghulta hai. Thande chai ke infusion ko separating funnel mein dichloromethane ke saath hilao, do immiscible layers ko settle hone do, neeche wali organic layer nikaalo aur use evaporate karo: caffeine peeche bach jaata hai. Yeh differential extraction hai. Agar compound organic solvent ko sirf thoda zyada pasand karta hai, to ek baar hilane se kam nikalta hai, isliye continuous extraction mein wahi solvent solution mein se baar-baar recycle kiya jaata hai.

8.8.5 Chromatography

1906 mein botanist Mikhail Tsvet ne ek leaf extract ko solvent ke saath chalk powder ke column mein se guzaara; pigments rangeen bands mein alag ho gaye, aur unhone is method ka naam chromatography ('colour writing') rakha. Tab se har version mein ek stationary phase (ek solid, ya solid par tika liquid) hota hai jo components ko rokta hai aur ek mobile phase (liquid ya gas) jo unhe saath le jaata hai. Jo components zyada mazbooti se pakde jaate hain woh dheere chalte hain, isliye alag ho jaate hain; yeh method compounds ko alag karta hai, purify karta hai aur unki purity check karta hai.

Adsorption chromatography is baat ka use karti hai ki components silica gel ya alumina par alag-alag had tak adsorb hote hain. Column chromatography mein adsorbent stopcock wali glass tube mein bhara jaata hai. Mixture top par daala jaata hai aur eluant neeche behta hai; sabse zyada adsorb hone wala component top ke paas rehta hai, baaki alag-alag speed se neeche chalte hain aur ek-ek karke collect kiye jaate hain.

Thin layer chromatography (TLC) mein glass plate (chromaplate) par lagbhag 0.2 mm moti adsorbent layer hoti hai. Mixture ka ek spot neeche se lagbhag 2 cm upar base line par lagaya jaata hai, aur plate thode eluant wale band jar mein khadi ki jaati hai. Jaise-jaise eluant upar chadhta hai, har component ek characteristic distance tak chalta hai, jise uske retardation factor ki tarah record kiya jaata hai:

R_{f} = \frac{x}{y}

(8.11)

jahan

$x$: substance dwara base line se tay ki gayi distance
$y$: solvent front dwara base line se tay ki gayi distance

Coloured spots seedhe dikhte hain. Colourless spots ko reagent spray karke (ninhydrin amino acids ko purple kar deta hai), plate ko kuch iodine crystals wale band jar mein rakh kar (brown spots) ya, fluoresce karne wale compounds ke liye, ultraviolet light mein dekha ja sakta hai. Fig. 8.17 mein ek run follow karo, phir TLC lab try karo.

4/4Rf = x / y, dono base line se naape gaye.

Fig. 8.17Har component solvent ki chadhaai ke apne hisse par ruk jaata hai.NCERT Fig. 8.12 · p. 283

Ek jar mein thode solvent ke saath TLC plate. Step 1: pencil base line par ek spot. Step 2: solvent front kuch upar chadh gaya hai aur spot tootna shuru hota hai. Step 3: run ke end ke paas alag-alag heights par teen spots. Step 4: front mark kiya gaya hai aur arrows beech wale spot tak x aur front tak y dikhate hain, Rf = x/y ke saath.

InteractiveTLC lab

Ek drawn TLC plate, base line aur eluant polarity ke liye ek slider ke saath, pure hexane se hexane–ethyl acetate mixture tak. Buttons teen mixtures mein se ek chunte hain. Develop solvent ko plate par upar chadhata hai; spots ek simple adsorption model se tay heights tak chadhte hain, zyada polar compounds neeche rehte hain. Phir ek table har spot ke liye x, y aur Rf list karti hai, aur ek message batata hai ki spots achhe se alag hue ya nahi. Values sirf samjhane ke liye hain, measured data nahi.

Fig. 8.18Ek mixture aur eluant ki polarity chuno, Develop dabao, phir har spot ka Rf naapo.

Partition chromatography is par depend karti hai ki components ek stationary liquid aur mobile phase ke beech kaise bat jaate hain. Paper chromatography mein stationary phase khaas paper ke cellulose mein ruka paani hota hai; solvent capillary action se spot lagi strip par upar chadhta hai, aur developed strip (chromatogram) ko TLC plate ki tarah padha jaata hai.

Try karke dekho

Black sketch pen ke colours alag karo

Tumhe chahiye

filter paper ya coffee filter ki ek strip
ek black water-based sketch pen
ek lamba glass aur paani
ek pencil aur ek ruler

Pehle guess karo

Jab paani paper par upar chadhega to black dot ka kya hoga?

ek hi black spot rahegacoloured bands mein bat jaayegapoori tarah gayab ho jaayega

Filter-paper strip ke neeche se 2 cm upar pencil se line kheencho aur us par black sketch-pen ink ka ek chhota dot lagao.
Strip ko thode paani wale lambe glass mein latkao, taaki paani line ke neeche paper ko chhuye.
Jab paani lagbhag top tak pahunch jaaye, strip nikaalo aur front mark karo. Har colour ke liye x aur paani ke liye y naapo, aur har Rf nikaalo.

Table 8.8

Colour	x (cm)	Rf

Tumhari entries apne aap save hoti hain.

Quick check

4 mein se 1

Naphthalene ko common salt se alag karne ka sabse achha method kaun sa hai?

sublimationsteam distillationfractional distillationdifferential extraction

Answer dekho

Answer (A) sublimation

Naphthalene sublime hota hai aur salt nahi. Distillations ke liye liquids chahiye.

8.9 Organic Compounds ka Qualitative Analysis

Har organic compound mein carbon hota hai aur lagbhag sab mein hydrogen; kai mein oxygen, nitrogen, sulphur, halogens ya phosphorus bhi hote hain. Qualitative analysis batata hai ki inmein se kaun se maujood hain.

8.9.1 Carbon aur Hydrogen ka Detection

Compound ko dry copper(II) oxide ke saath garam karo. Carbon oxidise hokar carbon dioxide banta hai, jo lime water ko milky kar deta hai ( $C O_{2}$ + $C a (O H)_{2}$ → $C a C O_{3} ↓$ + $H_{2} O$ ), aur hydrogen paani banta hai, jo safed anhydrous copper(II) sulphate ko blue kar deta hai ( $C u S O_{4}$ + $5 H_{2} O$ → $C u S O_{4} \cdot 5 H_{2} O$ ):

\begin{aligned} C + 2 C u O \\ \overset{Δ}{\to} 2 C u + C O_{2} \\ 2 H + C u O \\ \overset{Δ}{\to} C u + H_{2} O \end{aligned}

(8.12)

8.9.2 Doosre Elements ka Detection

Organic compounds mein nitrogen, sulphur, halogens aur phosphorus covalently bonded hote hain, isliye yeh aam ionic tests nahi dete. Lassaigne's test mein compound ka thoda hissa ignition tube mein sodium ke chhote tukde ke saath fuse kiya jaata hai, jo inhe ionic sodium salts mein badal deta hai (X = Cl, Br ya I):

\begin{aligned} N a + C + N \\ \overset{Δ}{\to} N a C N \\ 2 N a + S \\ \overset{Δ}{\to} N a_{2} S \\ N a + X \overset{Δ}{\to} N a X \end{aligned}

(8.13)

Laal-garam tube ko distilled water mein daala jaata hai, boil kiya jaata hai aur filter kiya jaata hai. Filtrate, yaani sodium fusion extract, mein har element ka test kiya jaata hai (Fig. 8.19).

4/4HNO₃ ke saath boil karo, AgNO₃ daalo: AgCl white, AgBr pale yellow, AgI yellow.

Fig. 8.19Ek fusion, phir har element ke liye alag test.NCERT p. 284

Ek flow diagram. Step 1: compound aur sodium ignition tube mein fuse kiye jaate hain, paani mein daale jaate hain aur filter karke sodium fusion extract milta hai. Step 2: nitrogen test, pehle iron(II) sulphate phir sulphuric acid, Prussian blue deta hai. Step 3: sulphur tests black lead sulphide ya violet colour dete hain. Step 4: nitric acid ke baad silver nitrate white, pale yellow ya yellow halide precipitates deta hai.

(A) Nitrogen. Extract ko iron(II) sulphate ke saath boil karo, phir concentrated sulphuric acid se acidify karo: Prussian blue colour ya precipitate nitrogen dikhata hai. Is colour ke peeche do steps hain: cyanide ions $F e^{2 +}$ ko gher kar complex ion $[F e (C N)_{6}]^{4 -}$ banate hain, aur garam acid iron(II) ke kuch hisse ko iron(III) mein oxidise kar deta hai, jo in complex ions ke saath milkar gehra neela solid iron(III) hexacyanidoferrate(II) banata hai:

\begin{aligned} 6 C N^{-} + F e^{2 +} \\ \overset{}{\to} [F e (C N)_{6}]^{4 -} \\ 3 [F e (C N)_{6}]^{4 -} \\ + 4 F e^{3 +} \\ \overset{}{\to} F e_{4} [F e (C N)_{6}]_{3} \end{aligned}

(8.14)

(B) Sulphur. Acetic acid se acidify karne par extract lead acetate ke saath black precipitate deta hai, $S^{2 -}$ + $P b^{2 +}$ → $P b S ↓$ , aur sodium nitroprusside ke saath violet colour, jo $[F e (C N)_{5} N O S]^{4 -}$ se aata hai. N aur S dono wala compound sodium thiocyanate de sakta hai: tab extract iron(III) ke saath blood red ho jaata hai, $[F e (S C N)]^{2 +}$ banakar, aur Prussian blue nahi deta. Extra sodium ise tod deta hai, $N a S C N$ + $2 N a$ → $N a C N$ + $N a_{2} S$ .

(C) Halogens. Extract ko nitric acid se acidify karo, boil karo, phir silver nitrate daalo, $X^{-}$ + $A g^{+}$ → $A g X ↓$ . Aqueous ammonia mein ghulne wala white precipitate chlorine dikhata hai; pale yellow aur thoda-sa ghulne wala, bromine; yellow aur na ghulne wala, iodine. Pehle nitric acid ke saath boil karna cyanide aur sulphide ko khatam kar deta hai, warna woh AgCN aur $A g_{2} S$ ki tarah precipitate hokar galat result dete.

(D) Phosphorus. Compound ko sodium peroxide jaise oxidising agent ke saath garam karo taaki phosphorus phosphate ban jaaye. Nitric acid ke saath boil karo, jo phosphate ko phosphoric acid bana deta hai ( $N a_{3} P O_{4}$ + $3 H N O_{3}$ → $H_{3} P O_{4}$ + $3 N a N O_{3}$ ), aur ammonium molybdate daalo: ammonium phosphomolybdate ka yellow colour ya precipitate phosphorus dikhata hai, $H_{3} P O_{4}$ + $12 (N H_{4})_{2} M o O_{4}$ + $21 H N O_{3}$ → $(N H_{4})_{3} P O_{4} \cdot 12 M o O_{3}$ + $21 N H_{4} N O_{3}$ + $12 H_{2} O$ .

Table 8.9Lassaigne's test ek nazar mein

Element	Extract par test	Positive result
N	$F e S O_{4}$ , boil; conc. $H_{2} S O_{4}$	Prussian blue
S	acetic acid + lead acetate	black PbS
S	sodium nitroprusside	violet
N aur S	$F e^{3 +}$	blood red
Cl, Br, I	$H N O_{3}$ , boil; $A g N O_{3}$	white, pale yellow, yellow
P	$N a_{2} O_{2}$ ke baad: $H N O_{3}$ + ammonium molybdate	yellow

8.10 Quantitative Analysis

Quantitative analysis compound mein har element ka mass per cent nikalta hai. Jaisa tumne Unit 1 mein dekha, yeh percentages empirical formula dete hain, aur molar mass ke saath molecular formula. Har classical method ek element ko aisi cheez mein badalta hai jise taula ya naapa ja sake.

8.10.1 Carbon aur Hydrogen

Ek taula hua sample garam copper(II) oxide ke upar dry oxygen mein poori tarah jalaya jaata hai, taaki uska saara carbon carbon dioxide ban jaaye aur saara hydrogen paani:

\begin{aligned} C_{x} H_{y} + (x + \frac{y}{4}) O_{2} \\ ⟶ x C O_{2} + \frac{y}{2} H_{2} O \end{aligned}

(8.15)

Gases pehle anhydrous calcium chloride ki taul li gayi U-tube se guzarti hain, jo paani absorb karti hai, phir concentrated potassium hydroxide solution wali tube se, jo carbon dioxide absorb karti hai (Fig. 8.20). Order zaroori hai, kyunki KOH solution paani bhi le leta. Har tube ke mass mein badhotri se paani aur carbon dioxide ke masses milte hain.

3/3KOH solution mein carbon dioxide ka mass, m₂, jud jaata hai.

Fig. 8.20Paani aur carbon dioxide ko isi order mein taulo.NCERT Fig. 8.14 · p. 286

Ek combustion tube jismein left se oxygen aati hai, ek sample boat aur copper(II) oxide, neeche se garam, uske baad do U-tubes. Step 1: sample jalta hai aur labelled arrows CO₂ aur H₂O ko right le jaate hain. Step 2: pehli U-tube, anhydrous calcium chloride, highlight hoti hai kyunki usmein paani judta hai. Step 3: doosri, potassium hydroxide solution, highlight hoti hai kyunki usmein carbon dioxide judti hai.

DerivationCarbon aur hydrogen ka per cent

Maano mass $m$ ka sample $m_{1}$ g paani aur $m_{2}$ g carbon dioxide deta hai. $C O_{2}$ ke har mole (44 g) mein 12 g carbon hota hai, aur $H_{2} O$ ke har mole (18 g) mein 2 g hydrogen.

% C = \frac{12 m_{2} \times 100}{44 m}

(8.16)

% H = \frac{2 m_{1} \times 100}{18 m}

(8.17)

Example 8.20NCERT Problem 8.20 · p. 286

Carbon aur hydrogen ka percentage

Complete combustion par ek organic compound ke 0.290 g ne 0.660 g carbon dioxide aur 0.270 g paani diya. Carbon aur hydrogen ke percentages, oxygen (jo akela doosra element hai) ka percentage aur empirical formula nikaalo.

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

8.10.2 Nitrogen

(i) Dumas method. Compound ko carbon dioxide ke atmosphere mein copper(II) oxide ke saath garam kiya jaata hai. Carbon aur hydrogen $C O_{2}$ aur paani ban jaate hain, aur nitrogen $N_{2}$ ki tarah free ho jaata hai:

\begin{aligned} C_{x} H_{y} N_{z} \\ + (2 x + \frac{y}{2}) C u O \\ \to x C O_{2} + \frac{y}{2} H_{2} O \\ + \frac{z}{2} N_{2} \\ + (2 x + \frac{y}{2}) C u \end{aligned}

(8.18)

Nitrogen oxides garam copper gauze ke upar $N_{2}$ mein reduce ho jaate hain, aur gases nitrometer mein KOH solution ke upar collect ki jaati hain, jo $C O_{2}$ absorb kar leta hai aur nitrogen ko naapne ke liye chhod deta hai. Gas water vapour se saturated hoti hai, isliye uska apna pressure atmospheric pressure minus aqueous tension hota hai.

DerivationDumas method se nitrogen ka per cent

Compound ka mass $m$ , $T_{1}$ aur atmospheric pressure $p$ (mm Hg) par $V_{1}$ mL nitrogen deta hai; $T_{1}$ par aqueous tension $f$ hai. Volume ko STP (273 K, 760 mm Hg) par badlo, jahan 22 400 mL $N_{2}$ ka mass 28 g hota hai:

\begin{aligned} p_{1} & = p - f \\ V & = \frac{p_{1} V_{1}}{760} \times \frac{273}{T_{1}} \end{aligned}

(8.19)

% N = \frac{28 V \times 100}{22 400 m}

(8.20)

Seedhe $p V = n R T$ use karne se bhi same result milta hai.

Example 8.21NCERT Problem 8.21 · p. 287

Dumas method

Ek compound ke 0.200 g ne 42.2 mL nitrogen diya, jo 290 K aur 740 mm Hg par paani ke upar collect kiya gaya. 290 K par aqueous tension 14.5 mm Hg hai. Nitrogen ka percentage nikaalo.

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

(ii) Kjeldahl's method. Concentrated sulphuric acid ke saath digestion se nitrogen ammonium sulphate mein badal jaata hai, sodium hydroxide se ammonia ki tarah nikalta hai, aur standard acid ki naapi hui extra quantity mein absorb hota hai; bache hue acid ko back-titrate karke pata chalta hai ki ammonia ne kitna acid use kiya.

Step by step

Kjeldahl's method ke chaar stages

Step 1 / 4

Digest karo

Conc. $H_{2} S O_{4}$ (+ $C u S O_{4}$ , $K_{2} S O_{4}$ ) ke saath garam karo: N → $(N H_{4})_{2} S O_{4}$ .

1 / 4

DerivationKjeldahl's method se nitrogen ka per cent

Maano $m$ g compound ka ammonia molarity $M_{a}$ wale $H_{2} S O_{4}$ ke $V_{a}$ mL mein absorb hota hai, aur bache acid ke liye molarity $M_{b}$ wala NaOH $V_{b}$ mL lagta hai. Millimoles mein:

\begin{aligned} n (N H_{3}) & = 2 M_{a} V_{a} \\ - M_{b} V_{b} \end{aligned}

(8.21)

$N H_{3}$ ke har millimole mein 14 mg nitrogen hota hai, isliye $n$ mmol mein aur $m$ g mein lene par:

% N = \frac{1.4 n (N H_{3})}{m}

(8.22)

Agar acid aur alkali ki molarity same $M$ ho, to $n = M (2 V_{a} - V_{b})$ . HCl jaise monobasic acid ke saath, $n = M_{a} V_{a} - M_{b} V_{b}$ .

Kjeldahl's method nitro aur azo groups ke nitrogen aur pyridine jaisi rings ke nitrogen ke liye kaam nahi karta, kyunki in conditions mein woh nitrogen ammonium sulphate mein nahi badalta.

Example 8.22NCERT Problem 8.22 · p. 288

Kjeldahl's method

Ek compound ke 0.60 g se nikle ammonia ne 0.50 M $H_{2} S O_{4}$ ke 15.0 mL ko theek neutralise kiya. Nitrogen ka percentage nikaalo.

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

8.10.3 Halogens

Carius method. Ek taula hua sample fuming nitric acid aur silver nitrate ke saath ek sealed hard-glass Carius tube mein furnace ke andar garam kiya jaata hai. Carbon aur hydrogen oxidise ho jaate hain, aur halogen silver halide, AgX, banata hai, jise filter, wash, dry karke taula jaata hai. AgX ke ek mole mein X ka ek mole hota hai:

\begin{aligned} % X & = \frac{A_{X}}{M_{A g X}} \\ \times \frac{m_{1}}{m} \times 100 \end{aligned}

(8.23)

jahan

$A_{X}$: halogen ka atomic mass
$M_{A g X}$: AgX ka molar mass
$m_{1}$: bane hue AgX ka mass

8.10.4 Sulphur

Ek taula hua sample Carius tube mein fuming nitric acid ya sodium peroxide ke saath garam kiya jaata hai, jo uske sulphur ko sulphuric acid mein oxidise kar deta hai. Extra barium chloride barium sulphate ko precipitate karta hai, jise filter, wash, dry karke taula jaata hai. $B a S O_{4}$ ke ek mole (233 g) mein 32 g sulphur hota hai:

% S = \frac{32 m_{1}}{233 m} \times 100

(8.24)

Example 8.23NCERT Problems 8.23, 8.24 · p. 289

Iodine aur sulphur ke Carius estimations

(i) Ek organic iodo compound ke 0.156 g ne 0.235 g AgI diya (Ag = 108, I = 127). (ii) Ek doosre compound ke 0.180 g ne 0.350 g $B a S O_{4}$ diya. (i) mein iodine aur (ii) mein sulphur ka percentage nikaalo.

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

8.10.5 Phosphorus

Sample ko fuming nitric acid ke saath garam kiya jaata hai, jo phosphorus ko phosphoric acid mein oxidise kar deta hai. Ise ammonia aur ammonium molybdate se ammonium phosphomolybdate, $(N H_{4})_{3} P O_{4} \cdot 12 M o O_{3}$ (1877 g/mol, ek P), ki tarah precipitate kiya jaata hai, ya magnesia mixture se $M g N H_{4} P O_{4}$ ki tarah, jo jalaane par $M g_{2} P_{2} O_{7}$ (222 g/mol, do P = 62 g) deta hai:

\begin{aligned} % P & = \frac{31 m_{1}}{1877 m} \times 100 \\ ya & \frac{62 m_{1}}{222 m} \times 100 \end{aligned}

(8.25)

8.10.6 Oxygen

Oxygen aksar fark se nikaala jaata hai: %O = 100 − (baaki sab percentages ka sum). Ise seedhe bhi naapa ja sakta hai. Compound ko nitrogen mein garam karke decompose kiya jaata hai, aur products laal-garam coke (1373 K) ke upar se guzarte hain, jo saari oxygen ko carbon monoxide mein badal deta hai. Phir garam iodine pentoxide CO ko $C O_{2}$ mein oxidise karta hai aur iodine chhodta hai:

\begin{aligned} 2 C + O_{2} \\ \overset{1373 K}{\to} 2 C O \\ I_{2} O_{5} + 5 C O \\ \overset{}{\to} I_{2} + 5 C O_{2} \end{aligned}

(8.26)

Pehli equation ko 5 se aur doosri ko 2 se multiply karne par dikhta hai ki nikla hua $O_{2}$ ka har mole (32 g) aakhir mein $C O_{2}$ ke do moles (88 g) ban jaata hai. Isliye agar $m$ g compound $m_{1}$ g $C O_{2}$ deta hai, to %O = $32 m_{1} \times 100 / (88 m)$ ; iske bajaye iodine ko titrate bhi kiya ja sakta hai.

Modern laboratories automatic CHN elemental analysers use karti hain, jinhe sirf 1–3 mg sample chahiye aur jo kuch hi minutes mein carbon, hydrogen aur nitrogen bata dete hain. Estimation lab mein classical calculations ki practice karo.

InteractiveEstimation lab

Buttons chhe methods mein se ek chunte hain: carbon aur hydrogen ke liye combustion, Dumas, Kjeldahl, halogen ke liye Carius, sulphur ke liye Carius, aur magnesium pyrophosphate ki tarah phosphorus. Number boxes mein sample ka mass aur naape gaye masses, volumes, temperature, pressures ya titre daale jaate hain. Lab calculation ka har step formula aur element ke percentage ke saath dikhata hai. New problem button random realistic data bharta hai aur answer tab tak chhupa rakhta hai jab tak tum apna answer type karke Check nahi dabaate.

Fig. 8.21Ek method chuno, data daalo, aur calculation ko step by step follow karo; New problem ek practice question deta hai jise tum khud solve karo.

Quick check

3 mein se 1

Complete combustion par ek compound ke 0.20 g ne 0.44 g

C O_{2}

diya. Carbon ka percentage kya hai?

Answer dekho

Answer 60 %

% C = 12 \times 0.44 \times 100 / (44 \times 0.20) = 60 %

Recap

Summary aur key points

Carbon sp³ (tetrahedral), sp² (trigonal planar) ya sp (linear) hota hai; zyada s character chhote, mazboot bonds deta hai (Section 8.2.1).
π bond ko parallel p orbitals chahiye: yeh geometry fix karta hai, rotation rokta hai aur ek reactive site hai (Section 8.2.2).
Structures complete, condensed ya bond-line formulas ki tarah, aur 3-D mein wedges aur dashes ke saath draw kiye jaate hain (Section 8.3).
Compounds acyclic ya cyclic (alicyclic, aromatic) hote hain aur functional group ke hisaab se homologous series mein aate hain (Section 8.4).
IUPAC names: sabse lambi chain, lowest locants, alphabetical prefixes; principal functional group suffix deta hai (Section 8.5).
Isomers ka molecular formula same hota hai: structural (chain, position, functional group, metamers) ya stereoisomers (Section 8.6).
Heterolysis carbocations (planar) ya carbanions (pyramidal) deta hai; homolysis free radicals deta hai (Section 8.7.1).
Nucleophiles electron pair donate karte hain, electrophiles accept karte hain; curved arrows pairs ko track karte hain (Section 8.7.2, Section 8.7.3).
Inductive effect σ bonds ko polarise karta hai aur lagbhag teen bonds ke andar fade ho jaata hai (Section 8.7.5).
Resonance aur R effect π aur lone-pair electrons ko delocalise karte hain; hybrid kisi bhi contributor se zyada stable hota hai (Section 8.7.6, Section 8.7.7).
Electromeric effect ek temporary π shift hai; α C–H bonds ka hyperconjugation carbocations aur alkenes ko stabilise karta hai (Section 8.7.8, Section 8.7.9).
Organic reactions substitutions, additions, eliminations ya rearrangements hote hain (Section 8.7.10).
Purification mein sublimation, crystallisation, distillation, extraction aur chromatography use hote hain, $R_{f} = x / y$ ke saath (Section 8.8).
Lassaigne's test N, S aur halogens detect karta hai; combustion, Dumas, Kjeldahl aur Carius methods mass percentages dete hain (Section 8.9, Section 8.10).

Apna NCERT exercise question dhoondo

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 Exercises 8.1	Practice Q1
NCERT Exercises 8.2	Practice Q2
NCERT Exercises 8.3	Practice Q3
NCERT Exercises 8.4	Practice Q4
NCERT Exercises 8.5	Practice Q5
NCERT Exercises 8.6	Practice Q6
NCERT Exercises 8.7	Practice Q7
NCERT Exercises 8.8	Practice Q8
NCERT Exercises 8.9	Practice Q9
NCERT Exercises 8.10	Practice Q10
NCERT Exercises 8.11	Practice Q11
NCERT Exercises 8.12	Practice Q12
NCERT Exercises 8.13	Practice Q13
NCERT Exercises 8.14	Practice Q14
NCERT Exercises 8.15	Practice Q15
NCERT Exercises 8.16	Practice Q16
NCERT Exercises 8.17	Practice Q17
NCERT Exercises 8.18	Practice Q18
NCERT Exercises 8.19	Practice Q19
NCERT Exercises 8.20	Practice Q20
NCERT Exercises 8.21	Practice Q21
NCERT Exercises 8.22	Practice Q22
NCERT Exercises 8.23	Practice Q23
NCERT Exercises 8.24	Practice Q24
NCERT Exercises 8.25	Practice Q25
NCERT Exercises 8.26	Practice Q26
NCERT Exercises 8.27	Practice Q27
NCERT Exercises 8.28	Practice Q28
NCERT Exercises 8.29	Practice Q29
NCERT Exercises 8.30	Practice Q30
NCERT Exercises 8.31	Practice Q31
NCERT Exercises 8.32	Practice Q32
NCERT Exercises 8.33	Practice Q33
NCERT Exercises 8.34	Practice Q34
NCERT Exercises 8.35	Practice Q35
NCERT Exercises 8.36	Practice Q36
NCERT Exercises 8.37	Practice Q37
NCERT Exercises 8.38	Practice Q38
NCERT Exercises 8.39	Practice Q39
NCERT Exercises 8.40	Practice Q40

Quick check: MCQs

8.2.2 π Bonds ki Kuch Khaas Features4 questions
8.3.2 Organic Molecules ka Three-Dimensional Representation4 questions
8.4.2 Homologous Series4 questions
8.5.4 Substituted Benzene Compounds ki Nomenclature4 questions
8.6.2 Stereoisomerism4 questions
8.7.2 Substrate aur Reagent4 questions
8.7.6 Resonance Structure3 questions
8.7.10 Organic Reactions aur Mechanisms ke Types4 questions
8.8.5 Chromatography4 questions
8.9.2 Doosre Elements ka Detection3 questions
8.10.6 Oxygen3 questions

Glossary · 15

Key terms (English + Hindi)

carbanionकार्बऋणायनkarbrinayan: Negative charge wala carbon jis par ek lone pair hai; aksar sp³ hybridised aur pyramidal.Chapter mein pehli baar yahan aaya →
carbocationकार्बधनायनkarbdhanayan: Positive charge wala carbon jiske paas sirf chhe valence electrons hain; yeh sp² hybridised aur planar hota hai.Chapter mein pehli baar yahan aaya →
catenationश्रृंखलनshrinkhalan: Kisi element, khaas kar carbon, ke atoms ki aapas mein judkar chains aur rings banane ki ability.Chapter mein pehli baar yahan aaya →
chromatography: Mixture ko alag karna, is basis par ki uske components mobile phase ke saath stationary phase par alag-alag speed se chalte hain.Chapter mein pehli baar yahan aaya →

electromeric effect: Reagent ke attack ke dauraan π electron pair ka multiple bond ke ek atom par temporary, poora shift.Chapter mein pehli baar yahan aaya →
electrophileइलेक्ट्रॉनरागीelectronragi: Aisa reagent jo electron-rich centre se electron pair accept karta hai.

free radicalमुक्त मूलकmukt moolak: Unpaired electron wala atom ya group, jo bond ke homolytic cleavage se banta hai.Chapter mein pehli baar yahan aaya →
functional groupप्रकार्यात्मक समूहprakaryatmak samooh: Atom ya atoms ka group jo compound ko uske characteristic reactions deta hai, jaise –OH ya –COOH.

homologous seriesसमजातीय श्रेणीsamjatiya shreni: Same functional group wale compounds ki family jiske lagaatar members mein ek $C H_{2}$ unit ka fark hota hai.
hyperconjugationअतिसंयुग्मनatisanyugman: C–H bond ke σ electrons ka padosi π system ya empty p orbital mein delocalisation.Chapter mein pehli baar yahan aaya →

inductive effectप्रेरणिक प्रभावpreranik prabhav: Polar bond se σ bonds ke along aage badhta permanent polarisation; lagbhag teen bonds mein fade ho jaata hai.Chapter mein pehli baar yahan aaya →
isomerismसमावयवताsamavayavata: Same molecular formula wale alag-alag compounds ka hona; aise compounds isomers hain.

nucleophileनाभिकरागीnabhikragi: Aisa reagent jo electron-poor centre ko electron pair donate karta hai.

resonanceअनुनादanunaad: Molecule ko aise Lewis structures ke hybrid ki tarah describe karna jo sirf electrons ki positions mein alag hain.
retardation factor: $R_{f}$ : substance dwara tay ki gayi distance ko solvent front ki distance se divide karna, dono base line se.Chapter mein pehli baar yahan aaya →

8.1 General Introduction

8.2 Carbon ki Tetravalence: Organic Compounds ke Shapes

8.2.1 Carbon Compounds ke Shapes

8.2.2 π Bonds ki Kuch Khaas Features

8.3 Organic Compounds ke Structural Representations

8.3.1 Complete, Condensed aur Bond-line Structural Formulas

8.3.2 Organic Molecules ka Three-Dimensional Representation

8.4 Organic Compounds ka Classification

8.4.1 Functional Group

8.4.2 Homologous Series

8.5 Organic Compounds ki Nomenclature

8.5.1 Nomenclature ka IUPAC System

8.5.2 Alkanes ki IUPAC Nomenclature

8.5.3 Functional Group(s) wale Organic Compounds ki Nomenclature

8.5.4 Substituted Benzene Compounds ki Nomenclature

8.6 Isomerism

8.6.1 Structural Isomerism

8.6.2 Stereoisomerism

8.7 Organic Reaction Mechanism ke Fundamental Concepts

8.7.1 Covalent Bond ka Fission

8.7.2 Substrate aur Reagent

8.7.3 Organic Reactions mein Electron Movement

8.7.4 Covalent Bonds mein Electron Displacement Effects

8.7.5 Inductive Effect

8.7.6 Resonance Structure

8.7.7 Resonance Effect

8.7.8 Electromeric Effect (E effect)

8.7.9 Hyperconjugation

8.7.10 Organic Reactions aur Mechanisms ke Types

8.8 Organic Compounds ki Purification ke Methods

8.8.1 Sublimation

8.8.2 Crystallisation

8.8.3 Distillation

8.8.4 Differential Extraction

8.8.5 Chromatography

8.9 Organic Compounds ka Qualitative Analysis

8.9.1 Carbon aur Hydrogen ka Detection

8.9.2 Doosre Elements ka Detection

8.10 Quantitative Analysis

8.10.1 Carbon aur Hydrogen

8.10.2 Nitrogen

8.10.3 Halogens

8.10.4 Sulphur

8.10.5 Phosphorus

8.10.6 Oxygen

Apna NCERT exercise question dhoondo

Quick check: MCQs

Tum chapter ke end tak pahunch gaye