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State Board · Class 11 · Chemistry

Lesson 14: Basic Principles of Organic Chemistry

7 concepts · 56 practice questions with answers · Free to read

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Classification of organic compounds and functional groups

Link to Std 9–10 : Std 9–10 introduced carbon and functional groups; here we make it systematic.

From daily life

Sugar, oil, vinegar and soap in the kitchen are all organic compounds; their behaviour depends on the functional group.

Simple definition

Organic compounds contain carbon and hydrogen (with O, N, S, halogens). A functional group is the atom or group deciding chemical behaviour. A homologous series has the same group and differs by –CH₂–.

ClassificationClassification of organic compoundsOrganic compoundsOpen chain (aliphatic)Closed chain (cyclic)straightCH₃CH₂CH₃branched(CH₃)₃CHhomocyclicC₆H₁₂heterocyclicC₅H₅NaromaticC₆H₆alicyclicC₆H₁₂Saturated: single bondsUnsaturated: C=C, C≡C

In the figure : Classification chart.

Rule 1: Special features of carbon — Tetravalence, catenation, multiple bonds and isomerism give millions of compounds.

  • Easy : CH₄, C₂H₆, C₃H₈ — a series differing by CH₂ (14 u).
  • Medium : C=C and C≡C multiple bonds.
  • Exam-type : C₄H₁₀ has 2 isomers.

Rule 2: Common functional groups — –OH alcohol, –CHO aldehyde, >C=O ketone, –COOH carboxylic acid, –COOR ester, –NH₂ amine, –X halide, –C≡N nitrile.

  • Easy : C₂H₅OH: alcohol.
  • Medium : CH₃COOH: carboxylic acid.
  • Exam-type : CH₃COCH₃: ketone.

Rule 3: Saturated and unsaturated — Saturated: only single bonds (alkanes). Unsaturated: C=C or C≡C (alkenes, alkynes). Aromatic: benzene ring.

  • Easy : C₃H₈ is saturated; C₃H₆ is unsaturated.
  • Medium : C₆H₁₂ is cyclic saturated.
  • Exam-type : C₆H₆ is aromatic.

Calculation: formula and molar mass for n = 5 in CₙH₂ₙ₊₂

  • Step 1: C₅H₁₂.
  • Step 2: M = 5 × 12 + 12 × 1 = 72 g/mol.
  • Step 3: The next member C₆H₁₄ = 72 + 14 = 86.

CET level (MCQ)

  • CET 1) –CHO is a? 1) ketone 2) aldehyde 3) acid 4) ester — Answer: 2).
  • CET 2) Difference in a homologous series? 1) CH 2) CH₂ 3) CH₃ 4) C — Answer: 2).

Real-life use: Medicines, plastics, fuels and dyes are all organic.

Remember

Tetravalence + catenation; the functional group decides behaviour; homologues differ by CH₂.

Common mistake

Treating CO₂ and carbonates as organic; they are inorganic.

CET trap

A ketone has carbon on both sides of C=O; an aldehyde has H on one side.

Shortcut

Memory clue: "the group decides the behaviour".

Spot the difference

GroupNameExample
–OHalcoholC₂H₅OH
–CHOaldehydeCH₃CHO
>C=OketoneCH₃COCH₃
–COOHacidCH₃COOH

Practice questions

Easy
  1. 1What is an organic compound?
  2. 2What is catenation?
  3. 3What is a functional group?
Medium
  1. 4What is a homologous series?
  2. 5Molar mass of C₅H₁₂?
  3. 6Aldehyde versus ketone.
Challenging
  1. 7Saturated versus unsaturated.
  2. 8Write four functional groups.
Show answers (try them yourself first)
  1. 1A compound containing C and H.
  2. 2Carbon's ability to form chains.
  3. 3The group deciding properties.
  4. 4A series growing by CH₂.
  5. 572 g/mol.
  6. 6Aldehyde has H, ketone has two carbons.
  7. 7Single bonds / multiple bonds.
  8. 8–OH, –CHO, –COOH, –NH₂.

IUPAC nomenclature

From daily life

Like a unique account number, each compound has one IUPAC name.

Simple definition

Name = prefix + root + suffix. Root = number of C in the longest chain (meth 1, eth 2, prop 3, but 4, pent 5, hex 6, hept 7, oct 8, non 9, dec 10). Suffix = principal functional group.

IUPAC nameIUPAC name: 3-methylpentan-2-olC1C2C3C4C5OHC-2CH₃C-3Root: pent (5 C) + an (single) ; suffix: -ol (OH)Prefix: 3-methyl (CH₃ on carbon 3)Numbering: lowest number to the principal group (OH) ⇒ 2Name = prefix + root + suffix

In the figure : Analysing 3-methylpentan-2-ol.

Rule 1: Rules — 1) Longest chain; 2) lowest number to the principal group; 3) branches as prefixes in alphabetical order; 4) di, tri are ignored in alphabetising.

  • Easy : CH₃CH(CH₃)CH₃: 2-methylpropane.
  • Medium : CH₃CH₂CH₂OH: propan-1-ol.
  • Exam-type : CH₃COCH₂CH₃: butan-2-one.

Rule 2: Priority of groups — –COOH > –COOR > –CONH₂ > –CN > –CHO > >C=O > –OH > –NH₂ > C=C > C≡C > –X, –R.

  • Easy : HOCH₂CH₂COOH: 3-hydroxypropanoic acid.
  • Medium : CH₂=CHCH₂OH: prop-2-en-1-ol.
  • Exam-type : CH₃CHClCH₃: 2-chloropropane.

Rule 3: Cyclic and aromatic — Cyclic: prefix 'cyclo'; benzene derivatives: methylbenzene (toluene), hydroxybenzene (phenol).

  • Easy : C₆H₁₂: cyclohexane.
  • Medium : C₆H₅NH₂: aniline / benzenamine.
  • Exam-type : C₆H₅COOH: benzoic acid.

Steps: naming CH₃–CH(OH)–CH(CH₃)–CH₂–CH₃

  • Step 1: Longest chain has 5 C ⇒ pent.
  • Step 2: OH is on C-2 from the left ⇒ -2-ol; CH₃ is on C-3.
  • Step 3: Name = 3-methylpentan-2-ol.

CET level (MCQ)

  • CET 1) Name of (CH₃)₂CHCH₂CH₃? 1) 2-methylbutane 2) 3-methylbutane 3) pentane 4) 2-ethylpropane — Answer: 1).
  • CET 2) CH₃COCH₂CH₃? 1) butanal 2) butan-2-one 3) butan-1-ol 4) butanoic acid — Answer: 2).

Real-life use: Precise communication in pharma and chemical industry.

Remember

Longest chain; lowest number to the principal group; alphabetical branches; meth-eth-prop-but.

Common mistake

Choosing the visually straight chain instead of the longest.

CET trap

di, tri and tetra are ignored when alphabetising, but iso and neo count.

Shortcut

Memory clue: "Meth-Eth-Prop-But" (Monkeys Eat Peeled Bananas…).

Spot the difference

FormulaName
CH₃CH₂CH₂CH₃butane
(CH₃)₃CH2-methylpropane
CH₃CH₂OHethanol
CH₃COOHethanoic acid

Practice questions

Easy
  1. 1What is an IUPAC name?
  2. 2Root names for 1 to 5 carbons.
  3. 3Name CH₃CH₂CH₂OH.
Medium
  1. 4Name CH₃CHClCH₃.
  2. 5Write the priority order of groups.
  3. 6Name CH₃COCH₂CH₃.
Challenging
  1. 7Name C₆H₁₂.
  2. 8Name HOCH₂CH₂COOH.
Show answers (try them yourself first)
  1. 1The unique systematic name of a compound.
  2. 2meth, eth, prop, but, pent.
  3. 3propan-1-ol.
  4. 42-chloropropane.
  5. 5–COOH > –CHO > >C=O > –OH > –NH₂.
  6. 6butan-2-one.
  7. 7cyclohexane.
  8. 83-hydroxypropanoic acid.

Isomerism (structural and stereo)

From daily life

The same letters form different words; the same formula gives different compounds.

Simple definition

Isomers share a molecular formula but differ in structure or properties. Structural: chain, position, functional, metamerism, tautomerism. Stereoisomers: geometrical (cis/trans) and optical.

IsomerismTypes of isomerism (example C₄H₁₀O)Chain(CH₃)₂CHCH₂OH ↔ CH₃(CH₂)₃OHPositionCH₃CH₂CH(OH)CH₃ ↔ CH₃(CH₂)₃OHFunctionalC₂H₅OC₂H₅ ↔ C₄H₉OHMetamerismCH₃OC₃H₇ ↔ C₂H₅OC₂H₅TautomerismCH₃CHO ↔ CH₂=CHOHStereoisomerism: geometrical (cis/trans) and optical (chiral C*)e.g. but-2-ene cis/trans; d/l of 2-butanol

In the figure : Types of isomerism.

Live figure

Isomerism (structural and stereo)

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Rule 1: Structural isomers — Chain: different branching; position: group or bond position; functional: different group; metamerism: different alkyl groups around O; tautomerism: H shift (keto–enol).

  • Easy : C₄H₁₀O: 4 alcohols + 3 ethers = 7 isomers.
  • Medium : C₅H₁₂: 3 isomers.
  • Exam-type : CH₃CHO ⇌ CH₂=CHOH (tautomers).

Rule 2: Geometrical isomerism — Because rotation about C=C is restricted and each carbon bears two different groups. cis = same side, trans = opposite.

  • Easy : but-2-ene: cis and trans.
  • Medium : but-1-ene: none (two H on CH₂).
  • Exam-type : trans usually has the higher melting point.

Rule 3: Optical isomerism — A chiral C* (four different groups) gives two mirror-image enantiomers, d/l. n chiral centres ⇒ 2ⁿ stereoisomers.

  • Easy : 2-butanol: one C*, 2 enantiomers.
  • Medium : n = 3 ⇒ 8.
  • Exam-type : Racemic mixture: equal d + l ⇒ no net rotation.

Counting isomers of C₄H₁₀O

  • Step 1: Alcohols: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol = 4.
  • Step 2: Ethers: CH₃–O–C₃H₇, CH₃–O–CH(CH₃)₂, C₂H₅–O–C₂H₅ = 3.
  • Step 3: Total 4 + 3 = 7 (excluding stereoisomers).

CET level (MCQ)

  • CET 1) Which has a chiral carbon? 1) CH₃CH₂OH 2) CH₃CH(OH)C₂H₅ 3) (CH₃)₂CHOH 4) CH₃CH₂CH₃ — Answer: 2).
  • CET 2) 3 chiral centres give how many stereoisomers? 1) 3 2) 6 3) 8 4) 9 — Answer: 3).

Real-life use: In drugs, often only one enantiomer is active (the thalidomide lesson).

Remember

chain, position, functional, meta, tauto; cis/trans; C* ⇒ 2ⁿ.

Common mistake

Saying but-1-ene shows geometrical isomerism.

CET trap

Meso compounds have C* yet are optically inactive, so fewer than 2ⁿ.

Shortcut

Memory clue: "four different = chiral".

Spot the difference

TypeExample
chainbutane / 2-methylpropane
positionpropan-1-ol / propan-2-ol
functionalC₂H₅OH / CH₃OCH₃
geometricalcis / trans-but-2-ene

Practice questions

Easy
  1. 1What are isomers?
  2. 2Draw the isomers of C₅H₁₂.
  3. 3Condition for geometrical isomerism.
Medium
  1. 4What is C*?
  2. 5Isomers of C₂H₆O.
  3. 6What is tautomerism?
Challenging
  1. 7What is a racemic mixture?
  2. 8When does the 2ⁿ rule fail?
Show answers (try them yourself first)
  1. 1Same formula, different structure.
  2. 2n-, iso-, neopentane.
  3. 3Each C of C=C has two different groups.
  4. 4A chiral carbon.
  5. 5C₂H₅OH and CH₃OCH₃.
  6. 6Isomers by H shift.
  7. 7An equal mixture of d and l.
  8. 8With meso compounds.

Electronic effects: inductive, resonance, hyperconjugation

From daily life

In tug-of-war the rope moves to the stronger team; electrons likewise drift toward the more electronegative atom.

Simple definition

Inductive effect: shift of electron density through σ-bonds. Resonance: delocalisation of π or lone-pair electrons. Hyperconjugation: interaction of α C–H σ electrons with a π bond.

Electronic effectsElectronic effectsInductive effect (I): σ-bond polarisation–I : NO₂ > CN > F > Cl > Br > I > OH+I : (CH₃)₃C– > (CH₃)₂CH– > CH₃CH₂– > CH₃–Resonance (R): delocalisation of π electrons+R: –OH, –NH₂, –X ; –R: –NO₂, –CN, –CHO, –COOHHyperconjugation: stability grows with α–H countCH₃⁺0 α-HCH₃CH₂⁺3 α-H(CH₃)₂CH⁺6 α-H(CH₃)₃C⁺9 α-H

In the figure : The three effects and hyperconjugation.

Rule 1: Inductive effect — –I (withdrawing): NO₂ > CN > F > Cl > Br > I > OH. +I (donating): alkyl; (CH₃)₃C– > (CH₃)₂CH– > CH₃CH₂– > CH₃–. Falls quickly with distance.

  • Easy : CF₃COOH > CCl₃COOH > CH₂ClCOOH > CH₃COOH (acidity).
  • Medium : (CH₃)₃C⁺ is most stable (+I).
  • Exam-type : ClCH₂CH₂COOH is weaker than ClCH₂COOH.

Rule 2: Resonance (R/M) — +R: –OH, –NH₂, –X (lone pair to ring); –R: –NO₂, –CN, –CHO, –COOH (pull electrons). The hybrid is more stable.

  • Easy : Phenol is acidic: phenoxide is resonance stabilised.
  • Medium : Aniline: –NH₂ +R ⇒ more density at o/p.
  • Exam-type : The C2–C3 bond in butadiene is short.

Rule 3: Hyperconjugation — 'No-bond resonance'; more α-H ⇒ more stability: CH₃⁺ (0) < CH₃CH₂⁺ (3) < (CH₃)₂CH⁺ (6) < (CH₃)₃C⁺ (9).

  • Easy : Propene is more stable than ethene.
  • Medium : Toluene is an o/p director.
  • Exam-type : Alkene stability rises with substitution.

Why is CCl₃COOH stronger than CH₃COOH?

  • Step 1: The –I effect of three Cl atoms.
  • Step 2: The negative charge on COO⁻ is spread out ⇒ the anion is stabilised.
  • Step 3: Stable conjugate base ⇒ stronger acid (larger Ka).

CET level (MCQ)

  • CET 1) Strongest acid? 1) CH₃COOH 2) ClCH₂COOH 3) Cl₂CHCOOH 4) Cl₃CCOOH — Answer: 4).
  • CET 2) How many α-H in (CH₃)₃C⁺? 1) 3 2) 6 3) 9 4) 12 — Answer: 3).

Real-life use: Acid-base strength, dyes, drug design.

Remember

–I: NO₂ > CN > F > Cl; +I: tertiary > secondary; α-H: 0, 3, 6, 9.

Common mistake

Treating alkyl as –I; alkyl donates electrons.

CET trap

–I order is F > Cl > Br > I, but halogens also show +R.

Shortcut

Memory clue: "more H, more stable" (hyperconjugation).

Spot the difference

EffectPathDistance
Inductiveσfalls off
Resonanceπ / lplong range
Hyperconj.σ(C–H) → πadjacent

Practice questions

Easy
  1. 1What is the inductive effect?
  2. 2Order of –I groups.
  3. 3What is resonance?
Medium
  1. 4Explain hyperconjugation.
  2. 5Why is CCl₃COOH strong?
  3. 6Why is phenol acidic?
Challenging
  1. 7Why is (CH₃)₃C⁺ the most stable?
  2. 8Two +R groups.
Show answers (try them yourself first)
  1. 1Shift of electron density through σ-bonds.
  2. 2NO₂ > CN > F > Cl > Br > I.
  3. 3Delocalisation of π electrons.
  4. 4C–H σ electrons conjugate with π.
  5. 5–I of three Cl atoms.
  6. 6Phenoxide is resonance stabilised.
  7. 79 α-H and +I.
  8. 8–OH, –NH₂.

Reaction intermediates: carbocation, carbanion, free radical, carbene

From daily life

Like a station on a journey, an intermediate is a short-lived stop in a reaction.

Simple definition

An intermediate is a short-lived species formed during a reaction: carbocation (R₃C⁺), carbanion (R₃C⁻), free radical (R₃C•), carbene (:CR₂).

IntermediatesReaction intermediatesR₃C⁺carbocationsp², planar, 120°; 6 electronsR₃C⁻carbanionsp³, pyramidal; lone pairR₃C•free radicalplanar-ish; one unpaired e⁻:CH₂carbenesp² (singlet); 6 e⁻, neutralStability: 3° > 2° > 1° (cation/radical); carbanion is reversed

In the figure : The four intermediates and their shapes.

Rule 1: Carbocation — sp², planar, 6 electrons; stability 3° > 2° > 1° > CH₃⁺; allyl and benzyl are extra stable (resonance). Rearrangement is possible.

  • Easy : (CH₃)₃C⁺ is tertiary.
  • Medium : A 1° → 2° hydride shift.
  • Exam-type : C₆H₅CH₂⁺ is benzyl.

Rule 2: Carbanion — sp³ pyramidal with a lone pair; stability CH₃⁻ > 1° > 2° > 3° (alkyl donates, so reversed). –I/–R groups stabilise it.

  • Easy : ⁻CH₂NO₂ is stable (–R).
  • Medium : Acetylide ion (sp, 50 % s).
  • Exam-type : Grignard reagents behave like carbanions.

Rule 3: Free radical and carbene — Radical: one unpaired electron, stability 3° > 2° > 1°, formed by homolysis. Carbene: neutral, 6 electrons, :CH₂ (singlet sp²).

  • Easy : Cl₂ → 2Cl• (light).
  • Medium : CHCl₃ + KOH ⇒ :CCl₂.
  • Exam-type : The benzyl radical is stable.

Why is a carbocation planar?

  • Step 1: C has three σ-bonds and an empty p orbital.
  • Step 2: Three σ-bonds spread to 120° ⇒ planar (sp²).
  • Step 3: The empty p orbital lies perpendicular ⇒ attack from either face.

CET level (MCQ)

  • CET 1) Hybridisation of a carbocation? 1) sp 2) sp² 3) sp³ 4) sp³d — Answer: 2).
  • CET 2) Shape of a carbanion? 1) planar 2) pyramidal 3) linear 4) octahedral — Answer: 2).

Real-life use: SN1, E1, radical chains, Grignard synthesis.

Remember

C⁺: sp², 6 e⁻; C⁻: sp³, lone pair; C•: 1 e⁻; :C: neutral 6 e⁻.

Common mistake

Treating carbanion stability like carbocation; it is reversed.

CET trap

Benzyl and allyl cations can beat 3° through resonance.

Shortcut

Memory clue: "cation flat, anion pyramid".

Spot the difference

SpeciesHybridisationShape
R₃C⁺sp²planar
R₃C⁻sp³pyramidal
R₃C•sp²nearly planar

Practice questions

Easy
  1. 1What is an intermediate?
  2. 2Stability order of carbocations.
  3. 3Stability order of carbanions.
Medium
  1. 4What is a carbene?
  2. 5Why is a carbocation planar?
  3. 6How is a radical formed?
Challenging
  1. 7Why is the benzyl cation stable?
  2. 8In which reaction does a carbene arise?
Show answers (try them yourself first)
  1. 1A short-lived species.
  2. 23° > 2° > 1° > CH₃⁺.
  3. 3CH₃⁻ > 1° > 2° > 3°.
  4. 4Neutral carbon with 6 electrons.
  5. 5sp² with an empty p.
  6. 6By homolysis.
  7. 7Resonance.
  8. 8CHCl₃ + KOH.

Reaction types, electrophiles and nucleophiles

From daily life

A magnet pulls iron; electrophiles are pulled to electron-rich sites.

Simple definition

Substitution, addition, elimination, rearrangement. An electrophile (E⁺) accepts an electron pair; a nucleophile (Nu⁻) donates one. Fission: homolytic (radicals) or heterolytic (ions).

Reaction typesBond fission and reaction typesHomolysisA–B → A• + B•free radicalsHeterolysisA–B → A⁺ + B⁻ionsSubstitutionCH₄ + Cl₂ → CH₃Cl + HClAdditionC₂H₄ + Br₂ → C₂H₄Br₂EliminationC₂H₅Br → C₂H₄ + HBrRearrangementCH₃CH₂CH₂⁺ → (CH₃)₂CH⁺Reagents: electrophile E⁺ (seeks e⁻) and nucleophile Nu⁻ (donates a pair)

In the figure : Bond fission and reaction types.

Rule 1: Types of reactions — Substitution: one group replaces another; addition: adds across a multiple bond; elimination: loses a small molecule (HX, H₂O); rearrangement: atoms rearranged.

  • Easy : CH₄ + Cl₂ → CH₃Cl + HCl: substitution.
  • Medium : C₂H₄ + HBr → C₂H₅Br: addition.
  • Exam-type : C₂H₅OH → C₂H₄ + H₂O: elimination.

Rule 2: E⁺ and Nu⁻ — E⁺: H⁺, NO₂⁺, Br⁺, BF₃, AlCl₃, carbocations. Nu⁻: OH⁻, CN⁻, RO⁻, NH₃, H₂O, Cl⁻ (lone pairs).

  • Easy : CH₃Br + OH⁻ → CH₃OH + Br⁻ (Nu attack).
  • Medium : Benzene + NO₂⁺ (E⁺ attack).
  • Exam-type : Nu⁻ attacks the δ⁺ carbon of C=O.

Rule 3: Curly arrows — A full-headed arrow shows movement of an electron pair; a half-headed (fish-hook) arrow shows a single electron. Arrows go from electron-rich to electron-poor.

  • Easy : A–B → A⁺ + :B⁻ (full arrow toward B).
  • Medium : A–B → A• + B• (two half arrows).
  • Exam-type : Count the arrows and balance charges.

Example: CH₃–Br + OH⁻ → CH₃OH + Br⁻ (SN2)

  • Step 1: C–Br is polar ⇒ C δ⁺, Br δ⁻.
  • Step 2: OH⁻ (Nu) attacks C from the back.
  • Step 3: Br⁻ leaves in the same step (transition state).

CET level (MCQ)

  • CET 1) Which is a nucleophile? 1) BF₃ 2) NH₃ 3) H⁺ 4) NO₂⁺ — Answer: 2).
  • CET 2) C₂H₅OH → C₂H₄ + H₂O is? 1) substitution 2) addition 3) elimination 4) rearrangement — Answer: 3).

Real-life use: Drug synthesis, polymers, fuel reforming.

Remember

Substitution, addition, elimination, rearrangement; E⁺ accepts e⁻, Nu⁻ donates e⁻.

Common mistake

Calling BF₃ or AlCl₃ nucleophiles; they are electrophiles (Lewis acids).

CET trap

H₂O and NH₃ are neutral yet nucleophiles (lone pairs).

Shortcut

Memory clue: "Nu seeks the nucleus (+); E seeks electrons".

Spot the difference

ItemElectrophileNucleophile
e⁻ pairacceptsdonates
ExampleNO₂⁺, BF₃OH⁻, NH₃
Attackse⁻ -समृद्धe⁻ -गरीब

Practice questions

Easy
  1. 1List four types of reactions.
  2. 2What is an electrophile?
  3. 3Examples of nucleophiles.
Medium
  1. 4Homolytic and heterolytic fission.
  2. 5Type of C₂H₅OH → C₂H₄.
  3. 6What does a curly arrow show?
Challenging
  1. 7Why is H₂O a nucleophile?
  2. 8Number of steps in SN2.
Show answers (try them yourself first)
  1. 1Substitution, addition, elimination, rearrangement.
  2. 2A species accepting an electron pair.
  3. 3OH⁻, CN⁻, NH₃.
  4. 4Homolytic: radicals; heterolytic: ions.
  5. 5Elimination.
  6. 6Movement of an electron pair.
  7. 7A lone pair on O.
  8. 8One.

Detection of elements and quantitative analysis

From daily life

Like a test for adulterated milk, we test whether a compound has N, S or Cl.

Simple definition

Lassaigne's extract: fuse the compound with Na ⇒ ionic NaCN, Na₂S, NaX; then test. Quantitative: C, H (combustion), N (Dumas/Kjeldahl), halogens and S (Carius).

LassaigneLassaigne's test: Na fusion ⇒ NaCN, Na₂S, NaXNitrogenFeSO₄ + NaOH, boil, HCl ⇒ Prussian blueSulphurNitroprusside ⇒ violet; Pb(OAc)₂ ⇒ blackHalogenHNO₃ + AgNO₃ ⇒ AgCl white, AgBr pale yellow, AgI yellowN and S bothFeCl₃ ⇒ blood red (Fe(SCN)²⁺)Dumas: %N = (28/22400) × (V/w) × 100 (V mL N₂ at STP, w g)Kjeldahl: %N = 1.4 × M × V × 2 / w (acid M, V mL)Carius: %Cl = (35.5/143.5) × (m₂/m₁) × 100

In the figure : Lassaigne tests and formulae.

Live figure

Detection of elements and quantitative analysis

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Rule 1: Qualitative tests — N: Prussian blue; S: violet (nitroprusside) or black PbS; X: AgCl white, AgBr pale yellow, AgI yellow; N + S: blood red.

  • Easy : 3Na₄[Fe(CN)₆] + 4Fe³⁺ → Fe₄[Fe(CN)₆]₃ (blue).
  • Medium : S²⁻ + Pb²⁺ → PbS (black).
  • Exam-type : Cl⁻ + Ag⁺ → AgCl (white; dissolves in NH₃).

Rule 2: Dumas and Kjeldahl — Dumas: %N = (28/22400) × (V/w) × 100 (V mL N₂ at STP). Kjeldahl: %N = 1.4 × N × V / w (acid normality N, V mL).

  • Easy : 0.3 g gives 50 mL N₂ at STP: %N = 20.83.
  • Medium : 0.5 g, 30 mL of 0.1 N acid: %N = 8.4.
  • Exam-type : Kjeldahl fails for nitro and azo compounds.

Rule 3: Carius and combustion — Carius: %Cl = (35.5/143.5) × (m₂/m₁) × 100 (mass m₂ of AgCl). Combustion: %C = (12/44) × (m(CO₂)/m) × 100; %H = (2/18) × (m(H₂O)/m) × 100.

  • Easy : 0.4 g gives 0.287 g AgCl: %Cl = 17.75.
  • Medium : 0.22 g CO₂ from 0.1 g sample: %C = 60.
  • Exam-type : 0.09 g H₂O from 0.1 g: %H = 10.

Calculation: 0.1 g sample gives 0.22 g CO₂ and 0.09 g H₂O

  • Step 1: %C = (12/44) × (0.22/0.1) × 100 = 60.
  • Step 2: %H = (2/18) × (0.09/0.1) × 100 = 10.
  • Step 3: The remaining 30 % is O (if present).

CET level (MCQ)

  • CET 1) Test for N? 1) Prussian blue 2) black 3) yellow ppt 4) violet — Answer: 1).
  • CET 2) 0.1 g sample, 0.22 g CO₂: %C? 1) 40 2) 50 3) 60 4) 70 — Answer: 3).

Real-life use: Composition of food/drug samples, adulteration tests.

Remember

N = Prussian blue; S = violet/black; X = AgX; Dumas 28/22400; Carius 35.5/143.5.

Common mistake

Forgetting to boil with HNO₃ before the halogen test (to remove S²⁻ and CN⁻).

CET trap

With both N and S the test gives blood-red Fe(SCN)²⁺, not blue.

Shortcut

Memory clue: "blue N, black S".

Spot the difference

ElementTestResult
NFeSO₄/FeCl₃blue
SPb(OAc)₂black
ClAgNO₃white

Practice questions

Easy
  1. 1What is Lassaigne's extract?
  2. 2Write the test for N.
  3. 3Test for S.
Medium
  1. 4Write the Dumas formula.
  2. 5Where does Kjeldahl fail?
  3. 60.4 g gives 0.287 g AgCl: %Cl?
Challenging
  1. 70.3 g, 50 mL N₂: %N?
  2. 8Why boil with HNO₃?
Show answers (try them yourself first)
  1. 1Extract from fusing with Na.
  2. 2FeSO₄ + NaOH, HCl ⇒ blue.
  3. 3Pb(OAc)₂ ⇒ black PbS.
  4. 4(28/22400)(V/w) × 100.
  5. 5Nitro and azo compounds.
  6. 617.75 %.
  7. 720.83 %.
  8. 8To remove S²⁻ and CN⁻.

Quick revision of the whole lesson

ConceptKey formula / fact
Classificationopen/cyclic; functional groups
IUPACprefix + root + suffix
Isomerismchain, position, group; cis/trans; C*
Effects–I, +I, R, hyperconjugation
IntermediatesC⁺ sp², C⁻ sp³, C•, :C
ReagentsE⁺ accepts e⁻; Nu⁻ donates e⁻
LassaigneN blue; S black; X AgX
Dumas28/22400
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🎯 Most important points for the exam

  • Classification and functional groups.
  • IUPAC nomenclature.
  • Isomerism.
  • Electronic effects.
  • Intermediates.
  • Reaction types, E⁺, Nu⁻.
  • Detection and estimation of elements.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.