Organic Chemistry – Some Basic Principles and Techniques Class 11: IUPAC, Effects and Analysis

What Organic Chemistry – Some Basic Principles and Techniques is about

Weightage: This chapter is Unit VIII, Organic Chemistry: Some basic Principles and Techniques, which the syllabus gives 11 of the 70 theory marks in the school annual exam.Source: CBSE Chemistry (043) syllabus 2026-27, Class XI course structure, Unit VIII Organic Chemistry: Some basic Principles and Techniques

Key concepts

Tetravalence of carbon, hybridisation and structural representations

NCERT section 8.2 Tetravalence of Carbon: Shapes of Organic Compounds; 8.3 Structural Representations of Organic Compounds

Carbon has four valence electrons and forms four covalent bonds (tetravalence). The type of hybridisation decides the shape and bond properties:

• sp³ (only single bonds): tetrahedral, 109.5°, as in CH₄.
• sp² (one double bond): trigonal planar, 120°, as in ethene.
• sp (one triple bond or two double bonds): linear, 180°, as in ethyne.

The more s character in the hybrid orbital (sp³ 25%, sp² 33%, sp 50%), the closer the electrons are held to the nucleus, so the carbon is more electronegative, and bonds are shorter and stronger. This is why the H of ethyne is slightly acidic.

A π bond needs parallel p orbitals on neighbouring atoms, so the atoms around a C=C bond lie in one plane and rotation is restricted. π electrons are exposed and are the reactive site of alkenes and alkynes.

Ways of drawing structures:

• Complete (Lewis/dash) formula: every bond shown, e.g. H–C–C–O–H with all hydrogens.
• Condensed formula: groups written together, CH₃CH₂OH or CH₃(CH₂)₂CH₃.
• Bond-line formula: a zig-zag line where every corner and line end is a carbon, hydrogens on carbon are not shown, and heteroatoms (O, N, Cl) are written explicitly. Cyclic compounds are drawn as polygons.
• Three-dimensional (wedge-dash): a solid wedge for a bond towards you, a dashed wedge for a bond going away, and plain lines in the plane of the paper.

Worked example

For CH₂=CH–CH₂–C≡CH, state the hybridisation of each carbon (numbered from the left) and the number of σ and π bonds. Draw its bond-line formula in words. (3 marks)

Try it first, then show the solutionSolution

1. C1 and C2 (double bond): sp²; C3 (only single bonds): sp³; C4 and C5 (triple bond): sp. (1)
2. Bonds: C–H σ = 2 + 1 + 2 + 1 = 6; C–C σ = 4; total σ = 10; π = 1 (C=C) + 2 (C≡C) = 3. (1)
3. Bond-line: a double line between the first two points, a single line to the third point, then a triple line ending at the fifth point (the C≡C–H end drawn straight because sp carbons are linear). (1)

Answer: C1, C2 sp²; C3 sp³; C4, C5 sp; 10 σ and 3 π bonds

Classification of organic compounds and functional groups

NCERT section 8.4 Classification of Organic Compounds (8.4.1 Functional Group, 8.4.2 Homologous Series)

Organic compounds are classified by their skeleton:
1. Acyclic (open-chain) or aliphatic: straight or branched chains (ethane, isobutane, acetaldehyde).
2. Cyclic (closed-chain):

• Alicyclic: rings of carbon that behave like aliphatic compounds (cyclopropane, cyclohexane, cyclohexene).
• Aromatic: benzene and related compounds. Benzenoid aromatics contain benzene rings (benzene, naphthalene, aniline); non-benzenoid aromatics (such as tropolone) are aromatic without a benzene ring.
• Heterocyclic: rings containing atoms other than carbon (O, N, S), such as furan, thiophene and pyridine. They may be aromatic.

A functional group is an atom or group of atoms that decides the characteristic reactions of the compound: –OH (alcohol), –CHO (aldehyde), C=O (ketone), –COOH (carboxylic acid), –NH₂ (amine), –X (halide) and so on.

A homologous series is a family of compounds with the same functional group, in which successive members differ by a –CH₂– unit (14 u). Members have similar chemical properties and a gradation in physical properties. Examples: alkanes, alkenes, alkynes, alcohols (CH₃OH, C₂H₅OH, C₃H₇OH …).

Worked example

Classify (a) cyclohexanol (b) pyridine (c) naphthalene (d) propanal. Name the functional group in (a) and (d), and write the next higher homologue of (d). (3 marks)

Try it first, then show the solutionSolution

1. (a) Alicyclic (ring of carbons with aliphatic behaviour); functional group –OH (alcohol). (1)
2. (b) Heterocyclic aromatic (ring contains N). (c) Benzenoid aromatic (two fused benzene rings). (1)
3. (d) Acyclic; functional group –CHO (aldehyde). Next homologue: butanal, CH₃CH₂CH₂CHO. (1)

Answer: (a) alicyclic, –OH (b) heterocyclic aromatic (c) benzenoid aromatic (d) acyclic, –CHO; next homologue butanal

IUPAC nomenclature of organic compounds

NCERT section 8.5 Nomenclature of Organic Compounds (8.5.1 to 8.5.4)

An IUPAC name has three parts: prefix(es) + word root + suffix(es).

• Word root gives the number of carbons in the parent chain: meth (1), eth (2), prop (3), but (4), pent (5), hex (6), hept (7), oct (8), non (9), dec (10).
• Primary suffix gives saturation: -ane (all single), -ene (C=C), -yne (C≡C).
• Secondary suffix names the main functional group.

Rules for alkanes:
1. Choose the longest carbon chain (parent).
2. Number it so that substituents get the lowest locants (lowest set at the first point of difference).
3. Name substituents (methyl, ethyl, isopropyl, tert-butyl …) with their locants; use di, tri, tetra for repeated ones.
4. Arrange substituent names in alphabetical order (ignore di, tri, sec-, tert-, but consider iso and neo).
5. If two chains are equally long, choose the one with more substituents.

Compounds with functional groups: the chain must contain the principal functional group, which gets the lowest possible locant. Priority order (highest first): –COOH > –SO₃H > –COOR > –COCl > –CONH₂ > –CN > –CHO > >C=O > –OH > –NH₂ > C=C > C≡C. Groups such as –R, –X (halo), –NO₂ (nitro) and –OR (alkoxy) are always prefixes. Lower-priority groups also become prefixes (–OH as hydroxy, C=O as oxo).

Examples: CH₃CH(CH₃)CH₂CH₃ is 2-methylbutane; CH₃COCH₂CH(OH)CH₃ is 4-hydroxypentan-2-one; CH₂=CHCH₂OH is prop-2-en-1-ol; CH₃CH(Cl)COOH is 2-chloropropanoic acid.

Substituted benzenes: one substituent takes a common or prefix name (chlorobenzene, nitrobenzene; phenol, aniline, toluene are accepted). For two or more, number the ring to give the lowest locants; older terms ortho (1,2), meta (1,3) and para (1,4) are still used.

Worked example

Write the IUPAC names of (a) (CH₃)₂CHCH(C₂H₅)CH₂CH₂CH₃ (b) CH₃COCH₂COOH (c) HOCH₂CH₂CH=CH₂. (3 marks)

Try it first, then show the solutionSolution

1. (a) Longest chain: from a CH₃ of the isopropyl group through C(H)(C₂H₅) and on to the propyl end = 6 carbons (hexane). Numbering from the isopropyl end gives substituents at C2 (methyl) and C3 (ethyl): 3-ethyl-2-methylhexane (alphabetical order: ethyl before methyl). (1)
2. (b) Principal group –COOH (C1); chain of 4 carbons; C=O at C3 becomes 'oxo': 3-oxobutanoic acid. (1)
3. (c) –OH outranks C=C; number from the CH₂OH end: OH on C1, double bond from C3: but-3-en-1-ol. (1)

Answer: (a) 3-ethyl-2-methylhexane (b) 3-oxobutanoic acid (c) but-3-en-1-ol

Isomerism

NCERT section 8.6 Isomerism (8.6.1 Structural Isomerism, 8.6.2 Stereoisomerism)

Isomers are compounds with the same molecular formula but different properties.

Structural isomerism (different arrangement of atoms):
1. Chain isomerism: different carbon skeletons. C₅H₁₂: pentane, 2-methylbutane, 2,2-dimethylpropane.
2. Position isomerism: same skeleton, different position of a substituent, functional group or multiple bond. Propan-1-ol and propan-2-ol; but-1-ene and but-2-ene.
3. Functional group isomerism: different functional groups. C₂H₆O: ethanol (CH₃CH₂OH) and methoxymethane (CH₃OCH₃); C₃H₆O: propanal and propanone.
4. Metamerism: different alkyl groups on either side of a functional group. C₄H₁₀O: methoxypropane (CH₃OC₃H₇) and ethoxyethane (C₂H₅OC₂H₅).

Stereoisomerism (same structure and bonding, different arrangement in space): divided into geometrical isomerism (cis and trans forms around a C=C bond) and optical isomerism (mirror-image molecules that rotate polarised light). NCERT only introduces them here; geometrical isomers of alkenes are studied in Hydrocarbons.

Worked example

For the molecular formula C₄H₁₀O, write (a) two position isomers that are alcohols, (b) one chain isomer of butan-1-ol, (c) two metamers, and (d) say how an alcohol and an ether of this formula are related. (3 marks)

Try it first, then show the solutionSolution

1. (a) Butan-1-ol, CH₃CH₂CH₂CH₂OH, and butan-2-ol, CH₃CH₂CH(OH)CH₃. (1)
2. (b) 2-Methylpropan-1-ol, (CH₃)₂CHCH₂OH. (½)
3. (c) Ethoxyethane, C₂H₅OC₂H₅, and 1-methoxypropane, CH₃OCH₂CH₂CH₃. (1)
4. (d) They are functional group isomers. (½)

Answer: (a) butan-1-ol and butan-2-ol (b) 2-methylpropan-1-ol (c) ethoxyethane and 1-methoxypropane (d) functional group isomers

Bond fission, reactive intermediates, nucleophiles and electrophiles

NCERT section 8.7 Fundamental Concepts in Organic Reaction Mechanism (8.7.1 to 8.7.3)

In a reaction the substrate (the molecule whose carbon forms a new bond) is attacked by a reagent.

Fission of a covalent bond:

• Heterolytic cleavage: the shared pair goes to one atom, giving ions. If carbon loses the pair, a carbocation (C⁺, sp², planar, six electrons) forms; if carbon keeps it, a carbanion (C⁻, sp³, pyramidal) forms.
• Homolytic cleavage: each atom takes one electron, giving free radicals (sp², with an unpaired electron). It happens in the gas phase or non-polar solvents, with heat or light.

Stability of carbocations: 3° > 2° > 1° > CH₃⁺, because alkyl groups release electrons (inductive effect and hyperconjugation) and spread the positive charge. Free radicals follow the same order. Carbanions follow the opposite order (CH₃⁻ > 1° > 2° > 3°).

Nucleophiles ('nucleus-loving') are electron-pair donors that attack electron-poor centres: OH⁻, CN⁻, Cl⁻, CH₃O⁻, and neutral molecules with lone pairs (H₂O, NH₃, R–NH₂).
Electrophiles ('electron-loving') are electron-pair acceptors that attack electron-rich centres: H⁺, NO₂⁺, carbocations, and neutral molecules with an electron-poor atom (BF₃, AlCl₃, the carbon of C=O, SO₃).

Curved arrows show electron movement: a full-headed arrow moves an electron pair, from a lone pair or a bond towards an atom or a new bond position; a half-headed (fish-hook) arrow moves a single electron.

Worked example

Classify each as a nucleophile or an electrophile: (a) NH₃ (b) BF₃ (c) CH₃O⁻ (d) NO₂⁺ (e) CH₃CH₂⁺ (f) H₂O. Arrange (CH₃)₃C⁺, CH₃⁺, (CH₃)₂CH⁺ and CH₃CH₂⁺ in decreasing stability. (3 marks)

Try it first, then show the solutionSolution

1. Nucleophiles (electron-pair donors): NH₃, CH₃O⁻, H₂O. (1)
2. Electrophiles (electron-pair acceptors): BF₃, NO₂⁺, CH₃CH₂⁺. (1)
3. (CH₃)₃C⁺ (3°) > (CH₃)₂CH⁺ (2°) > CH₃CH₂⁺ (1°) > CH₃⁺: more alkyl groups mean more electron release and more hyperconjugation. (1)

Answer: Nucleophiles: NH₃, CH₃O⁻, H₂O; electrophiles: BF₃, NO₂⁺, CH₃CH₂⁺; 3° > 2° > 1° > methyl

Electron displacement effects: inductive, resonance, electromeric and hyperconjugation

NCERT section 8.7.4 to 8.7.9 Electron Displacement Effects (inductive, resonance structure, resonance effect, electromeric effect, hyperconjugation)

Inductive effect (I): a permanent shift of σ electrons along a chain caused by an electronegative or electropositive atom. It weakens quickly with distance and is negligible beyond three carbons.

• −I groups (electron-withdrawing): –NO₂ > –CN > –COOH > –F > –Cl > –Br > –I > –OH > –C₆H₅.
• +I groups (electron-releasing): alkyl groups, (CH₃)₃C– > (CH₃)₂CH– > CH₃CH₂– > CH₃–.

Example: chloroethanoic acid is stronger than ethanoic acid because –Cl pulls electron density and stabilises the carboxylate ion.

Resonance: when a molecule can be drawn as two or more structures differing only in the positions of electrons (π electrons or lone pairs), the real molecule is the resonance hybrid, lower in energy than any single form. The difference is the resonance energy. Rules: same positions of nuclei, same number of unpaired electrons; more covalent bonds, octets complete, and negative charge on the more electronegative atom mean a more important contributor. Benzene and the carboxylate ion are classic examples.

Resonance (mesomeric) effect (R or M): a permanent polarity produced by the interaction of π electrons with lone pairs or with neighbouring π bonds, in a conjugated system (alternate single and multiple bonds).

• +R effect: the group donates electrons into the conjugated system: –OH, –OR, –NH₂, –NHR, halogens (–X). This raises electron density at ortho and para positions of benzene.
• −R effect: the group withdraws electrons: –NO₂, –CN, –CHO, >C=O, –COOH.

Electromeric effect (E): a temporary complete shift of a π electron pair to one atom of a multiple bond, only in the presence of an attacking reagent; it disappears when the reagent is removed. +E: the pair moves towards the atom to which the reagent attaches; −E: away from it.

Hyperconjugation: delocalisation of the σ electrons of a C–H bond on a carbon next to a carbocation, a free radical or a π system, into the empty p orbital or π system. It is a 'no-bond resonance'. The more α-hydrogens, the greater the stabilisation: (CH₃)₃C⁺ has nine α-H, CH₃CH₂⁺ has three. It also explains the extra stability of more substituted alkenes.

Worked example

Arrange 2-chlorobutanoic acid, 3-chlorobutanoic acid, 4-chlorobutanoic acid and butanoic acid in decreasing order of acid strength. Which effect is responsible, and why does its influence change along the series? (2 marks)

Try it first, then show the solutionSolution

1. The Cl atom withdraws electrons by the −I effect and stabilises the carboxylate ion, but the inductive effect weakens rapidly with distance: it is strongest when Cl is on C2, next to –COOH, and almost gone by C4. (1)
2. Order: 2-chlorobutanoic > 3-chlorobutanoic > 4-chlorobutanoic > butanoic acid (no Cl at all). (1)

Answer: 2-chloro > 3-chloro > 4-chloro > butanoic acid; the −I effect of Cl stabilises the anion and fades with distance.

Types of organic reactions and methods of purification

NCERT section 8.7.10 Types of Organic Reactions; 8.8 Methods of Purification of Organic Compounds

Types of reactions: substitution (one atom or group replaced by another), addition (atoms added across a multiple bond), elimination (atoms removed to form a multiple bond) and rearrangement (atoms reorganise within the molecule).

Purification methods (the choice depends on the nature of the compound and impurities):
1. Sublimation: for solids that change directly to vapour (camphor, naphthalene, anthracene, benzoic acid) and are mixed with non-sublimable impurities.
2. Crystallisation: dissolve in a solvent in which the compound is sparingly soluble cold but quite soluble hot; impurities stay in solution. Coloured impurities are removed with activated charcoal. Repeated crystallisation is used when solubilities are similar.
3. Distillation:

• Simple distillation: liquids with a large difference in boiling points (chloroform 334 K and aniline 457 K), or a liquid from a non-volatile impurity.
• Fractional distillation: liquids with close boiling points, using a fractionating column (crude oil in refineries).
• Distillation under reduced pressure: liquids that decompose at or below their boiling point (glycerol from spent lye in soap industry); lowering pressure lowers the boiling point.
• Steam distillation: steam-volatile substances immiscible with water (aniline, essential oils). The mixture boils when p(substance) + p(water) = atmospheric pressure, i.e. below 373 K.

4. Differential extraction: shaking an aqueous solution with an immiscible organic solvent in which the compound is more soluble (separating funnel).
5. Chromatography (Tswett): separation based on different rates of movement between a stationary phase and a mobile phase.

• Adsorption chromatography: column chromatography and thin layer chromatography (TLC) on silica gel or alumina. In TLC, the retardation factor $R_f=\frac{\text{distance moved by the substance}}{\text{distance moved by the solvent front}}$. Colourless spots are seen under UV light or with iodine vapour or a spraying reagent (ninhydrin for amino acids).
• Partition chromatography: paper chromatography, where water trapped in the paper is the stationary phase.

Worked example

Suggest a method to purify or separate: (a) camphor containing sand (b) glycerol from spent lye (c) aniline from a reaction mixture (d) a mixture of two dyes. In a TLC run, a spot moves 3.0 cm while the solvent front moves 7.5 cm. Find its Rf. (3 marks)

Try it first, then show the solutionSolution

1. (a) Sublimation. (b) Distillation under reduced pressure (glycerol decomposes at its normal boiling point). (1)
2. (c) Steam distillation (aniline is steam volatile and immiscible with water). (d) Chromatography (paper or TLC). (1)
3. Rf = 3.0 ÷ 7.5 = 0.40. (1)

Answer: (a) sublimation (b) reduced-pressure distillation (c) steam distillation (d) chromatography; Rf = 0.40

Qualitative analysis: detecting elements (Lassaigne's test)

NCERT section 8.9 Qualitative Analysis of Organic Compounds (8.9.1, 8.9.2)

Carbon and hydrogen: heat the compound with copper(II) oxide. Carbon is oxidised to CO₂, which turns lime water milky; hydrogen gives water, which turns anhydrous copper sulphate blue.

Lassaigne's test for N, S, halogens and P: the compound is fused with sodium metal, converting covalent N, S and X into ionic NaCN, Na₂S, NaX (and NaSCN if both N and S are present). The fused mass is boiled with water to give the sodium fusion extract.

• Nitrogen: boil the extract with iron(II) sulphate, then acidify with concentrated H₂SO₄. A Prussian blue colour (iron(III) hexacyanidoferrate(II), Fe₄[Fe(CN)₆]₃·xH₂O) shows nitrogen.
• Sulphur: (a) acidify with acetic acid and add lead acetate: a black precipitate of PbS; (b) add sodium nitroprusside: a violet colour.
• Both N and S: NaSCN gives a blood-red colour with Fe³⁺ (no Prussian blue, because there is no free cyanide).
• Halogens: acidify with nitric acid and add silver nitrate: white precipitate soluble in ammonia (AgCl, chlorine), pale yellow sparingly soluble (AgBr, bromine), yellow insoluble (AgI, iodine). If N or S is present, the extract is first boiled with concentrated HNO₃ to decompose cyanide and sulphide, which would otherwise interfere.
• Phosphorus: heat with an oxidising agent (sodium peroxide) to convert P to phosphate; boil with nitric acid and add ammonium molybdate: a yellow colour or precipitate.

Worked example

(a) Why is the sodium fusion extract acidified with acetic acid, and not with sulphuric acid, before lead acetate is added in the test for sulphur? (b) What is the role of concentrated H₂SO₄ in the test for nitrogen? (c) Will hydrazine (NH₂NH₂) give a positive Lassaigne's test for nitrogen? Explain. (3 marks)

Try it first, then show the solutionSolution

1. (a) Sulphuric acid would supply sulphate ions, which form a white precipitate of PbSO₄ with lead acetate and confuse the result; acetic acid adds no interfering anion, so a black precipitate can only be PbS. (1)
2. (b) The extract is alkaline, so iron hydroxides form on boiling with FeSO₄; concentrated H₂SO₄ dissolves them and gives the acidic medium in which Fe³⁺ (from partial oxidation of Fe²⁺) combines with hexacyanidoferrate(II) to give Prussian blue. (1)
3. (c) No: hydrazine has no carbon, so NaCN cannot form on fusion, and the Prussian blue test fails. (1)

Answer: (a) to avoid a white PbSO₄ precipitate (b) to dissolve iron hydroxides and give the acid medium for Prussian blue (c) no, there is no carbon to form CN⁻

Quantitative analysis: percentages of elements

NCERT section 8.10 Quantitative Analysis (8.10.1 to 8.10.6)

1. Carbon and hydrogen (Liebig's method): a known mass of compound is burnt in excess oxygen with copper(II) oxide. Water is absorbed in anhydrous CaCl₂ and CO₂ in KOH solution; the increases in mass give the amounts.
$\%C=\frac{12\times m_{C{O}_2}\times 100}{44\times m}$, $\%H=\frac{2\times m_{H_{2}O}\times 100}{18\times m}$
2. Nitrogen, Dumas method: the compound is heated with CuO in a CO₂ atmosphere; nitrogen is released as N₂ gas and collected over KOH solution (which absorbs CO₂). Volume converted to STP:
$\%N=\frac{28\times V_{N_2}(\text{mL at STP})\times 100}{22400\times m}$
3. Nitrogen, Kjeldahl's method: the compound is heated with concentrated H₂SO₄ (converting N to ammonium sulphate), then with NaOH; the ammonia is passed into a known volume of standard acid, and the unused acid is titrated with standard alkali. Moles of NH₃ = moles of N. It does not work for nitro and azo compounds or for N in rings (pyridine).
4. Halogens (Carius method): heated with fuming HNO₃ and AgNO₃ in a sealed Carius tube; the silver halide is weighed.
$\%X=\frac{\text{atomic mass of X}\times m_{AgX}\times 100}{\text{molar mass of AgX}\times m}$ (AgCl 143.5, AgBr 188, AgI 235).
5. Sulphur: heated with sodium peroxide or fuming HNO₃ in a Carius tube to give H₂SO₄, precipitated as BaSO₄ (233):
$\%S=\frac{32\times m_{BaS{O}_4}\times 100}{233\times m}$
6. Phosphorus: oxidised to phosphoric acid, precipitated as ammonium phosphomolybdate or as MgNH₄PO₄, ignited to Mg₂P₂O₇ (222): $\%P=\frac{62\times m_{M{g}_2{P}_2{O}_7}\times 100}{222\times m}$.
7. Oxygen: usually found by difference, % O = 100 − (sum of all other percentages).

Worked example

In Liebig's method, 0.45 g of an organic compound gave 0.66 g of CO₂ and 0.27 g of H₂O. The compound contains only C, H and O. Find the percentage of each element and its empirical formula. (3 marks)

Try it first, then show the solutionSolution

1. % C = 12 × 0.66 × 100 ÷ (44 × 0.45) = 40.0%; % H = 2 × 0.27 × 100 ÷ (18 × 0.45) = 6.67%. (1)
2. % O = 100 − 40.0 − 6.67 = 53.33%. (1)
3. Moles in 100 g: C 3.33, H 6.67, O 3.33 → ratio 1 : 2 : 1 → empirical formula CH₂O. (1)

Answer: C 40.0%, H 6.67%, O 53.3%; empirical formula CH₂O

Chemistry in real life: Organic Chemistry – Some Basic Principles and Techniques

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Electronic Effects, Reactive Intermediates and Types of Organic Reactions
• IUPAC Nomenclature and Structural Isomerism

Organic Chemistry – Some Basic Principles and Techniques formula sheet

Retardation factor (TLC/paper)

$R_f=\frac{\text{distance moved by substance}}{\text{distance moved by solvent front}}$

Always less than 1; no unit.

Carbon and hydrogen (Liebig)

$\%C=\frac{12\times m_{C{O}_2}\times 100}{44\times m},\%H=\frac{2\times m_{H_{2}O}\times 100}{18\times m}$

m = mass of compound taken.

Nitrogen (Dumas)

$\%N=\frac{28\times V_{N_2}\times 100}{22400\times m}$

V at STP in mL; correct for water vapour and temperature first if given.

Nitrogen (Kjeldahl)

$\%N=\frac{14\times n_{N{H}_3}\times 100}{m}=\frac{1.4\times M\times 2\left(V-\frac{V_1}{2}\right)}{m}$

M = molarity of H₂SO₄, V = volume of acid (mL), V₁ = volume of NaOH of the same molarity used for back titration.

Halogens (Carius)

$\%X=\frac{A_X\times m_{AgX}\times 100}{M_{AgX}\times m}$

AgCl 143.5, AgBr 188, AgI 235.

Sulphur

$\%S=\frac{32\times m_{BaS{O}_4}\times 100}{233\times m}$

BaSO₄ precipitate.

Phosphorus

$\%P=\frac{62\times m_{M{g}_2{P}_2{O}_7}\times 100}{222\times m}$

Or 31 × m × 100 ÷ (1877 × mass) for ammonium phosphomolybdate.

Organic Chemistry – Some Basic Principles and Techniques: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Organic Chemistry – Some Basic Principles and Techniques

The question patterns to expect, and the method that works for each.

Tricks to remember Organic Chemistry – Some Basic Principles and Techniques

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Monkeys Eat Peeled Bananas

  Meth, Eth, Prop, But for 1-4 carbons; then Greek numbers (pent, hex, hept, oct, non, dec).

  Use it when: Word roots.

• Carboxylic Sulphonic Esters Acid-chlorides Amides Nitriles Aldehydes Ketones Alcohols Amines

  Functional group priority from highest to lowest: COOH > SO₃H > COOR > COCl > CONH₂ > CN > CHO > C=O > OH > NH₂.

  Use it when: Choosing the suffix in polyfunctional compounds.

• More alkyl, more happy cation

  Carbocations and radicals: 3° > 2° > 1° > methyl. Carbanions: the reverse.

  Use it when: Stability orders.

• NIC on halogen colours

  AgCl white, AgBr pale yellow, AgI yellow: whiteness falls as size rises.

  Use it when: Lassaigne's halogen test.

• Liebig 12/44 and 2/18

  Fraction of C in CO₂ is 12/44; fraction of H in H₂O is 2/18.

  Use it when: Quantitative C and H numericals.

Organic Chemistry – Some Basic Principles and Techniques: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Organic Chemistry – Some Basic Principles and Techniques

The slips examiners see most often, and what to write instead.

• ✗Mistake: Not choosing the longest chain containing the functional group

  Students pick the longest chain overall.

  ✓Do this: The principal functional group must be in the parent chain, even if a longer chain exists elsewhere.

  Usually costs: 1 mark.

• ✗Mistake: Writing substituents in numerical, not alphabetical, order

  Locants are used to order the names.

  Wrong working

  2-methyl-3-ethylhexane

  Right working

  3-ethyl-2-methylhexane

  ✓Do this: Alphabetise the substituent names (ignore di, tri).

  Usually costs: ½-1 mark.

• ✗Mistake: Confusing inductive and electromeric effects

  Both describe electron shifts.

  ✓Do this: Inductive: permanent, σ electrons, partial shift. Electromeric: temporary, π electrons, complete shift, only when a reagent attacks.

  Usually costs: 1 mark.

• ✗Mistake: Testing for halogens without boiling the extract with HNO₃

  The step seems unnecessary.

  ✓Do this: If N or S is present, CN⁻ and S²⁻ also precipitate with Ag⁺; boil with HNO₃ first.

  Usually costs: 1 mark.

• ✗Mistake: Using Kjeldahl's method for nitrobenzene

  Limitations are forgotten.

  ✓Do this: Kjeldahl fails for nitro and azo compounds and ring N (pyridine); use Dumas.

  Usually costs: 1 mark.

• ✗Mistake: Using molar mass of AgX wrongly

  Atomic mass of X is placed in the denominator.

  Wrong working

  %Cl = 143.5 × m(AgCl) × 100 ÷ (35.5 × m)

  Right working

  %Cl = 35.5 × m(AgCl) × 100 ÷ (143.5 × m)

  ✓Do this: Fraction of X in AgX = A(X) ÷ M(AgX), which is less than 1.

  Usually costs: The answer.

NCERT exercise guide: Organic Chemistry – Some Basic Principles and Techniques

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Organic Chemistry – Some Basic Principles and Techniques

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The IUPAC name of CH₃–CH=CH–CHO is:
   (a) but-2-enal (b) but-3-enal (c) but-2-en-1-ol (d) 1-formylprop-1-ene

   Hint

   CHO carbon is C1.

   Answer

   (a) but-2-enal

2. Q2 · 1 mark · MCQExam level

   Which of the following is an electrophile?
   (a) NH₃ (b) H₂O (c) AlCl₃ (d) CN⁻

   Hint

   Electron-pair acceptor.

   Answer

   (c) AlCl₃

3. Q3 · 2 marks · Very short answerExam level

   Differentiate between the principle of adsorption chromatography and partition chromatography. (2 marks)

   Hint

   What holds the substance back?

   Answer

   Adsorption: components are adsorbed to different extents on a solid stationary phase (silica gel, alumina) (1). Partition: components distribute differently between a liquid stationary phase (water in paper) and a liquid mobile phase (1).

4. Q4 · 3 marks · Short answerHarder exam level

   Write the bond-line formulas of (a) 2,3-dimethylbutane (b) cyclohexanol (c) pent-2-yne. (3 marks)

   Hint

   Corners and ends are carbons.

   Answer

   (a) A four-point zig-zag with a short branch at the second and third points (1). (b) A hexagon with –OH on one corner (1). (c) A line of five carbons with a straight triple bond between the second and third (sp carbons linear) (1).

5. Q5 · 3 marks · Short answerHarder exam level

   Write all structural isomers of C₃H₆O that are aldehydes, ketones or alcohols with a C=C, naming each and the isomerism type relating propanal and propanone. (3 marks)

   Hint

   Three or four possibilities.

   Answer

   Propanal CH₃CH₂CHO; propanone CH₃COCH₃; prop-2-en-1-ol CH₂=CHCH₂OH (2). Propanal and propanone are functional group isomers (1). (Enols such as prop-1-en-2-ol are unstable and change into the carbonyl compounds, so they are not counted.)

6. Q6 · 2 marks · Very short answerHarder exam level

   Explain why –NO₂ group decreases electron density in benzene, while –OH increases it. Name the effects. (2 marks)

   Hint

   +R and −R.

   Answer

   –OH donates its lone pair into the ring by +R (resonance) effect, raising electron density at ortho and para positions (1). –NO₂ withdraws π electrons from the ring by −R (and −I) effect, lowering electron density (1).

7. Q7 · 2 marks · Very short answerHarder exam level

   Suppose 0.50 g of an organic compound treated by Kjeldahl's method gives ammonia that neutralises 10 mL of 1.0 M H₂SO₄. Calculate the percentage of nitrogen. (2 marks)

   Hint

   2 NH₃ per H₂SO₄.

   Answer

   H₂SO₄ = 0.010 mol, so NH₃ = 0.020 mol and N = 0.28 g (1). %N = 0.28 ÷ 0.50 × 100 = 56% (1).

8. Q8 · 2 marks · Very short answerHarder exam level

   A mixture of aniline (b.p. 457 K) and chloroform (b.p. 334 K) and another of ethanol (b.p. 351 K) and water (b.p. 373 K) are to be separated. Suggest the method for each and justify. (2 marks)

   Hint

   Gap in boiling points.

   Answer

   Aniline–chloroform: simple distillation, because the boiling points differ greatly (123 K) (1). Ethanol–water: fractional distillation, because the boiling points are close (1).

9. Q9 · 3 marks · Short answerHarder exam level

   Arrange in decreasing order of acid strength and explain: FCH₂COOH, ClCH₂COOH, BrCH₂COOH, CH₃COOH. (3 marks)

   Hint

   Compare −I strengths.

   Answer

   FCH₂COOH > ClCH₂COOH > BrCH₂COOH > CH₃COOH (1). Halogens withdraw electrons by −I, stabilising the carboxylate ion; F is the most electronegative and has the strongest −I (1). CH₃COOH has no halogen, so it is weakest (1).

10. Q10 · 3 marks · Short answerHOTS / competency

    Suppose a student's Lassaigne extract gives a black precipitate with lead acetate but no Prussian blue with FeSO₄ and H₂SO₄, and no precipitate with AgNO₃ after boiling with HNO₃. (a) Which elements are present and absent? (b) Name one compound that could give these results. (c) Give a second test for the element present. (3 marks)

    Hint

    PbS is black.

    Answer

    (a) Sulphur present; nitrogen and halogens absent (1). (b) Diethyl sulphide, (C₂H₅)₂S, or thiophene (1). (c) Sodium nitroprusside gives a violet colour (1).

11. Q11 · 5 marks · Long answerHOTS / competency

    (a) Explain hyperconjugation using the ethyl carbocation, CH₃CH₂⁺. (b) How many hyperconjugation structures can (CH₃)₂CH⁺ show? (c) Use this to explain the stability order of carbocations. (d) Why can't the methyl carbocation (CH₃⁺) be stabilised in this way? (5 marks)

    Hint

    Count α-C–H bonds.

    Answer

    (a) In CH₃CH₂⁺, a C–H σ bond of the CH₃ group (α to the positive carbon) overlaps with the empty p orbital of the sp² carbocation; the σ electrons partly delocalise, spreading the positive charge (2). (b) Six (two CH₃ groups × three C–H) (1). (c) More α-C–H bonds mean more hyperconjugation and more charge delocalisation: 3° (9) > 2° (6) > 1° (3) > CH₃⁺ (0) (1). (d) CH₃⁺ has no carbon next to the positive carbon, so it has no α-C–H bonds to donate electrons (1).

12. Q12 · 1 mark · Assertion-reasonHOTS / competency

    Assertion (A): Kjeldahl's method cannot be used to estimate nitrogen in nitrobenzene.
    Reason (R): The nitrogen of a nitro group is not converted to ammonium sulphate on heating with concentrated H₂SO₄.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

    Hint

    What does Kjeldahl rely on?

    Answer

    (a)

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What is the hybridisation of carbon in ethyne?Show answer

   sp.

2. 2.Give an example of a heterocyclic aromatic compound.Show answer

   Pyridine, furan or thiophene.

3. 3.By how much do successive members of a homologous series differ?Show answer

   By a CH₂ unit (14 u).

4. 4.Name CH₃CH₂CH(OH)CH₃.Show answer

   Butan-2-ol.

5. 5.Name the type of isomerism between ethanol and methoxymethane.Show answer

   Functional group isomerism.

6. 6.What is the shape and hybridisation of a carbocation?Show answer

   Trigonal planar, sp².

7. 7.Give two +R groups.Show answer

   –OH and –NH₂ (also –OR, halogens).

8. 8.Which purification method is used for glycerol from spent lye?Show answer

   Distillation under reduced pressure.

9. 9.What colour indicates nitrogen in Lassaigne's test?Show answer

   Prussian blue.

10. 10.Name the method in which nitrogen is collected as N₂ gas.Show answer

    Dumas method.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you give the hybridisation of every carbon in a molecule with single, double and triple bonds?
• ☐Can you draw bond-line formulas for butan-2-ol and 2-methylpropanal?
• ☐Can you name a compound with two functional groups using priority rules?
• ☐Can you write all structural isomers of C₅H₁₂ and name them?
• ☐Can you compare stability of 1°, 2° and 3° carbocations using hyperconjugation?
• ☐Can you list −I and +R groups and explain their effect on acidity?
• ☐Can you choose the right distillation method for a given mixture?
• ☐Can you calculate % N by Dumas and Kjeldahl, and % X by Carius?

Organic Chemistry – Some Basic Principles and Techniques: frequently asked questions