Chapters

The general organic chemistry of NCERT Class 11 Unit 8 (Organic Chemistry: Some Basic Principles and Techniques): shapes and hybridisation, structural formulas, bond fission, reactive intermediates, electrophiles and nucleophiles, the inductive, resonance and electromeric effects, hyperconjugation and the types of reactions (sections 8.2, 8.3 and 8.7), with aromaticity from Unit 9, plus what JEE Main adds beyond the book (marked Beyond NCERT). Naming is in IUPAC, isomers in Isomerism, purification and analysis in Practical Organic Chemistry.

Formula sheet

Every rule and order in the chapter on one screen, for the last pass. Details and examples are in the sections below.

Quantity	Rule or order	Watch
Hybridisation of C	σ bonds + lone pairs: 4 → sp³, 3 → sp², 2 → sp	Count σ only
s character	sp 50%, sp² 33%, sp³ 25%	More s: shorter bond, more electronegative C
σ bonds in a molecule	atoms − 1 + rings	Every multiple bond has exactly one σ
π bonds	one per C=X, two per C≡X
Carbocation	3° > 2° > 1° > CH₃⁺	Allyl, benzyl beat 1°; vinyl, aryl worst
Free radical	3° > 2° > 1° > CH₃•	Same reasons as cations
Carbanion	CH₃⁻ > 1° > 2° > 3°	sp > sp² > sp³; −R, −I groups help
Hyperconjugation	number of α-H	CH₃⁺ has none; (CH₃)₃C⁺ has 9
−I (pulls)	NO₂ > CN > COOH > F > Cl > Br > I > OR > C₆H₅	Dies out after about 3 bonds
+I (pushes)	O⁻ > COO⁻ > 3° alkyl > 2° > 1° > CH₃
+R	–X, –OH, –OR, –OCOR, –NH₂, –NHR, –NR₂, –NHCOR	Lone pair next to the π system
−R	–COOH, –CHO, >C=O, –CN, –NO₂	π bond to a more electronegative atom
Hückel rule	cyclic, planar, fully conjugated, (4n + 2) π e⁻	4n π: antiaromatic
Acid strength	the more stable the conjugate base, the stronger the acid

0 · What JEE Main asks from this chapter

113 questions in JEE Main 2019–2026, about 1 per paper; 10 are numerical answers. Grouped by type (counts are approximate):

Type	Share	Typical question
Stability of carbocations, carbanions, radicals, alkenes	~27%	Order: C₆H₅CH₂⁺ > CH₃CH₂⁺ > CH₂=CH⁺ > HC≡C⁺ (2021)
Inductive, resonance, electromeric, hyperconjugation (statements, resonance structures, bond lengths)	~22%	Relative stability of the contributing structures of CH₂=CH–CHO (2024)
Acidity, basicity, nucleophilicity	~17%	Strongest acid: CH(CN)₃, CHI₃, CHBr₃ or CHCl₃ (2019)
Aromaticity	~16%	Which is not aromatic: cyclopentadienyl cation (2019)
Hybridisation, σ/π counts, structural formulas	~10%	σ + π bonds in 2-formylhex-4-enoic acid: 22 (2024)
Fission, reagents, reaction types	~9%	Match: Williamson synthesis → nucleophilic substitution (2026)

Most are easy one-step orders. Draw the intermediate or the conjugate base before you rank anything.

1 · Shapes of carbon compounds

Carbon's tetravalence and shapes come from hybridisation of its s and p orbitals:

Carbon	s character	Angle, shape (example)
sp³	25%	109.5°, tetrahedral (CH₄)
sp²	33%	120°, trigonal planar (C₂H₄)
sp	50%	180°, linear (C₂H₂)

• More s character: the orbital is closer to the nucleus, so the bond is shorter and stronger, and the carbon is more electronegative: sp > sp² > sp³.
• π bond: sideways overlap of two parallel p orbitals. Both p orbitals stand perpendicular to the molecular plane, so all six atoms of H₂C=CH₂ lie in one plane, and rotation about C=C is restricted.
• The π cloud lies above and below the plane, loosely held: π bonds are the most reactive centres of a molecule.
• Bond lengths: C–C 154 pm, C=C 134 pm, C≡C 120 pm; benzene's C–C 139 pm.

Counting rule: for a C, count its σ bonds plus lone pairs (each multiple bond is one σ): 4 is sp³, 3 is sp², 2 is sp.

Example (NCERT Problem 8.1) · HC≡C–CH=CH–CH₃: σC–C 4, σC–H 6, πC=C 1, πC≡C 2. CH₂=C=CHCH₃: σC–C 3, σC–H 6, π 2.

Example (NCERT Problems 8.2, 8.3) · CH₃Cl sp³. (CH₃)₂CO sp³, sp². CH₃CN sp³, sp. HCONH₂ sp². CH₃CH=CHCN sp³, sp², sp², sp. H₂C=O sp², trigonal planar; CH₃F sp³, tetrahedral; HCN sp, linear.

Example (2021) · CH₂=C=CH–CH₃: C1 sp², C2 sp (two double bonds), C3 sp², C4 sp³. In CH₂=CH–CH₂–CH₂–C≡CH (exercise 8.36), the C2–C3 bond is sp²–sp³.

Example (2024) · 2-Formylhex-4-enoic acid, HOOC–CH(CHO)–CH₂–CH=CH–CH₃ (C₇H₁₀O₃, 20 atoms, no ring): σ = 20 − 1 = 19; π = 2 C=O + 1 C=C = 3. Total 22.

Example (2025) · O=CH–CH=CH–C≡C–CH=CH–C≡N: sp carbons 3 (two in C≡C, one in C≡N), sp² carbons 5. Hexa-1,3-dien-5-yne (C₆H₆): σ 11 + π 4 = 15.

Primary to quaternary carbon (NCERT Unit 9): a carbon bonded to one other carbon is 1°, two is 2°, three is 3°, four is 4° (quaternary). Terminal carbons are always 1°. In (CH₃)₂CH–CH₂–CH(CH₃)₂ only the middle CH₂ is 2° (2024).

2 · Writing structures

• Complete (dash) formula: every bond as a dash; lone pairs may or may not be shown.
• Condensed formula: dashes left out, identical groups as subscripts: CH₃(CH₂)₆CH₃; CH₃CH₃, H₂C=CH₂, HC≡CH, CH₃OH.
• Bond-line formula: only zigzag lines. Each end and each corner is a carbon with enough H's to make four bonds; heteroatoms (O, N, Cl) are written. HOCH(CN)₂ is an O–H on a corner carrying two –C≡N arms (2024).
• 3-D (wedge–dash): a solid wedge points towards you, a dashed wedge away from you, a plain line lies in the plane.
• Molecular models: framework (bonds only, atom size ignored); ball-and-stick (atoms and bonds; springs for C=C give ball-and-spring); space-filling (relative atom size by van der Waals radius, bonds not shown).

3 · Bond fission and reactive intermediates

Reaction: substrate + attacking reagent → [intermediate] → product(s). The substrate is the reactant that supplies carbon to the new bond; the other is the reagent. If both supply carbon, the one we watch is the substrate. A mechanism is the step-by-step account of electron movement, energetics and rates.

Heterolytic cleavage: the shared pair stays with one fragment. Gives ions: a carbocation (C with a sextet, + charge) or a carbanion (C with a lone pair, − charge). Reactions through heterolysis are ionic, heteropolar or polar.

Homolytic cleavage: one electron goes to each atom (a half-headed fish-hook arrow). Gives free radicals (an unpaired electron, neutral). Reactions through homolysis are free radical, homopolar or non-polar.

Intermediate	Carbon, shape	Electrons on C · acts as
Carbocation	sp², empty p orbital; trigonal planar	6 · electrophile
Carbanion	sp³, lone pair; distorted tetrahedral (pyramidal)	8 · nucleophile, base
Free radical	sp² (nearly planar); planar	7 · neither

Stability (NCERT): alkyl groups stabilise carbocations and radicals by inductive effect and hyperconjugation.

Free radicals follow the same order: methyl < ethyl < isopropyl < tert-butyl.

More on intermediates · Beyond NCERT

• A carbocation is stabilised by anything that gives it electrons: +I, hyperconjugation, +R (a lone pair or π bond next to C⁺), aromaticity. CH₂⁺–CH=CH–OCH₃ beats CH₂⁺–CH=CH–F and the others: the OCH₃ lone pair reaches C⁺ through the C=C (2025).
• A positive charge on an sp or sp² carbon is very unstable: C₆H₅CH₂⁺ > CH₃CH₂⁺ > CH₂=CH⁺ > HC≡C⁺ (2021).
• Carbanions are the reverse: alkyl groups (+I) destabilise them, so CH₃⁻ > 1° > 2° > 3° (2026). They are stabilised by more s character (HC≡C⁻ > CH₂=CH⁻ > CH₃CH₂⁻, 2026), by −I and −R groups, and by resonance (allyl, benzyl carbanions are resonance-stabilised, not inductively).
• Para groups on a benzyl carbanion: −R groups (CHO, NO₂) stabilise it most, −I halogens a little, +R groups (OCH₃) destabilise it. p-CHO > p-Br > H > p-CH₃ > p-OCH₃ (2026).
• Rearrangement: a carbocation shifts a neighbouring H⁻ or CH₃⁻ (1,2-shift) if that makes a more stable cation. 3-Methylbut-1-ene + HCl gives 2-chloro-2-methylbutane through a 1,2-H shift.
• Triphenylmethyl radical and cation are stabilised over three rings; tropylium and cyclopropenyl cations are aromatic.

4 · Electrophiles and nucleophiles

• Nucleophile (Nu:) "nucleus seeking": brings an electron pair to an electron-poor site. Anions with a lone pair (HO⁻, NC⁻, HS⁻, C₂H₅O⁻, H₂N⁻, R₃C:⁻) and neutral molecules with a lone pair (H₂O, R₃N, R₂NH, ROH).
• Electrophile (E⁺) "electron seeking": takes an electron pair from an electron-rich site. Carbocations, BF₃ and AlCl₃ (incomplete octet), ⁺NO₂, and neutral molecules with a δ+ carbon: the C of C=O, C≡N and C–X.
• An electrophilic centre is an electron-poor atom of the substrate: the starred C in CH₃CHO, CH₃CN, H₃C*–I (NCERT Problem 8.13).
• Curved arrows move an electron pair: from where it is (a π bond or a lone pair) to where it goes (an adjacent bond position or atom). A fish-hook moves one electron.

Example (NCERT Problem 8.12) · Nucleophiles: HS⁻, C₂H₅O⁻, (CH₃)₃N:, H₂N:⁻ (an unshared pair to give). Electrophiles: BF₃, Cl⁺, CH₃–C⁺=O, ⁺NO₂ (a site with only six electrons).

Nucleophilicity · Beyond NCERT

• An anion is a better nucleophile than its neutral parent: OH⁻ > H₂O, RO⁻ > ROH, NH₂⁻ > NH₃ (2022).
• Across a period it follows basicity: CH₃⁻ > NH₂⁻ > OH⁻ > F⁻.
• The anion of a strong acid is weak: H₂O < CH₃SO₃⁻ < CH₃COO⁻ < OH⁻ (2019).
• In water and alcohols, down a group it rises (bigger, less solvated): I⁻ > Br⁻ > Cl⁻ > F⁻.

5 · Electron displacement effects

Effects in the ground state, caused by an atom or group, polarise a bond permanently: the inductive and resonance effects (and hyperconjugation). An effect that appears only when a reagent approaches is temporary: the electromeric (polarisability) effect.

Inductive effect

A polar bond (C–Cl in CH₃CH₂Cl: δ+ C, δ− Cl) draws electrons from the next σ bond too, giving C2 a smaller δδ+. Polarisation of a σ bond caused by a neighbouring polar σ bond is the inductive effect. It falls off fast and is negligible after three bonds.

• Electron-withdrawing (−I) relative to H: halogens, –NO₂, –CN, –COOH, –COOR, –OAr (–OC₆H₅).
• Electron-donating (+I): alkyl groups (–CH₃, –C₂H₅).

Example (NCERT Problems 8.14, 8.15) · More polar: C–Br (vs C–H), C–O (vs C–N), C–O (vs C–S). In CH₃CH₂CH₂Br the effect is least at the bond farthest from Br (C3–H).

Example (NCERT exercise 8.17) · Acidity Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH: more −I chlorines stabilise the carboxylate more. CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH: more +I alkyl destabilises it.

Example (NCERT exercise 8.9) · O₂NCH₂CH₂O⁻ is more stable than CH₃CH₂O⁻: −NO₂ (−I) spreads the negative charge.

Resonance

When one Lewis structure can't describe a molecule, the real structure is a resonance hybrid of canonical (contributing) structures. These are hypothetical; none exists on its own. Each contributes in proportion to its stability.

• Benzene: all six C–C bonds are 139 pm, between C–C (154 pm) and C=C (134 pm): a hybrid of the two Kekulé structures.
• Nitromethane: both N–O bonds are the same length, between N–O and N=O.
• Resonance energy: the hybrid is lower in energy than any contributor. The difference between the actual structure and the lowest-energy contributor is the resonance (stabilisation) energy. The more important contributors, the larger it is; it is largest when the contributors are equivalent.

Rules for contributors: the same positions of nuclei and the same number of unpaired electrons; only electrons move. A contributor is more stable when it has:

1. more covalent bonds;
2. every atom with an octet (H a duplet);
3. less separation of opposite charges;
4. a negative charge on the more electronegative atom and a positive charge on the more electropositive one;
5. more dispersal of charge.

Example (NCERT Problem 8.18) · In CH₃COOCH₃, structures with C⁺ having only six electrons, or with charge separation that puts + on the wrong atom, are minor contributors.

Example (NCERT Problem 8.16) · CH₃COO⁻ has two equivalent contributors (the − charge on either O), so both C–O bonds are equal.

Resonance (mesomeric) effect

The polarity produced by the interaction of two π bonds, or of a π bond with a lone pair on an adjacent atom (asked 2024). It is passed along a conjugated chain. Written R or M.

• +R: electrons move away from the group into the conjugated system, raising the electron density at certain positions (aniline: ortho, para). Groups: –X (halogen), –OH, –OR, –OCOR, –NH₂, –NHR, –NR₂, –NHCOR.
• −R: electrons move towards the group (nitrobenzene). Groups: –COOH, –CHO, >C=O, –CN, –NO₂.
• Conjugated system: alternate single and multiple bonds (buta-1,3-diene, aniline, nitrobenzene). Its π electrons are delocalised; it often behaves abnormally.

Using resonance on bond lengths · Beyond NCERT

• A bond with partial double-bond character is shorter. In CH₂=CH–Cl the C–Cl bond is shorter than in CH₃Cl; in Cl–CH=CH–NO₂ it is shorter still, because the −R NO₂ pulls the Cl lone pair further in (2020).
• Methyl acetate: C=O (α) < C(O)–O (β, partly double) < O–CH₃ (γ) (2026).
• C=O length: tropone (C=O pulled to C⁺–O⁻, aromatic ring) > acetate (bond order 1.5) > acetone (2026).
• Conjugation raises the dipole moment: CH₃CH=CHCHO > CH₃CH₂CH₂CHO (2025).

Electromeric effect (E)

A temporary effect of compounds with a multiple bond, only in the presence of an attacking reagent: the complete transfer of a shared π pair to one of the two atoms, on the reagent's demand. It vanishes when the reagent is removed. Shown by a curved arrow.

• +E: the π pair goes to the atom to which the reagent attaches. An electrophile such as H⁺ adding to C=C.
• −E: the π pair goes to the atom the reagent does not attach to. A nucleophile such as CN⁻ attacking C=O: the pair moves to O.
• When the inductive and electromeric effects oppose, the electromeric effect wins.

Example (2024, 2026) · False: "H⁺ shows a −E effect" (an electrophile causes +E). False: "an unsaturated system shows +E with a nucleophile and −E with an electrophile" (the reverse).

Hyperconjugation

A general stabilising interaction: the σ electrons of a C–H bond on an alkyl group directly attached to an unsaturated atom, or to an atom with an empty p orbital, partly delocalise into it. It is permanent.

• In CH₃CH₂⁺, one C–H bond of CH₃ lines up with the empty p orbital and shares its electron density, spreading the positive charge.
• More alkyl groups on C⁺ means more hyperconjugation and a more stable cation. (CH₃)₃C⁺ has nine C–H bonds for it; in CH₃⁺ the empty p orbital is perpendicular to the C–H bonds, so there is none (NCERT Problem 8.19).
• Also in alkenes and alkylarenes (propene: CH₃ next to C=C). It can be seen as no-bond resonance: H⁺ ⁻CH₂–CH=CH₂ written as a contributor of propene.
• The overlap is of a C(sp³)–H σ bond with the empty 2p orbital (2021: "C(sp²)–H" is wrong).

Counting it · Beyond NCERT

• Hyperconjugative structures = number of α-H: H's on the carbons attached to C⁺, the radical carbon or the C=C.
• Alkene stability rises with α-H: (CH₃)₂C=C(CH₃)₂ (12) > (CH₃)₂C=CHCH₃ (9) > cis-but-2-ene (6) > propene (3) (2026). Trans beats cis for the same count (less crowding).

6 · Acidity and basicity · Beyond NCERT

An acid is stronger when its conjugate base is more stable. Rank the anions by charge on a more electronegative atom, resonance, −I and −R groups, s character and aromaticity.

• Carboxylic acid > phenol > water > alcohol > alkyne > alkene > alkane.
• −I groups raise acidity, more so when closer and more of them: FCH₂COOH > ClCH₂COOH > CH₃COOH; 2-chlorobutanoic > 3- > 4-.
• Carbon acids: HC≡CH > CH₂=CH₂ > CH₃CH₃ (s character; NCERT Unit 9). CH(CN)₃ is the strongest acid of CH(CN)₃, CHI₃, CHBr₃, CHCl₃: three –CN spread the charge by resonance (2019).
• α-H next to C=O is acidic: the carbanion is an enolate. Between two C=O (1,3-dicarbonyl) it is far more acidic; three esters on one C is more acidic still (2020).
• Aromatic anion: cyclopentadiene is a strong carbon acid because its anion is aromatic (6π); it beats toluene, cyclopropene and butane (2021).
• Conjugate bases: the weaker the acid, the stronger its base: RO⁻ > OH⁻ > CH₃COO⁻ > Cl⁻ (2024).
• Amines (Class 12): alkyl groups raise basicity; a lone pair in resonance (aniline) or on sp² N (pyridine) lowers it.

7 · Aromaticity

From NCERT Unit 9: a ring is aromatic if it is planar, has its π electrons completely delocalised round the ring, and has (4n + 2) π electrons (n = 0, 1, 2…; the Hückel rule). Benzene (6π) is the parent.

Aromatic, antiaromatic, non-aromatic · Beyond NCERT

Class	Condition	Examples
Aromatic	cyclic, planar, fully conjugated, 2, 6, 10… π e⁻	benzene, naphthalene, anthracene; cyclopropenyl cation (2π); cyclopentadienyl anion, tropylium cation (6π); pyrrole, furan, thiophene, pyridine; cyclooctatetraene dianion (10π)
Antiaromatic	cyclic, planar, conjugated, 4, 8… π e⁻	cyclobutadiene, cyclopentadienyl cation, cyclopropenyl anion
Non-aromatic	not planar, or an sp³ carbon breaks the ring	cyclooctatetraene (tub), cycloheptatriene, cyclopentadiene, [10]annulene (H's clash)

• Stability: aromatic > non-aromatic > antiaromatic.
• In pyrrole, furan, thiophene a heteroatom lone pair is one of the six π electrons. In pyridine the N lone pair lies in an sp² orbital outside the ring, so pyridine is a base and pyrrole hardly is.
• Count π electrons of a ring: each ring C=C 2; a ring C⁻ (lone pair) 2; a ring C⁺ 0; a heteroatom lone pair 2 (only one pair per atom).
• Tropolone and tropone are aromatic: the ring C=O is pulled to C⁺–O⁻, leaving a tropylium-like 6π ring. The C=O π electrons are not part of the aromatic six (2023).
• Benzenoid compounds contain a benzene ring (phenanthrene, naphthalene, aniline); cyclooctatetraene is not benzenoid (2022).

Example (2019) · Cyclopentadienyl anion, pyridine and pyrrole are aromatic; the cyclopentadienyl cation (4π) is not.

Example (2022) · [6]annulene (benzene) aromatic, [8]annulene non-aromatic (tub), cis-[10]annulene not aromatic because it can't be planar. The assertion said aromatic, so it is false; the reason (planarity is required) is true.

8 · Types of organic reactions

NCERT names four: substitution, addition, elimination and rearrangement. Each can be nucleophilic, electrophilic or free radical depending on the attacking species.

Example (NCERT exercise 8.14) · CH₃CH₂Br + HS⁻ → CH₃CH₂SH: nucleophilic substitution. (CH₃)₂C=CH₂ + HCl → (CH₃)₂CClCH₃: electrophilic addition. CH₃CH₂Br + HO⁻ → CH₂=CH₂: elimination. (CH₃)₃CCH₂OH + HBr → (CH₃)₂CBrCH₂CH₃: rearrangement (then substitution).

Example (2026) · Williamson synthesis: nucleophilic substitution. Friedel–Crafts reaction: electrophilic substitution. Bromination of vinylbenzene: electrophilic addition. Chlorination of toluene in light: free radical substitution.

9 · Traps

Trap	What goes wrong	Do this
Carbanion order	Using the carbocation order	Reverse it: CH₃⁻ > 1° > 2° > 3°
Hyperconjugation count	Counting all H's	Only α-H; CH₃⁺ has none
Hyperconjugation type	Calling it temporary	Permanent; only the electromeric effect is temporary
+E / −E	Swapping them	+E: electrons go to where the reagent attaches
Halogens	Calling them only −I	−I and +R: deactivating but o/p-directing
Aromatic count	Counting both lone pairs of furan's O	One pair joins the ring
Pyridine vs pyrrole	Assuming both N pairs are in the ring	Pyridine's is outside: it is the base
Non- vs antiaromatic	Calling cyclooctatetraene antiaromatic	It is tub-shaped: non-aromatic
σ count	Counting multiple bonds as several σ	σ = atoms − 1 + rings
Resonance energy	Measuring from the average	From the lowest-energy contributor
Acidity	Judging the acid, not the anion	Draw the conjugate base and rank its stability
Resonance structures	Moving atoms	Only electrons move; nuclei stay

10 · Numbers worth knowing

• Bond lengths: C–C 154 pm, C=C 134 pm, C≡C 120 pm, benzene C–C 139 pm.
• s character: sp 50%, sp² 33%, sp³ 25%. Angles 180°, 120°, 109.5°.
• Hückel numbers: 2, 6, 10, 14 π electrons aromatic; 4, 8 antiaromatic.
• α-H counts: CH₃CH₂⁺ 3, (CH₃)₂CH⁺ 6, (CH₃)₃C⁺ 9.

11 · For the boards only

JEE Main doesn't ask this, but the CBSE paper can.

• Berzelius proposed a "vital force" for organic compounds. Wöhler (1828) made urea from ammonium cyanate: NH₄CNO → NH₂CONH₂.
• Kolbe synthesised acetic acid (1845) and Berthelot methane (1856), settling that organic compounds can be made in the laboratory.
• Carbon's unique property of forming chains with itself is catenation.
• Molecular models are made of wood, plastic or metal; computer graphics are also used.