Chapters

The isomerism part of NCERT Class 11 Unit 8 (Organic Chemistry: Some Basic Principles and Techniques, section 8.6), with the conformations and cis–trans isomerism that NCERT gives in Unit 9 (Hydrocarbons), plus the optical and geometrical isomerism JEE Main asks beyond the book (marked Beyond NCERT).

Formula sheet

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

Quantity	Formula	Watch
Degree of unsaturation	DU = (2C + 2 + N − H − X)/2	O and S don't count; a ring or a π bond = 1
Geometrical isomers, unsymmetrical	2ⁿ (n stereogenic C=C)	Each C=C needs two different groups on both ends
Geometrical isomers, symmetrical	2ⁿ⁻¹ + 2ᵖ⁻¹	p = n/2 (n even), (n + 1)/2 (n odd)
Optical isomers, unsymmetrical	2ⁿ (n chiral centres)	No meso form
Symmetrical, n even	active 2ⁿ⁻¹, meso 2^(n/2 − 1)	Tartaric acid: 2 + 1 = 3
Symmetrical, n odd	active 2ⁿ⁻¹ − 2^((n−1)/2), meso 2^((n−1)/2)	Total 2ⁿ⁻¹
C=C and chiral C together	multiply when independent	CH₃CH(OH)CH=CHCH₃: 2 × 2 = 4
Specific rotation	[α] = α / (l × c)	l in dm, c in g mL⁻¹
Optical purity (ee)	observed rotation / pure rotation × 100	12.6/30 = 42%
Ethane barrier	about 12.5 kJ mol⁻¹	Staggered to eclipsed

0 · What JEE Main asks from this chapter

58 questions in JEE Main 2019–2026, about 1 in every 3 papers; 26 of them are numerical answers. Grouped by type:

Type	Share	Typical question
Count chiral centres, optical or stereoisomers	~43%	Optically active C₄H₈Cl₂ from 2-chlorobutane + Cl₂: 6 (2024)
Name the type of structural isomerism (incl. tautomerism)	~19%	Match: propanone/propanal → functional (2024)
Count structural isomers or substitution products	~14%	Monochlorination products of 2-methylbutane, with stereo: 6 (2024)
Geometrical isomerism	~14%	Which shows it: 4-methylpent-2-ene (2020)
Conformations	~10%	Order of energy of butane conformers (2021)

Numerical answers are small integers. Write every isomer down; don't trust a formula until symmetry has been checked.

1 · Types of isomerism

Isomerism: two or more compounds with the same molecular formula but different properties. NCERT's chart, with JEE's additions in grey:

Structural isomers have the same molecular formula but atoms linked differently.

1. Chain isomers: different carbon skeletons. C₅H₁₂: pentane, 2-methylbutane (isopentane), 2,2-dimethylpropane (neopentane). C₄H₁₀ has 2, C₅H₁₂ 3, C₆H₁₄ 5, C₇H₁₆ 9, C₁₀H₂₂ 75.
2. Position isomers: the same skeleton, the substituent or functional group in a different place. C₃H₈O: propan-1-ol and propan-2-ol.
3. Functional group isomers: different functional groups. C₃H₆O: propanal and propanone.
4. Metamers: different alkyl groups on either side of the same functional group. C₄H₁₀O: methoxypropane (CH₃OC₃H₇) and ethoxyethane (C₂H₅OC₂H₅).

Stereoisomers have the same constitution and sequence of bonds but differ in the relative positions of atoms or groups in space: geometrical and optical isomerism.

JEE's usual pairs · Beyond NCERT

Pair	Isomerism
Alcohol and ether (butan-2-ol, diethyl ether)	functional
Aldehyde and ketone; acid and ester	functional
1°, 2°, 3° amines of one formula (butanamine, diethylamine)	functional
Cyanide and isocyanide (butyl cyanide, butyl isocyanide)	functional
Alkene and cycloalkane (but-1-ene, cyclobutane)	functional (ring–chain)
Pentan-2-one and pentan-3-one; hexan-2-one and hexan-3-one	metamers (JEE's key, though also position)
Ethers, ketones, 2° amines, esters with groups swapped	metamers
Cyclopentane and methylcyclobutane	chain (ring size)
But-1-ene and but-2-ene; o- and p-nitrophenol	position
2-Methylpropene and but-1-ene	chain

Metamerism needs a polyvalent group (–O–, –CO–, –NH–, –COO–) with carbon on both sides. Isomers must have the same formula: butanamine (C₄H₁₁N) and N-methylethanamine (C₃H₉N) are not isomers at all.

Example (2026) · True: propanal and propanone are functional isomers; ethoxyethane and methoxypropane are metamers; pentane and 2,2-dimethylpropane are chain isomers. False: but-2-ene is not optically active; but-1-ene and but-2-ene are position, not functional, isomers.

2 · Tautomerism · Beyond NCERT

Tautomers are structural isomers in rapid equilibrium by the shift of an H atom (usually from an α-carbon to O). Keto–enol tautomerism needs at least one α-H on an sp³ carbon next to C=O.

• CH₃COCH₃ ⇌ CH₂=C(OH)CH₃: under 0.1% enol.
• 1,3-Dicarbonyls have much more enol: the enol C=C is conjugated with the other C=O, and an intramolecular H-bond makes a six-membered ring. Acetylacetone (CH₃COCH₂COCH₃) has far more enol than acetone (2021).
• Order of enol content: CH₃COCH₂COCH₃ > CH₃COCH₂COOC₂H₅ > CH₃COCH₂CONH₂ > CH₃COCH₃ (2019). The ester and amide groups give their own lone pair to the C=O, so it pulls less.
• An enol that is aromatic wins outright: cyclohexane-1,3,5-trione exists as phloroglucinol (benzene-1,3,5-triol) (2022).
• No α-H, no tautomerism: 2,2,6,6-tetramethylcyclohexanone, benzaldehyde, (CH₃)₃CCHO (2026).
• Other tautomers JEE names: amide ⇌ imidic acid (ethanamide ⇌ hydroxyethanimine, 2023), nitro ⇌ aci-nitro.

3 · Conformations

From NCERT Unit 9. Rotation about a C–C σ bond doesn't break it, so the molecule can take endless arrangements that interconvert. These are conformations (conformers, rotamers). Bond lengths and angles stay the same in all of them.

• Rotation is not completely free: a barrier of 1–20 kJ mol⁻¹ comes from repulsion between adjacent bonds, the torsional strain.
• The angle between the bonds on the front and back carbons is the dihedral (torsional) angle.
• Ethane: staggered (H's as far apart as possible, dihedral 60°) is most stable; eclipsed (H's as close as possible, 0°) is least stable; anything between is skew. The difference is about 12.5 kJ mol⁻¹, small enough that collisions supply it at room temperature, so the conformers cannot be isolated.
• Sawhorse projection: the C–C bond drawn as a slanting line, front carbon at the lower end, rear carbon at the upper end, three bonds at 120° on each.
• Newman projection: looking straight down the C–C bond; the front carbon is a point, the rear carbon a circle, three bonds at 120° on each.

In the eclipsed drawing the rear H's are drawn a few degrees off so they can be seen; in fact they sit directly behind the front ones.

Butane (about the C2–C3 bond) · Beyond NCERT

Conformer	CH₃–CH₃ dihedral	Energy
Anti (staggered)	180°	lowest
Gauche (staggered)	60°	low (steric strain between CH₃'s)
Partially eclipsed	120°	high
Fully eclipsed	0°	highest

• Increasing energy: anti < gauche < partially eclipsed < fully eclipsed (2021). From fully eclipsed to anti, the dihedral angle rises and the torsional strain falls (2026).
• What sets the stability of an acyclic conformer: torsional strain, steric interactions, and electrostatic forces (dipoles, H-bonds). Angle strain does not: the bond angles don't change (2019).
• In the staggered form of any ethane derivative (1,1,1-trichloroethane too) the dihedral angle between neighbouring groups is 60° (2021).
• Cyclohexane: chair (no angle or torsional strain) > twist-boat > boat > half-chair. Large groups prefer equatorial positions.

4 · Geometrical (cis–trans) isomerism

From NCERT Unit 9. Rotation about C=C is restricted (turning it would break the π bond), so groups on the two carbons are fixed in space.

• Condition: each doubly bonded carbon carries two different groups: XYC=CXY, XYC=CXZ, XYC=CZW. If either carbon has two identical groups (CH₂=CBr₂, (CH₃)₂C=CHC₂H₅), there is none.
• cis: identical groups on the same side; trans: on opposite sides. Same structure, different configuration, so different m.p., b.p., dipole moment and solubility.
• cis-But-2-ene has μ = 0.33 D; trans-but-2-ene has μ ≈ 0 because the two C–CH₃ dipoles cancel. The cis form is more polar, so it has the higher b.p.
• The trans form packs better in a crystal, so it has the higher m.p., and it is more stable (less steric strain).
• cis and trans-but-2-ene don't interconvert at room temperature.

Example (NCERT Problem 9.11) · Of (CH₃)₂C=CHC₂H₅, CH₂=CBr₂, C₆H₅CH=CHCH₃ and CH₃CH=CClCH₃, only the last two show cis–trans isomerism.

Example (2025) · Incorrect: "propene shows geometrical isomerism" (one carbon has two H's), "2-methylbut-2-ene has two geometrical isomers" (one carbon has two CH₃'s), "the trans isomer has the lower m.p." Correct: cis-but-2-ene has the higher dipole moment.

E/Z and rings · Beyond NCERT

• When all four groups differ, cis/trans fails. Rank the two groups on each carbon by the CIP rules: higher atomic number first; on a tie, compare the next atoms out; a double bond counts its atom twice. Higher-ranked groups on the same side: Z; opposite: E.
• Rings: a disubstituted ring (1,2-dimethylcyclopropane, 1,3-dimethylcyclopentane) has cis and trans forms, because ring bonds can't rotate either.
• A C=C inside a ring smaller than eight atoms can only be cis.

Example (2020) · 4-Methylpent-2-ene, (CH₃)₂CH–CH=CH–CH₃, shows geometrical isomerism; 2-methylpent-2-ene and the pent-1-enes don't.

5 · Optical isomerism · Beyond NCERT

NCERT names optical isomerism here only as a type of stereoisomerism; Class 12 Haloalkanes gives the basics. JEE asks this in detail.

• Chiral: not superimposable on its mirror image. The usual cause is a chiral (asymmetric) carbon: sp³ C with four different groups.
• Enantiomers: non-superimposable mirror images. Same physical properties (m.p., b.p., solubility) except the direction of rotating plane-polarised light: (+) dextro, (−) laevo, equal angles.
• Diastereomers: stereoisomers that are not mirror images (cis/trans pairs; isomers with two or more chiral centres differing at some of them). Different physical properties.
• Meso: has chiral centres but a plane of symmetry, so it is achiral and optically inactive (internal compensation). meso-Tartaric acid.
• Racemic mixture: 50 : 50 enantiomers, inactive by external compensation. An sp² carbocation or a planar C=O attacked equally from both faces gives a racemic product.
• Chirality without a chiral carbon: allenes abC=C=Cab with a ≠ b on each end; biphenyls with bulky ortho groups that block rotation.
• Molecules with a plane or centre of symmetry are achiral even when they look crowded.

R/S (CIP): rank the four groups 1 → 4. With group 4 pointing away, 1 → 2 → 3 clockwise is R, anticlockwise S. In a Fischer projection (horizontal bonds come towards you): if group 4 is on a horizontal bond, reverse the answer. Swapping any two groups inverts the configuration; turning the projection 180° in the plane keeps it; 90° inverts it.

Example (2024) · 2-Chlorobutane has one chiral carbon (H, Cl, CH₃, C₂H₅): 2 optical isomers.

Example (2022) · Chiral alcohols C₄H₁₀O: only butan-2-ol is chiral → 2 (its pair).

Example (2024) · 2-Chlorobutane + Cl₂ → C₄H₈Cl₂. Chlorine at C1 gives 1,2-dichlorobutane (chiral: 2); at C2, 2,2- (achiral); at C3, 2,3- (meso + a pair: 2 active); at C4, 1,3- (chiral: 2). Optically active: 2 + 2 + 2 = 6.

Example (2025) · CH₃CH(OH)CH=CHCH₃ has one chiral carbon and one stereogenic C=C: 2 × 2 = 4 stereoisomers. 5-Phenylpent-4-en-2-ol: also 4.

Example (2022) · Dimethylcyclopentane: 1,2- gives cis (meso) + trans pair = 3; 1,3- gives cis (meso) + trans pair = 3; 1,1- has none. Total 6. 1,2-Dimethylcyclopropane: 3 (2021).

Example (2022) · A mixture rotates +12.6°; the pure (+) isomer rotates +30°. Optical purity = 12.6/30 × 100 = 42%.

Example (2026) · Of n-propyl, isopropyl, sec-butyl and neopentyl chloride, only sec-butyl chloride (C₄H₉Cl, 92.5) is chiral. % C = 48/92.5 × 100 = 52%.

Example (2020) · 3-Methylbut-1-ene + HCl does not give a racemic product: the 2° carbocation shifts an H to become 3°, and the product 2-chloro-2-methylbutane is achiral. Butanone + HCN and but-1-ene + HBr do give racemates.

6 · Counting isomers · Beyond NCERT

1. Find the DU: rings plus π bonds.
2. List skeletons from the longest chain down, then move the group along each.
3. Delete repeats by symmetry, then add stereoisomers if asked.

Example (2022) · Monobromo C₅H₁₂, no stereo: pentane 3 (C1, C2, C3) + 2-methylbutane 4 + neopentane 1 = 8.

Example (2024) · Monochlorination of 2-methylbutane, with stereo: ClCH₂CH(CH₃)C₂H₅ (chiral, 2) + (CH₃)₂CClC₂H₅ (1) + (CH₃)₂CHCHClCH₃ (chiral, 2) + (CH₃)₂CHCH₂CH₂Cl (1) = 6.

Example (2022) · Monochlorination of methylcyclohexane, with stereo: 1-chloro (1) + 2-chloro (2 chiral C: 4) + 3-chloro (4) + 4-chloro (cis/trans: 2) + chloromethyl (1) = 12.

Example (2021) · Acyclic pentenes with geometrical isomers: pent-1-ene, cis- and trans-pent-2-ene, 2-methylbut-1-ene, 3-methylbut-1-ene, 2-methylbut-2-ene = 6.

Example (2025) · Benzene derivatives C₉H₁₂: propylbenzene, isopropylbenzene, three ethyltoluenes (o, m, p), three trimethylbenzenes (1,2,3; 1,2,4; 1,3,5) = 8.

7 · Traps

Trap	What goes wrong	Do this
Symmetrical molecule	Writing 2ⁿ for 2,3-dichlorobutane	Check for a meso form: 3, not 4
"Stereoisomers" of 1,1-dimethylcyclopentane	Counting it as 1	It has none; count only cis/trans and chiral forms
Geometrical isomerism	Missing two identical groups on one carbon	Check both carbons; propene and 2-methylbut-2-ene have none
m.p. vs b.p. of cis/trans	Swapping them	cis: higher b.p. (polar); trans: higher m.p. (packing)
Metamers vs position	Calling pentan-2-one/pentan-3-one position only	JEE keys call them metamers
"Isomers"	Pairing butanamine with N-methylethanamine	Check the formula first: C₄H₁₁N vs C₃H₉N
Tautomerism	Assuming every ketone has an enol	Needs an α-H on sp³ carbon
Racemic product	Forgetting carbocation shifts	A shift can make the product achiral
Conformer stability	Naming angle strain for butane	Acyclic: torsional + steric (+ electrostatic)
Fischer projection	Reading R/S with H on a horizontal bond	Reverse the answer

8 · Numbers worth knowing

• Chain isomers: C₄H₁₀ 2, C₅H₁₂ 3, C₆H₁₄ 5, C₇H₁₆ 9, C₁₀H₂₂ 75.
• C₄H₈ (alkenes) 3 structural (4 with cis/trans); C₅H₁₀ (alkenes) 5 structural (6 with cis/trans).
• Ethane barrier 12.5 kJ mol⁻¹; σ-rotation barriers 1–20 kJ mol⁻¹.
• cis-But-2-ene μ = 0.33 D; trans ≈ 0.
• Tartaric acid: 3 stereoisomers (d, l, meso).

9 · For the boards only

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

• NCERT's flow chart of isomerism (structural: chain, position, functional group, metamerism; stereo: geometrical, optical) is a favourite two-mark diagram.
• NCERT's cardboard model: two cardboards joined by two nails can't be rotated against each other, like a C=C.