Class 11Chemistry · Organic ChemistryFull chapter

Organic Chemistry — Some Basic Principles and Techniques

The whole chapter in one place — read it, then test yourself. Clear notes, key facts, a practice quiz, and worked NCERT solutions & PYQs.

Classification of Organic Compounds

Quick answer Organic compounds are first classified by their carbon skeleton as acyclic or cyclic, and then grouped into homologous series based on the functional group present.

Classification by carbon skeleton. Organic compounds are divided into two broad classes depending on how the carbon atoms are linked together.

  • Acyclic (open-chain) compounds, also called aliphatic compounds: carbon atoms form a straight or branched chain with no ring, e.g. CH3–CH2–CH2–CH3 (n-butane) or CH3–CH(CH3)–CH3 (isobutane, i.e. 2-methylpropane).
  • Cyclic (closed-chain / ring) compounds: carbon atoms are joined to form one or more rings. These are further divided into homocyclic (carbocyclic) compounds, in which every ring atom is carbon, and heterocyclic compounds, in which the ring contains at least one atom other than carbon (commonly N, O or S), e.g. pyridine and furan.
  • Homocyclic compounds are further split into alicyclic compounds (ring behaves like an open-chain compound, e.g. cyclohexane, cyclopropane) and aromatic compounds (planar, cyclic, conjugated ring systems showing special stability, e.g. benzene and its derivatives).

Classification by functional group. A functional group is an atom or group of atoms in a molecule that determines its characteristic chemical behaviour, e.g. –OH (alcohol), –CHO (aldehyde), >C=O (ketone), –COOH (carboxylic acid), –NH2 (amine), –X (haloalkane), –O– (ether), and the carbon–carbon multiple bonds C=C and C≡C. Compounds containing the same functional group and differing from the next member by a –CH2– unit form a homologous series. Members of a homologous series can be represented by the same general formula, have similar chemical properties, and show a gradual (graded) change in physical properties such as boiling point.

Worked example. Consider the first three members of the alcohol series: CH3OH (methanol), CH3CH2OH (ethanol) and CH3CH2CH2OH (propan-1-ol). Each differs from the previous one by exactly one –CH2– unit (mass 14 u): going from CH3OH (32 u) to C2H5OH (46 u) adds 14 u, and going to C3H7OH (60 u) adds another 14 u. All three share the same functional group (–OH), fit the general formula CnH2n+1OH, and show a steady rise in boiling point, confirming they belong to one homologous series.

General formula of alkanes CₙH₂ₙ₊₂ Saturated open-chain hydrocarbons
General formula of alkenes CₙH₂ₙ One C=C double bond
General formula of alkynes CₙH₂ₙ₋₂ One C≡C triple bond
Homologous increment –CH₂– = 14 u Constant mass difference between successive members
Remember
  • Acyclic (aliphatic) compounds have an open chain; cyclic compounds have a ring, further split into homocyclic (alicyclic/aromatic) and heterocyclic.
  • A functional group is the reactive atom/group that fixes a compound's chemical behaviour.
  • A homologous series is a family of compounds with the same functional group, same general formula, and successive members differing by –CH₂– (14 u).
  • Homologous series show similar chemical properties but a gradual change in physical properties (e.g. boiling point rises with size).
  • Aromatic compounds are cyclic, planar and conjugated, showing special stability distinct from alicyclic rings.

Structural Representation of Organic Compounds

Quick answer Organic molecules can be drawn as complete Lewis (dot) structures, condensed formulas, bond-line (skeletal) formulas, or three-dimensional wedge–dash structures.

Because organic molecules can be large, chemists use several shorthand ways to represent the same structure.

  • Complete structural (Lewis) formula: every atom and every bond (including C–H bonds) is drawn explicitly, showing all shared electron pairs.
  • Condensed structural formula: bonds to hydrogen (and sometimes to other atoms) are omitted; atoms are simply grouped next to the carbon they are attached to, e.g. CH3CH2OH for ethanol or CH3CH(CH3)2 for isobutane.
  • Bond-line (skeletal) formula: the most compact representation. Each line represents a carbon–carbon bond, each line end and each vertex (corner) represents one carbon atom, and hydrogen atoms attached to carbon are not shown at all — they are assumed to complete carbon's valency of four. Atoms other than carbon and hydrogen (O, N, X, etc.) are always written explicitly, along with any hydrogens attached to them.
  • Three-dimensional (wedge–dash) representation: used to show the actual spatial arrangement of atoms around a carbon. A plain line lies in the plane of the paper, a bold wedge represents a bond coming out of the plane towards the viewer, and a dashed (hashed) wedge represents a bond going behind the plane away from the viewer.

Worked example. Convert the condensed formula of isobutane, CH3–CH(CH3)–CH3, into its bond-line formula and verify the hydrogen count. The skeleton has 4 carbons: a central carbon joined to three terminal carbons (a 'Y' shape in bond-line notation, with no atom symbols shown for C or H). Applying the valency rule (carbon must form 4 bonds in total): the central carbon already has 3 C–C bonds shown, so it carries 4 − 3 = 1 hydrogen; each terminal carbon has only 1 C–C bond shown, so each carries 4 − 1 = 3 hydrogens. Total H = 1 + 3 + 3 + 3 = 10, which matches the molecular formula C4H10 exactly.

Valency used for H-count in skeletal formulas C = 4, H = 1, O = 2, N = 3, halogen(X) = 1
Implicit hydrogens on a skeletal carbon H(implicit) = 4 − (number of bonds shown at that carbon)
Remember
  • Complete structural formulas show every atom and bond; condensed formulas hide C–H bonds by grouping atoms.
  • In a bond-line formula, every line end/vertex is a carbon atom and hydrogens on carbon are never drawn explicitly.
  • Heteroatoms (O, N, halogens, etc.) and hydrogens attached to them are always shown explicitly in a bond-line formula.
  • Hydrogens implied at a skeletal carbon = 4 minus the number of bonds already drawn at that carbon.
  • Wedge–dash notation shows 3-D arrangement: plain line = in-plane, bold wedge = towards viewer, dashed wedge = away from viewer.

The IUPAC System of Nomenclature

Quick answer An IUPAC name is built from a word root (carbon count of the longest chain), a primary suffix (degree of saturation), and secondary prefixes/suffixes for substituents and functional groups, assigned using the lowest-locant rule.

IUPAC (systematic) names are constructed in a definite sequence: word root + primary suffix + secondary suffix, together with substituent prefixes and locants.

  • Word root indicates the number of carbons in the longest continuous chain (the parent chain): meth- (1C), eth- (2C), prop- (3C), but- (4C), pent- (5C), hex- (6C), hept- (7C), oct- (8C), non- (9C), dec- (10C).
  • Primary suffix shows saturation: -ane (all single bonds), -ene (one or more C=C), -yne (one or more C≡C).
  • Secondary suffix names the principal characteristic (functional) group, e.g. -ol (alcohol), -al (aldehyde), -one (ketone), -oic acid (carboxylic acid), -amine (amine). When several functional groups are present, only the senior-most one is expressed as a suffix; the rest are named as substituent prefixes (e.g. oxo- for a ketone, hydroxy- for –OH).
  • Substituent (side-chain) prefixes such as methyl-, ethyl-, chloro-, bromo- are cited in alphabetical order; multiplying prefixes (di-, tri-, tetra-) are used for repeated identical substituents but are ignored while alphabetising.
  • Numbering (locants) is chosen so that the principal characteristic group gets the lowest possible locant; if there is a choice, the set of locants for substituents as a whole must be as low as possible (lowest locant rule).

Worked example. Name CH3–CH(CH3)–CH2–CH2–OH. Step 1: the longest chain containing the –OH carbon has 4 carbons, so the word root is 'but-'. Step 2: all bonds are single, and –OH is the principal group, so the suffix is '-an-1-ol' once numbered. Step 3: number from the end nearer –OH so that it gets the lowest locant: C1(CH2OH)–C2(CH2)–C3(CH with a methyl branch)–C4(CH3). Step 4: the methyl substituent is at C3. Final IUPAC name: 3-methylbutan-1-ol. As a second example, CH3–CH2–CO–CH3 is a 4-carbon chain with a ketone group; numbering from the end nearer the C=O gives it locant 2, so the name is butan-2-one.

Word roots meth-(1C), eth-(2C), prop-(3C), but-(4C), pent-(5C), hex-(6C), hept-(7C), oct-(8C), non-(9C), dec-(10C)
Seniority order of functional groups –COOH > –SO₃H > –COOR > –COX > –CONH₂ > –C≡N > –CHO > >C=O > –OH > –NH₂ > –O– (ether)
Example IUPAC names CH₃CH(CH₃)CH₂CH₂OH = 3-methylbutan-1-ol; CH₃CH₂COCH₃ = butan-2-one
Remember
  • IUPAC name = word root (chain length) + primary suffix (saturation: -ane/-ene/-yne) + secondary suffix (principal functional group).
  • Only the senior-most functional group is shown as a suffix; others become substituent prefixes (e.g. hydroxy-, oxo-).
  • Substituent prefixes are listed alphabetically; di-/tri-/tetra- are multiplying prefixes and are not counted in alphabetising.
  • Numbering must give the lowest locant to the principal characteristic group first, then the lowest locant set to substituents.
  • Common seniority order (high to low): carboxylic acid > sulphonic acid > ester > acid halide > amide > nitrile > aldehyde > ketone > alcohol > amine > ether.

Isomerism: Structural and Stereoisomerism

Quick answer Structural isomers share a molecular formula but differ in how atoms are connected (chain, position, functional, metamerism, ring-chain); stereoisomers share the same connectivity but differ in spatial arrangement (geometrical or optical).

Compounds with the same molecular formula but different structures/arrangements are called isomers. Isomerism is divided into two broad types.

  • Structural isomerism (different connectivity of atoms):
    • Chain isomerism — differ in the arrangement of the carbon skeleton, e.g. n-butane and 2-methylpropane (isobutane), both C4H10.
    • Position isomerism — same carbon skeleton and same functional group but the group/substituent is at a different position, e.g. propan-1-ol and propan-2-ol, both C3H8O.
    • Functional isomerism — same molecular formula but different functional groups, e.g. ethanol (CH3CH2OH) and methoxymethane/dimethyl ether (CH3OCH3), both C2H6O.
    • Metamerism — same functional group but different alkyl groups distributed on either side of that group, e.g. diethyl ether (C2H5–O–C2H5) and methyl propyl ether (CH3–O–C3H7), both C4H10O.
    • Ring-chain isomerism — a cyclic compound and an open-chain compound share the same molecular formula, e.g. cyclopropane and propene, both C3H6.
  • Stereoisomerism (same connectivity, different spatial arrangement):
    • Geometrical (cis–trans) isomerism — arises from restricted rotation about a C=C double bond; e.g. but-2-ene exists as cis (identical groups on the same side) and trans (identical groups on opposite sides) forms.
    • Optical isomerism (basic idea) — arises when a carbon is attached to four different groups (a chiral centre); the molecule and its mirror image are non-superimposable, giving a pair of optical isomers called enantiomers.

Worked example. Classify the relationship in each pair: (i) CH3CH2CH2OH and CH3CH(OH)CH3 — same formula C3H8O, same –OH group, but the –OH is on C1 in one and C2 in the other → position isomers. (ii) CH3CH2CH2OH and CH3–O–CH2CH3 (methoxyethane) — both C3H8O, but one has –OH and the other has –O– → functional isomers. (iii) For but-2-ene, CH3–CH=CH–CH3, since rotation about the C=C bond is restricted, the two methyl groups can lie on the same side (cis-but-2-ene) or on opposite sides (trans-but-2-ene) of the double bond, giving a pair of geometrical isomers.

Chain isomer pair C₄H₁₀: CH₃CH₂CH₂CH₃ (n-butane) / (CH₃)₃CH (isobutane)
Functional isomer pair C₂H₆O: CH₃CH₂OH (ethanol) / CH₃OCH₃ (dimethyl ether)
Ring-chain isomer pair C₃H₆: cyclopropane / CH₂=CH–CH₃ (propene)
Remember
  • Isomers have the same molecular formula but different structures or spatial arrangements.
  • Chain, position, functional, metamerism and ring-chain isomerism are common types of structural isomerism.
  • Functional isomers have different functional groups altogether (e.g. an alcohol vs an ether of the same formula).
  • Geometrical (cis-trans) isomerism needs restricted rotation, typically about a C=C bond, with two different groups on each doubly-bonded carbon.
  • Optical isomerism (basic idea) arises from a chiral carbon attached to four different groups, giving non-superimposable mirror-image forms.

Electronic Displacement Effects I: Inductive and Resonance (Mesomeric) Effects

Quick answer The inductive effect is a permanent, distance-decaying shift of sigma-bond electrons caused by electronegativity differences; the resonance (mesomeric) effect is a permanent delocalisation of pi/lone-pair electrons over a conjugated system.

Inductive effect (I effect). When an electronegative atom or group is attached to a carbon chain, it pulls the shared sigma-bond electrons slightly towards itself, and this small polarity is relayed, with rapidly decreasing strength, along the rest of the chain. This is a permanent effect present in the ground state of the molecule.

  • Groups that withdraw electron density (relative to H) show a −I effect: examples in decreasing strength include –NO2, –CN, –COOH, halogens (–F > –Cl > –Br > –I), –OH.
  • Groups that release electron density (relative to H) show a +I effect: alkyl groups, with the order tertiary > secondary > primary > methyl in electron-donating ability.
  • The inductive effect explains trends in acid/base strength: electron-withdrawing (−I) groups near –COOH stabilise the conjugate base (carboxylate ion) and increase acid strength; electron-releasing (+I) groups do the opposite.

Worked example (inductive effect). Arrange CH3COOH, ClCH2COOH, Cl2CHCOOH and Cl3CCOOH in decreasing order of acid strength. Each chlorine attached near the –COOH group withdraws electron density through the −I effect, which stabilises the carboxylate anion formed after loss of H+ and makes the acid stronger. Since Cl3CCOOH has three −I chlorines (highest withdrawal), followed by two in Cl2CHCOOH, one in ClCH2COOH, and none in CH3COOH: Cl3CCOOH > Cl2CHCOOH > ClCH2COOH > CH3COOH.

Resonance (mesomeric) effect (R/M effect). When a molecule can be represented by two or more valid Lewis structures (differing only in the position of pi-bond or lone-pair electrons, never in the position of the nuclei), the actual molecule is a resonance hybrid of all these contributing structures and is more stable than any single structure. Groups that donate electrons into the pi system by resonance show a +R (+M) effect (e.g. –OH, –NH2, –OR, halogens attached to an unsaturated system); groups that withdraw electrons from the pi system show a −R (−M) effect (e.g. –NO2, –CHO, >C=O, –COOH, –CN).

Worked example (resonance). The carboxylate ion, CH3COO, can be drawn with the negative charge on either oxygen, giving two equivalent resonance structures. The true structure is a hybrid in which the negative charge and the pi bond are equally delocalised over both C–O bonds (each C–O bond has equal, intermediate bond length/order). This delocalisation stabilises the carboxylate ion far more than an alkoxide ion (RO, which has no such resonance), which is why carboxylic acids are much stronger acids than alcohols.

Acid strength order (inductive effect) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH
+I order of alkyl groups tertiary-alkyl > secondary-alkyl > primary-alkyl > –CH₃ > –H
Resonance structures of acetate ion CH₃–C(=O)–O⁻ ↔ CH₃–C(–O⁻)=O Two equivalent contributing structures forming one resonance hybrid
Remember
  • Inductive effect is a permanent sigma-electron shift that weakens rapidly with distance along the chain; −I groups withdraw, +I (alkyl) groups donate electron density.
  • Resonance structures differ only in electron position, not nuclear position; the real molecule is a more stable resonance hybrid of all valid contributing structures.
  • +R groups (e.g. –OH, –NH₂, halogens on a pi system) donate electron density by resonance; −R groups (e.g. –NO₂, –CHO, –COOH) withdraw electron density by resonance.
  • More −I substituents near –COOH increase acid strength by stabilising the conjugate base (e.g. Cl₃CCOOH is a much stronger acid than CH₃COOH).
  • Resonance delocalisation in the carboxylate ion (equal charge on both oxygens) is why carboxylic acids are far more acidic than alcohols.

Electronic Displacement Effects II: Electromeric Effect, Hyperconjugation, and Fission of Covalent Bonds

Quick answer The electromeric effect is a temporary complete pi-electron shift triggered only by an attacking reagent; hyperconjugation is delocalisation of adjacent C–H sigma electrons that stabilises carbocations, radicals and alkenes; covalent bonds break either homolytically (free radicals) or heterolytically (ions).

Electromeric effect (E effect). This is a temporary effect seen only when a multiple bond (C=C, C=O, etc.) is attacked by a reagent: the pi-bond electron pair is completely transferred to one of the two atoms at the demand of the attacking reagent, and the effect disappears the moment the reagent is removed. It is denoted +E when electrons move towards the atom to which the new bond forms and −E when they move away from it. Unlike the permanent inductive and resonance effects, the electromeric effect operates only during a reaction, at the very moment a reagent approaches.

Worked example (electromeric effect). During the addition of HBr to propene, CH3–CH=CH2, the approaching H+ (electrophile) triggers a complete shift of the pi-electron pair of C=C towards the carbon that will bond to H, generating the more stable secondary carbocation CH3+CH–CH3, which is then attacked by Br to complete Markovnikov addition. This electron shift happens only because HBr is attacking; it is not present before the reagent approaches.

Hyperconjugation (no-bond resonance). A C–H (or C–C) sigma bond on a carbon directly attached to a carbocation, a free radical, or a double bond can align with the empty/half-filled/pi orbital and delocalise its electron density into it, lending extra stability. The more such adjacent (alpha) C–H bonds available, the greater the stabilisation. This explains the observed stability order of carbocations and free radicals.

Worked example (hyperconjugation). Compare the stability of a primary carbocation CH3+CH2 (ethyl cation) with a tertiary carbocation (CH3)3C+. The ethyl cation has 3 alpha C–H bonds available for hyperconjugation (from the adjacent CH3 group), giving 3 hyperconjugative (no-bond resonance) structures, whereas the tertiary cation has 9 alpha C–H bonds (three CH3 groups), giving 9 such structures. More hyperconjugative structures mean greater delocalisation of the positive charge and greater stability, so the tertiary carbocation is far more stable: overall order 3° > 2° > 1° > methyl cation.

Fission of a covalent bond. A covalent bond can break in two ways.

  • Homolytic fission (homolysis): the bonding electron pair splits equally, one electron going to each atom, producing two neutral species with an unpaired electron called free radicals. It is favoured by non-polar conditions, heat, ultraviolet light, or the presence of peroxides, e.g. Cl–Cl breaking under UV light into two chlorine free radicals, Cl.
  • Heterolytic fission (heterolysis): the bonding electron pair goes entirely to the more electronegative atom, producing a cation and an anion. It is favoured in polar solvents, e.g. the C–Cl bond of chloromethane, CH3–Cl, breaking to give a methyl carbocation CH3+ and a chloride ion Cl.

Worked example (fission). Show homolytic and heterolytic fission of the C–Cl bond in CH3Cl. Homolysis (typically induced by heat/UV, symmetrical bond-breaking shown with single-barbed/'fish-hook' arrows): CH3–Cl → CH3 (methyl free radical) + Cl (chlorine free radical). Heterolysis (typically favoured in a polar medium, both electrons move to the more electronegative Cl): CH3–Cl → CH3+ (methyl carbocation, electron-deficient, sp2, planar) + Cl (chloride ion, complete octet).

Homolytic fission (general) A–B → A• + B• Free radicals; one electron to each fragment
Heterolytic fission (general) A–B → A⁺ + B⁻ Both electrons go to the more electronegative atom B
Fission of chloromethane CH₃–Cl →(homolysis) CH₃• + Cl• ; CH₃–Cl →(heterolysis) CH₃⁺ + Cl⁻
Carbocation stability order 3° > 2° > 1° > CH₃⁺
Remember
  • The electromeric effect is a temporary, complete pi-electron shift that occurs only in the presence of an attacking reagent, unlike the permanent inductive and resonance effects.
  • +E shifts electrons towards the atom forming the new bond; −E shifts them away from it.
  • Hyperconjugation delocalises adjacent C–H sigma electrons into an empty/pi orbital; more alpha C–H bonds means greater stabilisation.
  • Carbocation/free-radical stability order from hyperconjugation and inductive donation together: tertiary > secondary > primary > methyl.
  • Homolytic fission gives two neutral free radicals (favoured by heat/UV/non-polar conditions); heterolytic fission gives a cation and an anion (favoured in polar media).

Key facts & terms

Every formula in this chapter, in one place — screenshot it before your exam.

CₙH₂ₙ₊₂
General formula of alkanes
CₙH₂ₙ
General formula of alkenes
CₙH₂ₙ₋₂
General formula of alkynes
–CH₂– = 14 u
Homologous increment
C = 4, H = 1, O = 2, N = 3, halogen(X) = 1
Valency used for H-count in skeletal formulas
H(implicit) = 4 − (number of bonds shown at that carbon)
Implicit hydrogens on a skeletal carbon
meth-(1C), eth-(2C), prop-(3C), but-(4C), pent-(5C), hex-(6C), hept-(7C), oct-(8C), non-(9C), dec-(10C)
Word roots
–COOH > –SO₃H > –COOR > –COX > –CONH₂ > –C≡N > –CHO > >C=O > –OH > –NH₂ > –O– (ether)
Seniority order of functional groups
CH₃CH(CH₃)CH₂CH₂OH = 3-methylbutan-1-ol; CH₃CH₂COCH₃ = butan-2-one
Example IUPAC names
C₄H₁₀: CH₃CH₂CH₂CH₃ (n-butane) / (CH₃)₃CH (isobutane)
Chain isomer pair
C₂H₆O: CH₃CH₂OH (ethanol) / CH₃OCH₃ (dimethyl ether)
Functional isomer pair
C₃H₆: cyclopropane / CH₂=CH–CH₃ (propene)
Ring-chain isomer pair
Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH > CH₃COOH
Acid strength order (inductive effect)
tertiary-alkyl > secondary-alkyl > primary-alkyl > –CH₃ > –H
+I order of alkyl groups
CH₃–C(=O)–O⁻ ↔ CH₃–C(–O⁻)=O
Resonance structures of acetate ion
A–B → A• + B•
Homolytic fission (general)
A–B → A⁺ + B⁻
Heterolytic fission (general)
CH₃–Cl →(homolysis) CH₃• + Cl• ; CH₃–Cl →(heterolysis) CH₃⁺ + Cl⁻
Fission of chloromethane
3° > 2° > 1° > CH₃⁺
Carbocation stability order

Test yourself

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0 correct · 0/12 answered
Q1 Classification of organic compounds easy

Which of the following is a heterocyclic compound?

Q2 Homologous series easy

What is the general formula of alkynes?

Q3 Structural representation easy

In a bond-line (skeletal) structural formula, what does each line end or vertex represent?

Q4 IUPAC nomenclature medium

What is the correct IUPAC name of CH3-CH2-CO-CH3?

Q5 Structural isomerism medium

Which pair of compounds represents chain isomers?

Q6 Structural isomerism medium

The molecular formula C2H6O corresponds to which situation?

Q7 Stereoisomerism medium

Geometrical (cis-trans) isomerism arises mainly due to which of the following?

Q8 Inductive effect medium

What is the correct decreasing order of acid strength for CH3COOH, ClCH2COOH, Cl2CHCOOH and Cl3CCOOH, based on the inductive effect?

Q9 Resonance effect medium

Which statement about resonance structures is correct?

Q10 Electromeric effect medium

The electromeric effect is best described as:

Q11 Hyperconjugation hard

Hyperconjugation explains the greater stability of which of the following?

Q12 Fission of covalent bonds easy

Homolytic fission of a covalent bond produces which of the following species?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Write the IUPAC name of CH3-CH(CH3)-CH2-CH2-OH.IUPAC nomenclature

Step 1: Identify the longest chain containing the principal group. The longest continuous chain that includes the carbon bearing –OH has 4 carbons, so the word root is 'but-'.

Step 2: Identify saturation and principal group. All carbon-carbon bonds are single bonds (primary suffix '-ane'), and the principal characteristic group is –OH, giving the secondary suffix '-ol'.

Step 3: Number the chain to give the lowest locant to –OH: numbering from the end nearer –OH, C1 bears –OH, C2 is CH2, C3 bears a methyl substituent, C4 is CH3.

Step 4: Assemble the name. The methyl substituent is at C3, and –OH (as '-ol') is at C1.

IUPAC name: 3-methylbutan-1-ol.

2 How many structural isomers are possible for the molecular formula C4H10O? Draw them and classify each as an alcohol or an ether.Structural isomerism

Step 1: Split by functional group. C4H10O can exist either as an alcohol (–OH) or as an ether (–O–).

Step 2: List all alcohols (C4H9OH) — 4 isomers.

  • CH3-CH2-CH2-CH2-OH (butan-1-ol)
  • CH3-CH2-CH(OH)-CH3 (butan-2-ol)
  • (CH3)2CH-CH2-OH (2-methylpropan-1-ol)
  • (CH3)3C-OH (2-methylpropan-2-ol)

Step 3: List all ethers (R-O-R') — 3 isomers.

  • CH3-CH2-O-CH2-CH3 (diethyl ether / ethoxyethane)
  • CH3-O-CH2-CH2-CH3 (1-methoxypropane)
  • CH3-O-CH(CH3)2 (2-methoxypropane)

Step 4: Total count. 4 alcohol isomers + 3 ether isomers = 7 structural isomers in all. The alcohol/ether pairs (e.g. butan-1-ol vs diethyl ether) are functional isomers of one another, while the isomers within the alcohol set are chain or position isomers of each other, and the two propyl-methyl ethers are metamers.

3 Draw the resonance structures of the acetate ion (CH3COO⁻) and use them to explain why acetic acid is a stronger acid than ethanol.Resonance effect

Step 1: Draw the two contributing structures of the carboxylate ion formed when acetic acid loses a proton: CH3-C(=O)-O⁻ ↔ CH3-C(-O⁻)=O. These differ only in which oxygen carries the negative charge and which one carries the double bond; the carbon and oxygen nuclei do not move.

Step 2: Form the resonance hybrid. The actual acetate ion is a hybrid of these two equivalent structures, so the negative charge and the pi-bond character are equally shared (delocalised) over both oxygen atoms; both C-O bonds become identical in length, intermediate between a single and a double bond.

Step 3: Compare with the ethoxide ion. When ethanol loses a proton, the resulting ethoxide ion, C2H5-O⁻, has no analogous resonance structure — the negative charge stays localised entirely on one oxygen.

Step 4: Conclusion. Because the acetate ion is resonance-stabilised (delocalised charge) while the ethoxide ion is not, the acetate ion is much more stable than the ethoxide ion. A more stable conjugate base corresponds to a stronger acid, so acetic acid is a far stronger acid than ethanol.

4 Explain, with a suitable example, the difference between the inductive effect and the electromeric effect.Inductive vs electromeric effect

Step 1: State the key difference. The inductive effect is a permanent, weak electron shift through sigma bonds present in the molecule at all times (in its ground state), and it decreases rapidly with distance. The electromeric effect is a temporary, complete shift of pi electrons that appears only at the instant an attacking reagent approaches a multiple bond, and vanishes once the reagent is removed.

Step 2: Illustrate the inductive effect. In CH3-CH2-Cl, the electronegative chlorine permanently pulls sigma-bond electron density towards itself, making the adjacent carbon atoms slightly electron deficient even when no reagent is reacting with the molecule.

Step 3: Illustrate the electromeric effect. In the addition of HBr to propene, CH3-CH=CH2, the pi electrons of the C=C bond are still shared equally between the two carbons until H⁺ approaches; only at that moment do the pi electrons shift completely onto one carbon (the one not bonding to H⁺), generating the more stable secondary carbocation CH3-CH⁺-CH3. This electron shift did not exist before HBr approached and would not exist without it.

Step 4: Conclusion. The inductive effect is permanent and operates through sigma bonds over the whole molecule; the electromeric effect is temporary, complete, and operates only at a multiple bond under the demand of an attacking reagent.

5 Classify the following pairs of compounds as chain, position, or functional isomers: (i) neopentane and n-pentane, (ii) butan-1-ol and butan-2-ol, (iii) diethyl ether and butan-1-ol.Structural isomerism

(i) Neopentane, (CH3)4C, and n-pentane, CH3CH2CH2CH2CH3: both have the molecular formula C5H12 and no functional group other than C-H/C-C bonds; they differ only in the branching of the carbon skeleton. This is chain isomerism.

(ii) Butan-1-ol, CH3CH2CH2CH2OH, and butan-2-ol, CH3CH2CH(OH)CH3: both have the molecular formula C4H10O and the same functional group (–OH), but the –OH is attached at C1 in one and at C2 in the other. This is position isomerism.

(iii) Diethyl ether, CH3CH2-O-CH2CH3, and butan-1-ol, CH3CH2CH2CH2OH: both have the molecular formula C4H10O, but one is an ether (–O–) and the other is an alcohol (–OH) — different functional groups altogether. This is functional isomerism.

6 Show the homolytic and heterolytic fission of the C-Cl bond in chloromethane (CH3Cl). Name the species produced in each case and state the conditions that favour each type of fission.Fission of covalent bonds

Homolytic fission: CH3-Cl → CH3• + Cl•. The bonding electron pair splits equally, one electron staying with the carbon and one going to chlorine, producing a neutral methyl free radical (CH3•) and a neutral chlorine free radical (Cl•), each with one unpaired electron. This type of fission is favoured by high temperature, ultraviolet light, or non-polar (gas-phase/solvent-free) conditions.

Heterolytic fission: CH3-Cl → CH3⁺ + Cl⁻. Both bonding electrons move entirely to the more electronegative chlorine atom, producing an electron-deficient methyl carbocation (CH3⁺), which is sp2 hybridised and planar, and a chloride ion (Cl⁻), which has a complete octet. This type of fission is favoured by polar solvents, which can stabilise the separated positive and negative species by solvation.

Conclusion: homolysis gives two neutral radicals and is typically induced thermally/photochemically, while heterolysis gives a cation-anion pair and is typically favoured in a polar medium.

Previous-year board questions 4

Q1 Define the inductive effect. CBSE 2019 1 mark

The inductive effect is the permanent, partial displacement of sigma-bond electrons along a carbon chain towards a more electronegative atom or group, caused by the electronegativity difference between bonded atoms; its magnitude decreases rapidly as the distance from the electronegative atom/group increases.

Q2 Write the IUPAC names of the following compounds: (i) CH3-CH2-CHO (ii) (CH3)2CH-COOH CBSE 2020 2 marks

(i) CH3-CH2-CHO: the longest chain has 3 carbons with an aldehyde group at the terminal carbon, which by convention is C1. The name is propanal.

(ii) (CH3)2CH-COOH: the longest chain including the carboxylic acid carbon has 3 carbons (–COOH carbon is C1), with a methyl substituent at C2. The name is 2-methylpropanoic acid.

Q3 Explain the resonance effect with a suitable example. Also explain why carboxylic acids are more acidic than alcohols in terms of this effect. CBSE 2022 3 marks

Resonance effect: when a molecule or ion can be represented by two or more valid Lewis structures that differ only in the position of pi-bond or lone-pair electrons (not in the position of the nuclei), the true structure is a single, more stable resonance hybrid of all such contributing structures. For example, the acetate ion, CH3COO⁻, is represented by two equivalent structures, CH3-C(=O)-O⁻ ↔ CH3-C(-O⁻)=O, which delocalise the negative charge equally over both oxygen atoms.

Why carboxylic acids are more acidic than alcohols: when a carboxylic acid loses a proton, the resulting carboxylate ion is resonance-stabilised, with its negative charge spread over two oxygen atoms, making it markedly more stable. In contrast, when an alcohol loses a proton, the resulting alkoxide ion (RO⁻) has no such resonance and keeps its negative charge localised entirely on one oxygen. Since a more stable conjugate base means a stronger acid, carboxylic acids are considerably more acidic than alcohols.

Q4 (a) What is hyperconjugation? Explain, with an example, how it accounts for the relative stability of carbocations. (b) Differentiate between homolytic and heterolytic fission of a covalent bond, giving one example of each. (c) Draw the structures of all the possible isomers of molecular formula C4H10O and identify the type of isomerism shown by any one pair. CBSE 2023 5 marks

(a) Hyperconjugation: hyperconjugation (no-bond resonance) is the delocalisation of electrons of a C-H (or C-C) sigma bond on a carbon directly attached to a carbocation, a free radical, or a carbon of a double bond, into the adjacent empty or pi orbital. Consider the tertiary carbocation (CH3)3C⁺: it has 9 adjacent (alpha) C-H bonds, each of which can hyperconjugate with the empty p orbital on the positive carbon, giving 9 hyperconjugative structures and strong charge delocalisation. A primary carbocation such as CH3-CH2⁺ has only 3 such alpha C-H bonds, so it is delocalised, and hence stabilised, far less. This is why the stability order is tertiary > secondary > primary > methyl carbocation.

(b) Homolytic vs heterolytic fission: in homolytic fission, the bonding electron pair of a covalent bond splits equally, one electron going to each atom, producing two neutral free radicals, e.g. Cl-Cl → Cl• + Cl• under ultraviolet light. In heterolytic fission, both bonding electrons go entirely to the more electronegative atom, producing a cation and an anion, e.g. CH3-Cl → CH3⁺ + Cl⁻ in a polar medium. Homolysis is favoured by heat/UV light/non-polar conditions, whereas heterolysis is favoured by polar solvents that can stabilise the ions formed.

(c) Isomers of C4H10O: there are 7 structural isomers in total. The 4 alcohols are: CH3CH2CH2CH2OH (butan-1-ol), CH3CH2CH(OH)CH3 (butan-2-ol), (CH3)2CHCH2OH (2-methylpropan-1-ol), and (CH3)3COH (2-methylpropan-2-ol). The 3 ethers are: CH3CH2-O-CH2CH3 (diethyl ether), CH3-O-CH2CH2CH3 (1-methoxypropane), and CH3-O-CH(CH3)2 (2-methoxypropane). Taking the pair butan-1-ol and diethyl ether: both share the formula C4H10O but have different functional groups (–OH vs –O–), so this pair shows functional isomerism.

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