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Chemistry — Class 12 — L23: Nomenclature and General Principles

NIOS Code 313 · Module 7 · Chemistry of Organic Compounds

Notes extracted from NIOS Chemistry Course (313), Lesson 23 — Nomenclature and General Principles (313_Chemistry_Eng_Lesson23.pdf). Content covers sections 23.1–23.6.
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Overview — Module 7: Organic Foundations

Organic chemistry is the chemistry of carbon compounds (excluding simple oxides, carbonates, cyanides and carbides). Carbon’s unique catenation builds chains, rings and networks — fuels, foods, polymers, drugs and dyes. This lesson covers classification of hydrocarbons, IUPAC naming, bond fission and electronic effects, reaction types, isomerism (including R/S and D/L), and qualitative/quantitative organic analysis.

Section 1: Classification of Hydrocarbons (23.1)

Open-chain (aliphatic): saturated alkanes (C–C single) and unsaturated alkenes (C=C) / alkynes (C≡C).

Closed-chain (cyclic): homocyclic (only C in ring) — alicyclic (cycloalkanes etc.) or aromatic (benzene and derivatives); heterocyclic (N, O, S in ring).

Organic Compound Classification Open chainalkanes · alkenes · alkynes Homocyclicalicyclic · aromatic HeterocyclicN/O/S in ring Catenation of carbon → millions of organic compounds
Broad map of organic structural types before functional-group chemistry.

Section 2: IUPAC Nomenclature (23.2)

Word root gives carbon count (meth–dec); suffix gives saturation: -ane, -ene, -yne. Branched compounds use alkyl prefixes (methyl, ethyl, isopropyl, tert-butyl…).

Key rules: (1) longest continuous chain including multiple bond; if tie, more substituents; (2) lowest numbers for unsaturation then substituents; (3) alphabetical prefixes (ignore di/tri); (4) multiple identical groups di/tri/tetra. Cyclic: cyclo- + hydrocarbon name; number substituents for lowest set.

Word root + -ane/-ene/-yne  |  Longest chain · lowest locants
e.g. CH₃CH=CH₂ = propene · CH₃–CH(CH₃)–CH₃ = 2-methylpropane
IUPAC Naming Steps 1. Longest 2. Number 3. Prefixes 4. Suffix Include C=C/C≡C in parent · lowest sum of locants · alphabetise substituents Build structure from name: reverse the same rules
Systematic naming and structure drawing are inverse skills for exams.

Functional groups (halo, hydroxy, etc.) get prefixes or priority suffixes in later lessons; L23 focuses on hydrocarbon parents and general organic principles.

Section 3: Bond Fission, Reagents & Electronic Effects (23.3.1–23.3.3)

Homolytic fission: equal electron share → free radicals (·CH₃). Initiated by heat/light. Heterolytic fission: unequal share → carbocation (C⁺) + carbanion (C⁻).

Electrophiles (E⁺): electron-deficient — H⁺, NO₂⁺, Br⁺, BF₃. Attack high electron density. Nucleophiles (Nu): electron-rich — OH⁻, CN⁻, NH₃, H₂O. Attack δ⁺ carbon.

Homo → radicals  |  Hetero → C⁺ / C⁻  |  E⁺ seeks e⁻ · Nu donates e⁻
Mechanism language for all organic reactions

Inductive effect: permanent polarisation along σ chain. −I (withdrawing): NO₂, CN, halogens… +I (releasing): alkyl groups (tert-butyl strongest). Explains acid strength trends.

Electromeric: temporary complete transfer of π pair under reagent attack (C=O, C=C).

Resonance: molecule as hybrid of canonical forms (benzene equal C–C bonds 139 pm; carboxylate; nitro).

Hyperconjugation: σ–π conjugation (“no-bond resonance”), e.g. propene methyl hydrogens with C=C — stabilises alkenes/carbocations.

Steric hindrance: bulky groups block reagent approach (Hofmann/Meyer).

Electronic Effects Snapshot −INO₂, CN, X +Ialkyl groups Resonanceπ delocalisation Hyperconj.σ–π no-bond I permanent · E temporary · steric = bulk blocking
Electron displacement effects control reactivity and acid–base strength.

Section 4: Types of Organic Reactions (23.3.5–23.3.8)

Substitution: replace atom/group. Aliphatic: nucleophilic (R–X + OH⁻ → R–OH). Aromatic: electrophilic (benzene + HNO₃/H₂SO₄ → nitrobenzene).

Addition: to C=C/C≡C (π bond opens). Br₂ decolourises; H₂/Ni, HX add. Alkynes take two moles of H₂ to alkanes.

Elimination: remove small molecule from adjacent carbons → double bond (ethanol + H₂SO₄, 403 K → ethene + H₂O).

Rearrangement: skeleton change (1-chlorobutane + AlCl₃ → 2-chlorobutane).

R–X + Nu⁻ → R–Nu  |  C=C + Br₂ → dibromide  |  alcohol −H₂O → alkene
Substitution · addition · elimination — core organic reaction map

Section 5: Isomerism (23.4)

Structural: chain (n-butane / isobutane); position (propan-1-ol / propan-2-ol); functional (ethanol / methoxymethane); metamerism (1-methoxypropane / ethoxyethane).

Stereo: geometrical cis–trans (restricted rotation about C=C); optical when chiral carbon (four different groups) → enantiomers, d(+) / l(−), racemic (±) inactive.

Absolute configuration: D/L from Fischer projection (OH right = D for glyceraldehyde reference). R/S by Cahn–Ingold–Prelog priorities: lowest priority away; 1→2→3 clockwise = R, anticlockwise = S. Optical rotation (+/−) is independent of R/S or D/L labels.

Isomerism Overview Structuralchain · position · functionmetamerism Stereocis–trans · opticalR/S · D/L Same formula · different connectivity or 3D arrangement
Structural vs stereoisomerism — both appear constantly in organic exams.
Chiral C → enantiomers  |  CIP: clockwise = R · anticlockwise = S
d/l = observed rotation · R/S and D/L = absolute arrangement

Section 6: Qualitative Analysis (23.5)

C and H: heat with CuO → CO₂ (lime water milky) and H₂O (anhydrous CuSO₄ → blue).

Lassaigne’s test: fuse with Na → NaCN, Na₂S, NaX extracted in water. N: FeSO₄ + conc. H₂SO₄ → Prussian blue. N+S: blood-red Fe(CNS)₃. S: sodium nitroprusside violet or PbS black. Halogens: AgNO₃ after HNO₃ — white AgCl, pale yellow AgBr, yellow AgI (solubility in NH₄OH differs). Not given by NH₂NH₂, NH₂OH, or diazonium salts (no stable CN⁻).

Lassaigne Extract Tests NPrussian blue N+Sblood red Sviolet / PbS XAgX ppt Na fusion converts covalent N/S/X → ionic form in extract
Classic qualitative scheme for N, S and halogens in organic compounds.

Section 7: Quantitative Analysis (23.6)

Liebig (C, H): burn with CuO; absorb CO₂ in KOH, H₂O in CaCl₂.

%C = (12/44)×(m_CO₂/m)×100  |  %H = (2/18)×(m_H₂O/m)×100
Liebig combustion · m = mass of compound

Carius (halogens, S): fuming HNO₃ + AgNO₃ → AgX; S → BaSO₄. %Cl = 35.5/143.5 × (m_AgCl/m)×100; %S = 32/233 × (m_BaSO₄/m)×100.

Phosphorus: → H₃PO₄ → ammonium phosphomolybdate or Mg₂P₂O₇.

Nitrogen: Dumas (CuO, collect N₂ over KOH) or Kjeldahl (digest to (NH₄)₂SO₄, liberate NH₃, titrate).

%N (Dumas) ∝ V(N₂ at STP)  |  22400 mL N₂ = 28 g
Correct volume to STP · KOH absorbs CO₂

Exam Connections and Chapter Summary

This chapter is the grammar of organic chemistry. Exam favourites: IUPAC naming and structure writing; homo/hetero fission; E⁺/Nu lists; −I/+I order; reaction type identification; isomer classification; R/S or cis–trans assignment; Lassaigne observations; Liebig/Carius percentage calculations.

Leads into L24 hydrocarbons and later functional-group lessons. Master naming + mechanism vocabulary + analysis formulas and every later organic chapter becomes easier.

Intext-style drills: name branched alkenes; classify E⁺/Nu; identify −I groups; product of CH₂=CH₂ + HBr; type of isomerism for C₂H₆O pairs; cis–trans for CHF=CHF; R/S sketch; Lassaigne colour codes; compute %C from CO₂ mass. Liebig numbers: %C = (12/44)×(m_CO₂/m)×100; %H = (2/18)×(m_H₂O/m)×100. Carius: %Cl = 35.5/143.5 × (m_AgCl/m)×100; %S = 32/233 × (m_BaSO₄/m)×100.

MCQ Quiz — L23 Nomenclature and General Principles

0 / 10 correct

Flashcards — L23

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Golden Rules — L23 Nomenclature and General Principles

Most exam-important points from this chapter:

Classification & naming

Aliphatic / cyclic / aromatic / hetero. IUPAC: longest chain, lowest locants, prefixes, -ane/-ene/-yne.

Fission & reagents

Homo → radicals; hetero → ions. E⁺ attacks e⁻-rich sites; Nu attacks δ⁺ carbon.

Electronic effects

−I/+I permanent; electromeric temporary; resonance & hyperconjugation delocalise; steric blocks approach.

Reactions & isomers

Sub / add / elim / rearrange. Structural vs stereo; cis–trans; optical; R/S and D/L ≠ rotation sign.

Analysis

CuO detects C/H. Lassaigne for N/S/X. Liebig %C/%H; Carius X/S; Dumas/Kjeldahl N.

alk-ane / -ene / -yne
Longest chain · lowest number
Homo / hetero · free radical / ion
E⁺ / Nu⁻
−I / +I · resonance
Sub · add · elim · rearrange
Chain · pos · func · stereo
R/S · D/L · cis/trans
%C, %H Liebig · Carius · Kjeldahl

Section 1: Classification & IUPAC

NIOS Chemistry 313, Module 7 — Nomenclature and General Principles (sections 23.1–23.6).

Hydrocarbon Classes

Open-chain (aliphatic): alkanes, alkenes, alkynes

Cyclic: alicyclic · aromatic · heterocyclic (heteroatom in ring)

Catenation → vast carbon chemistry · organic = C compounds (not CO₂, carbonates, cyanides)

IUPAC Hydrocarbon Naming

Word root (meth–dec) + suffix (-ane/-ene/-yne)

Rules: longest chain (include multiple bond) · lowest numbers for substituents/unsaturation · alphabetical prefixes · di/tri/tetra · bis for en-type names

Alkyl R = CnH2n+1 · methyl, ethyl, isopropyl, tert-butyl

Section 2: Mechanisms & Effects

Bond Fission & Reagents

Homolytic: equal share → free radicals (·CH₃)

Heterolytic: unequal → carbocation (C⁺) / carbanion (C⁻)

Electrophile E⁺: electron-seeking (H⁺, NO₂⁺, Br⁺, BF₃)

Nucleophile Nu: electron-rich (OH⁻, CN⁻, NH₃, H₂O)

Electronic Effects

Inductive (−I/+I): permanent σ-bond polarisation · −I: NO₂>CN>F>Cl… · +I: tert-butyl > isopropyl > ethyl > methyl

Electromeric: temporary complete e⁻ transfer in C=C, C=O under reagent attack

Resonance: hybrid of canonical forms · benzene equal C–C 139 pm

Hyperconjugation: no-bond resonance (σ–π conjugation, e.g. propene)

Steric hindrance: bulky groups block approach of reagent

Reaction Types

Substitution: Nu on R–X · E⁺ on benzene (nitration HNO₃/H₂SO₄)

Addition: to C=C / C≡C · Br₂ decolourises · HX, H₂/Ni

Elimination: remove small molecule → double bond (alcohol + H₂SO₄ → alkene)

Rearrangement: skeleton change (1-chlorobutane → 2-chlorobutane with AlCl₃)

Section 3: Isomerism & Analysis

Isomerism Map

Structural: chain · position · functional · metamerism

Stereo: geometrical (cis–trans; restricted rotation) · optical (chiral C → enantiomers d/l, ± racemic)

Absolute config: D/L (Fischer; OH right = D) · R/S (CIP priority; 1→2→3 clockwise = R)

Qualitative & Quantitative Analysis

C,H: CuO heat → CO₂ (lime water) + H₂O (CuSO₄ blue)

Lassaigne: Na fusion → NaCN, Na₂S, NaX · Prussian blue (N) · nitroprusside violet (S) · AgNO₃ (halogens)

Liebig: %C = (12/44)×(mass CO₂/mass sample)×100 · %H = (2/18)×(mass H₂O/mass sample)×100

Carius: halogen → AgX; S → BaSO₄ · Kjeldahl/Dumas: nitrogen estimation

Section 2: Definitions

Catenation: Ability of C to form long chains, rings, networks.

Free radical: Neutral species with unpaired electron from homolysis.

Chiral carbon: Carbon attached to four different groups; optical activity.

Enantiomers: Non-superimposable mirror images; opposite optical rotation.

Lassaigne's extract: Na fusion extract for detecting N, S, halogens.

Section 3: Visual Map

L23 Map — Organic Foundations IUPAC names E⁺ / Nu · effects Rxn types Isomers Homo/hetero fission · −I/+I · resonance · hyperconjugation R/S · D/L · cis/trans · Liebig · Carius · Kjeldahl

Section 5: Q&A (12 Questions)

Q1: What is catenation?

Ability of carbon to form long chains, rings and networks of C–C bonds.

Q2: Longest chain rule for naming?

Select longest continuous C chain including multiple bond if present; name as derivative of that alkane/ene/yne.

Q3: Homolytic vs heterolytic fission?

Homo: equal e⁻ → free radicals. Hetero: unequal → carbocation + carbanion.

Q4: Electrophile vs nucleophile?

E⁺ seeks e⁻ (NO₂⁺, H⁺). Nu donates e⁻ (OH⁻, CN⁻, NH₃).

Q5: Order of −I effect?

NR₃⁺ > NO₂ > CN > F > Cl > Br > I > OH > … > H

Q6: Example of addition reaction?

CH₂=CH₂ + Br₂ → CH₂BrCH₂Br (Br₂ colour disappears).

Q7: Chain isomers of C₄H₁₀?

n-Butane and 2-methylpropane (isobutane).

Q8: cis–trans requirement?

Restricted rotation (C=C or ring) with two different groups on each double-bonded carbon.

Q9: R vs S configuration?

CIP priorities 1→2→3 clockwise = R; anticlockwise = S (lowest priority away).

Q10: Lassaigne test for nitrogen?

Na fusion → NaCN · FeSO₄ + acid → Prussian blue.

Q11: Liebig %C formula?

%C = (12/44)×(mass CO₂/mass compound)×100

Q12: Carius method estimates?

Halogens as AgX; sulphur as BaSO₄ (with HNO₃/BaCl₂).

Section 6: Tips & Exam Hacks

Memory Aids

  • Suffixes: "ane single · ene double · yne triple"
  • E⁺/Nu: "Electro seeks · nucleo attacks"
  • −I: "NO₂ and CN pull hard"
  • R/S: "Clockwise R · anticlockwise S"
  • D/L: "OH right = D (Fischer)"

Exam Tips

  • Number chain for lowest locants of unsaturation then substituents
  • Alphabetise substituents ignoring di/tri
  • Br₂ test for unsaturation
  • d/l is rotation; R/S and D/L are absolute config
  • Always acidify Lassaigne extract before AgNO₃ for halogens

Section 8: Quick Reference

• Classification: aliphatic / alicyclic / aromatic / hetero

• IUPAC: longest chain · lowest number · prefixes · -ane/-ene/-yne

• Homo → radicals · hetero → ions · E⁺ / Nu

• −I/+I · electromeric · resonance · hyperconjugation · steric

• Sub / add / elim / rearrange

• Structural + stereo (geo + optical) · R/S · D/L

• Qualitative: CuO; Lassaigne · Quantitative: Liebig, Carius, Dumas, Kjeldahl

PYQ — Previous Year Questions

Extracted from NIOS Chemistry (313) board exam papers in your PDF. Chapter L23 — Nomenclature and General Principles only. Use Model Answer for marking points; Explanation for concept clarity.

L23 — Nomenclature and General Principles

8 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A

Section A — MCQ / Objective (from papers)

PYQ1. The pair of complexes which shows linkage isomerism is — (A) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br   (B) [Co(NH3)5SCN]2+ and [Co(NH3)5NCS]2+   (C) [Co(NH3)6]3+ and [Cr(C2O4)3]3–   (D) [Cr(H2O)5Cl]Cl2

1 mark · Q11 · 313/MAY/205A

Model Answer

Model approach (select the best option):

  • (A) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br
  • (B) [Co(NH3)5SCN]2+ and [Co(NH3)5NCS]2+
  • (C) [Co(NH3)6]3+ and [Cr(C2O4)3]3–
  • (D) [Cr(H2O)5Cl]Cl2·H2O and [Cr(H2O)4Cl2]Cl·2H2O g§

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L23. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205A · Q11.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ2. The IUPAC name of CH3CH2SH is — (A) methanethiol   (B) ethanethiol   (C) ethyl sulphur hydride   (D) ethane CH3CH2SH

1 mark · Q12 · 313/MAY/205A

Model Answer

Model approach (select the best option):

  • (A) methanethiol
  • (B) ethanethiol
  • (C) ethyl sulphur hydride
  • (D) ethane CH3CH2SH

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L23. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205A · Q12.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ3. Match the items in Column—I with Column—II : Column—I Column—II Nucleophile Electrophile CH CH Carbocation (iii) Free radical

2 marks · Q24 · 313/MAY/205A

Model Answer

Answer using key concepts from L23 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L23 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q24 · 2 mark(s) · L23.

PYQ4. The pair of complexes which shows linkage isomerism is — (A) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br   (B) [Co(NH3)5SCN]2+ and [Co(NH3)5NCS]2+   (C) [Co(NH3)6]3+ and [Cr(C2O4)3]3–   (D) [Cr(H2O)5Cl]Cl2

1 mark · Q4 · 313/MAY/205B

Model Answer

Model approach (select the best option):

  • (A) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br
  • (B) [Co(NH3)5SCN]2+ and [Co(NH3)5NCS]2+
  • (C) [Co(NH3)6]3+ and [Cr(C2O4)3]3–
  • (D) [Cr(H2O)5Cl]Cl2·H2O and [Cr(H2O)4Cl2]Cl·2H2O g§

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L23. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205B · Q4.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ5. The IUPAC name of CH3CH2SH is — (A) methanethiol   (B) ethanethiol   (C) ethyl sulphur hydride   (D) ethane CH3CH2SH

1 mark · Q15 · 313/MAY/205B

Model Answer

Model approach (select the best option):

  • (A) methanethiol
  • (B) ethanethiol
  • (C) ethyl sulphur hydride
  • (D) ethane CH3CH2SH

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L23. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205B · Q15.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ6. The IUPAC name of CH3CH2SH is — (A) methanethiol   (B) ethanethiol   (C) ethyl sulphur hydride   (D) ethane CH3CH2SH

1 mark · Q4 · 313/MAY/205C

Model Answer

Model approach (select the best option):

  • (A) methanethiol
  • (B) ethanethiol
  • (C) ethyl sulphur hydride
  • (D) ethane CH3CH2SH

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L23. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205C · Q4.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ7. The pair of complexes which shows linkage isomerism is — (A) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br   (B) [Co(NH3)5SCN]2+ and [Co(NH3)5NCS]2+   (C) [Co(NH3)6]3+ and [Cr(C2O4)3]3–   (D) [Cr(H2O)5Cl]Cl2

1 mark · Q14 · 313/MAY/205C

Model Answer

Model approach (select the best option):

  • (A) [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br
  • (B) [Co(NH3)5SCN]2+ and [Co(NH3)5NCS]2+
  • (C) [Co(NH3)6]3+ and [Cr(C2O4)3]3–
  • (D) [Cr(H2O)5Cl]Cl2·H2O and [Cr(H2O)4Cl2]Cl·2H2O g§

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L23. Recall the core definition or formula from notes, then match it to one option. Paper: 313/MAY/205C · Q14.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

PYQ8. The temporary electron displacement which takes place in compounds containing multiple covalent bonds developing +ve and –ve charges within the molecule is known as — (A) resonance   (B) electromeric effect   (C) inductive effect   (D) hyperconjugation

1 mark · Q8 · 313/TUS/105A

Model Answer

Model approach (select the best option):

  • (A) resonance
  • (B) electromeric effect
  • (C) inductive effect
  • (D) hyperconjugation

Eliminate options that contradict definitions/equations from the chapter notes. NIOS awards full mark for the single correct choice.

Explanation

This MCQ belongs to L23. Recall the core definition or formula from notes, then match it to one option. Paper: 313/TUS/105A · Q8.

Tip: For numerical MCQs, write the formula first, substitute values, then pick the option.

Problem Solving — L23 Nomenclature and General Principles

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). If the question says draw, a labelled pencil sketch is provided. Explanations open by default.

Question 1 of 6IUPAC

State the correct order of steps in IUPAC naming of a simple organic compound.

Solution — step by step with formulas

  1. Select longest chain; number to give lowest locants; name substituents + parent + principal functional group suffix.

Final answer: Longest chain → number → substituents + parent + suffix

Textbook formal language

IUPAC provides unique systematic names.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Find the main chain, number it smartly, then stick prefixes and ending.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Organic naming

Functional group priority decides suffix.

Linked to chapter notes (L23). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 2 of 6Isomerism

What are structural isomers? Example C₄H₁₀.

Solution — step by step with formulas

  1. Same molecular formula, different connectivity; n-butane and isobutane.

Final answer: n-butane & 2-methylpropane

Textbook formal language

Structural isomerism includes chain, position, functional isomers.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Same atoms, different wiring diagram.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Structural isomers

Stereoisomerism is different arrangement in space.

Linked to chapter notes (L23). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 3 of 6Homologous

State two features of a homologous series.

Solution — step by step with formulas

  1. Same functional group; differ by CH₂; gradual property change; same general formula.

Final answer: CH₂ difference; same functional group

Textbook formal language

Homologues prepared by similar methods.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Like a family: methanol, ethanol, propanol…

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Homologous series

Boiling points rise with molar mass.

Linked to chapter notes (L23). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 4 of 6Inductive

What is the +I effect? Give one group that shows +I.

Solution — step by step with formulas

  1. Electron-releasing inductive effect; alkyl groups.

Final answer: Alkyl +I

Textbook formal language

Inductive effect operates through σ bonds.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Alkyl groups push electrons slightly toward the chain.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Electronic effects

Affects acid strength of carboxylic acids.

Linked to chapter notes (L23). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 5 of 6Resonance

Why is benzene more stable than a hypothetical cyclohexatriene?

Solution — step by step with formulas

  1. Resonance delocalisation of π electrons (aromatic stabilisation).

Final answer: Resonance / aromatic stabilisation

Textbook formal language

Resonance hybrid lower in energy than contributing structures.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Electrons are smeared over the ring, not stuck in three double bonds.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Resonance

Equal C–C bond lengths in benzene.

Linked to chapter notes (L23). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 6 of 6Purification

State the principle of chromatography briefly.

Solution — step by step with formulas

  1. Differential adsorption/partition between stationary and mobile phases separates components.

Final answer: Different affinities for stationary vs mobile phase

Textbook formal language

Physical method of separation based on relative migration.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Components race differently on paper/column.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Chromatography idea

Used for dyes, amino acids, purity checks.

Linked to chapter notes (L23). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.