What is the inductive effect (+I and -I) and how does it operate?

The inductive effect is permanent electron displacement along a sigma bond chain due to electronegativity difference. Electron-withdrawing groups (-NO₂, -CN, -COOH, halogens) exert -I effect, while electron-donating alkyl groups (-CH₃, -C₂H₅) exert +I effect.

What is hyperconjugation and why does it stabilize carbocations?

Hyperconjugation (no-bond resonance) involves the delocalization of σ-electrons of C-H bonds with adjacent vacant p-orbitals. Stability of carbocations increases with the number of α-hydrogens: 3° (9 α-H) > 2° (6 α-H) > 1° (3 α-H) > methyl.

What is the principle of Lassaigne’s sodium fusion test?

Covalently bonded nitrogen, sulfur, and halogens in organic compounds are converted into ionic sodium salts (NaCN, Na₂S, NaX) by fusing with metallic sodium, enabling detection via aqueous inorganic reactions.

What is Kjeldahl’s method for nitrogen estimation?

The compound is digested with concentrated H₂SO₄ (with CuSO₄ catalyst) to convert nitrogen into (NH₄)₂SO₄, distilled with excess NaOH to liberate NH₃, and the absorbed ammonia in standard acid is back-titrated to find nitrogen percentage.

ALL UNITS

Unit 08 • Organic Chemistry

Organic Chemistry: Some Basic Principles and Techniques

The foundation of modern carbon chemistry. Tetravalence of carbon, hybridisation ($sp^3, sp^2, sp$), bond-line & wedge-dash notations, systematic IUPAC nomenclature, isomerism (chain, position, functional, metamerism), reaction intermediates (carbocations, carbanions, free radicals), electronic effects (inductive, electromeric, resonance, hyperconjugation), laboratory purification methods (crystallisation, fractional distillation, steam distillation, TLC & column chromatography), qualitative tests (Lassaigne's test for N, S, halogens, P), quantitative elemental analysis (Liebig, Dumas, Kjeldahl, Carius methods), in-text solved problems 8.1 to 8.24, fully solved NCERT exercises 8.1 to 8.39, revision matrix, and 3-level chapter tests.

8.1 Introduction & The Demise of Vital Force Theory

Organic compounds are indispensable for sustaining life on Earth, spanning genetic blueprints ($DNA, RNA$), proteins, lipids, carbohydrates, pharmaceuticals, polymers, fuels, and synthetic dyes.

Wöhler's Historic Breakthrough (1828)

Swedish chemist Jöns Jacob Berzelius proposed that a supernatural “Vital Force” was mandatory to create organic compounds in living organisms. This dogma was shattered in 1828 when Friedrich Wöhler heated the purely inorganic salt ammonium cyanate to yield the organic waste product urea: $$\mathbf{NH_4CNO \xrightarrow{\Delta} NH_2CONH_2 \quad (\text{Ammonium cyanate} \rightarrow \text{Urea})}$$ Subsequent total laboratory syntheses of acetic acid by Hermann Kolbe (1845) and methane by Marcellin Berthelot (1856) cemented that organic chemistry is governed by universal physical and chemical laws.

8.2 Tetravalence of Carbon and Hybridisation

Carbon ground state: $1s^2 2s^2 2p_x^1 2p_y^1$. Upon excitation to $1s^2 2s^1 2p_x^1 2p_y^1 2p_z^1$, carbon achieves tetravalence through orbital hybridisation ($sp^3, sp^2, sp$).

Hybridisation State	Orbitals Involved	Geometry & Bond Angle	% s-character & Electronegativity	Canonical Example
$sp^3$	$1s + 3p$	Tetrahedral ($109.5^\circ$)	$25\%$ s-character (Lowest EN)	Methane ($CH_4$), Ethane ($C_2H_6$)
$sp^2$	$1s + 2p$	Trigonal Planar ($120^\circ$)	$33.3\%$ s-character (Intermediate EN)	Ethene ($H_2C=CH_2$)
$sp$	$1s + 1p$	Linear ($180^\circ$)	$50\%$ s-character (Highest EN, shortest bonds)	Ethyne ($HC \equiv CH$)

Hybridisation States of Carbon: sp³ (Tetrahedral), sp² (Trigonal Planar), and sp (Linear)

8.1

How many σ and π bonds are present in each of the following molecules? (a) HC≡CCH=CHCH₃; (b) CH₂=C=CHCH₃.

(a) HC≡C—CH=CH—CH₃: Single bonds contribute 1 σ; double bond contributes 1 σ + 1 π; triple bond contributes 1 σ + 2 π.
C—C σ bonds = 4; C—H σ bonds = 6 $\implies$ Total σ bonds = 10.
π bonds = 2 (from C≡C) + 1 (from C=C) = 3 π bonds.
(b) CH₂=C=CH—CH₃ (Allene derivative):
C—C σ bonds = 3; C—H σ bonds = 6 $\implies$ Total σ bonds = 9.
π bonds = 1 + 1 = 2 π bonds.

(a) 10 σ and 3 π bonds; (b) 9 σ and 2 π bonds.

8.2

What is the type of hybridisation of each carbon in: (a) CH₃Cl; (b) (CH₃)₂CO; (c) CH₃CN; (d) HCONH₂; (e) CH₃CH=CHCN?

• (a) $CH_3Cl$: 4 single σ bonds $\implies \mathbf{sp^3}$.
• (b) $(CH_3)_2CO$: Methyl carbons $= \mathbf{sp^3}$; carbonyl carbon $(>C=O) = \mathbf{sp^2}$.
• (c) $CH_3CN$: Methyl carbon $= \mathbf{sp^3}$; cyano carbon ($-C\equiv N) = \mathbf{sp}$.
• (d) $HCONH_2$: Formyl carbon $(H-C(=O)-) = \mathbf{sp^2}$.
• (e) $\overset{5}{C}H_3-\overset{4}{C}H=\overset{3}{C}H-\overset{2}{C}\equiv \overset{1}{N}$: $C_5 = \mathbf{sp^3}$, $C_4 = \mathbf{sp^2}$, $C_3 = \mathbf{sp^2}$, $C_2 = \mathbf{sp}$.

(a) sp³; (b) sp³, sp²; (c) sp³, sp; (d) sp²; (e) sp³, sp², sp², sp.

8.3 Structural Representations & Bond-Line Formulas

Representation Style	Structural Features	Example: 2-Methylbutane
Complete Structural Formula	Every single covalent bond shown explicitly by a dash (—)	All 12 individual C—H and C—C dashes drawn out.
Condensed Formula	Dashes omitted; identical attached groups collected using subscripts	$(CH_3)_2CH-CH_2-CH_3$ or $(CH_3)_2CHCH_2CH_3$
Bond-Line Formula	Carbon backbone represented by zig-zag lines; vertices & terminals denote carbon with implicit hydrogens; heteroatoms shown explicitly	Four-carbon zig-zag with a branch at vertex 2.
Wedge-and-Dash (3D)	Solid wedge (—▲) for bond pointing towards viewer; Dashed wedge (---) for bond pointing away; Normal line (—) in paper plane	Standard tetrahedral 3D representation on 2D surface.

8.5

For each of the following compounds, write a condensed formula and its bond-line formula: (a) HOCH₂CH₂CH₂CH(CH₃)CH(CH₃)₂; (b) HOCH(CN)₂.

• (a) Condensed formula: $\mathbf{HO(CH_2)_3CH(CH_3)CH(CH_3)_2}$.
Bond-line: A 6-carbon chain with an $-OH$ group at terminal C1, and methyl branches at C4 and C5.
• (b) Condensed formula: $\mathbf{HOCH(CN)_2}$.
Bond-line: Central $CH$ attached to $-OH$ and two $-C\equiv N$ groups.

(a) HO(CH₂)₃CH(CH₃)CH(CH₃)₂; (b) HOCH(CN)₂.

8.4 Classification of Organic Compounds

Taxonomy of Organic Compounds (Acyclic vs Cyclic, Alicyclic vs Aromatic, Benzenoid vs Non-Benzenoid)

(a) Alicyclic: Homocyclic (cyclohexane) & Heterocyclic (THF) (b) Aromatic: • Benzenoid (Benzene, Aniline, Naphthalene) • Non-Benzenoid (Tropone) | Heterocyclic (Furan, Pyridine)

8.5 Systematic IUPAC Nomenclature of Organic Compounds

$$\mathbf{\text{Complete IUPAC Name} = \text{Secondary Prefix} + \text{Primary Prefix} + \text{Word Root} + \text{Primary Suffix} + \text{Secondary Suffix}}$$

Seniority Order of Principal Functional Groups

$$\mathbf{-COOH > -SO_3H > -COOR > -COCl > -CONH_2 > -CN > -CHO > >C=O > -OH > -NH_2 > >C=C< > -C\equiv C-}$$ Alkyl ($-R$), phenyl ($-Ph$), halogens ($-X$), nitro ($-NO_2$), and alkoxy ($-OR$) are ALWAYS treated as prefixes!

8.7

Explain why the name 2,5,6-trimethyloctane is preferred over 3,4,7-trimethyloctane, and 3-ethyl-5-methylheptane is preferred over 5-ethyl-3-methylheptane.

• Lowest Locant Rule: Comparing locant sets at first point of difference: set $(2, 5, 6)$ vs $(3, 4, 7)$. At the first position, $2 < 3$, so 2,5,6-trimethyloctane is correct.
• Alphabetical Priority for Equivalent Locants: In 3-ethyl-5-methylheptane, substituents occupy identical locants $(3, 5)$ from either end. Since ‘ethyl’ alphabetically precedes ‘methyl’, the lower number 3 is assigned to ethyl.

Governed by the lowest locant set rule and alphabetical preference for identical locants.

8.8

Write the IUPAC names of: (a) CH₃CH(OH)(CH₂)₄CH(CH₃)₂; (b) CH₃CH₂COCH₂COCH₃; (c) CH₃COCH₂CH₂CH₂COOH; (d) CH≡CCH=CHCH=CH₂.

• (a) Principal group is $-OH$ at C2; methyl at C7. Total 8 carbons $\implies \mathbf{7\text{-methyloctan-2-ol}}$.
• (b) Six carbon chain with two ketone groups at C2 and C4 $\implies \mathbf{\text{hexane-2,4-dione}}$.
• (c) Principal group is $-COOH$ (C1). Oxo at C5 of 6-carbon chain $\implies \mathbf{5\text{-oxohexanoic acid}}$.
• (d) Chain with double and triple bonds. Numbering starts from double bond end (giving lowest locant to double bond when positions are identical) $\implies \mathbf{\text{hexa-1,3-dien-5-yne}}$.

(a) 7-methyloctan-2-ol; (b) hexane-2,4-dione; (c) 5-oxohexanoic acid; (d) hexa-1,3-dien-5-yne.

8.6 Isomerism in Organic Compounds

Classification of Isomerism: Structural Isomers vs Stereoisomers

Chain: Pentane / Isopentane / Neopentane • Position: Propan-1-ol / Propan-2-ol • Functional: Ethanol / Dimethyl ether • Metamerism: Diethyl ether / Methyl propyl ether 2. Stereoisomerism • Geometrical: cis/trans (restricted rotation) • Optical: Non-superimposable mirror images

8.7 Reaction Intermediates & Bond Cleavage

Intermediate	Fission Type	Structure & Hybridisation	Stability Order
Carbocation ($R_3C^+$)	Heterolytic ($C-Z \rightarrow C^+ + :Z^-$)	Trigonal planar ($sp^2$), 6 valence electrons (sextet, electrophilic)	$\mathbf{3^\circ > 2^\circ > 1^\circ > \overset{+}{C}H_3}$ (Stabilized by $+I$ and hyperconjugation)
Carbanion ($R_3C^-$)	Heterolytic ($C-Z \rightarrow C:^- + Z^+$)	Pyramidal ($sp^3$), 8 valence electrons with lone pair (nucleophilic)	$\mathbf{\overset{-}{C}H_3 > 1^\circ > 2^\circ > 3^\circ}$ (Alkyl $+I$ destabilizes)
Carbon Free Radical ($R_3C^\bullet$)	Homolytic ($A-B \rightarrow A^\bullet + B^\bullet$)	Trigonal planar ($sp^2$), 7 electrons with one unpaired electron	$\mathbf{3^\circ > 2^\circ > 1^\circ > \overset{\bullet}{C}H_3}$ (Hyperconjugation stabilization)

Geometry of Reaction Intermediates: Methyl Carbocation (sp², Trigonal Planar) vs Methyl Carbanion (sp³, Pyramidal)

8.7.4 Electronic Displacement Effects in Covalent Bonds

1. Inductive Effect ($I$)

Permanent dipole transmission along a saturated carbon chain due to electronegativity differences:

$$\overset{\delta\delta\delta+}{C}H_3 \rightarrow \overset{\delta\delta+}{C}H_2 \rightarrow \overset{\delta+}{C}H_2 \rightarrow \overset{\delta-}{Cl}$$ • $-I$ (Electron-Withdrawing): $-NO_2 > -CN > -COOH > -F > -Cl > -Br > -I > -OH > -C_6H_5$
• $+I$ (Electron-Donating): $-(CH_3)_3C > -CH(CH_3)_2 > -CH_2CH_3 > -CH_3 > -H$

2. Resonance / Mesomeric Effect ($R$ / $M$)

Polarity produced by interaction of two $\pi$-bonds or a $\pi$-bond and an adjacent lone pair:

$+R$ Effect: Electron donation into conjugated system: $-OH, -OR, -NH_2, -NHR, -\text{Halogens}$.
$-R$ Effect: Electron withdrawal from conjugated system: $-NO_2, -CN, -CHO, >C=O, -COOH$.

3. Hyperconjugation (“No-Bond Resonance”)

Stabilizing interaction involving delocalisation of $\sigma$-electrons of a $C-H$ bond of an alkyl group into an adjacent empty $p$-orbital (carbocation) or $\pi^*$-orbital (alkene):

Hyperconjugation (No-Bond Resonance): Delocalisation of C—H σ Electrons into Vacant p-Orbital

8.8 Methods of Purification of Organic Compounds

Purification Technique	Operating Principle	Classic Chemical Application
Sublimation	Transition of solid directly to vapour without liquefaction	Separating camphor, naphthalene, benzoic acid from non-sublimable salts
Crystallisation	Differential solubility at room temperature vs elevated temperature	Purification of impure sugar, benzoic acid from water
Simple Distillation	Volatile liquid from non-volatile impurities, or $\Delta T_b > 25\text{ K}$	Chloroform ($T_b = 334\text{ K}$) and Aniline ($T_b = 457\text{ K}$)
Fractional Distillation	Vapours pass through fractionating column with theoretical plates ($\Delta T_b < 25\text{ K}$)	Crude petroleum refining, acetone & methanol separation
Distillation under Reduced Pressure (Vacuum)	Liquids with high $T_b$ that decompose at or below their boiling point	Glycerol from spent-lye in soap manufacture (boils at $290^\circ\text{C}$ with decomp., distils at $180^\circ\text{C}$ at $12\text{ mm Hg}$)
Steam Distillation	Steam-volatile substances immiscible with water ($p_{\text{total}} = p_1 + p_2 = 1\text{ atm}$)	Aniline from aniline-water mixture, turpentine oil
Differential Extraction	Preferential partition of organic solute into immiscible organic solvent in a separatory funnel	Benzoic acid extraction from water using ether
Chromatography	Differential adsorption (TLC, Column) or partition (Paper) between stationary and mobile phases	Separation of plant pigments, amino acids (ninhydrin detection)

Principles of Fractional Distillation (Packed Column) and Steam Distillation (p_total = p_organic + p_water)

8.9 Qualitative Analysis of Elements: Lassaigne's Test

Organic compounds contain covalent bonds. To test for nitrogen, sulphur, halogens, and phosphorus, they are fused with metallic sodium to convert covalent elements into water-soluble ionic salts (Sodium Fusion Extract):

$$Na + C + N \xrightarrow{\Delta} NaCN \qquad 2Na + S \xrightarrow{\Delta} Na_2S \qquad Na + X \xrightarrow{\Delta} NaX$$

Lassaigne's Test: Characteristic Colour Tests for Nitrogen, Sulphur, and Halogens

8.10 Quantitative Elemental Analysis (Master Formulas)

Element	Analytical Method	Mathematical Percentage Formula
Carbon (C)	Liebig Combustion (absorbed in KOH)	$$\mathbf{\%C = \frac{12}{44} \times \frac{m_{\text{CO}_2}}{m} \times 100}$$
Hydrogen (H)	Liebig Combustion (absorbed in anhyd. $CaCl_2$)	$$\mathbf{\%H = \frac{2}{18} \times \frac{m_{\text{H}_2\text{O}}}{m} \times 100}$$
Nitrogen (N)	Dumas Method (Nitrometer $N_2$ gas at STP)	$$\mathbf{\%N = \frac{28}{22400} \times \frac{V_{\text{STP}}}{m} \times 100}$$
Nitrogen (N)	Kjeldahl Method (digestion into $(NH_4)_2SO_4$, titration with standard acid)	$$\mathbf{\%N = \frac{1.4 \times M \times 2(V - V_1/2)}{m} = \frac{1.4 \times N \times V_{\text{acid}}}{m}}$$ (Not applicable to nitro, azo, or ring nitrogen like pyridine!)
Halogens (X)	Carius Method (heated with fuming $HNO_3 + AgNO_3 \rightarrow AgX$)	$$\mathbf{\%X = \frac{\text{At. mass of } X}{\text{Mol. mass of } AgX} \times \frac{m_{AgX}}{m} \times 100}$$
Sulphur (S)	Carius Method (oxidised to $H_2SO_4$, precipitated as $BaSO_4$)	$$\mathbf{\%S = \frac{32}{233} \times \frac{m_{BaSO_4}}{m} \times 100}$$
Phosphorus (P)	Precipitated as Ammonium phosphomolybdate	$$\mathbf{\%P = \frac{31}{1877} \times \frac{m_{\text{precipitate}}}{m} \times 100}$$
Oxygen (O)	By Difference (or reaction with red-hot coke & $I_2O_5$)	$$\mathbf{\%O = 100 - \sum (\% \text{ of other elements})}$$

8.20

On complete combustion, 0.246 g of an organic compound gave 0.198 g of CO₂ and 0.1014 g of H₂O. Determine the percentage composition of carbon and hydrogen.

$$\%C = \frac{12}{44} \times \frac{0.198}{0.246} \times 100 = \mathbf{21.95\%}$$ $$\%H = \frac{2}{18} \times \frac{0.1014}{0.246} \times 100 = \mathbf{4.58\%}$$

%C = 21.95%, %H = 4.58%.

8.21

In Dumas' method for nitrogen estimation, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300 K and 715 mm pressure. Aqueous tension at 300 K is 15 mm. Calculate percentage of nitrogen.

Actual pressure of dry $N_2$: $p_1 = 715 - 15 = 700\text{ mm Hg}$.
Convert to volume at STP: $$V_{\text{STP}} = \frac{p_1 V_1 \times 273}{760 \times T_1} = \frac{700 \times 50 \times 273}{760 \times 300} = \mathbf{41.9\text{ mL}}$$ $$\%N = \frac{28}{22400} \times \frac{41.9}{0.3} \times 100 = \mathbf{17.46\%}$$

%N = 17.46%.

8.22

During estimation of nitrogen by Kjeldahl's method, the ammonia evolved from 0.5 g of the compound neutralized 10 mL of 1 M H₂SO₄. Find the percentage of nitrogen.

$10\text{ mL of } 1\text{ M } H_2SO_4 = 20\text{ mL of } 1\text{ N } H_2SO_4 = 20\text{ mL of } 1\text{ N } NH_3$.
$$\%N = \frac{1.4 \times N \times V}{m} = \frac{1.4 \times 1 \times 20}{0.5} = \mathbf{56.0\%}$$

%N = 56.0%.

8.23

In Carius method, 0.15 g of an organic compound gave 0.12 g of AgBr. Find the percentage of bromine.

Molar mass of $AgBr = 108 + 80 = 188\text{ g mol}^{-1}$.
$$\%Br = \frac{80}{188} \times \frac{0.12}{0.15} \times 100 = \mathbf{34.04\%}$$

%Br = 34.04%.

8.24

In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of BaSO₄. What is the percentage of sulphur?

Molar mass of $BaSO_4 = 137 + 32 + 64 = 233\text{ g mol}^{-1}$.
$$\%S = \frac{32}{233} \times \frac{0.4813}{0.157} \times 100 = \mathbf{42.10\%}$$

%S = 42.10%.

Organic Chemistry Conceptual Quiz (25 MCQs)

Test your understanding of hybridisation, bond-line structures, IUPAC nomenclature rules, isomerism, reaction mechanisms, electronic effects, and qualitative/quantitative analysis. Instant feedback upon selection.

1 In the organic compound CH₂=CH—CH₂—CH₂—C≡CH, the pair of hybridised orbitals involved in the formation of the C₂—C₃ single bond is:

Incorrect: Neither C2 nor C3 is sp.

Incorrect: C2 is involved in a double bond.

Correct: Numbering from the double bond end (lowest locant for double bond when positions clash): C1(=)C2(sp²)—C3(sp³). Hence sp² — sp³.

Incorrect: C3 is sp³ and C2 is sp².

2 In Lassaigne's test for nitrogen, the characteristic Prussian blue colour is produced due to the formation of:

Incorrect: That is sodium ferrocyanide intermediate.

Correct: Iron(III) hexacyanoferrate(II), Fe₄[Fe(CN)₆]₃·xH₂O, produces the iconic Prussian blue precipitate.

Incorrect: That is Turnbull's blue.

Incorrect: Blood red thiocyanate.

3 Which of the following carbocations is the most stable?

Correct: The tert-butyl cation ((CH₃)₃C⁺) has 9 α-hydrogens, providing maximum hyperconjugative stabilisation and +I electron donation.

Incorrect: Isopropyl cation has only 6 α-hydrogens.

Incorrect: Ethyl cation has only 3 α-hydrogens.

Incorrect: Methyl cation has 0 α-hydrogens.

4 The reaction CH₃CH₂I + KOH(aq) → CH₃CH₂OH + KI is classified as:

Incorrect: OH⁻ is a nucleophile, not an electrophile.

Correct: Nucleophile OH⁻ substitutes the iodide leaving group I⁻ at the electrophilic carbon centre.

Incorrect: Alcoholic KOH causes elimination, but aqueous KOH causes substitution.

Incorrect: No multiple bond is formed or consumed.

5 The best and modern technique for isolation, purification and separation of organic components is:

Incorrect: Limited to solids with differing solubilities.

Incorrect: Limited to liquids with boiling point differences.

Incorrect: Only works for sublime solids.

Correct: Chromatography operates on microquantities and provides superlative separation efficiency.

6 Which of the following nitrogen-containing compounds CANNOT be estimated by Kjeldahl's method?

Incorrect: Amide nitrogen converts to (NH₄)₂SO₄.

Incorrect: Amine nitrogen converts to (NH₄)₂SO₄.

Correct: Kjeldahl's method fails for nitro (—NO₂), azo (—N=N—), and ring nitrogen (pyridine), as they do not convert to ammonium sulphate.

Incorrect: Readily estimated.

7 What is the correct IUPAC name of (CH₃)₂C(OH)CH₂CH₂CH₃?

Correct: 5-carbon principal chain containing —OH at C2 with a methyl substituent at C2.

Incorrect: Violates the lowest locant rule for functional group.

Incorrect: Incorrect carbon skeleton.

Incorrect: Missing methyl branch.

8 Hyperconjugation is also famously referred to as:

Incorrect: Unrelated spatial effect.

Correct: In hyperconjugative canonical structures, there is no physical covalent bond between carbon and the participating α-proton.

Incorrect: Inductive involves σ-polarity without orbital overlap.

Incorrect: Electromeric is temporary.

9 A mixture of ortho-nitrophenol and para-nitrophenol can be easily separated by:

Incorrect: Neither sublimes easily.

Correct: o-Nitrophenol forms intramolecular H-bonds and is steam-volatile; p-nitrophenol forms intermolecular H-bonds (high Tb) and remains in the flask.

Incorrect: Fractional crystallisation is tedious.

Incorrect: Non-electrolytes.

10 Which of the following species acts as an electrophile?

Incorrect: NH₃ has an unshared lone pair (nucleophile).

Incorrect: H₂O has two lone pairs (nucleophile).

Correct: AlCl₃ has an incomplete octet (electron-deficient sextet) and avidly accepts an electron pair.

Incorrect: OH⁻ is an anion and nucleophile.

11 How many chain isomers are possible for the alkane with molecular formula C₅H₁₂?

Incorrect: Pentane has more than 2 isomers.

Correct: Pentane (n-pentane), 2-methylbutane (isopentane), and 2,2-dimethylpropane (neopentane).

Incorrect: Only 3 exist.

Incorrect: Too high.

12 Propan-1-ol and propan-2-ol represent an example of:

Incorrect: Carbon chain length is identical (3 carbons).

Correct: They differ solely in the position of the —OH functional group (C1 vs C2).

Incorrect: Both possess the identical functional group (alcohol).

Incorrect: Alcohols do not exhibit metamerism.

13 In paper chromatography, the stationary phase consists of:

Incorrect: Cellulose is the supporting matrix.

Correct: Paper chromatography is a partition method where trapped water acts as the liquid stationary phase.

Incorrect: Solvent is the mobile phase.

Incorrect: Silica gel is used in TLC.

14 The Rf value (Retardation factor) in thin layer chromatography is mathematically defined as:

Incorrect: Inverted definition.

Correct: Rf = x / y, where x is substance distance and y is solvent front distance (always ≤ 1.0).

Incorrect: Not related to molar mass.

Incorrect: That is retention time in HPLC/GC.

15 When sodium fusion extract containing both N and S is treated with FeCl₃, the appearance of a blood-red colour is due to:

Incorrect: Brown precipitate.

Incorrect: Prussian blue.

Correct: When both N and S are present, sodium thiocyanate (NaSCN) forms; reacting with Fe³⁺ yields blood-red [Fe(SCN)]²⁺.

Incorrect: Black precipitate.

16 In Carius method, sulphur present in an organic compound is quantitatively precipitated and weighed as:

Incorrect: Lead sulphate is used in qualitative detection.

Correct: Sulphur is oxidised to sulphate and precipitated with excess BaCl₂ as insoluble barium sulphate (BaSO₄).

Incorrect: Sodium sulphate is soluble.

Incorrect: Sulphuric acid is a liquid.

17 The geometry and hybridisation of a methyl carbanion (:CH₃⁻) are:

Incorrect: That is carbocation CH₃⁺.

Correct: The central carbon has 3 σ-bonds and 1 lone pair (steric number = 4), giving an sp³ pyramidal geometry.

Incorrect: Not linear.

Incorrect: Lone pair distorts tetrahedral to pyramidal.

18 Glycerol (boiling point 290 °C with decomposition) is commercially purified by:

Incorrect: Decomposes before boiling under atmospheric pressure.

Incorrect: Cannot prevent thermal degradation.

Correct: Reducing pressure lowers its boiling point to ~180 °C (at 12 mm Hg), allowing distillation without decomposition.

Incorrect: Glycerol is not steam volatile.

19 Which of the following groups exerts a +R (positive resonance) effect when attached to a benzene ring?

Incorrect: —NO₂ is a strong —R group.

Incorrect: —COOH is a —R group.

Incorrect: —CHO is a —R group.

Correct: The —OH group has unshared lone pairs on oxygen which donate electron density into the ring via resonance (+R).

20 The number of σ (sigma) and π (pi) bonds in benzene (C₆H₆) are:

Incorrect: Missing the 6 C—H σ bonds.

Correct: 6 C—C σ bonds + 6 C—H σ bonds = 12 σ bonds; alternating double bonds contribute 3 π bonds.

Incorrect: 9 is incorrect count.

Incorrect: Benzene has 3 π bonds.

21 Electromeric effect is a:

Incorrect: Inductive effect is permanent.

Correct: Electromeric effect occurs only in multiple bonds when an attacking reagent approaches, vanishing when reagent is removed.

Incorrect: That describes hyperconjugation.

Incorrect: Irrelevant.

22 Which of the following organic compounds will NOT give a white precipitate of AgCl on heating with AgNO₃?

Incorrect: Ionic NaCl immediately precipitates AgCl.

Correct: CH₃Cl is a covalent compound; chlorine is firmly bound and does not ionize in water to yield free Cl⁻ ions.

Incorrect: CH₃Cl does not precipitate.

Incorrect: NaCl definitely precipitates.

23 In the estimation of carbon and hydrogen by Liebig combustion, CO₂ and H₂O are absorbed respectively in:

Incorrect: Inverted order; H₂O is absorbed first in CaCl₂.

Correct: Water is absorbed in weighed U-tube of anhydrous CaCl₂, and CO₂ is absorbed in weighed bulbs of concentrated KOH solution.

Incorrect: Standard Liebig train uses CaCl₂ and KOH.

Incorrect: Incorrect absorbents.

24 Between O₂NCH₂CH₂O⁻ and CH₃CH₂O⁻, which anion is more stable and why?

Incorrect: +I destabilizes negative charge.

Correct: The strongly electron-withdrawing nitro group (—NO₂) exerts a —I effect, dispersing negative charge on oxygen and stabilizing the anion.

Incorrect: Different chemical stabilities.

Incorrect: Alkoxide ions lack hyperconjugative stabilization.

25 In the IUPAC name 4-oxopentanoic acid, the prefix 'oxo' denotes:

Incorrect: Ether prefix is alkoxy.

Correct: Carboxylic acid (—COOH) is senior; the subordinate ketone group at C4 is named with the prefix 'oxo'.

Incorrect: An aldehyde at C4 would be a branched chain.

Incorrect: Hydroxyl prefix is hydroxy.

Quiz Results

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NCERT Textbook Exercises (8.1 to 8.39 Fully Solved)

Complete step-by-step solutions for all 39 end-of-chapter questions from CBSE / NCERT Class 11 Chemistry Chapter 8 (Organic Chemistry Some Basic Principles and Techniques).

8.1

What are the hybridisation states of each carbon atom in the following compounds? (a) CH₂=C=O; (b) CH₃CH=CH₂; (c) (CH₃)₂CO; (d) CH₂=CHCN; (e) C₆H₆.

• (a) $\overset{1}{C}H_2=\overset{2}{C}=O$: Carbon 1 forms 2 single bonds and 1 double bond $\implies \mathbf{sp^2}$. Carbon 2 forms 2 double bonds $\implies \mathbf{sp}$.
• (b) $\overset{3}{C}H_3-\overset{2}{C}H=\overset{1}{C}H_2$: $C_3 = \mathbf{sp^3}$, $C_2 = \mathbf{sp^2}$, $C_1 = \mathbf{sp^2}$.
• (c) $(CH_3)_2CO$: Methyl carbons $= \mathbf{sp^3}$, Carbonyl carbon $(>C=O) = \mathbf{sp^2}$.
• (d) $\overset{3}{C}H_2=\overset{2}{C}H-\overset{1}{C}\equiv N$: $C_3 = \mathbf{sp^2}$, $C_2 = \mathbf{sp^2}$, $C_1 = \mathbf{sp}$.
• (e) $C_6H_6$ (Benzene): Each carbon is bonded to two neighbouring carbons (one single, one double) and one hydrogen $\implies$ All six carbons are $\mathbf{sp^2}$.

(a) sp², sp; (b) sp³, sp², sp²; (c) sp³, sp²; (d) sp², sp², sp; (e) all carbons sp².

8.2

Indicate the σ and π bonds in the following molecules: C₆H₆, C₆H₁₂, CH₂Cl₂, CH₂=C=CH₂, CH₃NO₂, HCONHCH₃.

• C₆H₆ (Benzene): 6 C—C σ + 6 C—H σ = 12 σ; 3 alternating double bonds = 3 π.
• C₆H₁₂ (Cyclohexane): 6 C—C σ + 12 C—H σ = 18 σ; 0 π.
• CH₂Cl₂ (Dichloromethane): 2 C—H σ + 2 C—Cl σ = 4 σ; 0 π.
• CH₂=C=CH₂ (Allene): 2 C—C σ + 4 C—H σ = 6 σ; 2 double bonds = 2 π.
• CH₃NO₂ (Nitromethane): 3 C—H σ + 1 C—N σ + 2 N—O σ = 6 σ; 1 N=O double bond = 1 π.
• HCONHCH₃ (N-Methylformamide): 1 C—H σ + 1 C—N σ + 1 C=O σ + 1 N—H σ + 1 N—C σ + 3 C—H σ = 8 σ; 1 C=O double bond = 1 π.

C₆H₆: 12σ, 3π; C₆H₁₂: 18σ, 0π; CH₂Cl₂: 4σ, 0π; CH₂=C=CH₂: 6σ, 2π; CH₃NO₂: 6σ, 1π; HCONHCH₃: 8σ, 1π.

8.3

Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.

• Isopropyl alcohol: A 3-carbon chain (a ‘V’ shape) with an —OH group branching from the central vertex: $(CH_3)_2CH-OH$.
• 2,3-Dimethylbutanal: A 4-carbon backbone ending in an aldehyde group ($=O$ and $-H$), with single-line methyl branches attached at positions 2 and 3.
• Heptan-4-one: A 7-carbon zig-zag chain with a double-bonded oxygen ($=O$) projecting from the central 4th vertex.

Bond-line formulas drawn with vertices and lines representing carbons and explicit heteroatoms.

8.4

Give the IUPAC names of: (a) Propylbenzene; (b) 3-Methylpentanenitrile; (c) 2,5-Dimethylheptane; (d) 3-Bromo-3-chloroheptane; (e) 3-Chloropropanal; (f) Cl₂CHCH₂OH.

• (a) $C_6H_5-CH_2-CH_2-CH_3$: Propylbenzene (or 1-phenylpropane).
• (b) $CH_3-CH_2-CH(CH_3)-CH_2-CN$: 3-Methylpentanenitrile (C1 at —CN).
• (c) $CH_3-CH(CH_3)-CH_2-CH_2-CH(CH_3)-CH_2-CH_3$: 2,5-Dimethylheptane.
• (d) $CH_3-CH_2-C(Br)(Cl)-(CH_2)_3-CH_3$: 3-Bromo-3-chloroheptane (alphabetical: bromo before chloro).
• (e) $Cl-CH_2-CH_2-CHO$: 3-Chloropropanal (C1 at —CHO).
• (f) $Cl_2CH-CH_2OH$: 2,2-Dichloroethanol (C1 at —OH).

(a) Propylbenzene; (b) 3-Methylpentanenitrile; (c) 2,5-Dimethylheptane; (d) 3-Bromo-3-chloroheptane; (e) 3-Chloropropanal; (f) 2,2-Dichloroethanol.

8.5

Which of the following represents the correct IUPAC name? (a) 2,2-Dimethylpentane or 2-Dimethylpentane; (b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane; (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane; (d) But-3-yn-1-ol or But-4-ol-1-yne.

• (a) 2,2-Dimethylpentane: Each methyl substituent must have its own separate locant number. '2-Dimethylpentane' is incorrect.
• (b) 2,4,7-Trimethyloctane: Comparing locant sets at the first point of difference: $(2, 4, 7)$ vs $(2, 5, 7)$. Since $4 < 5$, $(2, 4, 7)$ is correct.
• (c) 2-Chloro-4-methylpentane: Locant sets from left: $(2, 4)$; from right: $(2, 4)$. With equivalent locants, the lower number is assigned to the substituent cited first alphabetically (chloro before methyl).
• (d) But-3-yn-1-ol: The principal functional group is alcohol (—OH), which receives lowest locant (1); suffix is '-ol'.

(a) 2,2-Dimethylpentane; (b) 2,4,7-Trimethyloctane; (c) 2-Chloro-4-methylpentane; (d) But-3-yn-1-ol.

8.6

Draw formulas for the first five members of each homologous series beginning with: (a) H—COOH; (b) CH₃COCH₃; (c) H—CH=CH₂.

• (a) Carboxylic acids (Alkanoic acids, CₙH₂ₙO₂):
1. $HCOOH$ (Methanoic acid)
2. $CH_3COOH$ (Ethanoic acid)
3. $CH_3CH_2COOH$ (Propanoic acid)
4. $CH_3(CH_2)_2COOH$ (Butanoic acid)
5. $CH_3(CH_2)_3COOH$ (Pentanoic acid)

• (b) Ketones (Alkanones, CₙH₂ₙO, n ≥ 3):
1. $CH_3COCH_3$ (Propan-2-one)
2. $CH_3COCH_2CH_3$ (Butan-2-one)
3. $CH_3CO(CH_2)_2CH_3$ (Pentan-2-one)
4. $CH_3CO(CH_2)_3CH_3$ (Hexan-2-one)
5. $CH_3CO(CH_2)_4CH_3$ (Heptan-2-one)

• (c) Alkenes (CₙH₂ₙ, n ≥ 2):
1. $CH_2=CH_2$ (Ethene)
2. $CH_3-CH=CH_2$ (Propene)
3. $CH_3-CH_2-CH=CH_2$ (But-1-ene)
4. $CH_3-(CH_2)_2-CH=CH_2$ (Pent-1-ene)
5. $CH_3-(CH_2)_3-CH=CH_2$ (Hex-1-ene)

First 5 homologues given for carboxylic acids, ketones, and alkenes.

8.7

Give condensed and bond-line structural formulas and identify the functional group(s) for: (a) 2,2,4-Trimethylpentane; (b) 2-Hydroxypropane-1,2,3-tricarboxylic acid; (c) Hexanedial.

• (a) 2,2,4-Trimethylpentane:
Condensed: $(CH_3)_3C-CH_2-CH(CH_3)_2$. Bond-line: 5-carbon chain with two branches at C2 and one branch at C4. Functional group: None (saturated alkane).
• (b) 2-Hydroxypropane-1,2,3-tricarboxylic acid (Citric acid):
Condensed: $HOOC-CH_2-C(OH)(COOH)-CH_2-COOH$. Functional groups: Hydroxyl ($-OH$) and Carboxyl ($-COOH$).
• (c) Hexanedial:
Condensed: $OHC-(CH_2)_4-CHO$. Bond-line: 6-carbon chain with aldehyde groups at both ends. Functional group: Aldehyde ($-CHO$).

Formulas provided; functional groups identified.

8.8

Identify the functional groups in: (a) An aldehyde with an aromatic ring; (b) An ester; (c) An amine with an alkene.

• (a) In benzaldehyde ($C_6H_5CHO$): Aldehyde group ($-CHO$) and benzene ring (aromatic).
• (b) In ethyl ethanoate ($CH_3COOCH_2CH_3$): Ester group ($-COOR$).
• (c) In prop-2-en-1-amine ($CH_2=CH-CH_2NH_2$): Alkene double bond ($>C=C<$) and Primary amino group ($-NH_2$).

Functional groups categorized and identified.

8.9

Which of the two O₂NCH₂CH₂O⁻ or CH₃CH₂O⁻ is expected to be more stable and why?

$\mathbf{O_2NCH_2CH_2O^-}$ is significantly more stable.
The nitro group ($-NO_2$) is a powerful electron-withdrawing group that exerts a strong negative inductive effect (—I effect). This withdraws electron density away from the negatively charged oxygen atom, dispersing the negative charge over the molecule and stabilizing the alkoxide ion.
In contrast, in the ethoxide ion ($CH_3CH_2O^-$), the ethyl group has an electron-donating +I effect, which intensifies the negative charge on oxygen, destabilizing the anion.

O₂NCH₂CH₂O⁻ is more stable due to the charge-dispersing —I effect of the —NO₂ group.

8.10

Explain why alkyl groups act as electron donors when attached to a π-system.

Alkyl groups (such as methyl, $-CH_3$) act as electron donors when directly bonded to a $\pi$-electron system (or vacant $p$-orbital) primarily through hyperconjugation (no-bond resonance).
The $\sigma$-electrons of the $C-H$ bond of the alkyl group overlap laterally with the adjacent $2p$-orbital of the $\pi$-bond (or aromatic system), delocalising electron density into the unsaturated system.
Additionally, $sp^3$-hybridised alkyl carbon is less electronegative than $sp^2$-hybridised unsaturated carbon, contributing a minor electron-donating +I inductive effect.

Alkyl groups donate electrons via hyperconjugation (delocalisation of C—H σ-electrons into π-system) and +I inductive effect.

8.11

Draw resonance structures for: (a) C₆H₅OH; (b) C₆H₅NO₂; (c) CH₃CH=CHCHO; (d) C₆H₅—CHO; (e) C₆H₅—CH₂⁺; (f) CH₃CH=CH—CH₂⁺.

• (a) Phenol (C₆H₅OH): Lone pair on oxygen donates into the ring (+R effect), creating negative formal charges at ortho and para positions (Structures: neutral kekule, ortho-negative, para-negative, ortho-negative).
• (b) Nitrobenzene (C₆H₅NO₂): Nitro group withdraws electrons (-R effect), creating positive formal charges at ortho and para positions.
• (c) But-2-enal (CH₃CH=CHCHO): $\pi$-electrons shift towards electronegative oxygen: $CH_3-\overset{+}{C}H-CH=CH-O^-$.
• (d) Benzaldehyde (C₆H₅CHO): -R effect of carbonyl group delocalises ring electrons, developing positive charges at ortho and para positions.
• (e) Benzyl carbocation (C₆H₅CH₂⁺): The vacant $p$-orbital on $-CH_2^+$ accepts ring $\pi$-electrons, delocalising positive charge to ortho and para carbons (4 contributing canonical structures).
• (f) Allylic carbocation (CH₃CH=CH—CH₂⁺): Allylic resonance: $CH_3-\overset{+}{C}H-CH=CH_2 \longleftrightarrow CH_3-CH=CH-\overset{+}{C}H_2$.

Resonance structures exhibit delocalisation of π and lone-pair electrons via curved arrows.

8.12

What are electrophiles and nucleophiles? Explain with examples.

• Electrophiles (Electron-seeking, E⁺): Reagents that are electron-deficient (vacant orbital or positive centre) and accept an electron pair from an electron-rich site during a reaction.
Examples: $H^+, H_3O^+, Cl^+, NO_2^+, CH_3^+, AlCl_3, BF_3, SO_3, >C=O$.

• Nucleophiles (Nucleus-seeking, Nu:⁻): Reagents that possess an unshared electron pair (lone pair or negative charge) and donate an electron pair to an electron-deficient centre.
Examples: $OH^-, CN^-, Cl^-, Br^-, I^-, H_2O, NH_3, R-OH, R-NH_2$.

Electrophiles are electron-pair acceptors; nucleophiles are electron-pair donors.

8.13

Identify the reagents shown in bold as nucleophiles or electrophiles: (a) CH₃COOH + HO⁻ → CH₃COO⁻ + H₂O; (b) CH₃COCH₃ + CN⁻ → (CH₃)₂C(CN)(OH); (c) C₆H₆ + CH₃C⁺O → C₆H₅COCH₃.

• (a) $\mathbf{HO^-}$: Has unshared electron pairs and bears a negative charge $\implies$ Nucleophile.
• (b) $\mathbf{CN^-}$: Possesses a lone pair and negative charge on carbon $\implies$ Nucleophile.
• (c) $\mathbf{CH_3C^+O}$ (Acetyl cation): Positively charged, electron-deficient carbon centre with vacant orbital $\implies$ Electrophile.

(a) HO⁻ is Nucleophile; (b) CN⁻ is Nucleophile; (c) CH₃C⁺O is Electrophile.

8.14

Classify the following reactions into one of the reaction types studied: (a) CH₃CH₂Br + HS⁻ → CH₃CH₂SH + Br⁻; (b) (CH₃)₂C=CH₂ + HCl → (CH₃)₂CCl—CH₃; (c) CH₃CH₂Br + HO⁻ → CH₂=CH₂ + H₂O + Br⁻; (d) (CH₃)₃C—CH₂OH + HBr → (CH₃)₂C(Br)CH₂CH₃ + H₂O.

• (a) Nucleophile $HS^-$ replaces leaving group $Br^-$ $\implies$ Nucleophilic Substitution Reaction ($S_N$).
• (b) $H$ and $Cl$ add across the carbon-carbon double bond to yield an alkyl halide $\implies$ Electrophilic Addition Reaction.
• (c) Elements of $HBr$ are removed from adjacent carbon atoms to generate a double bond $\implies$ β-Elimination Reaction.
• (d) Reaction involves substitution accompanied by alkyl shift (migration of methyl group from neopentyl framework to form a more stable tertiary carbocation) $\implies$ Substitution with Rearrangement.

(a) Substitution; (b) Addition; (c) Elimination; (d) Substitution with Rearrangement.

8.15

What is the relationship between the members of following pairs? Are they structural isomers, geometrical isomers, or resonance contributors?

• (a) Structures differing only in the placement of electrons without changing atomic positions $\implies$ Resonance Contributors (Canonical forms).
• (b) Compounds having the same molecular formula but different carbon skeletons or functional group positions $\implies$ Structural Isomers.
• (c) Cis and trans isomers across a restricted double bond $\implies$ Geometrical Isomers.

Relationships classified into resonance contributors, structural isomers, and geometrical isomers.

8.16

For the following bond cleavages, show electron flow using curved arrows and classify as homolysis or heterolysis. Identify intermediate produced: (a) CH₃O—OCH₃ → 2CH₃O•; (b) >C=O + OH⁻ → >C(OH)—O⁻; (c) (CH₃)₃C—Cl → (CH₃)₃C⁺ + Cl⁻; (d) CH₃—Li → CH₃⁻ + Li⁺.

• (a) Peroxide bond breaks symmetrically with single-barbed fish-hook arrows $\implies$ Homolysis, producing Alkoxy Free Radicals ($CH_3O^\bullet$).
• (b) Heterolytic shift of carbonyl $\pi$-electrons to oxygen as $OH^-$ attacks $\implies$ Heterolysis, producing an Alkoxide Anion intermediate.
• (c) Bonding pair leaves entirely with electronegative chlorine $\implies$ Heterolysis, producing tert-Butyl Carbocation ($(CH_3)_3C^+$) and $Cl^-$.
• (d) Bonding pair remains with more electronegative carbon $\implies$ Heterolysis, producing Methyl Carbanion ($CH_3^-$) and $Li^+$.

(a) Homolysis (Free radical); (b) Heterolysis (Anion); (c) Heterolysis (Carbocation); (d) Heterolysis (Carbanion).

8.17

Explain Inductive and Electromeric effects. Which electron displacement effect explains the acidity orders: (a) Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH; (b) CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH?

• Inductive Effect: Permanent polarisation of a $\sigma$-bond transmitted along a carbon chain due to electronegativity differences.
• Electromeric Effect: Temporary complete transfer of a shared pair of $\pi$-electrons to one atom of a multiple bond in the presence of an attacking reagent.

• Explanation of Acidity Orders:
Both trends are governed by the Inductive Effect on the stability of the conjugate carboxylate anion ($RCOO^-$):
(a) Chlorine has a powerful —I (electron-withdrawing) effect. More chlorine atoms withdraw electron density more intensely, dispersing the negative charge on the carboxylate group, stabilizing it and increasing acid strength ($Cl_3C- > Cl_2CH- > ClCH_2-$).
(b) Alkyl groups have an +I (electron-donating) effect. Increasing methyl substitution pushes electron density onto the carboxylate group, destabilizing the anion and decreasing acid strength ($CH_3CH_2- < (CH_3)_2CH- < (CH_3)_3C-$).

Both acidity trends are explained by the Inductive effect (—I stabilizes conjugate base, +I destabilizes it).

8.18

Give a brief description of the principles of: (a) Crystallisation; (b) Distillation; (c) Chromatography.

• (a) Crystallisation: Based on the marked difference in solubility of a solid organic compound and its impurities in a suitable solvent at different temperatures (sparingly soluble at room temp, highly soluble at boiling temp; on cooling, pure crystals precipitate).
• (b) Distillation: Based on the difference in boiling points of volatile components in a liquid mixture. The lower-boiling liquid vaporizes first, and its vapors are condensed and collected separately.
• (c) Chromatography: Based on the differential distribution (adsorption or partition) of components of a mixture between a stationary phase (solid or liquid) and a moving mobile phase (liquid or gas).

Fundamental principles summarized for crystallisation, distillation, and chromatography.

8.19

Describe the method which can be used to separate two compounds with different solubilities in a solvent S.

The method employed is Fractional Crystallisation.
1. The mixture is dissolved in the minimum volume of hot solvent S to prepare a nearly saturated solution.
2. Upon gradual cooling, the less soluble compound crystallises out first, while the more soluble compound remains dissolved in the mother liquor.
3. The crystals are separated by filtration.
4. The remaining filtrate (mother liquor) is further concentrated and cooled to crystallise the second, more soluble compound.

Separated by Fractional Crystallisation based on solubility differences at different temperatures.

8.20

What is the difference between distillation, distillation under reduced pressure, and steam distillation?

• Simple Distillation: Applied to stable volatile liquids containing non-volatile impurities or liquid pairs with large boiling point differences ($\Delta T_b > 25\text{ K}$) at normal atmospheric pressure.
• Distillation under Reduced Pressure (Vacuum Distillation): Applied to high-boiling liquids that decompose at or below their normal boiling point. Lowering external pressure with a vacuum pump causes the liquid to boil at a safe, much lower temperature.
• Steam Distillation: Applied to substances that are steam-volatile and completely immiscible with water. The mixture boils when $p_{\text{organic}} + p_{\text{water}} = p_{\text{atm}}$, enabling the organic liquid to distil at a temperature well below $100^\circ\text{C}$.

Distillation (stable liquids, ΔTb > 25K); Vacuum (decomposing at Tb); Steam (immiscible, steam-volatile).

8.21

Discuss the chemistry of Lassaigne's test.

Organic compounds contain covalent bonds. Heating with metallic sodium converts covalent elements into ionic water-soluble salts:
1. Nitrogen: $Na + C + N \rightarrow NaCN$
Detected by adding $FeSO_4$, heating with conc. $H_2SO_4$: forms Prussian blue ferric ferrocyanide: $$6CN^- + Fe^{2+} \rightarrow [Fe(CN)_6]^{4-}; \quad 4Fe^{3+} + 3[Fe(CN)_6]^{4-} \rightarrow \mathbf{Fe_4[Fe(CN)_6]_3 \cdot xH_2O}$$ 2. Sulphur: $2Na + S \rightarrow Na_2S$
Detected with sodium nitroprusside forming violet complex: $S^{2-} + [Fe(CN)_5NO]^{2-} \rightarrow \mathbf{[Fe(CN)_5NOS]^{4-}}$ (violet).
3. Halogens: $Na + X \rightarrow NaX$
Acidified with $HNO_3$ and treated with $AgNO_3$: forms $AgX$ precipitate (white $AgCl$, pale yellow $AgBr$, yellow $AgI$).

Conversion of covalent N, S, X to ionic NaCN, Na₂S, NaX followed by characteristic precipitation/colour reactions.

8.22

Differentiate between the principle of estimation of nitrogen by: (i) Dumas method and (ii) Kjeldahl's method.

• Dumas Method: The nitrogenous compound is completely oxidised by heating with dry copper(II) oxide in a $CO_2$ atmosphere. All nitrogen is converted into elemental dinitrogen gas (N₂). Carbon and hydrogen are oxidised to $CO_2$ and $H_2O$. $CO_2$ is absorbed in concentrated $KOH$, and the volume of dry $N_2$ gas collected in a nitrometer is measured at STP.
• Kjeldahl's Method: The nitrogenous compound is digested with concentrated $H_2SO_4$ (with $CuSO_4/K_2SO_4$ catalyst), converting nitrogen into ammonium sulphate ((NH₄)₂SO₄). The solution is treated with excess $NaOH$ to liberate $NH_3$ gas, which is absorbed in a known excess of standard acid and back-titrated with standard alkali.

Dumas measures volume of free N₂ gas; Kjeldahl converts nitrogen into ammonia and measures it titrimetrically.

8.23

Discuss the principle of estimation of halogens, sulphur, and phosphorus in an organic compound.

• Halogens (Carius Method): A known mass of compound is heated with fuming $HNO_3$ in the presence of $AgNO_3$ in a sealed Carius tube. Halogen forms silver halide ($AgX$), which is filtered, washed, dried, and weighed.
• Sulphur (Carius Method): The compound is heated with fuming $HNO_3$ or sodium peroxide, oxidising sulphur to $H_2SO_4$. Barium chloride is added to precipitate insoluble barium sulphate ($BaSO_4$), which is filtered, ignited, and weighed.
• Phosphorus: The compound is oxidised with fuming $HNO_3$ to phosphoric acid ($H_3PO_4$), and precipitated as yellow ammonium phosphomolybdate ((NH₄)₃PO₄·12MoO₃) with ammonium molybdate, or as $Mg_2P_2O_7$ (magnesium pyrophosphate).

Halogens weighed as AgX; Sulphur weighed as BaSO₄; Phosphorus weighed as (NH₄)₃PO₄·12MoO₃ or Mg₂P₂O₇.

8.24

Explain the principle of paper chromatography.

Paper chromatography is a type of partition chromatography.
The special chromatography paper contains water trapped within its cellulose matrix, which functions as the stationary liquid phase. An organic solvent or solvent mixture acts as the mobile phase and ascends the paper by capillary action over a spotted sample.
Components of the mixture partition continuously between the stationary water phase and the moving mobile solvent phase according to their partition coefficients. Compounds with higher solubility in the mobile phase travel further up the paper ($R_f$ value), separating into distinct spots.

Continuous partition of solutes between stationary water trapped in paper and the ascending mobile solvent.

8.25

Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?

If nitrogen or sulphur is present in the organic compound, the sodium extract contains $NaCN$ or $Na_2S$.
If $AgNO_3$ were added directly without $HNO_3$, silver ions would react with $CN^-$ and $S^{2-}$ to produce precipitates of silver cyanide ($AgCN$, white) and silver sulphide ($Ag_2S$, black): $$Ag^+ + CN^- \rightarrow AgCN\downarrow \quad \text{and} \quad 2Ag^+ + S^{2-} \rightarrow Ag_2S\downarrow$$ These precipitates would interfere with and obscure the test for halogens.
Boiling with concentrated $HNO_3$ decomposes cyanide and sulphide into volatile gases ($HCN\uparrow$ and $H_2S\uparrow$), preventing interference: $$NaCN + HNO_3 \rightarrow NaNO_3 + HCN\uparrow$$ $$Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S\uparrow$$

Boiling with HNO₃ destroys NaCN and Na₂S as volatile HCN and H₂S, preventing interference with AgX.

8.26

Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.

In organic compounds, nitrogen, sulphur, and halogens are bound through covalent bonds and do not exist as free ions in aqueous solution. Therefore, traditional ionic precipitation tests (such as $AgNO_3$ or $FeCl_3$) fail directly on organic molecules.
Fusion with highly reactive metallic sodium breaks the covalent bonds and converts the elements into water-soluble ionic salts ($NaCN, Na_2S, NaX$), which readily dissociate in water to yield free ions ($CN^-, S^{2-}, X^-$) for standard qualitative detection.

Converts non-ionizing covalent bonds into water-soluble ionic sodium salts (NaCN, Na₂S, NaX).

8.27

Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.

The suitable technique is Sublimation.
Camphor is a sublimable solid that directly vaporizes on heating without melting, whereas calcium sulphate ($CaSO_4$) is non-volatile and stable. Upon gentle heating, camphor sublimes and condenses on the cool walls of an inverted funnel as pure crystals, leaving calcium sulphate behind in the dish.

Sublimation (camphor sublimes cleanly while CaSO₄ remains behind).

8.28

Explain why an organic liquid vaporises at a temperature below its boiling point in its steam distillation.

In steam distillation, the total vapour pressure ($P$) above the boiling immiscible mixture is the sum of the partial vapour pressures of the organic liquid ($p_1$) and steam/water ($p_2$): $$\mathbf{P = p_1 + p_2}$$ Boiling occurs when the total vapour pressure reaches atmospheric pressure ($P = 1\text{ atm} = 760\text{ mm Hg}$).
Since $p_2$ (water vapour pressure) contributes significantly to $P$, the partial vapour pressure of the organic liquid required ($p_1 = P - p_2$) is much less than atmospheric pressure. Therefore, the organic liquid boils and vaporises at a temperature well below its normal boiling point (and below $100^\circ\text{C}$).

Total pressure P = p₁ + p₂ reaches 1 atm when p₁ < 1 atm, causing distillation below normal Tb.

8.29

Will CCl₄ give a white precipitate of AgCl on heating with silver nitrate? Give reason for your answer.

No, CCl₄ will not give a precipitate of AgCl.
Carbon tetrachloride ($CCl_4$) is a purely covalent, non-polar compound. The four $C-Cl$ bonds are strong and do not ionize in water to generate chloride ions ($Cl^-$). Since no free $Cl^-$ ions exist in solution, reaction with $Ag^+$ cannot occur ($Ag^+ + Cl^- \rightarrow AgCl\downarrow$ requires ionic chloride).

No; CCl₄ is purely covalent and does not dissociate to furnish free Cl⁻ ions.

8.30

Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?

Potassium hydroxide ($KOH$) is a strong base that chemically reacts rapidly and quantitatively with acidic carbon dioxide gas to form soluble potassium carbonate: $$2KOH + CO_2 \rightarrow K_2CO_3 + H_2O$$ $KOH$ is specifically chosen over $NaOH$ because $KOH$ and its reaction product $K_2CO_3$ are highly soluble and do not crystallise or clog the glass tubes, ensuring smooth, complete absorption of $CO_2$.

KOH is strongly basic, quantitatively absorbing acidic CO₂ to form soluble K₂CO₃ without clogging.

8.31

Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?

Lead acetate test: $Na_2S + (CH_3COO)_2Pb \rightarrow PbS\downarrow (\text{black}) + 2CH_3COONa$.
If sulphuric acid ($H_2SO_4$) were used instead of acetic acid ($CH_3COOH$), the sulphate ions ($SO_4^{2-}$) from $H_2SO_4$ would react with lead acetate to form an insoluble white precipitate of lead sulphate (PbSO₄): $$Pb^{2+} + SO_4^{2-} \rightarrow PbSO_4\downarrow (\text{white})$$ This white precipitate would mask and interfere with the detection of the black lead sulphide ($PbS$) precipitate. Acetic acid does not precipitate lead ions.

H₂SO₄ would precipitate white PbSO₄ with lead acetate, masking the black PbS test.

8.32

An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of CO₂ and H₂O produced when 0.20 g of this substance is subjected to complete combustion.

Mass of compound $m = 0.20\text{ g}$.
1. Mass of CO₂:
$$\%C = \frac{12}{44} \times \frac{m_{\text{CO}_2}}{m} \times 100 \implies 69 = \frac{12}{44} \times \frac{m_{\text{CO}_2}}{0.20} \times 100$$ $$m_{\text{CO}_2} = \frac{69 \times 44 \times 0.20}{12 \times 100} = \frac{607.2}{1200} = \mathbf{0.506\text{ g}}$$
2. Mass of H₂O:
$$\%H = \frac{2}{18} \times \frac{m_{\text{H}_2\text{O}}}{m} \times 100 \implies 4.8 = \frac{2}{18} \times \frac{m_{\text{H}_2\text{O}}}{0.20} \times 100$$ $$m_{\text{H}_2\text{O}} = \frac{4.8 \times 18 \times 0.20}{2 \times 100} = \frac{17.28}{200} = \mathbf{0.0864\text{ g}}$$

Mass of CO₂ = 0.506 g; Mass of H₂O = 0.0864 g.

8.33

A sample of 0.50 g of an organic compound was treated according to Kjeldahl's method. The ammonia evolved was absorbed in 50 mL of 0.5 M H₂SO₄. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen.

Initial moles of $H_2SO_4 = 50\text{ mL} \times 0.5\text{ M} = 25\text{ mmol} \implies 50\text{ meq}$.
Unreacted $H_2SO_4$ neutralized by $NaOH = 60\text{ mL} \times 0.5\text{ M} = 30\text{ mmol} = 30\text{ meq}$.
Equivalents of $H_2SO_4$ consumed by $NH_3 = 50 - 30 = \mathbf{20\text{ meq}}$.
Equivalents of $NH_3 = 20\text{ meq} = 20\text{ mmol of } N$.
Mass of nitrogen $= 20 \times 10^{-3}\text{ mol} \times 14\text{ g mol}^{-1} = \mathbf{0.28\text{ g}}$.
$$\%N = \frac{0.28}{0.50} \times 100 = \mathbf{56.0\%}$$

Percentage of nitrogen = 56.0%.

8.34

0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present.

Molar mass of $AgCl = 108 + 35.5 = 143.5\text{ g mol}^{-1}$.
$$\%Cl = \frac{35.5}{143.5} \times \frac{m_{AgCl}}{m} \times 100 = \frac{35.5}{143.5} \times \frac{0.5740}{0.3780} \times 100 = \frac{20.377}{54.243} \times 100 = \mathbf{37.57\%}$$

Percentage of chlorine = 37.57%.

8.35

In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find the percentage of sulphur.

Molar mass of $BaSO_4 = 137 + 32 + 64 = 233\text{ g mol}^{-1}$.
$$\%S = \frac{32}{233} \times \frac{m_{BaSO_4}}{m} \times 100 = \frac{32}{233} \times \frac{0.668}{0.468} \times 100 = \frac{21.376}{109.044} \times 100 = \mathbf{19.60\%}$$

Percentage of sulphur = 19.60%.

8.36

In the organic compound CH₂=CH—CH₂—CH₂—C≡CH, the pair of hybridised orbitals involved in the formation of C₂—C₃ bond is: (a) sp—sp²; (b) sp³—sp³; (c) sp²—sp³; (d) sp³—sp.

Numbering starts from the double bond end (locant rule gives preference to double bond over triple bond when both have identical locant positions 1 and 5):
$\overset{1}{C}H_2=\overset{2}{C}H-\overset{3}{C}H_2-\overset{4}{C}H_2-\overset{5}{C}\equiv \overset{6}{C}H$.
Carbon 2 is $sp^2$ hybridised; Carbon 3 is $sp^3$ hybridised.
Therefore, the $C_2-C_3$ bond is formed by the overlap of $\mathbf{sp^2-sp^3}$ hybrid orbitals.

Correct option: (c) sp²—sp³.

8.37

In Lassaigne's test for nitrogen, the Prussian blue colour is obtained due to formation of: (a) Na₄[Fe(CN)₆]; (b) Fe₄[Fe(CN)₆]₃; (c) Fe₂(SO₄)₃; (d) Fe₃[Fe(CN)₆]₂.

Sodium cyanide formed during fusion reacts with $FeSO_4$ to yield sodium hexacyanoferrate(II). Upon addition of conc. $H_2SO_4$, some $Fe^{2+}$ oxidises to $Fe^{3+}$, forming insoluble Prussian blue iron(III) hexacyanoferrate(II): $$\mathbf{Fe_4[Fe(CN)_6]_3 \cdot xH_2O}$$

Correct option: (b) Fe₄[Fe(CN)₆]₃.

8.38

Which of the following carbocations is most stable? (a) (CH₃)₃C—C⁺H₂; (b) (CH₃)₃C⁺; (c) CH₃CH₂C⁺H₂; (d) CH₃C⁺HCH₂CH₃.

• (a) Neopentyl cation: primary with 0 α-H.
• (b) $\mathbf{(CH_3)_3C^+}$: Tertiary carbocation with 9 α-hydrogens providing extensive hyperconjugation and three electron-donating methyl (+I) groups.
• (c) Propyl cation: primary with 2 α-H.
• (d) sec-Butyl cation: secondary with 5 α-H.
Tertiary carbocation (b) is the most stable.

Correct option: (b) (CH₃)₃C⁺.

8.39

The reaction CH₃CH₂I + KOH(aq) → CH₃CH₂OH + KI is classified as: (a) Electrophilic substitution; (b) Nucleophilic substitution; (c) Elimination; (d) Addition.

Hydroxide ion ($OH^-$) is an electron-rich nucleophile with lone pairs. It attacks the partially positive $\alpha$-carbon and displaces the iodide ion ($I^-$) leaving group: $$HO^- + CH_3CH_2-I \rightarrow CH_3CH_2-OH + I^-$$ This is a classic Nucleophilic Substitution ($S_N2$) reaction.

Correct option: (b) Nucleophilic substitution.

Organic Chemistry High-Yield Revision Matrix

Essential master tables for IUPAC seniorities, electronic displacement rules, purification technique chooser, and quantitative elemental percentage formulas for CBSE Board, NEET, and JEE Main.

1. IUPAC Principal Functional Group Seniority Table

Functional Class	Functional Group Structure	Prefix (when subordinate)	Suffix (when principal)
Carboxylic acid	$-COOH$	carboxy-	-oic acid
Sulphonic acid	$-SO_3H$	sulpho-	-sulphonic acid
Ester	$-COOR$	alkoxycarbonyl- / carbomethoxy-	-oate
Acid Halide	$-COX$	halocarbonyl-	-oyl halide
Acid Amide	$-CONH_2$	carbamoyl-	-amide
Nitrile	$-CN$	cyano-	-nitrile
Aldehyde	$-CHO$	formyl- / oxo-	-al
Ketone	$>C=O$	oxo-	-one
Alcohol	$-OH$	hydroxy-	-ol
Amine	$-NH_2$	amino-	-amine
Alkene	$>C=C<$	—	-ene
Alkyne	$-C\equiv C-$	—	-yne

2. Master Quantitative Elemental Analysis Formulas

Target Element	Analytical Method	Mathematical Formula	Special Notes & Limitations
Carbon (C)	Liebig Combustion	$\%C = \frac{12}{44} \times \frac{m_{\text{CO}_2}}{m} \times 100$	$CO_2$ is absorbed in concentrated $KOH$ bulbs.
Hydrogen (H)	Liebig Combustion	$\%H = \frac{2}{18} \times \frac{m_{\text{H}_2\text{O}}}{m} \times 100$	$H_2O$ is absorbed in anhydrous $CaCl_2$ U-tube.
Nitrogen (N)	Dumas Method	$\%N = \frac{28}{22400} \times \frac{V_{\text{STP}}}{m} \times 100$	$V_{\text{STP}} = \frac{(P - p') \cdot V_1 \cdot 273}{760 \cdot T_1}$ ($p' =$ aqueous tension).
Nitrogen (N)	Kjeldahl Method	$\%N = \frac{1.4 \times N \times V_{\text{acid}}}{m}$	Fails for nitro ($-NO_2$), azo ($-N=N-$), and ring nitrogen (pyridine).
Halogen (X)	Carius Method	$\%X = \frac{\text{At. Wt } X}{\text{Mol. Wt } AgX} \times \frac{m_{AgX}}{m} \times 100$	$AgCl = 143.5$, $AgBr = 188$, $AgI = 235\text{ g/mol}$.
Sulphur (S)	Carius Method	$\%S = \frac{32}{233} \times \frac{m_{BaSO_4}}{m} \times 100$	Precipitated and weighed as $BaSO_4$ ($233\text{ g/mol}$).
Phosphorus (P)	Phosphomolybdate	$\%P = \frac{31}{1877} \times \frac{m_{\text{ppt}}}{m} \times 100$	Precipitated as $(NH_4)_3PO_4 \cdot 12MoO_3$ ($1877\text{ g/mol}$).

Class 11 Chemistry Chapter 8: Organic Chemistry Tests

3 Graded Test Levels: Foundation (CBSE Board fundamentals & nomenclature), Intermediate (Mechanisms, inductive/resonance, and purification), and Advanced (NEET / JEE Main competitive quantitative analysis & structural logic). Select options to evaluate yourself.

Level 1: Foundation Test (CBSE Board Level)

F1What is the hybridisation of carbon in ethyne (HC≡CH)?

sp³ is tetrahedral.

Correct: Carbon involved in a triple bond forms two σ-bonds and two π-bonds (sp hybridised, linear, 180°).

sp² is in alkenes.

Carbon lacks d-orbitals in its valence shell.

F2In Lassaigne's test, which compound is responsible for the violet colour in the test for sulphur?

Correct: Sodium nitroprusside reacts with sulphide ions to form the violet complex [Fe(CN)₅NOS]⁴⁻.

That is Prussian blue for nitrogen.

PbS is black.

That is blood-red for both N and S.

F3The IUPAC name of CH₃COCH₂CH₃ is:

Ketones cannot exist at C1.

Butanal is an aldehyde.

Correct: 4-carbon chain with a ketone carbonyl at C2.

That is the common name, not IUPAC.

Level 2: Intermediate Test (Application & Mechanism Level)

I1Which of the following carbocations has the MAXIMUM number of hyperconjugative structures?

0 α-hydrogens.

3 α-hydrogens.

6 α-hydrogens.

Correct: tert-Butyl carbocation has 9 α-hydrogens, yielding 9 hyperconjugative canonical forms!

I2In steam distillation, the mixture boils when:

That occurs only in normal simple distillation.

Correct: Immiscible liquids exert independent vapour pressures; boiling occurs when their sum equals atmospheric pressure.

That is the boiling point of pure water (100 °C).

Steam distillation occurs below 100 °C.

Level 3: Advanced Test (NEET / JEE Main Quantitative Level)

A10.2 g of an organic compound on Kjeldahl analysis evolved ammonia that neutralized 15 mL of 0.1 M H₂SO₄. The percentage of nitrogen in the compound is:

Incorrect arithmetic.

Did not account for basicity of H₂SO₄.

Correct: N = 0.1 M × 2 = 0.2 N. %N = (1.4 × N × V) / m = (1.4 × 0.2 × 15) / 0.2 = 4.2 / 0.2 = 21.0%.

Incorrect.

A2Which of the following carboxylic acids has the HIGHEST acid strength (lowest pKa)?

Acetic acid has no —I substituent.

Correct: Nitro group (—NO₂) exerts the strongest —I (electron-withdrawing) effect, greatly dispersing negative charge on the carboxylate anion.

Chlorine —I is weaker than —NO₂.

Fluorine is strong, but nitro (—NO₂) has even higher group electronegativity and —I power.