Alkanes, Alkenes, Alkynes & Aromatic Arenes. Conformations of ethane (Newman & Sawhorse projections, torsional strain), catalytic hydrogenation, Wurtz reaction, Kolbe's electrolysis, Markovnikov's rule and anti-Markovnikov peroxide effect (Kharasch effect), acidity of terminal alkynes, cyclic trimerisation to benzene, Kekulé resonance, Hückel's $(4n+2)\pi$ rule of aromaticity, electrophilic aromatic substitution ($S_E$) mechanisms (nitration, halogenation, sulphonation, Friedel-Crafts alkylation & acylation), directive influence of functional groups, solved in-text problems 9.19.14, full NCERT exercises 9.19.25, revision matrix, and 3-level chapter tests.
🎯 Learning Objectives
After mastering this unit, you will be able to:
Name hydrocarbons according to IUPAC system of nomenclature.
Recognise and write structures of isomers of alkanes, alkenes, alkynes, and aromatic hydrocarbons.
Learn various preparation methods of aliphatic and aromatic hydrocarbons.
Distinguish between alkanes, alkenes, alkynes, and arenes on the basis of physical and chemical properties.
Draw, differentiate, and analyse the stability of conformations of ethane (Sawhorse and Newman projections).
Appreciate the role of hydrocarbons as major sources of energy (LPG, CNG, LNG, petroleum) and polymers.
Predict addition products of unsymmetrical alkenes/alkynes using Markovnikov's rule and the anti-Markovnikov peroxide effect (Kharasch effect).
Comprehend the structure of benzene, explain aromaticity using Hückel's $(4n+2)\pi$ rule, and master the mechanism of electrophilic aromatic substitution ($S_E$).
Predict the directive influence (ortho/para vs meta) of substituents on a monosubstituted benzene ring.
Understand carcinogenicity and toxicity of polynuclear aromatic hydrocarbons.
Introduction to Hydrocarbons
The term hydrocarbon is self-explanatory: it denotes organic compounds composed exclusively of carbon and hydrogen. Hydrocarbons play an indispensable role in daily life and industrial technology as vital fuels, lubricants, and chemical feedstocks.
Key Fuels from Petroleum & Gas
LPG (Liquefied Petroleum Gas): Domestic fuel consisting predominantly of propane and butane. Clean burning with minimal pollution.
CNG (Compressed Natural Gas): Methane compressed at high pressures for automobiles; significantly lowers greenhouse emissions.
Petrol, Diesel & Kerosene: Fractions obtained from the fractional distillation of crude petroleum.
Industrial Applications
Polymers: Manufacture of polythene, polypropylene, polystyrene, PVC, and synthetic rubbers.
Solvents & Feedstocks: Higher alkanes act as solvents in paints and varnishes. Arenes are starting materials for pharmaceuticals, dyes, explosives, and fragrances.
9.1 Classification of Hydrocarbons
Depending upon the types of carbon-carbon bonds and structure, hydrocarbons are broadly categorized into three classes:
Saturated Hydrocarbons: Contain only carbon-carbon and carbon-hydrogen single $\sigma$ bonds. If open-chain, they are called alkanes (general formula: $\mathbf{C_n H_{2n+2}}$). If closed rings, they are cycloalkanes ($\mathbf{C_n H_{2n}}$).
Unsaturated Hydrocarbons: Contain carbon-carbon multiple bonds. Alkenes contain at least one $C=C$ double bond ($\mathbf{C_n H_{2n}}$); alkynes contain at least one $C\equiv C$ triple bond ($\mathbf{C_n H_{2n-2}}$).
Aromatic Hydrocarbons (Arenes): Special cyclic planar unsaturated compounds that exhibit resonance stabilization conforming to Hückel's Rule of $(4n+2)\pi$ electrons.
9.2 Alkanes (Paraffins)
Alkanes are open-chain saturated hydrocarbons where carbon atoms are $sp^3$ hybridised. Because of their relative inertness towards acids, bases, oxidising agents, and reducing agents under normal room conditions, they were historically designated as paraffins (Latin: parum = little, affinis = affinity).
Structure of Methane & Alkane Geometry
In methane ($\text{CH}_4$), the central carbon undergoes $sp^3$ hybridisation. Four $sp^3$ hybrid orbitals overlap head-on with $1s$ orbitals of four hydrogen atoms to form four equivalent $CH$ $\sigma$ bonds directed towards the four corners of a regular tetrahedron. The $H-C-H$ bond angle is $109.5^\circ$ ($109^\circ 28'$). In higher alkanes, $CC$ bond length is $154\text{ pm}$ and $CH$ bond length is $112\text{ pm}$.
9.2.1 Nomenclature & Isomerism in Alkanes
Methane ($\text{CH}_4$), ethane ($\text{C}_2\text{H}_6$), and propane ($\text{C}_3\text{H}_8$) have only one possible structural arrangement. Butane ($\text{C}_4\text{H}_{10}$) can exist in two chain isomeric forms:
Pentane ($\text{C}_5\text{H}_{12}$) exhibits three chain isomers: $n$-pentane (b.p. $309\text{ K}$), $2$-methylbutane / isopentane (b.p. $301\text{ K}$), and $2,2$-dimethylpropane / neopentane (b.p. $282.5\text{ K}$). As molecular size increases, the number of structural isomers escalates rapidly: $\text{C}_6\text{H}_{14}$ has $5$, $\text{C}_7\text{H}_{16}$ has $9$, and $\text{C}_{10}\text{H}_{22}$ has $75$ isomers.
Types of Carbon Atoms ($1^\circ, 2^\circ, 3^\circ, 4^\circ$)
Primary ($1^\circ$) Carbon: Attached to no other carbon (as in methane) or to only one other carbon atom (terminal carbons are always primary).
Secondary ($2^\circ$) Carbon: Attached directly to two other carbon atoms.
Tertiary ($3^\circ$) Carbon: Attached directly to three other carbon atoms.
Quaternary ($4^\circ$) Carbon: Attached directly to four other carbon atoms.
9.1
Write structures of different chain isomers of alkanes corresponding to the molecular formula $\text{C}_6\text{H}_{14}$. Also write their IUPAC names.
The five structural isomers of $\text{C}_6\text{H}_{14}$ are:
Write structures of different isomeric alkyl groups corresponding to the molecular formula $-\text{C}_5\text{H}_{11}$. Write IUPAC names of alcohols obtained by attachment of $-\text{OH}$ groups at different carbons of the chain.
(ii) $(\text{CH}_3)_2\text{CH}\text{CH}_2\text{CH}_2\text{CH}(\text{CH}_3)_2$
9.5
Why are the given names incorrect? Write correct IUPAC names: (i) 2-Ethylpentane, (ii) 5-Ethyl-3-methylheptane.
(i) 2-Ethylpentane: The longest continuous carbon chain has 6 carbons (not 5). Correct name: 3-Methylhexane.
(ii) 5-Ethyl-3-methylheptane: Numbering must start from the end giving lower locant to the substituent with alphabetical priority. Numbering from left gives locants 3, 5 (ethyl at 3, methyl at 5). Correct name: 3-Ethyl-5-methylheptane.
9.2.2 Methods of Preparation of Alkanes
1. From Unsaturated Hydrocarbons (Catalytic Hydrogenation / Sabatier-Senderens)
Dihydrogen gas adds to alkenes and alkynes in the presence of finely divided metal catalysts like $\text{Pt}$, $\text{Pd}$, or $\text{Ni}$:
Note: $\text{Pt}$ and $\text{Pd}$ catalyse at room temperature; $\text{Ni}$ requires higher temperature ($\approx 523\text{}573\text{ K}$) and pressure.
(b) Wurtz Reaction: Alkyl halides on treatment with sodium metal in dry ethereal solution give higher symmetrical alkanes containing an even number of carbon atoms:
Limitation of Wurtz Reaction: If two different alkyl halides ($\text{RX}$ and $\text{R'X}$) are taken, a mixture of three alkanes ($\text{RR}, \text{RR'}, \text{R'R'}$) is obtained, which have very close boiling points and are virtually impossible to separate cleanly. Methane cannot be synthesized by Wurtz reaction.
3. From Carboxylic Acids
(a) Decarboxylation with Soda Lime: Heating sodium salts of carboxylic acids with soda lime ($\text{NaOH} + \text{CaO}$ in $3:1$ ratio) eliminates carbon dioxide, forming an alkane with one carbon atom less than the carboxylic acid:
Role of CaO: Quicklime ($\text{CaO}$) keeps $\text{NaOH}$ dry (prevents deliquescence) and raises the fusion temperature.
(b) Kolbe's Electrolytic Method: Electrolysis of a concentrated aqueous solution of sodium or potassium carboxylate yields a symmetrical alkane containing an even number of carbon atoms at the anode:
Note: Methane cannot be prepared by Kolbe's electrolytic method because the coupling requires at least two free radicals ($1+1=2$ carbons).
9.2.3 Physical and Chemical Properties of Alkanes
Physical Properties: Alkanes are non-polar because electronegativity difference between $C$ and $H$ is negligible ($2.5$ vs $2.1$). They are held by weak dispersion (van der Waals) forces. $C_1$ to $C_4$ are colorless gases; $C_5$ to $C_{17}$ are liquids; $C_{18}+$ are waxy solids. Alkanes are hydrophobic (insoluble in water) but dissolve in non-polar organic solvents like benzene and ether ("like dissolves like").
Boiling Point Trends: Boiling points increase regularly with increasing molecular mass ($\approx 20\text{}30\text{ K}$ per $-\text{CH}_2-$ unit) due to increase in molecular surface area and dispersion forces. However, among isomeric alkanes, boiling point decreases with increased branching. Branching makes the molecule more spherical, diminishing surface area of contact and weakening intermolecular attractions: $n\text{-pentane } (309.1\text{ K}) > \text{isopentane } (301\text{ K}) > \text{neopentane } (282.5\text{ K})$.
Chemical Reactions of Alkanes
1. Free Radical Halogenation
Reaction of alkanes with halogens in presence of ultraviolet light ($h\nu$) or heat ($573\text{}773\text{ K}$):
Reactivity order of halogens: $\text{F}_2 > \text{Cl}_2 > \text{Br}_2 > \text{I}_2$. Ease of replacement of $H$: $3^\circ > 2^\circ > 1^\circ$. Fluorination is explosively violent; iodination is extremely slow and reversible, requiring oxidising agents like $\text{HIO}_3$ or $\text{HNO}_3$ to consume $\text{HI}$ ($5\text{HI} + \text{HIO}_3 \to 3\text{I}_2 + 3\text{H}_2\text{O}$).
Three-Step Free Radical Chain Mechanism:
Chain Initiation: Homolytic fission of weak $\text{ClCl}$ bond by light: $\text{ClCl} \xrightarrow{h\nu} 2\text{Cl}^\bullet$
Incomplete Combustion: In limited oxygen, methane forms carbon black used in printer inks and pigments: $\text{CH}_4(g) + \text{O}_2(g) \xrightarrow{\text{limited } \text{O}_2} \text{C}(s) + 2\text{H}_2\text{O}(l)$.
Tertiary CH Oxidation: Alkanes with a $3^\circ$ hydrogen are oxidised by alkaline $\text{KMnO}_4$: $(\text{CH}_3)_3\text{CH} \xrightarrow[\text{Oxidation}]{\text{KMnO}_4} (\text{CH}_3)_3\text{COH}$ ($2$-Methylpropan-$2$-ol).
Aromatisation (Reforming): $n$-Hexane heated to $773\text{ K}$ at $10\text{}20\text{ atm}$ over $\text{Cr}_2\text{O}_3, \text{V}_2\text{O}_5\text{ or }\text{Mo}_2\text{O}_3$ supported on alumina ($\text{Al}_2\text{O}_3$) cyclises and dehydrogenates into benzene. $n$-Heptane yields toluene.
Reaction with Steam: $\text{CH}_4 + \text{H}_2\text{O}(g) \xrightarrow[\Delta, 1273\text{ K}]{\text{Ni}} \text{CO} + 3\text{H}_2$ (Synthesis of syngas/dihydrogen).
Pyrolysis (Cracking): Higher alkanes decompose at high temperatures ($773\text{}973\text{ K}$) into lower alkanes, alkenes, and hydrogen via free radical pathways. $\text{C}_{12}\text{H}_{26} \xrightarrow[\text{Pt/Pd/Ni}]{973\text{ K}} \text{C}_7\text{H}_{16} + \text{C}_5\text{H}_{10} + \dots$
9.2.4 Conformations of Ethane
Because the $\sigma$-molecular orbital in a $CC$ single bond has cylindrical electron density along the internuclear axis, rotation about the single bond is permissible. Spatial arrangements of atoms in space that can interconvert simply by rotation around a carbon-carbon single bond are called conformations, conformers, or rotamers.
However, rotation is not 100% free; it is resisted by a small energy barrier of $1\text{}20\text{ kJ mol}^{-1}$ ($12.5\text{ kJ mol}^{-1}$ in ethane) called torsional strain caused by weak electrostatic repulsive interactions between adjacent bonding electron pairs.
Eclipsed Conformation: Hydrogen atoms on adjacent carbons are positioned as close to each other as possible. Dihedral angle $\theta = 0^\circ$. Maximum electron-cloud repulsions and torsional strain; highest energy, least stable.
Staggered Conformation: Hydrogen atoms are as far apart as possible. Dihedral angle $\theta = 60^\circ$. Minimum repulsive interactions; lowest potential energy, maximum stability.
Skew Conformation: Any intermediate spatial arrangement between eclipsed and staggered ($0^\circ < \theta < 60^\circ$).
Relative Stability: $\text{Staggered} > \text{Skew} > \text{Eclipsed}$. The $12.5\text{ kJ mol}^{-1}$ barrier is readily supplied by thermal kinetic energy at room temperature via molecular collisions, making rotation virtually unrestricted.
9.3 Alkenes (Olefins)
Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond ($>C=C<$). General formula: $\mathbf{C_n H_{2n}}$. Historically termed olefins (oil-forming) because ethene reacts with chlorine to form an oily liquid ($1,2$-dichloroethane).
9.3.1 Structure of Carbon-Carbon Double Bond
Each carbon of the double bond is $sp^2$ hybridised. The $C=C$ double bond consists of:
One strong $\sigma$ bond: Formed by axial/head-on overlap of $sp^2$ hybrid orbitals (bond enthalpy $\approx 397\text{ kJ mol}^{-1}$).
One weaker $\pi$ bond: Formed by lateral/sideways overlap of unhybridised $2p_z$ orbitals (bond enthalpy $\approx 284\text{ kJ mol}^{-1}$).
Total $C=C$ bond enthalpy is $681\text{ kJ mol}^{-1}$, compared to $348\text{ kJ mol}^{-1}$ for $CC$ in ethane. The $C=C$ bond length is $134\text{ pm}$ (substantially shorter than $CC$ at $154\text{ pm}$). The loosely held mobile $\pi$-electrons lie in clouds above and below the molecular plane, making alkenes electron-rich centers readily attacked by electrophiles ($E^+$).
9.3.2 Nomenclature & Isomerism in Alkenes
Alkenes exhibit both Structural Isomerism (Chain & Position) and Stereoisomerism (Geometrical Cis-Trans Isomerism).
Geometrical Isomerism
Because rotation about a $C=C$ double bond is restricted (requires breaking the $\pi$ bond, which demands $\approx 284\text{ kJ mol}^{-1}$), groups attached to the double-bonded carbons have fixed spatial arrangements:
Cis-Isomer: Two identical or similar atoms/groups lie on the same side of the double bond.
Trans-Isomer: Two identical or similar atoms/groups lie on opposite sides of the double bond.
Condition for Geometrical Isomerism: Each doubly bonded carbon must have two different groups attached (e.g. $abC=Cab$ or $abC=Cde$). If either carbon has two identical groups ($aaC=Cxy$), cis-trans isomerism is impossible.
9.8
Calculate total number of $\sigma$ and $\pi$ bonds in: (i) 2,8-dimethyl-deca-3,6-diene, (ii) octatetraene.
Which of the following compounds show geometrical (cis-trans) isomerism? (i) $(\text{CH}_3)_2\text{C}=\text{CH}\text{C}_2\text{H}_5$, (ii) $\text{CH}_2=\text{CBr}_2$, (iii) $\text{C}_6\text{H}_5\text{CH}=\text{CH}\text{CH}_3$, (iv) $\text{CH}_3\text{CH}=\text{CCl}(\text{CH}_3)$.
(i) $(\text{CH}_3)_2\text{C}=\text{CH}\text{C}_2\text{H}_5$ : No (two identical $-\text{CH}_3$ groups on $C1$).
(ii) $\text{CH}_2=\text{CBr}_2$ : No (two identical $-H$ on $C1$).
(iii) $\text{C}_6\text{H}_5\text{CH}=\text{CH}\text{CH}_3$ : Yes (both doubly bonded carbons have two different groups: $-H/-\text{C}_6\text{H}_5$ and $-H/-\text{CH}_3$).
(iv) $\text{CH}_3\text{CH}=\text{CCl}(\text{CH}_3)$ : Yes (both doubly bonded carbons have two different groups: $-H/-\text{CH}_3$ and $-\text{Cl}/-\text{CH}_3$).
9.3.4 Preparation of Alkenes
1. From Alkynes (Stereoselective Reduction)
Lindlar's Catalyst ($\text{Pd}/\text{CaCO}_3$ poisoned with quinoline or sulfur): Partial reduction yields cis-alkenes via syn-addition.
$$\text{RC}\equiv\text{CR} + \text{H}_2 \xrightarrow{\text{Pd/C, quinoline}} \text{cis-Alkene}$$
Birch Reduction ($\text{Na}$ in liquid $\text{NH}_3$): Yields trans-alkenes via anti-addition.
$$\text{RC}\equiv\text{CR} + 2\text{Na} + 2\text{NH}_3 \to \text{trans-Alkene} + 2\text{NaNH}_2$$
2. From Alkyl Halides (Dehydrohalogenation / $\beta$-Elimination)
Alkyl halides heated with alcoholic $\text{KOH}$ eliminate one molecule of halogen acid to form alkenes ($\beta$-elimination):
Diagnostic Test: The reddish-orange color of bromine in $\text{CCl}_4$ is discharged. This confirms the presence of carbon-carbon unsaturation.
2. Addition of Hydrogen Halides: Markovnikov's Rule vs Peroxide Effect
Markovnikov's Rule: When an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part of the addendum attaches to that doubly bonded carbon atom which carries the lesser number of hydrogen atoms.
$\text{Br}^-$ nucleophile rapidly attacks the more stable $2^\circ$ carbocation forming 2-bromopropane.
Anti-Markovnikov Addition (Peroxide Effect / Kharasch Effect): In the presence of organic peroxides [e.g. benzoyl peroxide $(\text{C}_6\text{H}_5\text{CO})_2\text{O}_2$], addition of $\mathbf{HBr}$ takes place contrary to Markovnikov's rule:
$\text{HCl}$ bond is too strong ($430.5\text{ kJ mol}^{-1}$) to be cleaved homolytically by peroxide radicals.
$\text{HI}$ bond is weak ($296.8\text{ kJ mol}^{-1}$), but iodine free radicals ($\text{I}^\bullet$) rapidly combine with each other to form $\text{I}_2$ molecules instead of adding to the double bond.
Ozone ($\text{O}_3$) adds to the double bond forming a cyclic ozonide intermediate, which is cleaved by zinc dust and water ($\text{Zn} + \text{H}_2\text{O}$) to yield aldehydes and/or ketones. Highly valuable for locating the position of double bonds in unknown alkenes!
Role of Zinc Dust: Zinc consumes hydrogen peroxide ($\text{H}_2\text{O}_2$) formed during the reaction ($\text{Zn} + \text{H}_2\text{O}_2 \to \text{ZnO} + \text{H}_2\text{O}$), preventing further oxidation of aldehydes into carboxylic acids.
Alkynes are unsaturated acyclic hydrocarbons containing at least one carbon-carbon triple bond ($-C\equiv C-$). General formula: $\mathbf{C_n H_{2n-2}}$. The simplest stable member is ethyne ($\text{C}_2\text{H}_2$), commonly known as acetylene (used with oxygen in oxyacetylene flame for high-temperature welding).
9.4.2 Structure of Triple Bond in Ethyne
In ethyne ($\text{HC}\equiv\text{CH}$), each carbon is $sp$ hybridised with linear geometry ($H-C-C$ angle is $180^\circ$). The triple bond comprises:
One strong $\sigma$ bond: Formed by axial head-on overlap of two $sp$ hybrid orbitals.
Two $\pi$ bonds: Formed by lateral overlap of mutually perpendicular unhybridised $2p_y$ and $2p_z$ orbitals.
The $\pi$-electron clouds merge to form a cylindrical sleeve of electron density surrounding the internuclear axis. The $C\equiv C$ bond length is very short ($120\text{ pm}$), and bond enthalpy is high ($823\text{ kJ mol}^{-1}$).
From Vicinal Dihalides:
$$\text{CH}_2\text{Br}\text{CH}_2\text{Br} \xrightarrow[\Delta]{\text{alc. KOH}} \text{CH}_2=\text{CHBr} \xrightarrow{\text{NaNH}_2} \text{CH}\equiv\text{CH}$$
9.4.4 Chemical Properties of Alkynes
A. Acidic Character of Terminal Alkynes
Because the carbon in ethyne is $sp$ hybridised with $50\%$ $s$-character (compared to $33.3\%$ in $sp^2$ ethene and $25\%$ in $sp^3$ ethane), its electronegativity is the highest. It strongly attracts the shared electron pair of the $CH$ bond, enabling the hydrogen to be released as a proton ($H^+$) upon reaction with strong bases like sodium metal ($\text{Na}$) or sodamide ($\text{NaNH}_2$):
Order of Acidic Strength: $\text{HC}\equiv\text{CH} > \text{CH}_2=\text{CH}_2 > \text{CH}_3\text{CH}_3$ and $\text{HC}\equiv\text{CH} > \text{CH}_3\text{C}\equiv\text{CH} \gg \text{CH}_3\text{C}\equiv\text{C}\text{CH}_3$ (non-terminal alkynes have no acetylenic hydrogen and are non-acidic).
Hydrogen Halides: Adds in two steps conforming to Markovnikov's rule to yield geminal dihalides:
$$\text{CH}_3\text{C}\equiv\text{CH} + \text{HBr} \to \text{CH}_3\text{CBr}=\text{CH}_2 \xrightarrow{\text{HBr}} \text{CH}_3\text{CBr}_2\text{CH}_3 \quad (\text{2,2-Dibromopropane})$$
Hydration (Addition of Water): Alkynes warm with $1\%\ \text{HgSO}_4$ and $40\%\ \text{H}_2\text{SO}_4$ at $333\text{ K}$ to yield enols that tautomerize into carbonyl compounds:
$$\text{HC}\equiv\text{CH} + \text{H}_2\text{O} \xrightarrow[\text{333 K}]{\text{Hg}^{2+} / \text{H}^+} [\text{CH}_2=\text{CHOH}] \xrightarrow{\text{tautomerises}} \text{CH}_3\text{CHO} \quad (\text{Ethanal / Acetaldehyde})$$
$$\text{CH}_3\text{C}\equiv\text{CH} + \text{H}_2\text{O} \xrightarrow{\text{Hg}^{2+} / \text{H}^+} [\text{CH}_3\text{C(OH)}=\text{CH}_2] \xrightarrow{\text{tautomerises}} \text{CH}_3\text{COCH}_3 \quad (\text{Propanone / Acetone})$$
C. Polymerisation of Ethyne
Linear Polymerisation: Produces polyacetylene $-(\text{CH}=\text{CH}-\text{CH}=\text{CH})_n-$. Conducts electricity under specialized doping conditions; used for lightweight polymer battery electrodes.
Cyclic Polymerisation (Aromatisation): Ethyne passed through a red hot iron tube at 873 K polymerises to form benzene ($3\text{C}_2\text{H}_2 \to \text{C}_6\text{H}_6$). Vital pathway from aliphatic to aromatic chemistry!
9.5 Aromatic Hydrocarbons (Arenes)
Aromatic hydrocarbons are known as arenes (Greek: aroma = sweet smelling). Compounds containing a benzene ring are termed benzenoids (e.g. benzene, toluene, naphthalene, biphenyl); those with aromatic stability lacking a benzene ring are termed non-benzenoids (e.g. azulene, tropylium cation, cyclopentadienyl anion).
9.5.2 Structure & Resonance Stability of Benzene
Michael Faraday isolated benzene in 1825 ($\text{C}_6\text{H}_6$). In 1865, Friedrich August Kekulé proposed a hexagonal ring with alternating single and double bonds. Because benzene only yields one monosubstituted derivative and one ortho-disubstituted derivative, Kekulé proposed an oscillating equilibrium between the two canonical structures.
Resonance & Orbital Description
All 6 carbon atoms are $sp^2$ hybridised, forming a planar hexagonal ring with $CCC$ bond angles of $120^\circ$.
Each carbon has one unhybridised $2p_z$ orbital perpendicular to the ring plane. These 6 $p$-orbitals overlap laterally to form a completely delocalised $\pi$-electron cloud in the shape of two continuous doughnut-shaped rings—one above and one below the plane of the carbon atoms.
X-ray diffraction confirms that all six $CC$ bond lengths are identical at $\mathbf{139\text{ pm}}$, exactly intermediate between a single bond ($154\text{ pm}$) and a double bond ($133\text{ pm}$).
Resonance energy of benzene is $\mathbf{150.6\text{ kJ mol}^{-1}}$ ($36\text{ kcal mol}^{-1}$), accounting for its extraordinary stability and strong resistance to addition reactions.
9.5.3 Hückel's Rule of Aromaticity
For any planar cyclic system to be aromatic, it must satisfy three criteria:
Planarity: Ring atoms must lie in the same geometric plane.
Complete Delocalisation: Conjugated cyclic ring of unhybridised $p$-orbitals.
Because benzene is an electron-rich $\pi$-system, it readily attracts electrophiles ($E^+$). Substitution preserves the highly stable aromatic resonance sextet, whereas addition destroys aromaticity.
General 3-Step Mechanism of $S_E$:
Step 1: Generation of Electrophile ($E^+$):
Nitration: $\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightleftharpoons \mathbf{NO_2^+} + \text{H}_3\text{O}^+ + 2\text{HSO}_4^-$ ($\text{H}_2\text{SO}_4$ acts as acid, $\text{HNO}_3$ acts as base!)
Step 2: Attack of $E^+$ & Formation of Arenium Ion ($\sigma$-Complex):
The electrophile attacks the ring, using two $\pi$-electrons to form a $CE$ bond at an $sp^3$ hybridised carbon. The resulting carbocation is delocalised over the remaining 5 carbon atoms and stabilised by three resonance structures.
Step 3: Loss of Proton to Restore Aromaticity:
A weak base (e.g. $\text{HSO}_4^-$ or $[\text{AlCl}_4]^-$) removes the proton from the $sp^3$ carbon, returning the electron pair to the ring to regenerate the stable $(4n+2)\pi$ aromatic system.
Combustion: Burns with a smoky/sooty flame due to high carbon content ($92.3\%$):
$$\text{C}_6\text{H}_6 + \frac{15}{2}\text{O}_2 \to 6\text{CO}_2 + 3\text{H}_2\text{O}$$
9.5.6 Directive Influence of Substituents
When a monosubstituted benzene undergoes further substitution, the incoming electrophile is directed to specific positions depending upon the substituent already attached:
Ortho & Para Directing Groups
Ring Activating: Donate electrons by resonance ($+R$ or $+M$), creating high electron density specifically at ortho and para positions:
Examples: $-\text{OH}, -\text{NH}_2, -\text{NHR}, -\text{OCH}_3, -\text{CH}_3, -\text{C}_2\text{H}_5$ (alkyls activate via hyperconjugation and $+I$).
Deactivating Ortho-Para Directors (Halogens): Halogens ($-\text{F}, -\text{Cl}, -\text{Br}, -\text{I}$) exhibit strong electron-withdrawing inductive effect ($-I$) which deactivates the ring overall, but their resonance ($+R$) effect directs the incoming electrophile to ortho/para positions.
Meta Directing Groups
Ring Deactivating: Withdraw electrons via $-R$ / $-M$ and $-I$ effects, placing partial positive charges at ortho and para positions. Consequently, the meta position remains comparatively electron-rich, directing electrophilic attack to meta:
Benzene and polynuclear aromatic hydrocarbons (PAHs) containing more than two benzene rings fused together are toxic and possess cancer-producing (carcinogenic) properties.
Such polynuclear hydrocarbons are formed during the incomplete combustion of organic matter like tobacco, coal, petroleum, and charbroiled foods. Once inhaled or ingested, they enter cells and undergo biochemical oxidation by cytochrome P450 enzymes into reactive epoxide metabolites that covalently bind to DNA bases, producing mutations and initiating cancer.
Prominent Carcinogenic Hydrocarbons:
$1,2$-Benzanthracene
$3$-Methylcholanthrene
$1,2$-Benzpyrene and $9,10$-Dimethyl-1,2-benzanthracene
Hydrocarbons: Concept Mastery Quiz (25 Questions)
Test your conceptual understanding of alkanes, alkenes, alkynes, conformational analysis, addition mechanisms, and electrophilic aromatic substitution with these 25 curated NCERT/JEE questions.
Q01
Which of the following compounds has the highest boiling point?
Correct Answer: For isomeric alkanes, boiling point decreases with increasing branching because branching makes the molecule more compact and spherical, decreasing surface area of contact and van der Waals forces. Hence, unbranched n-pentane has the highest b.p. (309.1 K).
Incorrect: For isomeric alkanes, boiling point decreases with increasing branching because branching makes the molecule more compact and spherical, decreasing surface area of contact and van der Waals forces. Hence, unbranched n-pentane has the highest b.p. (309.1 K).
Incorrect: For isomeric alkanes, boiling point decreases with increasing branching because branching makes the molecule more compact and spherical, decreasing surface area of contact and van der Waals forces. Hence, unbranched n-pentane has the highest b.p. (309.1 K).
Incorrect: For isomeric alkanes, boiling point decreases with increasing branching because branching makes the molecule more compact and spherical, decreasing surface area of contact and van der Waals forces. Hence, unbranched n-pentane has the highest b.p. (309.1 K).
Q02
Why is methane unable to be prepared by the Wurtz reaction?
Incorrect: Wurtz reaction couples two alkyl free radicals (R• + R• -> R-R), thereby producing an alkane with at least 2 carbon atoms (ethane). Methane contains only one carbon.
Correct Answer: Wurtz reaction couples two alkyl free radicals (R• + R• -> R-R), thereby producing an alkane with at least 2 carbon atoms (ethane). Methane contains only one carbon.
Incorrect: Wurtz reaction couples two alkyl free radicals (R• + R• -> R-R), thereby producing an alkane with at least 2 carbon atoms (ethane). Methane contains only one carbon.
Incorrect: Wurtz reaction couples two alkyl free radicals (R• + R• -> R-R), thereby producing an alkane with at least 2 carbon atoms (ethane). Methane contains only one carbon.
Q03
In the free radical chlorination of methane, the formation of ethane byproduct is explained by which mechanistic step?
Incorrect: In the termination phase, two methyl free radicals collide and combine: •CH3 + •CH3 -> CH3-CH3, resulting in the termination of the chain and formation of trace ethane.
Incorrect: In the termination phase, two methyl free radicals collide and combine: •CH3 + •CH3 -> CH3-CH3, resulting in the termination of the chain and formation of trace ethane.
Correct Answer: In the termination phase, two methyl free radicals collide and combine: •CH3 + •CH3 -> CH3-CH3, resulting in the termination of the chain and formation of trace ethane.
Incorrect: In the termination phase, two methyl free radicals collide and combine: •CH3 + •CH3 -> CH3-CH3, resulting in the termination of the chain and formation of trace ethane.
Q04
The torsional strain in ethane is primarily caused by:
Incorrect: Torsional strain in ethane arises from repulsive interactions between electron clouds of C-H sigma bonds on adjacent carbon atoms when they align in the eclipsed conformation.
Correct Answer: Torsional strain in ethane arises from repulsive interactions between electron clouds of C-H sigma bonds on adjacent carbon atoms when they align in the eclipsed conformation.
Incorrect: Torsional strain in ethane arises from repulsive interactions between electron clouds of C-H sigma bonds on adjacent carbon atoms when they align in the eclipsed conformation.
Incorrect: Torsional strain in ethane arises from repulsive interactions between electron clouds of C-H sigma bonds on adjacent carbon atoms when they align in the eclipsed conformation.
Q05
What is the dihedral angle between C-H bonds in the staggered conformation of ethane?
Incorrect: In staggered ethane, the dihedral angle theta between adjacent C-H bonds is exactly 60°, minimizing electron cloud repulsions.
Correct Answer: In staggered ethane, the dihedral angle theta between adjacent C-H bonds is exactly 60°, minimizing electron cloud repulsions.
Incorrect: In staggered ethane, the dihedral angle theta between adjacent C-H bonds is exactly 60°, minimizing electron cloud repulsions.
Incorrect: In staggered ethane, the dihedral angle theta between adjacent C-H bonds is exactly 60°, minimizing electron cloud repulsions.
Q06
Which of the following alkenes will NOT show geometrical (cis-trans) isomerism?
Incorrect: Propene (CH2=CH-CH3) has two identical hydrogen atoms attached to carbon-1. For cis-trans isomerism, each double-bonded carbon must hold two distinct groups.
Incorrect: Propene (CH2=CH-CH3) has two identical hydrogen atoms attached to carbon-1. For cis-trans isomerism, each double-bonded carbon must hold two distinct groups.
Correct Answer: Propene (CH2=CH-CH3) has two identical hydrogen atoms attached to carbon-1. For cis-trans isomerism, each double-bonded carbon must hold two distinct groups.
Incorrect: Propene (CH2=CH-CH3) has two identical hydrogen atoms attached to carbon-1. For cis-trans isomerism, each double-bonded carbon must hold two distinct groups.
Q07
Partial reduction of an alkyne with Lindlar's catalyst (Pd/CaCO3 poisoned with quinoline) produces:
Incorrect: Lindlar's catalyst facilitates syn-addition of H2 across the triple bond, stereoselectively yielding the cis-alkene.
Correct Answer: Lindlar's catalyst facilitates syn-addition of H2 across the triple bond, stereoselectively yielding the cis-alkene.
Incorrect: Lindlar's catalyst facilitates syn-addition of H2 across the triple bond, stereoselectively yielding the cis-alkene.
Incorrect: Lindlar's catalyst facilitates syn-addition of H2 across the triple bond, stereoselectively yielding the cis-alkene.
Q08
Reduction of but-2-yne with sodium in liquid ammonia (Birch reduction) stereoselectively yields:
Incorrect: Sodium in liquid ammonia conducts an anti-addition mechanism via a trans-radical anion, selectively yielding the trans-alkene.
Correct Answer: Sodium in liquid ammonia conducts an anti-addition mechanism via a trans-radical anion, selectively yielding the trans-alkene.
Incorrect: Sodium in liquid ammonia conducts an anti-addition mechanism via a trans-radical anion, selectively yielding the trans-alkene.
Incorrect: Sodium in liquid ammonia conducts an anti-addition mechanism via a trans-radical anion, selectively yielding the trans-alkene.
Q09
The dipole moment of trans-but-2-ene compared to cis-but-2-ene is:
Incorrect: In trans-but-2-ene, the two C-CH3 bond moments are oriented in exactly opposite directions (180°), canceling each other to give mu ≈ 0 D.
Incorrect: In trans-but-2-ene, the two C-CH3 bond moments are oriented in exactly opposite directions (180°), canceling each other to give mu ≈ 0 D.
Correct Answer: In trans-but-2-ene, the two C-CH3 bond moments are oriented in exactly opposite directions (180°), canceling each other to give mu ≈ 0 D.
Incorrect: In trans-but-2-ene, the two C-CH3 bond moments are oriented in exactly opposite directions (180°), canceling each other to give mu ≈ 0 D.
Q10
Addition of HBr to propene in the presence of benzoyl peroxide yields 1-bromopropane because:
Incorrect: In the peroxide effect, Br• adds first to CH2=CH-CH3 to generate the more stable secondary free radical (CH3-•CH-CH2Br), which abstracts H• to yield 1-bromopropane.
Correct Answer: In the peroxide effect, Br• adds first to CH2=CH-CH3 to generate the more stable secondary free radical (CH3-•CH-CH2Br), which abstracts H• to yield 1-bromopropane.
Incorrect: In the peroxide effect, Br• adds first to CH2=CH-CH3 to generate the more stable secondary free radical (CH3-•CH-CH2Br), which abstracts H• to yield 1-bromopropane.
Incorrect: In the peroxide effect, Br• adds first to CH2=CH-CH3 to generate the more stable secondary free radical (CH3-•CH-CH2Br), which abstracts H• to yield 1-bromopropane.
Q11
Why does the peroxide effect (anti-Markovnikov addition) NOT operate with HCl or HI?
Correct Answer: H-Cl bond enthalpy is high (430.5 kJ/mol) so chain propagation with RO• is endothermic. For HI, H-I is weak (296.8 kJ/mol), but iodine radicals combine into I2 rather than adding to the double bond.
Incorrect: H-Cl bond enthalpy is high (430.5 kJ/mol) so chain propagation with RO• is endothermic. For HI, H-I is weak (296.8 kJ/mol), but iodine radicals combine into I2 rather than adding to the double bond.
Incorrect: H-Cl bond enthalpy is high (430.5 kJ/mol) so chain propagation with RO• is endothermic. For HI, H-I is weak (296.8 kJ/mol), but iodine radicals combine into I2 rather than adding to the double bond.
Incorrect: H-Cl bond enthalpy is high (430.5 kJ/mol) so chain propagation with RO• is endothermic. For HI, H-I is weak (296.8 kJ/mol), but iodine radicals combine into I2 rather than adding to the double bond.
Q12
An alkene on ozonolysis followed by Zn/H2O cleavage yields methanal and propanone. The alkene is:
Incorrect: Joining the carbonyl groups with a double bond: (CH3)2C=O + O=CH2 -> (CH3)2C=CH2 (2-Methylpropene).
Correct Answer: Joining the carbonyl groups with a double bond: (CH3)2C=O + O=CH2 -> (CH3)2C=CH2 (2-Methylpropene).
Incorrect: Joining the carbonyl groups with a double bond: (CH3)2C=O + O=CH2 -> (CH3)2C=CH2 (2-Methylpropene).
Incorrect: Joining the carbonyl groups with a double bond: (CH3)2C=O + O=CH2 -> (CH3)2C=CH2 (2-Methylpropene).
Q13
What is the role of zinc dust in the reductive cleavage of alkene ozonides?
Incorrect: Cleavage of ozonide produces H2O2. Zinc reacts with H2O2 (Zn + H2O2 -> ZnO + H2O), preventing it from oxidising aldehydes into acids.
Incorrect: Cleavage of ozonide produces H2O2. Zinc reacts with H2O2 (Zn + H2O2 -> ZnO + H2O), preventing it from oxidising aldehydes into acids.
Correct Answer: Cleavage of ozonide produces H2O2. Zinc reacts with H2O2 (Zn + H2O2 -> ZnO + H2O), preventing it from oxidising aldehydes into acids.
Incorrect: Cleavage of ozonide produces H2O2. Zinc reacts with H2O2 (Zn + H2O2 -> ZnO + H2O), preventing it from oxidising aldehydes into acids.
Q14
The acidic character of terminal alkynes (e.g. ethyne) is primarily attributed to:
Correct Answer: An sp carbon has 50% s-character, making it significantly more electronegative than sp2 (33.3%) or sp3 (25%) carbons. It strongly draws the C-H electron pair, facilitating proton release.
Incorrect: An sp carbon has 50% s-character, making it significantly more electronegative than sp2 (33.3%) or sp3 (25%) carbons. It strongly draws the C-H electron pair, facilitating proton release.
Incorrect: An sp carbon has 50% s-character, making it significantly more electronegative than sp2 (33.3%) or sp3 (25%) carbons. It strongly draws the C-H electron pair, facilitating proton release.
Incorrect: An sp carbon has 50% s-character, making it significantly more electronegative than sp2 (33.3%) or sp3 (25%) carbons. It strongly draws the C-H electron pair, facilitating proton release.
Q15
Which of the following compounds reacts with sodium metal to evolve dihydrogen gas?
Incorrect: Only terminal alkynes (containing an sp C-H bond) are acidic enough to react with active metals like Na to evolve H2. But-2-yne is internal and lacks acidic hydrogen.
Incorrect: Only terminal alkynes (containing an sp C-H bond) are acidic enough to react with active metals like Na to evolve H2. But-2-yne is internal and lacks acidic hydrogen.
Incorrect: Only terminal alkynes (containing an sp C-H bond) are acidic enough to react with active metals like Na to evolve H2. But-2-yne is internal and lacks acidic hydrogen.
Correct Answer: Only terminal alkynes (containing an sp C-H bond) are acidic enough to react with active metals like Na to evolve H2. But-2-yne is internal and lacks acidic hydrogen.
Q16
When ethyne is passed through a red hot iron tube at 873 K, the product formed is:
Incorrect: Cyclic trimerisation of three ethyne molecules at 873 K over iron catalyst gives benzene: 3 C2H2 -> C6H6.
Incorrect: Cyclic trimerisation of three ethyne molecules at 873 K over iron catalyst gives benzene: 3 C2H2 -> C6H6.
Correct Answer: Cyclic trimerisation of three ethyne molecules at 873 K over iron catalyst gives benzene: 3 C2H2 -> C6H6.
Incorrect: Cyclic trimerisation of three ethyne molecules at 873 K over iron catalyst gives benzene: 3 C2H2 -> C6H6.
Q17
All carbon-carbon bond lengths in benzene are equal to 139 pm because:
Incorrect: Due to resonance and continuous lateral overlap of six 2pz orbitals, the 6 pi-electrons are uniformly delocalised over the ring, resulting in an intermediate bond length of 139 pm.
Correct Answer: Due to resonance and continuous lateral overlap of six 2pz orbitals, the 6 pi-electrons are uniformly delocalised over the ring, resulting in an intermediate bond length of 139 pm.
Incorrect: Due to resonance and continuous lateral overlap of six 2pz orbitals, the 6 pi-electrons are uniformly delocalised over the ring, resulting in an intermediate bond length of 139 pm.
Incorrect: Due to resonance and continuous lateral overlap of six 2pz orbitals, the 6 pi-electrons are uniformly delocalised over the ring, resulting in an intermediate bond length of 139 pm.
Q18
According to Hückel's Rule, a planar cyclic conjugated system is aromatic if it contains:
Incorrect: Hückel's rule specifies (4n + 2) pi electrons where n = 0, 1, 2, 3... (e.g. 2, 6, 10, 14 pi electrons).
Correct Answer: Hückel's rule specifies (4n + 2) pi electrons where n = 0, 1, 2, 3... (e.g. 2, 6, 10, 14 pi electrons).
Incorrect: Hückel's rule specifies (4n + 2) pi electrons where n = 0, 1, 2, 3... (e.g. 2, 6, 10, 14 pi electrons).
Incorrect: Hückel's rule specifies (4n + 2) pi electrons where n = 0, 1, 2, 3... (e.g. 2, 6, 10, 14 pi electrons).
Q19
In the nitration of benzene with conc. HNO3 and conc. H2SO4, the role of conc. H2SO4 is:
Incorrect: H2SO4 is a stronger acid than HNO3. It donates a proton to HNO3: HNO3 + 2 H2SO4 <=> NO2+ + H3O+ + 2 HSO4-.
Correct Answer: H2SO4 is a stronger acid than HNO3. It donates a proton to HNO3: HNO3 + 2 H2SO4 <=> NO2+ + H3O+ + 2 HSO4-.
Incorrect: H2SO4 is a stronger acid than HNO3. It donates a proton to HNO3: HNO3 + 2 H2SO4 <=> NO2+ + H3O+ + 2 HSO4-.
Incorrect: H2SO4 is a stronger acid than HNO3. It donates a proton to HNO3: HNO3 + 2 H2SO4 <=> NO2+ + H3O+ + 2 HSO4-.
Q20
During electrophilic aromatic substitution, the arenium ion (sigma-complex) is:
Incorrect: The arenium ion has one sp3 hybridized carbon where the electrophile added. This interrupts the continuous conjugated cyclic p-orbital system, rendering the intermediate non-aromatic.
Correct Answer: The arenium ion has one sp3 hybridized carbon where the electrophile added. This interrupts the continuous conjugated cyclic p-orbital system, rendering the intermediate non-aromatic.
Incorrect: The arenium ion has one sp3 hybridized carbon where the electrophile added. This interrupts the continuous conjugated cyclic p-orbital system, rendering the intermediate non-aromatic.
Incorrect: The arenium ion has one sp3 hybridized carbon where the electrophile added. This interrupts the continuous conjugated cyclic p-orbital system, rendering the intermediate non-aromatic.
Q21
When benzene is treated with 1-chloropropane in the presence of anhydrous AlCl3, the major product is:
Incorrect: The primary carbocation CH3-CH2-CH2+ initially generated rearranges via a 1,2-hydride shift into the more stable secondary carbocation CH3-CH(+)-CH3, which attacks benzene to give isopropylbenzene.
Correct Answer: The primary carbocation CH3-CH2-CH2+ initially generated rearranges via a 1,2-hydride shift into the more stable secondary carbocation CH3-CH(+)-CH3, which attacks benzene to give isopropylbenzene.
Incorrect: The primary carbocation CH3-CH2-CH2+ initially generated rearranges via a 1,2-hydride shift into the more stable secondary carbocation CH3-CH(+)-CH3, which attacks benzene to give isopropylbenzene.
Incorrect: The primary carbocation CH3-CH2-CH2+ initially generated rearranges via a 1,2-hydride shift into the more stable secondary carbocation CH3-CH(+)-CH3, which attacks benzene to give isopropylbenzene.
Q22
Although halogens are ortho- and para-directing, they are ring deactivating because:
Correct Answer: Halogens possess a strong -I effect which withdraws electron density from the entire ring (deactivating it towards electrophilic attack), but resonance (+R) donation specifically stabilizes ortho and para arenium ions.
Incorrect: Halogens possess a strong -I effect which withdraws electron density from the entire ring (deactivating it towards electrophilic attack), but resonance (+R) donation specifically stabilizes ortho and para arenium ions.
Incorrect: Halogens possess a strong -I effect which withdraws electron density from the entire ring (deactivating it towards electrophilic attack), but resonance (+R) donation specifically stabilizes ortho and para arenium ions.
Incorrect: Halogens possess a strong -I effect which withdraws electron density from the entire ring (deactivating it towards electrophilic attack), but resonance (+R) donation specifically stabilizes ortho and para arenium ions.
Q23
Which of the following substituents is META-directing and strongly deactivating?
Incorrect: -NO2 exerts powerful -I and -R electron-withdrawing effects, deactivating ortho and para positions and directing attack to the meta position.
Incorrect: -NO2 exerts powerful -I and -R electron-withdrawing effects, deactivating ortho and para positions and directing attack to the meta position.
Correct Answer: -NO2 exerts powerful -I and -R electron-withdrawing effects, deactivating ortho and para positions and directing attack to the meta position.
Incorrect: -NO2 exerts powerful -I and -R electron-withdrawing effects, deactivating ortho and para positions and directing attack to the meta position.
Q24
Chlorination of benzene in the presence of ultraviolet light (h*nu) without a Lewis acid yields:
Incorrect: Under UV light, three molecules of Cl2 undergo free-radical addition across the benzene ring to yield 1,2,3,4,5,6-hexachlorocyclohexane (BHC / gammaxane / lindane).
Correct Answer: Under UV light, three molecules of Cl2 undergo free-radical addition across the benzene ring to yield 1,2,3,4,5,6-hexachlorocyclohexane (BHC / gammaxane / lindane).
Incorrect: Under UV light, three molecules of Cl2 undergo free-radical addition across the benzene ring to yield 1,2,3,4,5,6-hexachlorocyclohexane (BHC / gammaxane / lindane).
Incorrect: Under UV light, three molecules of Cl2 undergo free-radical addition across the benzene ring to yield 1,2,3,4,5,6-hexachlorocyclohexane (BHC / gammaxane / lindane).
Q25
Polynuclear aromatic hydrocarbons (such as 1,2-benzpyrene) cause cancer because:
Incorrect: Metabolic oxidation of fused polynuclear hydrocarbons yields diol-epoxides that covalently react with amino groups of DNA purine bases, causing mutations that trigger cancer.
Correct Answer: Metabolic oxidation of fused polynuclear hydrocarbons yields diol-epoxides that covalently react with amino groups of DNA purine bases, causing mutations that trigger cancer.
Incorrect: Metabolic oxidation of fused polynuclear hydrocarbons yields diol-epoxides that covalently react with amino groups of DNA purine bases, causing mutations that trigger cancer.
Incorrect: Metabolic oxidation of fused polynuclear hydrocarbons yields diol-epoxides that covalently react with amino groups of DNA purine bases, causing mutations that trigger cancer.
NCERT End-of-Chapter Exercises (9.1 9.25) Solved
Thorough, step-by-step solutions to all 25 textbook exercises for Unit 9: Hydrocarbons, strictly aligned with official CBSE & NCERT marking guidelines.
9.1
How do you account for the formation of ethane during chlorination of methane?
Solution:
Chlorination of methane proceeds via a free radical chain mechanism consisting of initiation, propagation, and termination steps.
Termination: As the reaction progresses and reactant concentration decreases, the probability of free radicals colliding with each other increases. When two methyl free radicals ($\text{CH}_3^\bullet$) collide, they couple to form a molecule of ethane:
$$\text{CH}_3^\bullet + \text{CH}_3^\bullet \to \text{CH}_3\text{CH}_3 \quad (\text{Ethane})$$
This coupling reaction in the chain termination step accounts for the presence of trace amounts of ethane as an impurity/byproduct during the chlorination of methane.
$\text{CH}_3\text{CH}=\text{C}(\text{CH}_3)_2$: The longest continuous chain containing the double bond has 4 carbons. Numbering from right to give double bond the lowest locant ($C2$):
$\overset{4}{\text{C}}\text{H}_3\overset{3}{\text{C}}\text{H}=\overset{2}{\text{C}}(\text{CH}_3)\overset{1}{\text{C}}\text{H}_3 \implies$ 2-Methylbut-2-ene.
$\text{CH}_2=\text{CH}\text{C}\equiv\text{C}\text{CH}_3$: When both double and triple bonds are equidistant from ends, the double bond takes numbering priority:
$\overset{1}{\text{C}}\text{H}_2=\overset{2}{\text{C}}\text{H}\overset{3}{\text{C}}\equiv\overset{4}{\text{C}}\overset{5}{\text{C}}\text{H}_3 \implies$ Pent-1-en-3-yne.
$\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2$: Principal chain is 4 carbons with double bond at $C1$ and phenyl group at $C4$: 4-Phenylbut-1-ene.
2-Methylbut-1-en-3-yne: Double bond at 1, methyl at 2, triple bond at 3: 2-Methylbut-1-en-3-yne.
$\text{CH}_3(\text{CH}_2)_4\text{CH}(\text{CH}_2\text{CH}(\text{CH}_3)_2)(\text{CH}_2)_3\text{CH}_3$: Longest chain contains 10 carbons (decane). At carbon-5, there is a $2$-methylpropyl (isobutyl) substituent:
$\overset{10}{\text{C}}\text{H}_3\dots\overset{6}{\text{C}}\text{H}_2\overset{5}{\text{C}}\text{H}[\text{CH}_2\text{CH}(\text{CH}_3)_2]\overset{4}{\text{C}}\text{H}_2\dots\overset{1}{\text{C}}\text{H}_3 \implies$ 5-(2-Methylpropyl)decane (or $5$-isobutyldecane).
$\text{CH}_3\text{CH}=\text{CH}\text{CH}_2\text{CH}=\text{CH}\text{CH}(\text{C}_2\text{H}_5)\text{CH}_2\text{CH}=\text{CH}_2$: Longest chain containing all double bonds has 10 carbons (decatriene). Numbering from right gives lower locants ($1, 4, 7$ for double bonds, $3$ for ethyl group): 4-Ethyldeca-1,5,8-triene (or $3$-ethyldeca-$1,4,7$-triene depending on numbering direction).
9.3
For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated:
The longest continuous chain has 5 carbons with an ethyl group at carbon-3 and double bond at carbon-2.
Therefore, the structure is $\text{CH}_3\text{CH}=\text{C}(\text{C}_2\text{H}_5)_2$ and the IUPAC name is 3-Ethylpent-2-ene.
9.6
An alkene 'A' contains three CC, eight CH $\sigma$ bonds and one CC $\pi$ bond. 'A' on ozonolysis gives two moles of an aldehyde of molar mass $44\text{ u}$. Write IUPAC name of 'A'.
Solution:
An aldehyde has the formula $\text{RCHO}$. Its molar mass is $44\text{ u}$.
$\text{Molar mass of RCHO} = M(\text{R}) + 12 + 1 + 16 = M(\text{R}) + 29 = 44 \implies M(\text{R}) = 15\text{ u}$.
An alkyl radical with molar mass $15$ is a methyl group ($-\text{CH}_3$).
Therefore, the aldehyde is ethanal ($\text{CH}_3\text{CHO}$).
Since ozonolysis of 1 mole of 'A' produces 2 moles of $\text{CH}_3\text{CHO}$, alkene 'A' must be symmetrical:
$$\text{CH}_3\text{CH}=\text{O} + \text{O}=\text{CH}\text{CH}_3 \to \text{CH}_3\text{CH}=\text{CH}\text{CH}_3 \quad (\text{But-2-ene})$$
Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?
Solution:
Hex-2-ene is $\text{CH}_3\text{CH}=\text{CH}\text{CH}_2\text{CH}_2\text{CH}_3$:
cis-Hex-2-ene: Both alkyl groups ($-\text{CH}_3$ and $-\text{CH}_2\text{CH}_2\text{CH}_3$) are on the same side of the double bond.
trans-Hex-2-ene: The alkyl groups lie on opposite sides of the double bond.
Boiling Point Comparison:
The cis-isomer has a higher boiling point than the trans-isomer. In the cis-isomer, the dipole moments of the two $C\text{alkyl}$ bonds reinforce each other because they are inclined in the same direction, resulting in a net dipole moment ($\mu > 0$). This induces stronger dipole-dipole intermolecular attractions. In the trans-isomer, the two bond dipoles point in nearly opposite directions and largely cancel out, giving a lower dipole moment and weaker intermolecular forces.
9.10
Why is benzene extraordinarily stable though it contains three double bonds?
Solution:
Benzene is exceptionally stable because of cyclic $\pi$-electron resonance delocalisation:
The six $\pi$-electrons are not fixed between pairs of carbon atoms; instead, they are completely delocalised into a continuous cyclic ring forming two doughnut-shaped electron clouds above and below the planar hexagonal ring.
This delocalisation gives benzene an enormous resonance energy of $150.6\text{ kJ mol}^{-1}$ ($36\text{ kcal mol}^{-1}$).
Because all six $CC$ bonds have equal bond length ($139\text{ pm}$) and partial double-bond character, addition reactions would disrupt the closed aromatic sextet and forfeit this massive resonance stabilisation. Consequently, benzene resists addition and preferentially undergoes electrophilic substitution.
9.11
What are the necessary conditions for any system to be aromatic?
Solution:
For a compound to be aromatic, it must fulfill the following criteria:
Cyclic Structure: The molecule must be cyclic.
Planarity: Ring atoms must be coplanar (all ring atoms are $sp^2$ or $sp$ hybridised) to permit unhindered lateral overlap of $p$-orbitals.
Complete Conjugation: A continuous, uninterrupted loop of overlapping $p$-orbitals around the entire perimeter of the ring.
Hückel's Rule: The cyclic conjugated $\pi$-system must contain $(4n+2)\pi$ electrons, where $n = 0, 1, 2, 3 \dots$ ($2, 6, 10, 14 \dots$ $\pi$-electrons).
9.12
Explain why the following systems are not aromatic: (i) Cyclooctatetraene, (ii) Cyclopentadiene, (iii) Cycloheptatrienyl radical.
Solution:
Cyclooctatetraene ($\text{C}_8\text{H}_8$): Contains $8\ \pi$-electrons, which equals $4n$ ($n=2$), violating Hückel's $(4n+2)$ rule. Moreover, to minimize anti-aromatic destabilization, it adopts a non-planar tub-shaped conformation, preventing complete planar delocalisation. Hence, it is non-aromatic.
Cyclopentadiene: Contains an $sp^3$ hybridised $-\text{CH}_2-$ carbon atom in the ring. This breaks the continuous cyclic conjugation of $p$-orbitals, rendering the molecule non-aromatic.
Cycloheptatrienyl radical: Has 7 $\pi$-electrons. This is an odd number of electrons, which does not satisfy Hückel's $(4n+2)$ rule ($4n+2$ gives $2, 6, 10$). Hence, it is non-aromatic. (In contrast, the cycloheptatrienyl cation has $6\ \pi$-electrons and is aromatic).
9.13
How will you convert benzene into:
(i) $p$-Nitrobromobenzene (ii) $m$-Nitrochlorobenzene (iii) $p$-Nitrotoluene (iv) Acetophenone?
$m$-Nitrochlorobenzene:
Chlorine is $o/p$-directing, but $-\text{NO}_2$ is $m$-directing. Hence, nitrate first to direct the subsequent halogen to the meta position:
$$\text{Benzene} \xrightarrow[\text{conc. } \text{H}_2\text{SO}_4]{\text{conc. } \text{HNO}_3} \text{Nitrobenzene} \xrightarrow{\text{Cl}_2 / \text{Anhyd. } \text{AlCl}_3} m\text{-Nitrochlorobenzene}$$
$p$-Nitrotoluene:
Alkyl group is $o/p$-directing. Perform Friedel-Crafts alkylation first, then nitrate:
$$\text{Benzene} \xrightarrow{\text{CH}_3\text{Cl} / \text{Anhyd. } \text{AlCl}_3} \text{Toluene} \xrightarrow[\text{conc. } \text{H}_2\text{SO}_4]{\text{conc. } \text{HNO}_3, 303\text{ K}} p\text{-Nitrotoluene} \ (+ \text{ortho})$$
Acetophenone:
Friedel-Crafts acylation with acetyl chloride in the presence of anhydrous $\text{AlCl}_3$:
$$\text{Benzene} + \text{CH}_3\text{COCl} \xrightarrow{\text{Anhyd. } \text{AlCl}_3} \text{C}_6\text{H}_5\text{COCH}_3 + \text{HCl}$$
9.14
In the alkane $\text{H}_3\text{C}\text{CH}_2\text{C}(\text{CH}_3)_2\text{CH}_2\text{CH}(\text{CH}_3)_2$, identify $1^\circ, 2^\circ, 3^\circ, 4^\circ$ carbon atoms and give the number of H atoms bonded to each.
Solution:
Numbering the parent chain as $2,4,4$-trimethylhexane:
Primary ($1^\circ$) Carbon atoms: Attached to only 1 carbon atom. There are 5 primary carbon atoms (terminal carbons and the 3 methyl side groups).
Number of hydrogen atoms bonded to $1^\circ$ carbons = $5 \times 3 = \mathbf{15\text{ H atoms}}$.
Secondary ($2^\circ$) Carbon atoms: Attached to 2 carbons ($C2$ and $C5$ methylene groups). There are 2 secondary carbon atoms.
Number of hydrogen atoms bonded to $2^\circ$ carbons = $2 \times 2 = \mathbf{4\text{ H atoms}}$.
Tertiary ($3^\circ$) Carbon atoms: Attached to 3 carbons ($-\text{CH}(\text{CH}_3)_2$). There is 1 tertiary carbon atom.
Number of hydrogen atoms bonded to $3^\circ$ carbon = $1 \times 1 = \mathbf{1\text{ H atom}}$.
Quaternary ($4^\circ$) Carbon atoms: Attached to 4 carbons ($-\text{C}(\text{CH}_3)_2-$). There is 1 quaternary carbon atom.
Number of hydrogen atoms bonded to $4^\circ$ carbon = $\mathbf{0\text{ H atoms}}$.
9.15
What effect does branching of an alkane chain have on its boiling point?
Solution:
Branching of an alkane chain lowers its boiling point.
Scientific Reason: As branching increases, the molecular geometry becomes more spherical and compact. A sphere has the minimum surface area for a given volume. This diminished surface area of contact reduces the magnitude of intermolecular van der Waals (London dispersion) forces between adjacent molecules. Consequently, less thermal energy is required to overcome intermolecular attractions, resulting in a lower boiling point.
Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
Solution:
In absence of peroxide (Markovnikov's Rule - Ionic Mechanism):
$\text{HBr}$ ionises to generate electrophile $\text{H}^+$. Attack of $\text{H}^+$ on propene forms two possible carbocations:
$$\text{CH}_3\text{CH}=\text{CH}_2 + \text{H}^+ \to \text{CH}_3\overset{+}{\text{C}}\text{H}\text{CH}_3 \ (2^\circ, \text{more stable}) \quad \text{or} \quad \text{CH}_3\text{CH}_2\overset{+}{\text{C}}\text{H}_2 \ (1^\circ, \text{less stable})$$
The secondary carbocation is substantially more stable due to $+I$ inductive effect and 6 hyperconjugative structures. Nucleophilic attack of $\text{Br}^-$ on the $2^\circ$ carbocation yields 2-bromopropane as the major product.
In presence of peroxide (Peroxide / Kharasch Effect - Free Radical Mechanism):
Organic peroxides undergo homolysis to generate alkoxy/aryl free radicals, which react with $\text{HBr}$ to produce bromine free radicals ($\text{Br}^\bullet$):
$$(\text{C}_6\text{H}_5\text{COO})_2 \to 2\text{C}_6\text{H}_5\text{COO}^\bullet \to 2\text{C}_6\text{H}_5^\bullet + 2\text{CO}_2$$
$$\text{C}_6\text{H}_5^\bullet + \text{HBr} \to \text{C}_6\text{H}_6 + \text{Br}^\bullet$$
The $\text{Br}^\bullet$ radical attacks propene to form a free radical:
$$\text{CH}_3\text{CH}=\text{CH}_2 + \text{Br}^\bullet \to \text{CH}_3\overset{\bullet}{\text{C}}\text{H}\text{CH}_2\text{Br} \ (2^\circ \text{ radical, more stable})$$
The more stable secondary free radical abstracts a hydrogen atom from another $\text{HBr}$ molecule, yielding 1-bromopropane as the major product:
$$\text{CH}_3\overset{\bullet}{\text{C}}\text{H}\text{CH}_2\text{Br} + \text{HBr} \to \text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{Br}^\bullet$$
9.17
Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene?
Solution:
If $o$-xylene existed as a single static structure, ozonolysis would yield only two products. However, $o$-xylene is a resonance hybrid of two contributing Kekulé forms:
Contributing Structure I (methyl carbons separated by a double bond):
Ozonolysis cleaves the double bond between the two methyl-bearing carbons, giving: 1 mole of Butane-2,3-dione (dimethylglyoxal, $\text{CH}_3\text{COCOCH}_3$) + 2 moles of Glyoxal ($\text{CHOCHO}$).
Contributing Structure II (methyl carbons separated by a single bond):
Ozonolysis leaves the $CC$ single bond between methyl groups intact, giving: 2 moles of 2-Oxopropanal (methylglyoxal, $\text{CH}_3\text{COCHO}$) + 1 mole of Glyoxal ($\text{CHOCHO}$).
Conclusion: In practice, all three products (glyoxal, methylglyoxal, and dimethylglyoxal in $3:2:1$ ratio) are isolated. This conclusively proves that the two Kekulé structures oscillate/resonate, confirming Kekulé's dynamic model of oscillating double bonds.
9.18
Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
Solution:
Decreasing Order of Acidic Strength:
$$\mathbf{\text{Ethyne (HC}\equiv\text{CH)} > \text{Benzene (C}_6\text{H}_6\text{)} > \text{n-Hexane (C}_6\text{H}_{14}\text{)}}$$
Reasoning Based on Hybridisation and $s$-Character:
In ethyne, carbon is $sp$ hybridised ($50\%\ s$-character). Since $s$-orbitals are closer to the nucleus, $sp$-carbon has the highest electronegativity. It strongly polarises the $CH$ bond, enabling hydrogen to dissociate as a proton ($H^+$).
In benzene, carbon is $sp^2$ hybridised ($33.3\%\ s$-character), making it moderately electronegative.
In $n$-hexane, carbon is $sp^3$ hybridised ($25\%\ s$-character), having the lowest electronegativity, so the $CH$ bond is least polar and virtually non-acidic.
9.19
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Solution:
Benzene possesses a planar ring enveloped by two continuous doughnut-shaped clouds of delocalised six $\pi$-electrons. This high concentration of mobile electron density makes benzene an electron-rich species that repels electron-rich nucleophiles ($Nu^-$) while strongly attracting electron-deficient electrophiles ($E^+$).
Nucleophilic attack would require forcing an electron-rich group into an already electron-dense aromatic ring, which encounters massive electrostatic repulsion. Hence, benzene undergoes electrophilic substitution readily, but nucleophilic substitution only under extreme conditions with powerful electron-withdrawing substituents.
9.20
How would you convert the following compounds into benzene? (i) Ethyne, (ii) Ethene, (iii) Hexane.
Solution:
From Ethyne: Pass ethyne gas through a red hot iron or quartz tube at 873 K (cyclic trimerisation):
$$3\text{HC}\equiv\text{CH} \xrightarrow[\text{873 K}]{\text{Red hot Fe tube}} \text{C}_6\text{H}_6$$
Step 2: Dehydrohalogenate with alcoholic $\text{KOH}$ followed by $\text{NaNH}_2$ to ethyne: $\text{CH}_2\text{Br}\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH / NaNH}_2} \text{CH}\equiv\text{CH}$
Step 3: Trimerise ethyne over red hot iron tube at $873\text{ K}$ to benzene.
From Hexane: Pass $n$-hexane over $\text{Cr}_2\text{O}_3 / \text{V}_2\text{O}_5 / \text{Mo}_2\text{O}_3$ supported on $\text{Al}_2\text{O}_3$ at $773\text{ K}$ and $10\text{}20\text{ atm}$ pressure (aromatisation/reforming):
$$\text{CH}_3(\text{CH}_2)_4\text{CH}_3 \xrightarrow[\text{1020 atm}]{\text{Cr}_2\text{O}_3 / \text{Al}_2\text{O}_3, 773\text{ K}} \text{C}_6\text{H}_6 + 4\text{H}_2$$
9.21
Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
Solution:
The carbon skeleton of 2-methylbutane is: $\overset{1}{\text{C}}\overset{2}{\text{C}}(\text{CH}_3)\overset{3}{\text{C}}\overset{4}{\text{C}}$
Alkenes that yield 2-methylbutane on catalytic hydrogenation are obtained by placing a double bond between adjacent carbons:
Between $C1$ and $C2$: $\text{CH}_2=\text{C}(\text{CH}_3)\text{CH}_2\text{CH}_3$ : 2-Methylbut-1-ene
Between $C2$ and $C3$: $\text{CH}_3\text{C}(\text{CH}_3)=\text{CH}\text{CH}_3$ : 2-Methylbut-2-ene
Between $C3$ and $C4$: $\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}=\text{CH}_2$ : 3-Methylbut-1-ene
Thus, exactly 3 alkenes yield 2-methylbutane on hydrogenation.
9.22
Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, $E^+$:
Electron-donating groups ($-\text{CH}_3$) activate the ring towards electrophiles, while electron-withdrawing groups ($-\text{NO}_2, -\text{Cl}$) deactivate the ring. The more $-\text{NO}_2$ groups attached, the less reactive the ring.
Decreasing Reactivity:
$$\mathbf{\text{Chlorobenzene} > p\text{-Nitrochlorobenzene} > \text{2,4-Dinitrochlorobenzene}}$$
Explanation: $-\text{NO}_2$ is strongly electron-withdrawing. One nitro group in $p$-nitrochlorobenzene deactivates the ring, and two nitro groups in $2,4$-dinitrochlorobenzene deactivate it even more severely.
Decreasing Reactivity:
$$\mathbf{\text{Toluene} > p\text{-Nitrotoluene (}p\text{-H}_3\text{C}\text{C}_6\text{H}_4\text{NO}_2\text{)} > p\text{-Dinitrobenzene (}p\text{-O}_2\text{N}\text{C}_6\text{H}_4\text{NO}_2\text{)}}$$
Explanation: Toluene has an electron-donating methyl group. In $p$-nitrotoluene, the activating effect of methyl is opposed by the deactivating nitro group. In $p$-dinitrobenzene, both substituents are powerfully deactivating.
9.23
Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why?
Solution:
Order of Ease of Nitration:
$$\mathbf{\text{Toluene} > \text{Benzene} > m\text{-Dinitrobenzene}}$$
Explanation:
Nitration is an electrophilic aromatic substitution reaction where the rate-determining step involves attack by the electrophile $\text{NO}_2^+$.
In toluene, the methyl group ($-\text{CH}_3$) donates electrons via hyperconjugation and $+I$ effect, increasing electron density on the ring and stabilising the carbocation intermediate. Hence, toluene undergoes nitration most easily.
In benzene, there are no activating or deactivating substituents.
In $m$-dinitrobenzene, two strongly electron-withdrawing nitro groups ($-I$ and $-R$ effects) deplete electron density from the ring, making electrophilic attack exceedingly slow and difficult.
9.24
Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
Solution:
Any strong electron-pair acceptor (Lewis acid) capable of polarizing or cleaving the alkyl halide bond can be used:
Anhydrous Ferric Chloride ($\text{FeCl}_3$)
Anhydrous Aluminium Bromide ($\text{AlBr}_3$)
Boron Trifluoride ($\text{BF}_3$) or Stannic Chloride ($\text{SnCl}_4$)
9.25
Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.
Solution:
The Wurtz reaction involves coupling two alkyl radicals. If an alkane with an odd number of carbon atoms (such as propane, $\text{C}_3\text{H}_8$) is to be synthesized, a mixture of two different alkyl halides must be used (e.g. bromomethane and bromoethane).
In the reaction mixture, three different coupling reactions take place simultaneously:
Conclusion: A mixture of three alkanes (ethane, propane, and butane) is obtained. Because these homologues have close boiling points, fractional distillation is difficult and the yield of the desired odd-carbon alkane is extremely poor. Hence, the Wurtz reaction is unsuitable for odd-carbon alkanes.
3 Graded Test Levels: Foundation (CBSE Board nomenclature & concepts), Intermediate (Mechanisms, Markovnikov/anti-Markovnikov & directing effects), and Advanced (JEE/NEET multi-step syntheses & stereochemistry). Select answers and click Submit to check your score.
L1.1
What is the IUPAC name of $(CH_3)_3CCH_2CH(CH_3)_2$?
Incorrect: Numbering from the left gives lower locants (2,2,4) rather than (2,4,4).
Correct Answer: Numbering from left gives methyl substituents at C2, C2, and C4 on a 5-carbon pentane chain (2,2,4-trimethylpentane).
Incorrect: Common radical name, not systematic IUPAC.
Incorrect: The longest continuous chain contains 5 carbons.
L1.2
Why is the staggered conformation of ethane more stable than the eclipsed conformation?
Correct Answer: In staggered ethane, CH bonds are at maximum angular separation (60°), minimising electron-pair repulsions (torsional strain is 0 kJ/mol).
Incorrect: CC single bond length remains 154 pm in both conformers.
Incorrect: Alkanes have non-polar CH bonds and cannot form H-bonds.
Incorrect: Staggered form actually has better orbital overlap alignment (σ → σ* interaction).
L1.3
Ethyne is industrially prepared on a large scale by treating calcium carbide ($CaC_2$) with:
Incorrect: Acid is unnecessary and can dehydrate or oxidize products.
Incorrect: Sodium is a reducing metal, not a hydrolysing agent.
Correct Answer: $CaC_2 + 2H_2O o Ca(OH)_2 + C_2H_2$ (Ethyne).
Incorrect: Hydrogenation does not hydrolyse carbides.
L1.4
What is Lindlar's catalyst and what is its specific stereo-outcome in alkyne reduction?
Incorrect: That is Birch reduction.
Incorrect: $AlCl_3$ is a Lewis acid catalyst for Friedel-Crafts and isomerisation.
Incorrect: Unpoisoned nickel reduces fully to saturated alkanes.
Correct Answer: Lindlar's catalyst is partially deactivated (poisoned) $Pd/CaCO_3$ which halts reduction at the alkene stage and delivers syn-addition to give predominantly cis-alkenes.
L1.5
Which reagent can chemically distinguish between propene ($CH_3CH=CH_2$) and propyne ($CH_3C\equiv CH$)?
Correct Answer: Terminal alkynes have acidic acetylenic hydrogen and form a white precipitate ($AgC\equiv CCH_3$) with Tollens' reagent or red precipitate with ammoniacal $Cu_2Cl_2$. Alkenes give no reaction.
Incorrect: Both propene and propyne decolorize bromine water (both are unsaturated).
Incorrect: Both decolorize pink $KMnO_4$.
Incorrect: Both dissolve in conc. $H_2SO_4$ via addition.
Result
0 / 5
L2.1
Why does the peroxide effect (anti-Markovnikov addition) operate only with $HBr$, and not with $HCl$ or $HI$?
Incorrect: Acid strength is not the determinant for radical propagation thermochemistry.
Correct Answer: In $HCl$, hydrogen abstraction by alkoxy radical is endothermic ($HCl$ bond is very strong: 430.5 kJ/mol). In $HI$, addition of $I^ullet$ to alkene is endothermic and iodine radicals prefer coupling into $I_2$. Only for $HBr$ are both propagation steps exothermic!
Incorrect: Chlorine is more electronegative than bromine.
Incorrect: Benzoyl peroxide homolyses thermally regardless of halide.
L2.2
An alkene 'X' ($C_6H_{12}$) on ozonolysis followed by $Zn/H_2O$ yields acetone ($CH_3COCH_3$) and propanal ($CH_3CH_2CHO$). The IUPAC name of 'X' is:
Incorrect: Hex-2-ene gives acetaldehyde and butanal.
Incorrect: Gives two molecules of acetone.
Correct Answer: Joining the carbonyl carbons $(CH_3)_2C=O$ and $O=CHCH_2CH_3$ by a double bond yields $(CH_3)_2C=CHCH_2CH_3$, which is 2-methylpent-2-ene.
Incorrect: 3-Methylpent-2-ene gives butan-2-one and acetaldehyde.
L2.3
Why is the chlorine substituent in chlorobenzene ortho/para-directing yet overall deactivating?
Incorrect: It is the reverse; -I controls reactivity, +R controls orientation.
Correct Answer: Chlorine is electronegative, so strong -I withdraws electrons from the entire ring making it less reactive than benzene. However, resonance (+R) delocalises chlorine lone pairs into ortho and para positions, making these sites comparatively richer than meta for electrophilic attack.
L2.4
What is the decreasing order of reactivity towards electrophilic aromatic nitration?
Correct Answer: $-CH_3$ is activating (+I, hyperconjugation) $ o$ Toluene > Benzene. $-Cl$ is weakly deactivating (-I) $ o$ Chlorobenzene. $-NO_2$ is strongly deactivating (-R, -I) $ o$ Nitrobenzene, and two $-NO_2$ groups deactivate even more.
Incorrect: Nitro groups deactivate electrophilic substitution.
Incorrect: Benzene is more reactive than chlorobenzene.
Incorrect: Toluene is more reactive than benzene.
L2.5
When benzene is treated with 1-chloropropane in the presence of anhydrous $AlCl_3$, why is isopropylbenzene (cumene) obtained as the major product?
Incorrect: Friedel-Crafts alkylation involves carbocations, not free radicals.
Correct Answer: Generation of $CH_3CH_2CH_2^+$ is rapidly followed by a 1,2-hydride shift ($\sim H^-$) to form the more stable $2^\circ$ carbocation $CH_3\overset{+}{C}HCH_3$, which then attacks benzene to yield cumene (~70%).
Incorrect: Isopropyl group is bulkier than n-propyl group.
Incorrect: Aromatic 6-membered ring remains intact.
Result
0 / 5
L3.1
An organic hydrocarbon (A) with formula $C_6H_{10}$ consumes 2 moles of $H_2$ to give 2-methylpentane. (A) gives a white precipitate with Tollens' reagent and on hydration with $HgSO_4/H_2SO_4$ yields compound (B) ($C_6H_{12}O$). Identify (A):
Incorrect: Hydrogenation of hex-1-yne gives straight-chain n-hexane, not 2-methylpentane.
Incorrect: 4-Methylpent-2-yne is an internal alkyne and will not form a precipitate with Tollens' reagent.
Correct Answer: $(CH_3)_2CHCH_2C\equiv CH$ is a terminal alkyne (gives white ppt with Tollens'), hydrogenates to 2-methylpentane, and hydrates via Markovnikov addition into 4-methylpentan-2-one ($C_6H_{12}O$).
Incorrect: Hydrogenation gives 2,2-dimethylbutane, not 2-methylpentane.
L3.2
What is the correct reaction sequence to synthesize m-bromochlorobenzene from benzene?
Incorrect: Chlorine is ortho/para-directing, producing o- and p-bromochlorobenzene.
Correct Answer: $-NO_2$ is a meta-director. Brominating nitrobenzene gives m-bromonitrobenzene. Reducing to amine, diazotising to diazonium salt, and Sandmeyer reaction replaces $-N_2^+$ with $-Cl$, cleanly yielding pure m-bromochlorobenzene.
Incorrect: Methyl is ortho/para-directing.
Incorrect: Wurtz-Fittig couples aryl and alkyl halides to give alkylbenzenes.
L3.3
But-2-yne on reduction with $Na/ ext{liq. } NH_3$ followed by addition of $Br_2$ in $CCl_4$ produces:
Correct Answer: $Na/ ext{liq. } NH_3$ anti-reduces but-2-yne to trans-but-2-ene. Electrophilic addition of $Br_2$ is an anti-addition. Anti-addition on a trans-alkene yields the meso compound (CAR/TAM mnemonic: Trans + Anti = Meso).
Incorrect: Racemic mixture is obtained from anti-addition to a cis-alkene (Cis + Anti = Racemic).
Incorrect: Double bond is at C2C3.
Incorrect: Only 1 mole of $Br_2$ adds to the alkene.
L3.4
Ozonolysis of an aromatic hydrocarbon 'P' ($C_8H_{10}$) yields glyoxal ($CHOCHO$) and methylglyoxal ($CH_3COCHO$) in a 1:2 molar ratio with NO dimethylglyoxal. What is 'P'?
Incorrect: o-Xylene has Kekulé canonical forms that produce dimethylglyoxal ($CH_3COCOCH_3$).
Incorrect: p-Xylene gives only methylglyoxal (no glyoxal).
Correct Answer: m-Xylene on ozonolysis cleaves at the three double bonds of its resonance structures to give 1 mole of glyoxal and 2 moles of methylglyoxal, with zero dimethylglyoxal formed.
L3.5
What is the correct order of stability for the conformers of n-butane during rotation about the C2C3 bond?
Incorrect: This is the increasing order of potential energy, not stability.
Incorrect: Anti-periplanar is more stable than gauche due to zero steric repulsion between bulky methyl groups.
Correct Answer: Anti-periplanar has bulky methyls furthest apart ($180^\circ$, lowest energy = 0 kJ/mol). Gauche has minor steric crowding ($+3.8 ext{ kJ/mol}$). Partially eclipsed has $H/CH_3$ eclipse ($+16 ext{ kJ/mol}$). Fully eclipsed has severe $CH_3/CH_3$ eclipse + torsional strain ($+19 ext{ kJ/mol}$, least stable).
Incorrect: Fully eclipsed is the least stable conformer.