Hydrocarbons

What to remember

• Alkanes undergo free-radical substitution (chain mechanism: initiation, propagation, termination); ease of replacement of H is 3° > 2° > 1°. Alkenes and alkynes undergo electrophilic addition, following Markovnikov's rule, except for HBr with peroxide (anti-Markovnikov).
• Alkene preparation and reduction control stereochemistry: Lindlar catalyst gives cis-alkene from an alkyne, and Na in liquid NH₃ gives trans-alkene.
• Terminal alkynes are weakly acidic (sp carbon, 50 % s-character): they react with Na and with ammoniacal AgNO₃ or Cu₂Cl₂. Alkenes and internal alkynes do not.
• Benzene is aromatic (planar, cyclic, 6 delocalised π electrons, 4n + 2 with n = 1) and gives electrophilic substitution, not addition. –OH, –NH₂, –CH₃ and halogens direct ortho/para; –NO₂, –CN, –CHO, –COOH and –SO₃H direct meta.

Classification and alkanes

Hydrocarbons are saturated (alkanes, cycloalkanes), unsaturated (alkenes, alkynes) and aromatic. Alkanes are CₙH₂ₙ₊₂, alkenes CₙH₂ₙ, alkynes CₙH₂ₙ₋₂.

Nomenclature and isomerism. Alkane names use the longest chain with the lowest locants for substituents, in alphabetical order, for example 2-methylbutane and 2,2-dimethylpropane. Alkanes show chain isomerism: C₄H₁₀ has 2 isomers, C₅H₁₂ has 3 and C₆H₁₄ has 5.

Conformations of ethane. Rotation about the C–C σ bond gives different arrangements (conformations) without breaking any bond. The two extremes:

The energy difference is about 12.5 kJ mol⁻¹. This small barrier is called torsional strain, and the molecule rotates rapidly at room temperature. In a Newman projection, the front carbon is a point where three bonds meet and the rear carbon is a circle with three bonds. A sawhorse projection shows the C–C bond as a slanted line with three bonds at each end. In the staggered Newman view each front C–H bond lies midway between two rear C–H bonds; in the eclipsed view the rear C–H bonds hide behind the front ones.

Preparation of alkanes

Physical properties. C₁–C₄ are gases, C₅–C₁₇ are liquids and higher members are solids. Alkanes are non-polar and insoluble in water. The boiling point rises with molar mass. For isomers, branching lowers the boiling point because the molecule is more compact and has weaker van der Waals forces: pentane > 2-methylbutane > 2,2-dimethylpropane.

Chemical reactions

• Halogenation (free-radical substitution). CH₄ + Cl₂ → CH₃Cl + HCl in UV light, heat or at 520–670 K. Further products are CH₂Cl₂, CHCl₃ and CCl₄. Reactivity of halogens: F₂ > Cl₂ > Br₂ > I₂. Iodination is reversible, so it needs an oxidising agent such as HIO₃ or HNO₃ to remove HI.
• Mechanism (chlorination of methane):
• Initiation: Cl–Cl → 2Cl• (homolytic fission by light or heat).
• Propagation: CH₄ + Cl• → •CH₃ + HCl ; •CH₃ + Cl₂ → CH₃Cl + Cl•.
• Termination: Cl• + Cl• → Cl₂ ; •CH₃ + •CH₃ → CH₃CH₃ ; •CH₃ + Cl• → CH₃Cl.
• Combustion. CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = −890 kJ mol⁻¹. General: CₙH₂ₙ₊₂ + (3n + 1)/2 O₂ → nCO₂ + (n + 1)H₂O.
• Controlled oxidation. 2CH₄ + O₂ → 2CH₃OH (Cu, 523 K, 100 atm); CH₄ + O₂ → HCHO + H₂O (Mo₂O₃); 2CH₃CH₃ + 3O₂ → 2CH₃COOH + 2H₂O ((CH₃COO)₂Mn, heat); (CH₃)₃CH → (CH₃)₃COH with KMnO₄.
• Isomerisation. n-Butane → 2-methylpropane with anhydrous AlCl₃ and HCl.
• Aromatisation. n-Hexane → benzene + 4H₂ at 773 K, 10–20 atm, with Cr₂O₃/V₂O₅/Mo₂O₃ on alumina.
• Pyrolysis (cracking). Heating higher alkanes at high temperature in absence of air gives lower alkanes, alkenes and hydrogen, for example C₁₂H₂₆ → C₇H₁₆ + C₅H₁₀.

Alkenes

The double-bond carbons are sp² hybridised. The C=C bond (134 pm) is one σ and one π bond. The six atoms of ethene lie in one plane with H–C–H and H–C=C angles near 120°. Restricted rotation about C=C gives geometrical (cis-trans) isomerism when each carbon has two different groups.

Preparation

Reactions

• Addition of H₂: RCH=CH₂ + H₂ → RCH₂CH₃ (Pt, Pd or Ni).
• Addition of halogens: CH₂=CH₂ + Br₂ (in CCl₄) → BrCH₂CH₂Br. The reddish-orange colour of bromine is discharged. This is a test for unsaturation.
• Addition of HX (reactivity HI > HBr > HCl). Markovnikov's rule: in addition of an unsymmetrical reagent HX, the negative part goes to the carbon with fewer hydrogen atoms (the more substituted carbon). CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (2-bromopropane). Mechanism: (1) H⁺ adds to the double bond to form the more stable carbocation, the secondary (CH₃)₂CH⁺ rather than the primary CH₃CH₂CH₂⁺ (slow step); (2) Br⁻ attacks the carbocation (fast step).
• Peroxide (Kharasch) effect: with peroxide, HBr adds against Markovnikov's rule by a free-radical chain mechanism: CH₃CH=CH₂ + HBr → CH₃CH₂CH₂Br (1-bromopropane). It occurs only with HBr, not with HCl or HI. The Br• adds first, forming the more stable secondary free radical.
• Addition of H₂SO₄ and water: CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₃H (ethyl hydrogensulphate). Propene + conc. H₂SO₄ gives isopropyl hydrogensulphate, which on boiling with water gives propan-2-ol. Hydration with dilute H₂SO₄ also follows Markovnikov's rule: propene → propan-2-ol.
• Oxidation with Baeyer's reagent (cold, dilute, alkaline KMnO₄): CH₂=CH₂ → HOCH₂CH₂OH (ethane-1,2-diol). The pink colour is discharged, and this is a test for unsaturation. With acidic KMnO₄ or K₂Cr₂O₇ the C=C bond is cleaved: CH₃CH=CHCH₃ → 2CH₃COOH; (CH₃)₂C=CH₂ → CH₃COCH₃ + CO₂ + H₂O.
• Ozonolysis: the alkene reacts with O₃ to form an ozonide, and Zn/H₂O gives aldehydes or ketones: CH₃CH=CH₂ → CH₃CHO + HCHO. It locates the position of the double bond.
• Polymerisation: n CH₂=CH₂ → (–CH₂–CH₂–)ₙ polythene (high temperature, high pressure, catalyst); propene gives polypropene.

Alkynes

The triple-bond carbons are sp hybridised and the molecule is linear. C≡C is 120 pm, one σ and two π bonds.

Preparation. (1) Calcium carbide: CaC₂ + 2H₂O → HC≡CH + Ca(OH)₂ (CaC₂ is made from CaO and carbon). (2) Vicinal dihalide with alcoholic KOH twice, then NaNH₂: BrCH₂CH₂Br → CH₂=CHBr → HC≡CH.

Acidic character of terminal alkynes. The sp carbon is more electronegative (50 % s-character), so the C–H bond is more polar and the hydrogen can be removed as H⁺. Acidic strength: HC≡CH > CH₂=CH₂ > CH₃CH₃.

• 2HC≡CH + 2Na → 2HC≡C⁻Na⁺ + H₂ (sodium ethynide)
• HC≡CH + ammoniacal AgNO₃ → white precipitate of silver acetylide
• HC≡CH + ammoniacal Cu₂Cl₂ → red precipitate of copper acetylide

These tests identify terminal alkynes (R–C≡C–H). But-2-yne does not respond.

Addition reactions

• H₂: HC≡CH → CH₂=CH₂ → CH₃CH₃ (Pt, Pd or Ni).
• Halogens: HC≡CH + Br₂ → CHBr=CHBr → CHBr₂CHBr₂ (1,1,2,2-tetrabromoethane); Br₂ is decolourised.
• HX: HC≡CH + HBr → CH₂=CHBr → CH₃CHBr₂ (1,1-dibromoethane, Markovnikov).
• Water: HC≡CH + H₂O → [CH₂=CHOH] → CH₃CHO (dilute H₂SO₄, HgSO₄, 333 K). The unstable enol (vinyl alcohol) tautomerises to the aldehyde. Propyne gives propanone.

Polymerisation. Linear polymerisation gives polyacetylene. Cyclic polymerisation: 3HC≡CH → benzene when ethyne is passed through a red-hot iron tube at 873 K.

Aromatic hydrocarbons: benzene

Structure. Benzene is C₆H₆, a planar ring of six sp² carbon atoms with a delocalised π cloud above and below the ring. Benzene is the resonance hybrid of two structures with alternating single and double bonds. All six C–C bonds are equal, about 139 pm (between 154 pm for C–C and 134 pm for C=C). Resonance makes benzene unusually stable, so it prefers substitution over addition.

Aromaticity (Hückel rule). A compound is aromatic if it is planar, cyclic, fully conjugated, and has (4n + 2) π electrons, where n = 0, 1, 2, …

Preparation. (1) Cyclic polymerisation of ethyne. (2) Decarboxylation: C₆H₅COONa + NaOH → C₆H₆ + Na₂CO₃ (CaO, heat). (3) Reduction of phenol with zinc dust: C₆H₅OH + Zn → C₆H₆ + ZnO.

Electrophilic substitution. Mechanism in three steps: (1) generation of the electrophile; (2) attack of the π electrons on the electrophile to give a resonance-stabilised carbocation (arenium ion, sigma complex), the slow step; (3) loss of H⁺ to restore the aromatic ring.

NO₂⁺ forms as: HNO₃ + 2H₂SO₄ ⇌ NO₂⁺ + H₃O⁺ + 2HSO₄⁻.

Addition and other reactions. C₆H₆ + 3H₂ → C₆H₁₂ (cyclohexane; Ni, 453–473 K). C₆H₆ + 3Cl₂ → C₆H₆Cl₆ (benzene hexachloride, BHC) in UV light. Combustion: 2C₆H₆ + 15O₂ → 12CO₂ + 6H₂O, with a sooty flame because of the high carbon content.

Directive influence. A group already on the ring decides where the next substituent goes.

Activating groups donate electrons (+R, +I) and increase electron density at ortho and para positions. Deactivating meta directors withdraw electrons (−R, −I). Halogens withdraw by −I (deactivating) but their lone pairs donate by +R, which controls the position (ortho/para).

Carcinogenicity and toxicity. Benzene is toxic and is a carcinogen. Polynuclear hydrocarbons, which contain more than two fused benzene rings, are carcinogenic. They form by incomplete combustion of organic matter such as coal, petroleum and tobacco, and can damage DNA and cause cancer.

NEET traps

• Wurtz reaction gives alkanes with an even number of carbon atoms from one halide and cannot give methane; decarboxylation gives one carbon fewer.
• In Kolbe electrolysis the alkane forms at the anode, not at the cathode.
• Lindlar catalyst gives cis-alkene; Na/liquid NH₃ gives trans-alkene.
• The peroxide effect works only for HBr. It does not work for HCl or HI.
• Markovnikov addition goes through the more stable carbocation; anti-Markovnikov goes through the more stable free radical.
• Baeyer's reagent gives a diol and keeps the carbon skeleton; acidic KMnO₄ breaks the C=C bond.
• Only terminal alkynes give a precipitate with ammoniacal AgNO₃ or Cu₂Cl₂.
• Benzene does not decolourise bromine water or Baeyer's reagent, although it is unsaturated on paper.
• Halogens in benzene are ortho/para directing even though they are deactivating.
• Staggered ethane is more stable than eclipsed; the dihedral angle in staggered is 60°, in eclipsed 0°.

One-liners

• Alkane general formula is CₙH₂ₙ₊₂.
• The most stable conformation of ethane is staggered.
• The energy barrier between ethane conformations is about 12.5 kJ mol⁻¹.
• Methane burns with ΔH of −890 kJ mol⁻¹.
• Anhydrous AlCl₃/HCl isomerises n-butane to 2-methylpropane.
• n-Hexane gives benzene on aromatisation at 773 K.
• Ethene is made from ethanol with conc. H₂SO₄ at 443 K.
• Saytzeff rule gives the more substituted alkene.
• Baeyer's reagent is cold, dilute, alkaline KMnO₄.
• Ozonolysis of propene gives ethanal and methanal.
• Calcium carbide with water gives ethyne.
• Hydration of ethyne with Hg²⁺/H⁺ gives ethanal.
• Three molecules of ethyne give benzene.
• All C–C bonds in benzene are about 139 pm.
• The electrophile in nitration is NO₂⁺.

Practice questions

1. The Wurtz reaction of ethyl bromide with sodium metal in dry ether gives:

   1. butane
   2. hexane
   3. propane
   4. ethane
   Answer

   A. butane

   2CH₃CH₂Br + 2Na → CH₃CH₂CH₂CH₃ + 2NaBr; the two ethyl groups join.

2. Heating sodium propanoate with soda lime gives:

   1. methane
   2. ethane
   3. butane
   4. propane
   Answer

   B. ethane

   Decarboxylation removes one carbon as Na₂CO₃: CH₃CH₂COONa → CH₃CH₃.

3. On electrolysis of aqueous sodium acetate, the gas liberated at the cathode is:

   1. carbon dioxide
   2. methane
   3. ethane
   4. hydrogen
   Answer

   D. hydrogen

   Ethane and CO₂ form at the anode; H₂ and NaOH form at the cathode.

4. In the most stable conformation of ethane the dihedral angle between the nearest C–H bonds on the two carbons is:

   1. 180°
   2. 90°
   3. 60°
   4. 0°
   Answer

   C. 60°

   The staggered conformation has the nearest C–H bonds 60° apart; the eclipsed has 0°.

5. The correct order of boiling points of the isomers of C₅H₁₂ is:

   1. n-pentane > 2,2-dimethylpropane > 2-methylbutane
   2. 2,2-dimethylpropane > 2-methylbutane > n-pentane
   3. n-pentane > 2-methylbutane > 2,2-dimethylpropane
   4. 2-methylbutane > n-pentane > 2,2-dimethylpropane
   Answer

   C. n-pentane > 2-methylbutane > 2,2-dimethylpropane

   Branching reduces surface area and van der Waals forces, so more branching means a lower boiling point.

6. Iodination of methane is carried out in the presence of HIO₃ or HNO₃ in order to:

   1. oxidise HI and prevent the reverse reaction
   2. generate I• radicals faster
   3. reduce CH₃I to CH₄
   4. act as a solvent for iodine
   Answer

   A. oxidise HI and prevent the reverse reaction

   The reaction is reversible; HI would reduce CH₃I back to CH₄, so it is removed by oxidation.

7. The product of the reaction of propene with HBr in the absence of peroxide is:

   1. 1-bromopropane
   2. propan-2-ol
   3. 1,2-dibromopropane
   4. 2-bromopropane
   Answer

   D. 2-bromopropane

   Markovnikov addition: Br goes to the carbon with fewer hydrogen atoms.

8. The peroxide effect is shown by the addition of:

   1. HBr only
   2. HCl only
   3. HI only
   4. HCl, HBr and HI
   Answer

   A. HBr only

   Only HBr adds by the free-radical path in the presence of peroxide.

9. Reduction of but-2-yne with H₂ over Lindlar catalyst gives:

   1. butane
   2. cis-but-2-ene
   3. but-1-ene
   4. trans-but-2-ene
   Answer

   B. cis-but-2-ene

   Lindlar catalyst (poisoned Pd) stops at the cis-alkene.

10. Ozonolysis of CH₃CH=CHCH₃ followed by Zn/H₂O gives:

    1. ethanoic acid
    2. propanone
    3. methanal
    4. ethanal
    Answer

    D. ethanal

    Cleavage at the C=C bond gives two molecules of CH₃CHO.

11. Ethyne on hydration with dilute H₂SO₄ and HgSO₄ at 333 K gives:

    1. ethane-1,2-diol
    2. ethanal
    3. propanone
    4. ethanol
    Answer

    B. ethanal

    Vinyl alcohol first forms and tautomerises to CH₃CHO.

12. Which of the following gives a white precipitate with ammoniacal silver nitrate?

    1. CH₃C≡CCH₃
    2. CH₂=CHCH₃
    3. CH₃C≡CH
    4. CH₃CH₂CH₃
    Answer

    C. CH₃C≡CH

    Only terminal alkynes (with an acidic C–H) form silver acetylides.

13. The number of π electrons in naphthalene is:

    1. 10
    2. 14
    3. 6
    4. 8
    Answer

    A. 10

    Naphthalene has five double bonds in its fused rings, so 10 π electrons (4n + 2, n = 2).

14. The electrophile in the nitration of benzene is:

    1. NO₂•
    2. NO₂⁺
    3. NO₃⁻
    4. NO₂⁻
    Answer

    B. NO₂⁺

    Conc. HNO₃ with conc. H₂SO₄ forms the nitronium ion NO₂⁺.

15. Which of the following groups directs an incoming substituent to the meta position of benzene?

    1. –OH
    2. –CH₃
    3. –NH₂
    4. –NO₂
    Answer

    D. –NO₂

    –NO₂ is an electron-withdrawing meta-directing group; the others direct ortho/para.

16. Benzene reacts with CH₃COCl in the presence of anhydrous AlCl₃ to give:

    1. benzoic acid
    2. benzaldehyde
    3. acetophenone (C₆H₅COCH₃)
    4. toluene
    Answer

    C. acetophenone (C₆H₅COCH₃)

    Friedel–Crafts acylation introduces the CH₃CO group.

17. Given below are two statements: Statement I: The staggered conformation of ethane is more stable than the eclipsed conformation. Statement II: Torsional strain is greatest in the eclipsed conformation.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Repulsion between the C–H bonds is maximum when they are eclipsed.

18. Given below are two statements: Statement I: The Wurtz reaction cannot be used to prepare methane. Statement II: Heating sodium acetate with soda lime gives ethane.

    1. Both Statement I and Statement II are correct
    2. Statement I is correct but Statement II is incorrect
    3. Both Statement I and Statement II are incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Statement I is correct but Statement II is incorrect

    Wurtz coupling doubles the carbon number, so methane cannot form. Decarboxylation of CH₃COONa gives methane.

19. Given below are two statements: Statement I: Lindlar catalyst converts an alkyne into a trans-alkene. Statement II: Sodium in liquid ammonia converts an alkyne into a trans-alkene.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is incorrect but Statement II is correct
    4. Statement I is correct but Statement II is incorrect
    Answer

    C. Statement I is incorrect but Statement II is correct

    Lindlar gives the cis-alkene; Na/liquid NH₃ gives the trans-alkene.

20. Given below are two statements: Statement I: The addition of HBr to propene in the presence of peroxide proceeds by a free-radical mechanism. Statement II: 1-Bromopropane is the product of this reaction.

    1. Both Statement I and Statement II are incorrect
    2. Statement I is correct but Statement II is incorrect
    3. Statement I is incorrect but Statement II is correct
    4. Both Statement I and Statement II are correct
    Answer

    D. Both Statement I and Statement II are correct

    Br• adds first to give the more stable secondary radical, so Br ends up on the terminal carbon.

21. Given below are two statements: Statement I: Markovnikov addition of HX to propene goes through the less stable carbocation. Statement II: The peroxide effect is shown by HCl as well as HBr.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Markovnikov addition goes through the more stable carbocation, and the peroxide effect is shown only by HBr.

22. Given below are two statements: Statement I: Baeyer's reagent converts ethene into ethane-1,2-diol. Statement II: Acidic KMnO₄ oxidation of an alkene keeps the carbon skeleton unchanged.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Acidic KMnO₄ breaks the C=C bond.

23. Given below are two statements: Statement I: Benzene readily undergoes addition reactions like alkenes. Statement II: Benzene undergoes electrophilic substitution reactions.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Addition would destroy the stable delocalised π system, so benzene prefers substitution.

24. Which of the following statements are correct? 1. Ethyne is more acidic than ethene. 2. Ethyne forms a red precipitate with ammoniacal Cu₂Cl₂. 3. But-2-yne reacts with sodium to liberate hydrogen.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 2 only

    But-2-yne has no terminal C–H, so it does not react with sodium.

25. Which of the following statements are correct? 1. The electrophile in nitration of benzene is NO₂⁻. 2. The electrophile in sulphonation is SO₃. 3. Friedel–Crafts acylation uses CH₃COCl and anhydrous AlCl₃.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    C. 2 and 3 only

    The electrophile in nitration is NO₂⁺, not NO₂⁻.

26. Which of the following statements are correct? 1. In dehydrohalogenation the more substituted alkene is the major product. 2. Acid dehydration is easiest for primary alcohols. 3. Dehydrohalogenation uses alcoholic KOH.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1, 2 and 3
    4. 1 and 3 only
    Answer

    D. 1 and 3 only

    Acid dehydration is easiest for tertiary alcohols (3° > 2° > 1°).

27. Assertion (A): Terminal alkynes react with sodium to liberate hydrogen. Reason (R): An sp-hybridised carbon is more electronegative than an sp² or sp³ carbon, so the C–H bond is acidic.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    50 % s-character makes the sp carbon hold the bond pair more tightly, so H⁺ can be lost.

28. Assertion (A): Benzene undergoes electrophilic substitution rather than addition. Reason (R): Addition would destroy the stable delocalised π electron system.

    1. Both A and R are true but R is not the correct explanation of A
    2. Both A and R are true and R is the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true and R is the correct explanation of A

    Substitution keeps the aromatic ring intact.

29. Assertion (A): The Wurtz reaction can be used to prepare methane. Reason (R): The Wurtz reaction couples two alkyl groups from alkyl halides.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    Coupling two alkyl groups gives at least ethane, so methane cannot be formed.

30. Assertion (A): n-Hexane gives benzene when heated at 773 K with Cr₂O₃/V₂O₅/Mo₂O₃ on alumina. Reason (R): This reaction is called pyrolysis.

    1. Both A and R are true and R is the correct explanation of A
    2. A is true but R is false
    3. Both A and R are true but R is not the correct explanation of A
    4. A is false but R is true
    Answer

    B. A is true but R is false

    The reaction is aromatisation (cyclisation and dehydrogenation), not pyrolysis.

31. Assertion (A): Chlorobenzene reacts faster than benzene in electrophilic substitution. Reason (R): The chlorine atom is an ortho/para-directing group.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is false but R is true
    4. A is true but R is false
    Answer

    C. A is false but R is true

    Chlorine is deactivating (−I outweighs +R) so the reaction is slower, although it directs ortho/para.

32. Match List I with List II. List I: a. Wurtz reaction b. Kolbe electrolysis c. Decarboxylation d. Lindlar reduction List II: i. Electrolysis of the sodium salt of a carboxylic acid ii. Alkyl halide with sodium in dry ether iii. Sodium salt of an acid heated with soda lime iv. Poisoned Pd with H₂ converts an alkyne to a cis-alkene

    1. a-iii, b-i, c-ii, d-iv
    2. a-i, b-ii, c-iii, d-iv
    3. a-ii, b-i, c-iv, d-iii
    4. a-ii, b-i, c-iii, d-iv
    Answer

    D. a-ii, b-i, c-iii, d-iv

    Match each reaction with its reagent or condition.

33. Match List I with List II. List I: a. Propene + HBr (no peroxide) b. Propene + HBr (with peroxide) c. Ethyne + H₂O (Hg²⁺/H⁺) d. Propene + H₂O (dilute H₂SO₄) List II: i. 1-Bromopropane ii. 2-Bromopropane iii. Ethanal iv. Propan-2-ol

    1. a-ii, b-i, c-iii, d-iv
    2. a-ii, b-i, c-iv, d-iii
    3. a-i, b-ii, c-iii, d-iv
    4. a-ii, b-iii, c-i, d-iv
    Answer

    A. a-ii, b-i, c-iii, d-iv

    Markovnikov addition gives 2-bromopropane and propan-2-ol; peroxide gives 1-bromopropane; ethyne gives ethanal.

34. Match List I with List II. List I: a. Nitration b. Sulphonation c. Friedel–Crafts alkylation (CH₃Cl/AlCl₃) d. Friedel–Crafts acylation (CH₃COCl/AlCl₃) List II: i. CH₃CO⁺ ii. NO₂⁺ iii. SO₃ iv. CH₃⁺

    1. a-iii, b-ii, c-iv, d-i
    2. a-ii, b-iii, c-iv, d-i
    3. a-ii, b-iii, c-i, d-iv
    4. a-ii, b-iv, c-iii, d-i
    Answer

    B. a-ii, b-iii, c-iv, d-i

    Each reaction has its own electrophile.

35. Match List I with List II. List I: a. Benzene b. Naphthalene c. Anthracene d. Ethyne List II: i. 14 π electrons ii. 4 π electrons iii. 6 π electrons iv. 10 π electrons

    1. a-iii, b-iv, c-ii, d-i
    2. a-iv, b-iii, c-i, d-ii
    3. a-iii, b-iv, c-i, d-ii
    4. a-iii, b-i, c-iv, d-ii
    Answer

    C. a-iii, b-iv, c-i, d-ii

    Benzene 6, naphthalene 10, anthracene 14, ethyne 4 π electrons.

36. Match List I with List II. List I: a. Staggered ethane b. Eclipsed ethane c. Newman projection d. Sawhorse projection List II: i. C–C bond drawn as a slanted line ii. Dihedral angle 0° iii. Front carbon shown as a point and rear carbon as a circle iv. Dihedral angle 60°

    1. a-ii, b-iv, c-iii, d-i
    2. a-iv, b-iii, c-ii, d-i
    3. a-iv, b-ii, c-iii, d-i
    4. a-iv, b-ii, c-i, d-iii
    Answer

    C. a-iv, b-ii, c-iii, d-i

    Staggered 60°, eclipsed 0°; Newman uses a point and a circle; sawhorse uses a slanted C–C line.

37. Match List I with List II. List I: a. Aromatisation b. Isomerisation c. Controlled oxidation of methane to methanol d. Pyrolysis List II: i. Cu, 523 K, 100 atm ii. High temperature in the absence of air iii. Anhydrous AlCl₃ and HCl iv. Cr₂O₃/V₂O₅/Mo₂O₃, 773 K

    1. a-iv, b-iii, c-ii, d-i
    2. a-i, b-iii, c-iv, d-ii
    3. a-iii, b-iv, c-i, d-ii
    4. a-iv, b-iii, c-i, d-ii
    Answer

    D. a-iv, b-iii, c-i, d-ii

    Match each alkane reaction with its conditions.

38. An alkene on ozonolysis (Zn/H₂O) gives CH₃COCH₃ and CH₃CHO. The alkene is:

    1. 2-methylbut-2-ene
    2. 2-methylpropene
    3. pent-2-ene
    4. but-2-ene
    Answer

    A. 2-methylbut-2-ene

    (CH₃)₂C=CHCH₃ cleaves to CH₃COCH₃ and CH₃CHO.

39. A hydrocarbon of formula C₃H₄ gives a white precipitate with ammoniacal AgNO₃. On hydration with HgSO₄ and dilute H₂SO₄ it gives:

    1. propanoic acid
    2. propanone
    3. propan-2-ol
    4. propanal
    Answer

    B. propanone

    It is propyne; Markovnikov hydration gives an enol that becomes CH₃COCH₃.

40. The number of structurally different monochloro derivatives of 2-methylbutane is:

    1. 3
    2. 5
    3. 4
    4. 2
    Answer

    C. 4

    The distinct positions are the two equivalent CH₃ on C-2, C-2 (tertiary), C-3 and C-4: four products.

41. The major product of dehydrohalogenation of 2-bromobutane with alcoholic KOH is:

    1. butan-2-ol
    2. but-1-ene
    3. butane
    4. but-2-ene
    Answer

    D. but-2-ene

    By Saytzeff's rule the more substituted alkene is the major product.

42. An alkene reacts with HBr in the absence of peroxide to give 2-bromo-2-methylpropane. The alkene is:

    1. but-2-ene
    2. 2-methylpropene
    3. propene
    4. but-1-ene
    Answer

    B. 2-methylpropene

    H⁺ adds to the CH₂ carbon to give the tertiary carbocation (CH₃)₃C⁺, then Br⁻ adds.

43. Nitration of nitrobenzene with conc. HNO₃ and conc. H₂SO₄ gives mainly:

    1. 1,3-dinitrobenzene
    2. 1,2-dinitrobenzene
    3. 1,3,5-trinitrobenzene
    4. 1,4-dinitrobenzene
    Answer

    A. 1,3-dinitrobenzene

    The –NO₂ group is meta-directing and deactivating.

44. Reduction of but-2-yne with sodium in liquid ammonia gives:

    1. butane
    2. cis-but-2-ene
    3. trans-but-2-ene
    4. but-1-ene
    Answer

    C. trans-but-2-ene

    Na/liquid NH₃ gives trans-alkenes.