Class 11Chemistry · Organic ChemistryFull chapter

Hydrocarbons

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

Classification and Nomenclature of Hydrocarbons

Quick answer Hydrocarbons are compounds of only carbon and hydrogen; they are classified as saturated, unsaturated, or aromatic, and named using systematic IUPAC rules.

Hydrocarbons are organic compounds made up of only carbon and hydrogen atoms. They are broadly classified into saturated hydrocarbons (alkanes, containing only C-C and C-H single bonds), unsaturated hydrocarbons (alkenes with a C=C double bond and alkynes with a C≡C triple bond), and aromatic hydrocarbons (benzene and its derivatives, containing a stable ring of delocalised pi electrons). Saturated rings that are not aromatic are called alicyclic hydrocarbons (cycloalkanes).

IUPAC nomenclature of hydrocarbons follows a set procedure: (i) select the longest continuous carbon chain containing the maximum number of multiple bonds (if any) as the parent chain; (ii) number the chain from the end that gives the lowest locants to the multiple bond and/or substituents; (iii) name substituents as prefixes in alphabetical order with their locants; (iv) use the suffix -ane for alkanes, -ene for alkenes and -yne for alkynes.

Worked example: Name the compound CH3-CH(CH3)-CH(CH3)-CH2-CH3.

  • The longest chain has 5 carbons (pentane).
  • Numbering from the left gives methyl groups at C2 and C3 (locants 2,3); numbering from the right would give locants 3,4. Since {2,3} is lower, we number from the left.
  • Two identical substituents are named using the multiplying prefix di: 2,3-dimethyl.
  • The IUPAC name is therefore 2,3-dimethylpentane.
General formula of alkanes CₙH₂ₙ₊₂
General formula of alkenes CₙH₂ₙ
General formula of alkynes CₙH₂ₙ₋₂
General formula of arenes (benzene series) CₙH₂ₙ₋₆ for a single benzene ring, n ≥ 6
Remember
  • Hydrocarbons contain only C and H; classified as saturated (alkanes), unsaturated (alkenes, alkynes) and aromatic.
  • General formulas: alkanes CnH2n+2, alkenes CnH2n, alkynes CnH2n-2 (n = number of carbon atoms).
  • IUPAC naming: longest chain + lowest locants + alphabetically ordered substituent prefixes + appropriate suffix (-ane/-ene/-yne).
  • Chain isomerism (different carbon skeletons) and position isomerism (different location of multiple bond/substituent) are common among hydrocarbons.

Alkanes: Preparation and Properties

Quick answer Alkanes are prepared by Wurtz reaction, decarboxylation, Kolbe electrolysis and hydrogenation, and they mainly undergo free-radical substitution (halogenation) and combustion.

Alkanes can be prepared by several methods. In the Wurtz reaction, an alkyl halide is treated with sodium metal in dry ether to give a higher alkane with double the number of carbon atoms; a single alkyl halide always gives an alkane with an even number of carbon atoms. Alkanes are also obtained by decarboxylation of the sodium salt of a carboxylic acid heated with soda lime, by Kolbe's electrolytic method (electrolysis of an aqueous solution of a sodium/potassium carboxylate), and by catalytic hydrogenation of alkenes or alkynes over finely divided Ni, Pd or Pt.

Alkanes are largely unreactive towards common acids, bases and oxidising agents because of their strong, non-polar C-C and C-H sigma bonds, but they undergo free-radical substitution with halogens in sunlight or UV light, and they burn completely in excess oxygen (combustion) to give carbon dioxide and water with release of heat. They also undergo controlled oxidation, isomerisation, aromatisation and pyrolysis (cracking) under suitable conditions.

Worked example (mechanism of free-radical chlorination of methane):

  • Initiation: Cl2 absorbs UV light and undergoes homolytic fission to give two chlorine free radicals.
  • Propagation step 1: A chlorine radical abstracts a hydrogen atom from methane, generating a methyl free radical and HCl.
  • Propagation step 2: The methyl free radical reacts with another Cl2 molecule to give chloromethane and regenerate a chlorine radical, which continues the chain.
  • Termination: Two radicals combine (Cl• + Cl•, CH3• + Cl•, or CH3• + CH3•) to stop the chain; excess Cl2 allows further substitution to give CH2Cl2, CHCl3 and CCl4.
Wurtz reaction 2R-X + 2Na → (dry ether) R-R + 2NaX
Decarboxylation (soda lime) CH₃COONa + NaOH → (CaO, Δ) CH₄ + Na₂CO₃
Kolbe's electrolytic method 2CH₃COONa + 2H₂O → (electrolysis) CH₃-CH₃ + 2CO₂ + 2NaOH + H₂
Catalytic hydrogenation CH₂=CH₂ + H₂ → (Ni/Pd/Pt) CH₃-CH₃
Free-radical chlorination (overall) CH₄ + Cl₂ → (hv) CH₃Cl + HCl
Complete combustion of methane CH₄ + 2O₂ → CO₂ + 2H₂O
Remember
  • Wurtz reaction: 2R-X + 2Na (dry ether) gives R-R + 2NaX; gives alkanes with an even number of carbons from a single alkyl halide.
  • Decarboxylation (soda-lime) and Kolbe's electrolysis both convert carboxylic acid salts into alkanes.
  • Free-radical halogenation proceeds via initiation, propagation and termination steps and needs sunlight/UV light.
  • Alkanes undergo combustion to CO2 and H2O and can also undergo controlled oxidation, isomerisation, aromatisation, and pyrolysis (cracking).

Alkenes: Preparation and Properties

Quick answer Alkenes are prepared by dehydration of alcohols or dehydrohalogenation of alkyl halides and characteristically undergo electrophilic addition governed by Markovnikov's rule.

Alkenes are commonly prepared by dehydration of alcohols using concentrated H2SO4 with heating, and by dehydrohalogenation of alkyl halides using alcoholic KOH, in which the more substituted alkene is usually the major product (Saytzeff's rule). Alkenes can also be obtained by partial hydrogenation of alkynes: Lindlar's catalyst (Pd on BaSO4, poisoned with quinoline) gives the cis alkene, while sodium in liquid ammonia gives the trans alkene.

The C=C double bond makes alkenes reactive towards electrophilic addition. Addition of hydrogen halides (HX) follows Markovnikov's rule: the halide ion attaches to the carbon already carrying fewer hydrogen atoms, because the reaction proceeds through the more stable carbocation intermediate. In the presence of peroxides, HBr alone adds in the opposite sense (anti-Markovnikov), known as the peroxide (Kharasch) effect, since the mechanism switches from ionic to free-radical. Alkenes also add H2SO4, water (acid-catalysed hydration) and halogens; they decolourise cold, dilute, alkaline KMnO4 (Baeyer's test, forming a vicinal diol) and undergo ozonolysis, which cleaves the double bond to give two carbonyl compounds and is used to locate the position of the double bond.

Worked example (Markovnikov addition): Propene reacts with HBr in the absence of peroxide.

  • H+ adds first to the terminal (less substituted) carbon, generating the more stable secondary carbocation CH3-C+H-CH3 rather than the less stable primary carbocation.
  • Br- then attacks this secondary carbocation.
  • The major product is 2-bromopropane, CH3-CHBr-CH3, consistent with Markovnikov's rule.
Dehydration of ethanol CH₃CH₂OH → (conc. H₂SO₄, 443 K) CH₂=CH₂ + H₂O
Dehydrohalogenation (Saytzeff) CH₃-CH₂-CHBr-CH₃ + KOH(alc.) → CH₃-CH=CH-CH₃ (major) + KBr + H₂O
Markovnikov addition of HBr CH₃-CH=CH₂ + HBr → CH₃-CHBr-CH₃
Anti-Markovnikov (peroxide effect) CH₃-CH=CH₂ + HBr → (peroxide) CH₃-CH₂-CH₂Br
Baeyer's test (cold dil. alkaline KMnO4) 3CH₂=CH₂ + 2KMnO₄ + 4H₂O → 3HOCH₂-CH₂OH + 2MnO₂ + 2KOH
Ozonolysis of but-2-ene CH₃-CH=CH-CH₃ → (i. O₃ ii. Zn/H₂O) 2CH₃CHO
Remember
  • Dehydration of alcohols (conc. H2SO4) and dehydrohalogenation of alkyl halides (alc. KOH, Saytzeff's rule) are the main preparation routes.
  • Lindlar's catalyst gives cis-alkene from alkyne; Na/liquid NH3 gives trans-alkene.
  • Markovnikov's rule: H adds to the carbon with more H atoms already; the more stable carbocation intermediate decides the major product.
  • Peroxide effect (Kharasch effect) reverses Markovnikov addition, but only for HBr, via a free-radical mechanism.
  • Ozonolysis cleaves C=C to give two carbonyl compounds and reveals the original position of the double bond.

Alkynes: Preparation and Properties

Quick answer Alkynes, prepared mainly from calcium carbide or vicinal dihalides, show acidic terminal C-H bonds and undergo electrophilic/nucleophilic addition across the triple bond.

The most important preparation of ethyne (acetylene) is the reaction of calcium carbide with water. Alkynes are also prepared by double dehydrohalogenation of vicinal (or gem) dihalides using excess alcoholic KOH.

A unique feature of terminal alkynes (R-C≡C-H) is the acidic character of the hydrogen attached to the triple-bonded carbon, because the sp-hybridised carbon holds its bonding electrons closer to the nucleus. This hydrogen can be replaced by sodium (using Na or NaNH2) to form sodium acetylides, and terminal alkynes give a characteristic white precipitate with ammoniacal silver nitrate (and a red precipitate with ammoniacal cuprous chloride), which serves as a test to distinguish terminal from non-terminal alkynes. Alkynes undergo addition of H2, halogens, and hydrogen halides (following Markovnikov's rule for unsymmetrical alkynes) across the triple bond, and controlled acid-catalysed hydration in presence of Hg2+/H2SO4 converts them into carbonyl compounds via an unstable enol intermediate.

Worked example (hydration of ethyne):

  • Ethyne reacts with water in the presence of 40% H2SO4 and HgSO4 catalyst.
  • Markovnikov addition of water across the triple bond first gives the unstable enol, CH2=CH-OH.
  • The enol immediately tautomerises (keto-enol tautomerism) to the more stable carbonyl form.
  • The final product is acetaldehyde (ethanal), CH3CHO; higher terminal alkynes give methyl ketones by the same route.
Preparation of ethyne from calcium carbide CaC₂ + 2H₂O → HC≡CH + Ca(OH)₂
Preparation from vicinal dihalide CH₃-CHBr-CH₂Br + 2KOH(alc.) → CH₃-C≡CH + 2KBr + 2H₂O
Acidic character (sodium acetylide) HC≡CH + Na → HC≡C-Na + ½H₂
Test for terminal alkyne R-C≡C-H + [Ag(NH₃)₂]⁺ → R-C≡C-Ag↓(white) + NH₃ + NH₄⁺
Markovnikov hydration of ethyne HC≡CH + H₂O → (Hg²⁺/H₂SO₄) CH₃CHO
Remember
  • CaC2 + 2H2O gives C2H2 + Ca(OH)2, the standard preparation of ethyne.
  • Terminal alkynes are weakly acidic; they form acetylides with Na/NaNH2 and give a white precipitate with ammoniacal AgNO3 (test for terminal triple bond).
  • Addition reactions (H2, X2, HX, H2O) occur in two stages across the triple bond and follow Markovnikov's rule for unsymmetrical reagents.
  • Hg2+/H2SO4-catalysed hydration converts alkynes into aldehydes (from ethyne) or ketones (from higher alkynes) via an enol intermediate.

Benzene: Structure and Aromaticity

Quick answer Benzene's structure is a resonance hybrid of equivalent Kekulé forms with a delocalised pi system, and Hückel's (4n+2) rule sets the criterion for aromaticity.

Benzene, C6H6, is represented by two equivalent Kekulé structures with alternating single and double bonds; the true structure is a resonance hybrid of these (and other minor) structures. All six carbon-carbon bonds in benzene are of equal length (139 pm), intermediate between a pure C-C single bond (154 pm) and a pure C=C double bond (133 pm), because the six p-orbitals on the ring carbons overlap sideways to form a continuous, delocalised pi electron cloud above and below the planar hexagonal ring. This delocalisation makes benzene about 150 kJ mol-1 more stable than the hypothetical Kekulé structure with localised double bonds (its resonance/delocalisation energy), which explains why benzene prefers substitution reactions over the addition reactions typical of alkenes.

A cyclic, planar compound is called aromatic if it satisfies Hückel's rule: it must have a continuous, complete cyclic conjugation of p-orbitals and possess (4n+2) pi electrons, where n = 0, 1, 2, .... Benzene has 6 pi electrons (n = 1), so it is aromatic; other examples include naphthalene and the cyclopentadienyl anion.

Worked example (testing aromaticity):

  • Benzene: cyclic, planar, fully conjugated, with 6 pi electrons. Setting 4n+2 = 6 gives n = 1 (a whole number), so benzene is aromatic.
  • Cyclobutadiene: cyclic and conjugated, but has 4 pi electrons. Setting 4n+2 = 4 gives n = 0.5, which is not a whole number, so cyclobutadiene does not satisfy Hückel's rule and is not aromatic (it is in fact anti-aromatic and highly unstable).
Molecular formula of benzene C₆H₆
Hückel's rule Number of π electrons = 4n + 2 (n = 0, 1, 2, ...)
C-C bond length comparison C-C (154 pm) > C(benzene)-C (139 pm) > C=C (133 pm)
Remember
  • Benzene is a resonance hybrid of Kekulé structures; all C-C bonds are equal (139 pm), between single (154 pm) and double (133 pm) bond lengths.
  • Delocalisation of the pi electron cloud gives benzene extra stability (delocalisation/resonance energy, about 150 kJ/mol).
  • Hückel's rule: aromaticity needs a cyclic, planar, fully conjugated system with (4n+2) pi electrons.
  • Because of this extra stability, benzene undergoes electrophilic substitution (retaining aromaticity) far more readily than addition.

Electrophilic Substitution Reactions of Benzene

Quick answer Benzene undergoes electrophilic substitution — nitration, halogenation, and Friedel-Crafts alkylation/acylation — through a common arenium-ion mechanism, directed by ring substituents.

Because of its stable, delocalised pi system, benzene reacts with electrophiles by substitution rather than addition, so that aromaticity is preserved in the product. The general mechanism has three steps: (i) generation of the electrophile; (ii) attack of the electrophile on the pi cloud of the ring to form a resonance-stabilised, non-aromatic carbocation intermediate called the arenium ion (or sigma complex); and (iii) loss of a proton from the arenium ion, which restores the aromatic sextet of electrons.

In nitration, benzene is heated with a mixture of concentrated nitric acid and concentrated sulphuric acid (the nitrating mixture) at about 333 K; the electrophile is the nitronium ion, NO2+. In halogenation, benzene reacts with chlorine or bromine in presence of a Lewis acid catalyst such as anhydrous FeCl3 or AlCl3, which polarises the halogen molecule to generate the electrophile. In Friedel-Crafts alkylation, an alkyl halide with anhydrous AlCl3 introduces an alkyl group (though the reaction suffers from over-alkylation and carbocation rearrangement, and fails on strongly deactivated rings such as nitrobenzene); Friedel-Crafts acylation, using an acyl halide with anhydrous AlCl3, avoids rearrangement and over-substitution because the acyl group formed is deactivating.

Substituents already present on the ring control where the next electrophile attacks: electron-donating groups such as -OH, -NH2, -OR and alkyl groups activate the ring and direct incoming groups to the ortho and para positions; halogens are weakly deactivating but still ortho, para-directing; strongly electron-withdrawing groups such as -NO2, -CN, -COOH and -SO3H deactivate the ring and direct to the meta position.

Worked example (mechanism of nitration of benzene):

  • Step 1: HNO3 is protonated by H2SO4 and loses water to generate the electrophile, NO2+.
  • Step 2: NO2+ attacks the benzene pi cloud, forming a resonance-stabilised arenium ion in which the positive charge is delocalised over three ring carbons.
  • Step 3: HSO4- removes a proton from the sp3 carbon of the arenium ion, regenerating the aromatic ring and giving nitrobenzene.
Generation of nitronium ion HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻
Nitration of benzene C₆H₆ + HNO₃ → (conc. H₂SO₄, 333 K) C₆H₅NO₂ + H₂O
Halogenation of benzene C₆H₆ + Cl₂ → (anhyd. AlCl₃) C₆H₅Cl + HCl
Friedel-Crafts alkylation C₆H₆ + CH₃Cl → (anhyd. AlCl₃) C₆H₅CH₃ + HCl
Friedel-Crafts acylation C₆H₆ + CH₃COCl → (anhyd. AlCl₃) C₆H₅COCH₃ + HCl
Remember
  • Electrophilic substitution mechanism: electrophile generation, then arenium ion (sigma complex) formation, then deprotonation to restore aromaticity.
  • Nitration uses conc. HNO3/conc. H2SO4 (NO2+ electrophile); halogenation and Friedel-Crafts reactions need a Lewis acid catalyst (FeX3/AlCl3).
  • Friedel-Crafts alkylation is prone to polyalkylation/rearrangement and fails on strongly deactivated rings; acylation avoids these problems.
  • Ring substituents are classified as ortho, para-directors (activating, or halogens which are deactivating) or meta-directors (deactivating).

Key equations

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

CₙH₂ₙ₊₂
General formula of alkanes
CₙH₂ₙ
General formula of alkenes
CₙH₂ₙ₋₂
General formula of alkynes
CₙH₂ₙ₋₆
General formula of arenes (benzene series)
2R-X + 2Na → (dry ether) R-R + 2NaX
Wurtz reaction
CH₃COONa + NaOH → (CaO, Δ) CH₄ + Na₂CO₃
Decarboxylation (soda lime)
2CH₃COONa + 2H₂O → (electrolysis) CH₃-CH₃ + 2CO₂ + 2NaOH + H₂
Kolbe's electrolytic method
CH₂=CH₂ + H₂ → (Ni/Pd/Pt) CH₃-CH₃
Catalytic hydrogenation
CH₄ + Cl₂ → (hv) CH₃Cl + HCl
Free-radical chlorination (overall)
CH₄ + 2O₂ → CO₂ + 2H₂O
Complete combustion of methane
CH₃CH₂OH → (conc. H₂SO₄, 443 K) CH₂=CH₂ + H₂O
Dehydration of ethanol
CH₃-CH₂-CHBr-CH₃ + KOH(alc.) → CH₃-CH=CH-CH₃ (major) + KBr + H₂O
Dehydrohalogenation (Saytzeff)
CH₃-CH=CH₂ + HBr → CH₃-CHBr-CH₃
Markovnikov addition of HBr
CH₃-CH=CH₂ + HBr → (peroxide) CH₃-CH₂-CH₂Br
Anti-Markovnikov (peroxide effect)
3CH₂=CH₂ + 2KMnO₄ + 4H₂O → 3HOCH₂-CH₂OH + 2MnO₂ + 2KOH
Baeyer's test (cold dil. alkaline KMnO4)
CH₃-CH=CH-CH₃ → (i. O₃ ii. Zn/H₂O) 2CH₃CHO
Ozonolysis of but-2-ene
CaC₂ + 2H₂O → HC≡CH + Ca(OH)₂
Preparation of ethyne from calcium carbide
CH₃-CHBr-CH₂Br + 2KOH(alc.) → CH₃-C≡CH + 2KBr + 2H₂O
Preparation from vicinal dihalide
HC≡CH + Na → HC≡C-Na + ½H₂
Acidic character (sodium acetylide)
R-C≡C-H + [Ag(NH₃)₂]⁺ → R-C≡C-Ag↓(white) + NH₃ + NH₄⁺
Test for terminal alkyne
HC≡CH + H₂O → (Hg²⁺/H₂SO₄) CH₃CHO
Markovnikov hydration of ethyne
C₆H₆
Molecular formula of benzene
Number of π electrons = 4n + 2 (n = 0, 1, 2, ...)
Hückel's rule
C-C (154 pm) > C(benzene)-C (139 pm) > C=C (133 pm)
C-C bond length comparison
HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻
Generation of nitronium ion
C₆H₆ + HNO₃ → (conc. H₂SO₄, 333 K) C₆H₅NO₂ + H₂O
Nitration of benzene
C₆H₆ + Cl₂ → (anhyd. AlCl₃) C₆H₅Cl + HCl
Halogenation of benzene
C₆H₆ + CH₃Cl → (anhyd. AlCl₃) C₆H₅CH₃ + HCl
Friedel-Crafts alkylation
C₆H₆ + CH₃COCl → (anhyd. AlCl₃) C₆H₅COCH₃ + HCl
Friedel-Crafts acylation

Test yourself

Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.

0 correct · 0/12 answered
Q1 Classification easy

What is the general formula of alkynes?

Q2 Nomenclature easy

What is the IUPAC name of CH3-CH(CH3)-CH2-CH2-CH3?

Q3 Alkanes - preparation medium

Which statement correctly describes a limitation of the Wurtz reaction?

Q4 Alkanes - halogenation easy

In the free-radical chlorination of methane, which step produces chlorine free radicals from Cl2?

Q5 Alkenes - addition reactions medium

Propene reacts with HBr in the absence of peroxide. What is the major product and which rule explains it?

Q6 Alkenes - addition reactions medium

The anti-Markovnikov addition of HBr to alkenes in the presence of peroxides is known as:

Q7 Alkenes - ozonolysis medium

Ozonolysis of but-2-ene (CH3-CH=CH-CH3) followed by Zn/H2O workup gives:

Q8 Alkynes - preparation easy

Which reagent reacts with calcium carbide to prepare ethyne (acetylene)?

Q9 Alkynes - acidic character medium

Terminal alkynes (R-C≡C-H) give a characteristic white precipitate with which reagent, confirming the terminal triple bond?

Q10 Aromaticity medium

According to Hückel's rule, a planar cyclic molecule is aromatic if it contains:

Q11 Electrophilic substitution medium

In the nitration of benzene using concentrated HNO3 and concentrated H2SO4, the electrophile generated is:

Q12 Friedel-Crafts reaction hard

Friedel-Crafts alkylation of benzene fails when the ring already carries:

NCERT solutions & previous-year questions

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

NCERT questions 6

1 An alkene 'A' contains three C-C sigma bonds, eight C-H sigma bonds and one C-C pi bond. Ozonolysis of 'A' gives two moles of an aldehyde with molecular mass 44. Identify 'A' and justify.Alkenes - ozonolysis

An aldehyde of molecular mass 44 must be CH3CHO (ethanal): 12+1+12+3(1)+16 = 44. Since ozonolysis gives two moles of the same aldehyde, both carbons of the double bond must carry an identical -CH3 group attached, so the alkene is symmetrical: CH3-CH=CH-CH3 (but-2-ene).

Checking the bond count in but-2-ene: C-C sigma bonds are C1-C2, C2-C3 (sigma part of the double bond) and C3-C4, i.e. 3 C-C sigma bonds. C-H sigma bonds: C1 has 3H, C2 has 1H, C3 has 1H, C4 has 3H, totalling 8 C-H sigma bonds. There is exactly one C-C pi bond (the pi component of C2=C3). All conditions match.

Ozonolysis: CH3-CH=CH-CH3 → (i. O3 ii. Zn/H2O) 2CH3CHO.

Hence A is but-2-ene, CH3-CH=CH-CH3.

2 Draw the cis and trans structures of hex-2-ene, CH3-CH=CH-CH2-CH2-CH3. Which isomer would you expect to react faster in catalytic hydrogenation, and why?Alkenes - stability

Hex-2-ene: CH3-CH=CH-CH2-CH2-CH3.

  • cis-hex-2-ene: the CH3 group and the -CH2CH2CH3 chain lie on the same side of the C=C bond.
  • trans-hex-2-ene: the CH3 group and the -CH2CH2CH3 chain lie on opposite sides of the C=C bond.

The cis isomer suffers greater steric strain between the bulky groups on the same side, making it less stable (higher in energy, with a greater heat of hydrogenation) than the more stable, less strained trans isomer. Since the cis isomer is at a higher energy level to begin with, it releases more energy and reacts faster on catalytic hydrogenation (with Ni, Pd or Pt) to reach the same stable alkane product.

3 Write the structures of all the alkenes which yield 2-methylbutane, CH3-CH(CH3)-CH2-CH3, on hydrogenation.Alkenes - isomerism

2-Methylbutane has the skeleton CH3-CH(CH3)-CH2-CH3. The alkenes that give this alkane on addition of H2 are those with the same carbon skeleton and a double bond anywhere along the chain:

  • 2-methylbut-1-ene: CH2=C(CH3)-CH2-CH3
  • 2-methylbut-2-ene: CH3-C(CH3)=CH-CH3
  • 3-methylbut-1-ene: CH2=CH-CH(CH3)-CH3

Each of these, on catalytic hydrogenation (Ni/Pd/Pt, H2), adds H2 across the double bond to give 2-methylbutane.

4 How would you convert benzene into (i) p-nitrobromobenzene and (ii) m-nitrochlorobenzene? Explain the order of steps using directive influence of substituents.Aromatic - directive influence

(i) p-Nitrobromobenzene: Since -Br is a weakly deactivating but ortho, para-directing group, while -NO2 is meta-directing, bromination must be carried out first so that the second group (nitro) is directed to the para position.

  • Step 1: C6H6 + Br2 → (anhyd. FeBr3) C6H5Br + HBr (bromobenzene).
  • Step 2: C6H5Br + HNO3 → (conc. H2SO4, 333 K) p-O2N-C6H4-Br (major) + H2O, along with a minor ortho isomer that is separated by physical methods.

(ii) m-Nitrochlorobenzene: Since -NO2 is strongly deactivating and meta-directing, nitration must be carried out first, so the incoming chlorine is directed to the meta position.

  • Step 1: C6H6 + HNO3 → (conc. H2SO4, 333 K) C6H5NO2 + H2O (nitrobenzene).
  • Step 2: C6H5NO2 + Cl2 → (anhyd. AlCl3) m-Cl-C6H4-NO2 + HCl (m-nitrochlorobenzene).
5 Write the balanced equation for the complete combustion of butane, and calculate the volume of O2 (at STP, 22.4 L per mole) required to completely burn 2 moles of butane.Alkanes - combustion (numerical)

Balanced equation for complete combustion of butane:

2C4H10 + 13O2 → 8CO2 + 10H2O

From the equation, 2 mol of butane requires 13 mol of O2.

Volume of O2 required at STP = 13 mol × 22.4 L mol-1 = 291.2 L.

6 Why does the Friedel-Crafts reaction fail with nitrobenzene? What role does anhydrous AlCl3 play in this reaction, and why must it be anhydrous?Friedel-Crafts reaction

Anhydrous AlCl3 acts as a Lewis acid catalyst: it accepts a lone pair from the halogen of the alkyl/acyl halide, polarising the R-X (or RCO-X) bond and generating the electrophile (R+ or RCO+) needed to attack the benzene ring. AlCl3 must be anhydrous because water hydrolyses it, destroying its Lewis-acidic (electrophile-generating) activity.

The Friedel-Crafts reaction fails on nitrobenzene because the strongly electron-withdrawing -NO2 group deactivates the benzene ring so severely that the ring pi cloud cannot supply enough electron density to attack the electrophile, so no arenium ion intermediate can form under normal Friedel-Crafts conditions. (For a similar reason, aniline also fails, since its -NH2 lone pair forms a salt with AlCl3, converting it into a strongly deactivating, meta-directing -NH2-AlCl3 group and consuming the catalyst.)

Previous-year board questions 4

Q1 Write the stepwise mechanism of nitration of benzene. CBSE 2020 3 marks

Step 1 - Generation of the electrophile: Concentrated H2SO4 protonates HNO3, which then loses a molecule of water to generate the nitronium ion:

HNO3 + 2H2SO4 → NO2+ + H3O+ + 2HSO4-

Step 2 - Formation of the arenium ion: The electrophile NO2+ attacks the delocalised pi electron cloud of benzene, forming a resonance-stabilised, non-aromatic carbocation intermediate (the arenium ion or sigma complex), in which the positive charge is delocalised over three ring carbons.

Step 3 - Loss of proton: HSO4- removes a proton from the sp3 carbon of the arenium ion, restoring the aromatic sextet of electrons and giving nitrobenzene.

Overall: C6H6 + HNO3 → (conc. H2SO4, 333 K) C6H5NO2 + H2O.

Q2 An alkene on ozonolysis gives a mixture of ethanal and pentan-3-one. Write the structure and IUPAC name of the alkene, and write the ozonolysis reaction. CBSE 2019 2 marks

Ethanal, CH3CHO, arises from a =CH-CH3 fragment of the alkene. Pentan-3-one, CH3CH2-CO-CH2CH3, arises from a =C(C2H5)2 fragment. Joining these two fragments at the position of the original double bond gives the alkene:

CH3-CH=C(CH2CH3)2

Taking the longest chain through one of the ethyl groups (5 carbons) with the second ethyl group as a substituent at C3, the IUPAC name is 3-ethylpent-2-ene.

Ozonolysis reaction: CH3-CH=C(C2H5)2 → (i. O3 ii. Zn/H2O) CH3CHO + CH3CH2-CO-CH2CH3.

Q3 (i) Why is the Wurtz reaction not a good method for preparing alkanes containing an odd number of carbon atoms? (ii) Illustrate with a suitable equation. CBSE 2023 3 marks

(i) Reason: In the Wurtz reaction, 2R-X + 2Na → (dry ether) R-R + 2NaX, a single alkyl halide always couples two identical R groups, so the product alkane always has an even number of carbon atoms. To obtain an alkane with an odd number of carbon atoms, two different alkyl halides (R-X and R'-X) must be reacted together, but coupling then occurs randomly among all the radicals/species present, producing a statistical mixture of three products - R-R, R'-R' and R-R' - which are difficult to separate, giving only a poor yield of the desired odd-carbon alkane. Hence the Wurtz reaction is not a good method for odd-carbon alkanes.

(ii) Illustration: Using bromomethane and bromoethane together with sodium in dry ether, coupling occurs randomly and gives three products simultaneously:

  • 2CH3Br + 2Na → CH3-CH3 (ethane) + 2NaBr
  • 2C2H5Br + 2Na → CH3-CH2-CH2-CH3 (n-butane) + 2NaBr
  • CH3Br + C2H5Br + 2Na → CH3-CH2-CH3 (propane, desired) + 2NaBr

Propane (the odd-carbon alkane) is obtained only as one of three simultaneous products, in a low yield, alongside ethane and n-butane.

Q4 Complete the following reactions and write the IUPAC name of the product in (a): (a) CH3-CH=CH2 + HBr in the presence of peroxide; (b) C6H6 + CH3COCl in the presence of anhydrous AlCl3; (c) CaC2 + H2O. CBSE 2022 5 marks

(a) CH3-CH=CH2 + HBr → (peroxide) CH3-CH2-CH2Br. This is anti-Markovnikov addition (the Kharasch/peroxide effect), giving 1-bromopropane. IUPAC name of the product: 1-bromopropane.

(b) C6H6 + CH3COCl → (anhyd. AlCl3) C6H5-CO-CH3 + HCl. This is a Friedel-Crafts acylation, giving acetophenone (1-phenylethan-1-one).

(c) CaC2 + 2H2O → HC≡CH + Ca(OH)2. This gives ethyne (acetylene) gas and calcium hydroxide.

Part of Priodemy for School

Interactive Maths & Science — free with every school on Priodemy EduSuite. Explore more chapters and labs on the Priodemy for School hub.

Ask AI