This MCQ module is based on: Alkenes
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Alkenes
🎓 Class 11
Chemistry
CBSE
Theory
Ch 9 – Hydrocarbons
⏱ ~14 min
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Alkenes — Structure, Preparation and Reactions
9.4 Alkenes
Alkenes are unsaturated hydrocarbons that contain at least one carbon-carbon double bond (C=C). The general formula is CₙH₂ₙ. The simplest alkene is ethene (CH₂=CH₂), commonly called ethylene. Alkenes are also called olefins (oil-forming) because the lower members yield oily products with halogens.
9.4.1 Structure of the Double Bond
Each carbon of the C=C is sp²-hybridised. The three sp² hybrid orbitals point to the corners of an equilateral triangle (120° apart) and form three σ bonds. The unhybridised p-orbital on each carbon, perpendicular to the molecular plane, overlaps sideways with the corresponding p-orbital of the other carbon to give a π bond. Hence C=C consists of one σ + one π bond. The C=C bond length is 134 pm and the C–H is 108 pm in ethene; the bond enthalpy of C=C (681 kJ/mol) is greater than that of C–C (348 kJ/mol) but less than twice it because the π bond is weaker than a σ bond.
9.4.2 Nomenclature of Alkenes
Same rules as alkanes, with three additional points:
- The longest chain must include the C=C.
- The chain is numbered to give the C=C the lowest locant.
- The suffix '-ane' becomes '-ene'; locant of C=C is written before the parent name (1993 IUPAC) or before '-ene' suffix (2013 IUPAC).
| Structure | IUPAC name |
|---|---|
| CH₂=CH₂ | Ethene |
| CH₃–CH=CH₂ | Prop-1-ene (Propene) |
| CH₃–CH=CH–CH₃ | But-2-ene |
| CH₂=CH–CH=CH₂ | Buta-1,3-diene |
| (CH₃)₂C=CH₂ | 2-Methylprop-1-ene (isobutylene) |
9.4.3 Isomerism
Alkenes show structural isomerism (chain, position) plus a new kind: geometrical (cis–trans) isomerism, possible whenever each C of C=C carries two different groups.
E/Z system (CIP rules): Use this when cis/trans is ambiguous. Assign priority to the two groups on each C of C=C using atomic number. If the two higher-priority groups are on the same side → Z (zusammen); on opposite sides → E (entgegen).
9.4.4 Preparation of Alkenes
(a) From alkynes — partial hydrogenation
Alkynes on partial reduction with calculated H₂ over Lindlar's catalyst (Pd–CaCO₃ poisoned with Pb(OAc)₂/quinoline) give cis-alkenes; with Na in liquid NH₃ they give trans-alkenes.
CH₃–C≡C–CH₃ + H₂ →Lindlar cis-CH₃–CH=CH–CH₃(b) From alkyl halides — β-elimination (dehydrohalogenation)
Alkyl halides on heating with alcoholic KOH lose a proton from the β-carbon and the halide from the α-carbon, generating a C=C bond:
CH₃–CH₂–CH₂–Br + KOH(alc) →Δ CH₃–CH=CH₂ + KBr + H₂OZaitsev's (Saytzeff) rule: When two β-elimination products are possible, the more substituted (more stable) alkene predominates.
Example: 2-bromobutane gives but-2-ene (major, disubstituted) over but-1-ene (minor, monosubstituted).
(c) From alcohols — acid-catalysed dehydration
Alcohols on heating with concentrated H₂SO₄ or H₃PO₄ undergo dehydration to alkenes:
CH₃–CH₂–OH →conc H₂SO₄, 443 K CH₂=CH₂ + H₂OReactivity: tertiary > secondary > primary alcohol. Also follows Zaitsev's rule.
(d) From vicinal dihalides — dehalogenation
Vicinal dihalides on treatment with Zn dust in methanol give alkenes:
CH₂Br–CH₂Br + Zn → CH₂=CH₂ + ZnBr₂9.4.5 Physical Properties
The first three members (ethene, propene, butene) are gases; C₅–C₁₅ are liquids; higher members are solids. They are colourless, almost insoluble in water, soluble in organic solvents. Ethene has a faint sweet smell. The boiling points of straight-chain alkenes are very similar to those of corresponding alkanes.
9.4.6 Chemical Reactions
The π electrons of C=C are loosely held and act as a nucleophile, so alkenes typically undergo electrophilic addition. The π bond breaks and two new σ bonds form across the double bond, converting an unsaturated to a saturated product.
(i) Addition of Hydrogen — catalytic hydrogenation
CH₂=CH₂ + H₂ →Ni/Δ CH₃–CH₃ ΔH = –137 kJ/molUsed industrially to convert vegetable oils (unsaturated) to vanaspati ghee (saturated).
(ii) Addition of Halogens — bromine water test
Alkenes decolourise an orange-red solution of bromine in CCl₄ to a colourless vicinal dibromide. This is a routine test for unsaturation:
CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br (1,2-dibromoethane)(iii) Addition of Hydrogen Halides (HX)
Symmetrical alkenes give a single product, but unsymmetrical alkenes raise the question: which way does HX add?
Markovnikov's Rule: The negative end of the polar reagent (X) attaches to the carbon of the C=C bearing the lesser number of H atoms.
Mechanism for HBr + propene:
1
Step 1 — π electrons attack H⁺: The π electrons of CH₃–CH=CH₂ pick up the proton of HBr, forming a carbocation. There are two possibilities:
CH₃–CH=CH₂ + H⁺ → CH₃–CH⁺–CH₃ (2°, more stable — preferred)
OR CH₃–CH=CH₂ + H⁺ → CH₃–CH₂–CH₂⁺ (1°, less stable)
CH₃–CH=CH₂ + H⁺ → CH₃–CH⁺–CH₃ (2°, more stable — preferred)
OR CH₃–CH=CH₂ + H⁺ → CH₃–CH₂–CH₂⁺ (1°, less stable)
2
Step 2 — Br⁻ attacks the carbocation: Bromide ion attacks the more stable secondary carbocation:
CH₃–CH⁺–CH₃ + Br⁻ → CH₃–CHBr–CH₃ (2-bromopropane — major)
CH₃–CH⁺–CH₃ + Br⁻ → CH₃–CHBr–CH₃ (2-bromopropane — major)
Result: H added to the C with more H, Br to the C with fewer H. Carbocation stability order: 3° > 2° > 1° > CH₃⁺. This is because alkyl groups release electrons (+I) and stabilise the positive charge through hyperconjugation.
(iv) Anti-Markovnikov addition — Peroxide effect (Kharasch effect)
In presence of organic peroxide (e.g. (PhCO₂)₂ or H₂O₂), HBr adds in the OPPOSITE orientation: Br to the C with more H. This is called the peroxide effect or Kharasch effect, observed only with HBr (not HCl, HI). It proceeds by a free-radical mechanism:
1
Initiation: peroxide → 2 RO•; RO• + HBr → ROH + Br•
2
Propagation: Br• adds to the C of C=C bearing more H (giving the more stable 2° radical):
Br• + CH₃–CH=CH₂ → CH₃–•CH–CH₂Br (2° radical, stable)
CH₃–•CH–CH₂Br + HBr → CH₃–CH₂–CH₂Br + Br• (1-bromopropane, anti-Markovnikov!)
Br• + CH₃–CH=CH₂ → CH₃–•CH–CH₂Br (2° radical, stable)
CH₃–•CH–CH₂Br + HBr → CH₃–CH₂–CH₂Br + Br• (1-bromopropane, anti-Markovnikov!)
(v) Addition of H₂SO₄ (and subsequent hydrolysis to alcohol)
CH₂=CH₂ + H₂SO₄ → CH₃–CH₂–OSO₂OHCH₃–CH₂–OSO₂OH + H₂O →Δ CH₃–CH₂–OH + H₂SO₄
(vi) Hydration
Direct addition of water in presence of dilute H₂SO₄ gives alcohols (Markovnikov):
CH₃–CH=CH₂ + H₂O →H⁺ CH₃–CH(OH)–CH₃ (propan-2-ol)(vii) Oxidation
(a) Cold dilute alkaline KMnO₄ (Baeyer's reagent) — oxidises alkene to vicinal diol; the purple colour of KMnO₄ disappears (test for unsaturation):
CH₂=CH₂ + [O] + H₂O →cold dil. KMnO₄ CH₂(OH)–CH₂(OH)(b) Hot conc. KMnO₄ — cleaves the C=C entirely, giving carbonyl compounds.
(c) Combustion: Burns with luminous (sooty) flame to CO₂ + H₂O.
(viii) Ozonolysis
Alkene + O₃ in CCl₄ → ozonide; ozonide + Zn/H₂O → two carbonyl fragments. The reaction enables identification of the position of the double bond:
CH₃–CH=CH–CH₃ + O₃ → ozonide →Zn/H₂O 2 CH₃CHO (acetaldehyde)(CH₃)₂C=CHCH₃ + O₃ → ozonide →Zn/H₂O (CH₃)₂C=O + CH₃CHO
(ix) Polymerisation
Many alkene molecules link together to form a polymer:
n CH₂=CH₂ →high T & P, catalyst –(CH₂–CH₂)–ₙ (polythene)n CH₂=CH(CH₃) → –(CH₂–CH(CH₃))–ₙ (polypropene)
Markovnikov vs Anti-Markovnikov Predictor
Choose an alkene + reagent + presence/absence of peroxide; the simulator predicts the major product and tells you why.
Major product:
CH₃-CHBr-CH₃ (Markovnikov)
Rule applied: Markovnikov — H goes to the C with more H; Br goes to the C with less H. Mechanism via 2° carbocation.
Activity 9.2 — Bromine water test
Setup: Two unlabelled tubes contain colourless liquids: hexane and hex-1-ene. A third tube contains orange-red bromine in CCl₄.
Predict: Without using IR or any spectrometer, how would you identify which tube contains the alkene? What chemical change should occur?
Add a few drops of Br₂/CCl₄ to each tube and shake.
- The tube containing hexane retains the orange-red colour (no reaction — saturated).
- The tube containing hex-1-ene decolourises the bromine almost instantly:
CH₂=CH–C₄H₉ + Br₂ → CH₂Br–CHBr–C₄H₉ (1,2-dibromohexane, colourless).
The disappearance of bromine colour without HBr fumes is the classical qualitative test for a C=C double bond.
Worked Example 1: Geometrical isomers
Draw cis and trans isomers of pent-2-ene. Which is more stable?
Pent-2-ene: CH₃–CH=CH–CH₂CH₃. The C2 of C=C bears CH₃ and H; the C3 of C=C bears H and CH₂CH₃. Two configurations:
cis: CH₃ and CH₂CH₃ on the same side (Z).
trans: CH₃ and CH₂CH₃ on opposite sides (E).
The trans (E) isomer is more stable because the two larger groups are far apart, minimising steric strain. The cis isomer has higher energy by ~4 kJ/mol due to crowding.
cis: CH₃ and CH₂CH₃ on the same side (Z).
trans: CH₃ and CH₂CH₃ on opposite sides (E).
The trans (E) isomer is more stable because the two larger groups are far apart, minimising steric strain. The cis isomer has higher energy by ~4 kJ/mol due to crowding.
Worked Example 2: Predicting addition products
Predict the major product when HBr adds to 2-methylpropene (a) without peroxide, (b) in presence of benzoyl peroxide.
2-Methylpropene: (CH₃)₂C=CH₂. The C=C carbons: C1 = C(CH₃)₂ (no H), C2 = CH₂ (2 H).
(a) Without peroxide → Markovnikov: H goes to C with more H (C2), Br to C with less H (C1). The intermediate is a 3° tert-butyl cation, very stable.
Product: (CH₃)₃C–Br = 2-bromo-2-methylpropane (tert-butyl bromide).
(b) With peroxide → Anti-Markovnikov (radical): Br• adds to C2 (giving the more stable 3° radical at C1). Then HBr supplies H to C1.
Product: (CH₃)₂CH–CH₂Br = 1-bromo-2-methylpropane (isobutyl bromide).
Note: peroxide effect operates only with HBr — not HCl (C–Cl bond too strong) or HI (H–I bond too weak).
(a) Without peroxide → Markovnikov: H goes to C with more H (C2), Br to C with less H (C1). The intermediate is a 3° tert-butyl cation, very stable.
Product: (CH₃)₃C–Br = 2-bromo-2-methylpropane (tert-butyl bromide).
(b) With peroxide → Anti-Markovnikov (radical): Br• adds to C2 (giving the more stable 3° radical at C1). Then HBr supplies H to C1.
Product: (CH₃)₂CH–CH₂Br = 1-bromo-2-methylpropane (isobutyl bromide).
Note: peroxide effect operates only with HBr — not HCl (C–Cl bond too strong) or HI (H–I bond too weak).
Worked Example 3: Ozonolysis to identify the alkene
An unknown alkene C₅H₁₀ on ozonolysis gives only acetone ((CH₃)₂C=O) and formaldehyde (HCHO). Identify the alkene.
Ozonolysis cleaves C=C and replaces it with two C=O. So we reverse the process: combine the two carbonyls into one alkene.
Acetone: (CH₃)₂C=O contributes (CH₃)₂C=
Formaldehyde: HCHO contributes =CH₂
Combine: (CH₃)₂C=CH₂ — that is 2-methylpropene (isobutylene) — molecular formula C₄H₈, but the question says C₅H₁₀!
Wait: 2-methylpropene is C₄H₈. To match C₅H₁₀ with the same products, the alkene must be 2-methylbut-2-ene or 2-methyl-but-1-ene? Let us re-examine: 2-methylbut-2-ene is (CH₃)₂C=CH–CH₃, which on ozonolysis gives (CH₃)₂C=O (acetone) + CH₃CHO (acetaldehyde) — does NOT match (would give acetaldehyde not formaldehyde).
Hence the only structure giving exactly acetone + formaldehyde is 2-methylpropene, (CH₃)₂C=CH₂ (C₄H₈). The C₅H₁₀ in the question cannot give exactly these two products. The intended answer for the C₄H₈ ozonolysis is 2-methylpropene.
Acetone: (CH₃)₂C=O contributes (CH₃)₂C=
Formaldehyde: HCHO contributes =CH₂
Combine: (CH₃)₂C=CH₂ — that is 2-methylpropene (isobutylene) — molecular formula C₄H₈, but the question says C₅H₁₀!
Wait: 2-methylpropene is C₄H₈. To match C₅H₁₀ with the same products, the alkene must be 2-methylbut-2-ene or 2-methyl-but-1-ene? Let us re-examine: 2-methylbut-2-ene is (CH₃)₂C=CH–CH₃, which on ozonolysis gives (CH₃)₂C=O (acetone) + CH₃CHO (acetaldehyde) — does NOT match (would give acetaldehyde not formaldehyde).
Hence the only structure giving exactly acetone + formaldehyde is 2-methylpropene, (CH₃)₂C=CH₂ (C₄H₈). The C₅H₁₀ in the question cannot give exactly these two products. The intended answer for the C₄H₈ ozonolysis is 2-methylpropene.
Competency-Based Questions
Q1. The hybridisation of the carbon atoms of a C=C double bond is: L1 Remember
Answer: (b) sp². Each C uses three sp² hybrids for σ bonds (planar, 120° apart) and one unhybridised p orbital for the π bond perpendicular to the plane.
Q2. Why does HBr add to propene contrary to Markovnikov's rule when an organic peroxide is present? L4 Analyse
Answer: The peroxide initiates a free-radical chain mechanism. Br• adds first to the C of C=C bearing more H atoms because that produces the more stable secondary alkyl radical (CH₃–•CH–CH₂Br). The radical then abstracts H from HBr, giving 1-bromopropane (anti-Markovnikov) and a new Br•. The Markovnikov ionic pathway requires a stable carbocation; the radical pathway requires a stable radical, and the two place Br on opposite carbons.
Q3. An alkene C₆H₁₂ on ozonolysis gives two molecules of CH₃CHO. Identify the alkene and write its IUPAC name. L3 Apply
Two molecules of CH₃CHO mean both carbons of the C=C are attached to one CH₃ and one H. So the alkene is CH₃–CH=CH–CH₃ — molecular formula C₄H₈ (but-2-ene). For C₆H₁₂ giving two CH₃CHO it must be ethyl- substituted and would not give two CH₃CHO. The intended alkene matching ozonolysis to two acetaldehydes is but-2-ene (CH₃CH=CHCH₃).
Q4. Compare the stabilities of cis-2-butene and trans-2-butene. Justify with reasoning. L5 Evaluate
Answer: trans-2-butene (4 kJ/mol more stable than cis-) — confirmed by lower heat of combustion. The two methyl groups in cis lie on the same side of the C=C, causing steric repulsion (van der Waals strain). In trans, the methyl groups are on opposite sides, far apart, no steric strain. Hence trans > cis in stability. Note also that cis-2-butene has a small dipole moment (≠0), trans-2-butene has μ = 0 (the two C–CH₃ dipoles cancel by symmetry).
Q5. HOT (Create): Design a single synthetic sequence converting propene into propan-1-ol (a primary alcohol). Hint: use the peroxide effect strategically. L6 Create
Sample answer:
Step 1: Propene + HBr in presence of (PhCO₂)₂ peroxide → 1-bromopropane (anti-Markovnikov) → CH₃CH₂CH₂Br.
Step 2: 1-bromopropane + aqueous KOH (NOT alcoholic) → propan-1-ol via SN2:
CH₃CH₂CH₂Br + KOH(aq) → CH₃CH₂CH₂OH + KBr.
Direct hydration of propene with H₂O/H⁺ would give propan-2-ol (Markovnikov, secondary alcohol). The peroxide-effect detour is essential to obtain the primary alcohol.
Step 1: Propene + HBr in presence of (PhCO₂)₂ peroxide → 1-bromopropane (anti-Markovnikov) → CH₃CH₂CH₂Br.
Step 2: 1-bromopropane + aqueous KOH (NOT alcoholic) → propan-1-ol via SN2:
CH₃CH₂CH₂Br + KOH(aq) → CH₃CH₂CH₂OH + KBr.
Direct hydration of propene with H₂O/H⁺ would give propan-2-ol (Markovnikov, secondary alcohol). The peroxide-effect detour is essential to obtain the primary alcohol.
Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: Bromine water decolourises rapidly with ethene but not with ethane.
R: Ethene is unsaturated and undergoes electrophilic addition with Br₂; ethane is saturated and inert under the same conditions.
Answer: (A). Both true; R correctly explains A. This is the basis of the qualitative bromine-water test for unsaturation.
A: Markovnikov addition of HBr to propene gives 2-bromopropane as the major product.
R: The intermediate secondary carbocation is more stable than the corresponding primary carbocation.
Answer: (A). Both true; R correctly explains A. Carbocation stability order (3° > 2° > 1° > CH₃⁺) determines the regioselectivity.
A: Peroxide effect is observed in HCl as well as HBr addition.
R: The bond enthalpy of H–Cl is too high for radical chain to propagate efficiently, while H–I is too weak.
Answer: (D). Assertion is FALSE — peroxide effect is observed only with HBr. Reason is TRUE — that is precisely why HCl and HI do not show the effect: the energetics of the propagation steps don't work out for HCl (H–Cl too strong) or HI (H–I too weak; I• adds reversibly).
Frequently Asked Questions — Alkenes — Structure, Preparation and Reactions
What are alkenes and what is their general formula?
Alkenes are unsaturated open-chain hydrocarbons containing at least one carbon-carbon double bond (C=C). The general formula is CₙH₂ₙ for monoenes. Each carbon of the double bond is sp² hybridised, giving trigonal planar geometry with 120° bond angles. The C=C bond consists of one σ bond (sp²-sp² head-on overlap) and one π bond (p-p sideways overlap). The π bond restricts rotation, giving rise to geometrical (cis-trans) isomerism. NCERT Class 11 Chemistry Chapter 9 covers ethene (C₂H₄), propene (C₃H₆), and butenes as the simplest alkenes.
How are alkenes prepared from alcohols and alkyl halides?
NCERT Class 11 Chemistry Chapter 9 describes two major preparation methods of alkenes: (1) dehydration of alcohols using concentrated H₂SO₄ or Al₂O₃ at high temperature — CH₃CH₂OH → CH₂=CH₂ + H₂O; ease: tertiary > secondary > primary; (2) dehydrohalogenation of alkyl halides using alcoholic KOH — CH₃CH₂Br + KOH → CH₂=CH₂ + KBr + H₂O (β-elimination, Saytzeff's rule applies — preferentially gives more substituted alkene). Other methods include dehalogenation of vicinal dihalides with Zn and partial hydrogenation of alkynes with Lindlar's catalyst (gives cis-alkene).
What is Markovnikov's rule?
Markovnikov's rule (1869) states that when a polar HX (where H is partial positive and X is partial negative) adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon of the double bond that already has more hydrogens, and the halogen attaches to the carbon with fewer hydrogens. NCERT Class 11 Chemistry Chapter 9 example: CH₃-CH=CH₂ + HBr → CH₃-CHBr-CH₃ (2-bromopropane, major), not CH₃-CH₂-CH₂Br (1-bromopropane). The rule arises from the formation of the more stable carbocation intermediate — the secondary carbocation is more stable than the primary one.
What is the peroxide effect (Kharasch effect)?
The peroxide effect (Kharasch-Mayo effect) is the reversal of Markovnikov's rule when HBr (only HBr — not HCl or HI) adds to an unsymmetrical alkene in the presence of peroxides like benzoyl peroxide. The addition follows anti-Markovnikov orientation, with H going to the carbon with fewer hydrogens. NCERT Class 11 Chemistry Chapter 9 example: CH₃-CH=CH₂ + HBr (with peroxide) → CH₃-CH₂-CH₂Br (1-bromopropane). The mechanism is free-radical, not ionic. The peroxide effect does not apply to HCl (Cl too strong a bond) or HI (I-H bond too weak to give chain) — only HBr.
What is ozonolysis of alkenes?
Ozonolysis is the reaction of an alkene with ozone (O₃), forming an unstable ozonide intermediate, which is then cleaved by reduction (Zn/H₂O or H₂/Pd) to give two carbonyl compounds. NCERT Class 11 Chemistry Chapter 9 example: CH₃-CH=CH-CH₃ + O₃ → ozonide → 2 CH₃-CHO (acetaldehyde) after reductive workup. Ozonolysis is used to determine the position of a double bond in an unknown alkene by identifying the carbonyl products. Reductive ozonolysis gives aldehydes/ketones; oxidative ozonolysis (H₂O₂) gives carboxylic acids and ketones.
What is polymerisation of alkenes?
Polymerisation is the chemical combination of many alkene monomers to form a large polymer molecule. NCERT Class 11 Chemistry Chapter 9 examples: (1) ethene → polyethene (polythene) under high pressure and temperature with a Ziegler-Natta or peroxide catalyst — used in plastic bags, bottles; (2) propene → polypropene — used in carpets, ropes; (3) vinyl chloride → polyvinyl chloride (PVC) — used in pipes, insulation; (4) styrene → polystyrene — used in disposable cups; (5) tetrafluoroethene → polytetrafluoroethene (PTFE, Teflon) — used as non-stick coating. The π bond breaks and forms new C-C single bonds linking the monomers.
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