Master the four ways alkenes react: discover why the pi bond is the reactive hotspot, how Markovnikov's rule predicts which product dominates, and why making industrial ethanol needs heat, pressure, and an acid catalyst all at once.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
Margarine was invented in 1869 as a butter substitute. It is made from vegetable oil — which is liquid at room temperature — by reacting it with hydrogen gas in the presence of a nickel catalyst. The liquid oil becomes solid or semi-solid margarine. The oil and margarine have the same carbon skeletons and almost the same molecular formula — the only change is that some carbon-carbon double bonds have disappeared and been replaced with single bonds.
Before you read on, write down what you think happens to the hydrogen gas during this reaction. Where do the H atoms end up, and what happens to the double bond? You will return to this at the end of the lesson.
The pi bond is the reactive site — electron-rich, exposed, and ready to break
The C=C double bond is electron-rich and accessible — the pi bond sits exposed above and below the molecular plane, making it a magnet for electron-seeking reagents that add across it and convert it to a single bond.
An addition reaction is one in which two reactants combine to form a single product with no atoms lost. This is the defining reaction type of alkenes. The C=C double bond consists of two components:
When an addition reaction occurs, the pi bond breaks, the sigma bond remains intact, and the two carbons each form a new bond to the incoming reagent atoms. The product contains only single bonds — it is now saturated at those carbons.
| Reaction | Reagent added | Product type | What opens the C=C |
|---|---|---|---|
| Hydrogenation | H—H | Alkane (saturated) | H2 across C=C |
| Halogenation | X—X (Br2/Cl2) | Dihaloalkane | X2 across C=C |
| Hydrohalogenation | H—X | Monohalogenated alkane | HX across C=C |
| Hydration | H—OH | Alcohol | H2O across C=C |
Margarine and the bromine water test — two reactions, two industries
Hydrogenation adds hydrogen across a double bond to produce a saturated product — halogenation adds a halogen molecule — and both reactions are the foundation of major industrial processes you encounter every day.
Reactants: alkene + H2(g). Product: alkane (saturated). The catalyst provides a surface on which both the alkene and H2 adsorb, lowering the activation energy for H-H bond breaking and H addition to C.
Examples:
Industrial significance: vegetable oil hydrogenation (margarine production) — liquid unsaturated fats (containing C=C in long fatty acid chains) are partially hydrogenated using a nickel catalyst at ~200°C to reduce the number of double bonds and increase the melting point from liquid to semi-solid.
Reactants: alkene + X2 (Br2 or Cl2). Product: dihaloalkane — one halogen on each carbon of the former double bond (vicinal dihalide).
Examples:
Bromine water test for unsaturation: Bromine water is an aqueous solution of Br2 — orange/brown in colour. When an alkene (or alkyne) is added, Br2 reacts with the C=C by addition to produce a colourless dihaloalkane. The colour change — orange/brown → colourless — is the positive test result.
Two possible products — but one forms in greater amounts
When an asymmetrical reagent like HBr adds to an asymmetrical alkene, there are two possible products — Markovnikov's rule tells you which one forms predominantly, and understanding why requires thinking about where the electrons go.
Reactants: alkene + HX (HCl, HBr, or HI). Product: monohalogenated alkane. Conditions: HX gas or HX dissolved in organic solvent; no catalyst; room temperature.
For symmetrical alkenes (both C=C carbons have identical substituents), there is only one possible product and Markovnikov's rule is not needed. For unsymmetrical alkenes, two products are possible — the H can add to either carbon and X goes to the other. Markovnikov's rule determines the major product.
When HX adds to an unsymmetrical alkene, the H adds to the carbon of the C=C that already has MORE hydrogen atoms. The X (halogen) adds to the carbon with FEWER hydrogens (the more substituted carbon).
Memory aid: "The rich get richer" — the carbon already rich in H atoms gets the extra H.
Example — propene + HBr (unsymmetrical):
H → C1 (has 2H, more H). Br → C2 (has 1H, more substituted). Product: 2-bromopropane. The secondary carbocation at C2 is more stable → faster pathway → major product.
H → C2. Br → C1 (primary carbon). Product: 1-bromopropane. The primary carbocation at C1 is less stable → slower pathway → minor product.
Step-by-step method for applying Markovnikov's rule:
Industrial ethanol, Le Chatelier, and why three conditions must all be stated
Hydration of an alkene converts a hydrocarbon into an alcohol by adding water across the double bond — it is the industrial route to ethanol and to many other alcohols, and it is the reverse of the dehydration reaction you will meet in L12.
Reactants: alkene + H2O (steam). Product: alcohol. The H of water adds to one carbon of the C=C; the OH adds to the other. For unsymmetrical alkenes, Markovnikov's rule applies: H adds to the carbon with more H, OH adds to the more substituted carbon.
Examples:
The reaction is reversible: alkene + H2O ⇌ alcohol. High pressure is used industrially to drive the equilibrium toward the alcohol product (Le Chatelier — increasing pressure favours the side with fewer moles of gas).
"Bromine water decolourising proves the compound is an alkene." Wrong. Alkynes also decolourise bromine water. The test confirms C=C or C≡C unsaturation, not specifically an alkene.
"Markovnikov's rule applies to all alkenes." Wrong. For symmetrical alkenes (like but-2-ene), there is only one possible product regardless. The rule is only needed when the two C=C carbons are different.
"Hydration just needs a catalyst at room temperature." Wrong. Hydration requires H3PO4 OR H2SO4 catalyst, ~300°C, AND high pressure (~65 atm) — all three must be stated.
"The nickel catalyst in hydrogenation is consumed." Wrong. Ni/Pd/Pt is a heterogeneous catalyst — it provides a surface for the reaction to occur but is not consumed and does not appear in the balanced equation.
Problem: Write balanced equations and state all conditions for: (a) propene + hydrogen; (b) ethene + bromine; (c) but-1-ene + water.
(a) Hydrogenation of propene: Propene: CH3CH=CH2. H adds to each carbon of C=C. Product: CH3CH2CH3 (propane).
Equation: CH3CH=CH2 + H2 → CH3CH2CH3
Conditions: H2 gas | Ni catalyst | ~150–200°C | pressure vessel
(b) Halogenation of ethene: Ethene: CH2=CH2. Br adds to each carbon of C=C. Product: CH2BrCH2Br (1,2-dibromoethane).
Equation: CH2=CH2 + Br2 → CH2BrCH2Br
Conditions: Br2 in solution | no catalyst | room temperature | fume cupboard
(c) Hydration of but-1-ene: But-1-ene: CH3CH2CH=CH2. Markovnikov applies: C1 (=CH2) has 2H → H adds here; C2 (CH=) has 1H → OH adds here. Product: CH3CH2CH(OH)CH3 = butan-2-ol.
Equation: CH3CH2CH=CH2 + H2O → CH3CH2CH(OH)CH3
Conditions: H2O (steam) | H3PO4 catalyst | ~300°C | high pressure
Answer: (a) CH3CH=CH2 + H2 → CH3CH2CH3 (Ni, 150–200°C) | (b) CH2=CH2 + Br2 → CH2BrCH2Br (no catalyst, r.t.) | (c) CH3CH2CH=CH2 + H2O → CH3CH2CH(OH)CH3 (H3PO4, 300°C, high pressure)
Problem: (a) Write the major product of the reaction between but-1-ene and HBr. Name the product and explain why it is the major product using Markovnikov's rule. (b) Write the minor product and explain why it forms in smaller amounts.
Draw but-1-ene: CH3CH2CH=CH2. The C=C is between C1 (=CH2, two H atoms) and C2 (=CH-, one H atom and one CH2CH3 group).
Major product: H adds to C1 (more H). Br adds to C2 (fewer H, more substituted). Product: CH3CH2CHBrCH3 = 2-bromobutane.
Why it is the major product: When H+ adds to C1, the positive charge develops on C2. C2 has an ethyl group (CH2CH3) attached — a secondary carbocation, more stable due to alkyl group electron donation. The more stable carbocation forms more readily → this pathway is faster → 2-bromobutane is the major product.
Minor product: H adds to C2, Br adds to C1 (anti-Markovnikov). The carbocation on C1 is primary — less stable (no stabilising alkyl groups). This pathway is slower → 1-bromobutane (CH3CH2CH2CH2Br) forms in smaller amounts.
Answer: (a) Major: 2-bromobutane (CH3CH2CHBrCH3) — H to C1 (more H), Br to C2 (more substituted); secondary carbocation intermediate is more stable. (b) Minor: 1-bromobutane (CH3CH2CH2CH2Br) — primary carbocation at C1 is less stable, so this pathway is slower.
Problem (6 marks): A student adds an unknown compound X (molecular formula C4H8) to bromine water and observes the orange-brown colour disappears. The student concludes: "Compound X must be an alkene because it decolourised bromine water." (a) Evaluate the student's conclusion. (b) Compound X, treated with H2O in the presence of H3PO4 at 300°C and high pressure, produces a single alcohol. Identify the structural formula and IUPAC name of the alcohol. (c) If compound X had been but-2-ene, how would the alcohol product differ?
(a) Evaluate: The observation is correct — C4H8 decolourising Br2(aq) confirms unsaturation (C=C). However, the conclusion is incomplete: C4H8 is consistent with alkene (CnH2n) but also with cycloalkane (also CnH2n). Cycloalkanes do NOT decolourise bromine water. Alkynes are C4H6, so not applicable here. Additional evidence — e.g. reaction with H2/Ni producing a saturated product, or hydration producing an alcohol — would be needed to confirm C=C specifically.
(b) Alcohol from hydration of C4H8: The question states a SINGLE alcohol product. For but-1-ene (CH3CH2CH=CH2), Markovnikov hydration gives one major product: H → C1, OH → C2. Product: CH3CH2CH(OH)CH3 = butan-2-ol. Structural formula: CH3-CH(OH)-CH2-CH3. IUPAC name: butan-2-ol (secondary alcohol).
(c) But-2-ene comparison: But-2-ene (CH3CH=CHCH3) is symmetrical — both C=C carbons each have one H and one CH3. Addition of H2O in either direction gives the same compound: butan-2-ol. Markovnikov's rule is not needed — there is no ambiguity. But-1-ene also gives butan-2-ol as its Markovnikov product, but Markovnikov's rule IS needed to choose this over butan-1-ol. So both alkenes give butan-2-ol; the difference is that Markovnikov's rule is required for but-1-ene but not for but-2-ene.
Answer: (a) Partially correct — decolourisation confirms C=C or C≡C; C4H8 could be a cycloalkane (which doesn't react); more evidence needed. (b) Butan-2-ol — CH3CH(OH)CH2CH3. (c) But-2-ene also gives butan-2-ol (symmetrical alkene, Markovnikov's rule unnecessary), whereas but-1-ene requires Markovnikov's rule to select butan-2-ol over butan-1-ol.
Complete the Learn phase to unlock Practice.
For each reaction below, write the balanced equation, identify the reaction type, and state all conditions (reagent, catalyst, temperature, equipment).
A chemist carries out four experiments. For each observation, identify the reaction type, write a possible equation, and evaluate any conclusion made.
1. What is the major product when 2-methylpropene (CH2=C(CH3)2) reacts with HBr?
2. A student adds bromine water to compound P — it decolourises. Compound Q does not decolourise. Which conclusion is best supported?
3. Which set of conditions correctly describes the industrial hydration of ethene to produce ethanol?
4. Which statement about addition reactions of alkenes is correct?
5. Propene reacts with water under appropriate conditions to form an alcohol. Using Markovnikov's rule, what is the major product?
1. Explain why alkenes undergo addition reactions while alkanes typically undergo substitution. Refer to bond types in your answer. (3 marks)
2. Write the equation for the reaction of but-1-ene with HBr and identify the major product. Explain your answer using Markovnikov's rule and the relative stability of the carbocation intermediates. (4 marks)
3. Evaluate the use of the bromine water test to identify an alkene. In your answer, discuss what the test confirms, what it does not confirm, and what additional evidence would be needed to specifically identify an alkene rather than any unsaturated compound. (5 marks)
1. Ethene + Cl2: CH2=CH2 + Cl2 → CH2ClCH2Cl. Halogenation. No catalyst, room temperature, fume cupboard (Cl2 is toxic).
2. Propene + H2: CH3CH=CH2 + H2 → CH3CH2CH3. Hydrogenation. Ni catalyst, ~150–200°C, sealed/pressure apparatus.
3. But-1-ene + HCl: CH3CH2CH=CH2 + HCl → CH3CH2CHClCH3. Hydrohalogenation. Markovnikov: H to C1 (2H), Cl to C2 (1H, more substituted). Major product: 2-chlorobutane. No catalyst, room temperature, fume cupboard.
4. Ethene + H2O: CH2=CH2 + H2O → CH3CH2OH. Hydration. Product: ethanol. Conditions: H2O (steam), H3PO4 catalyst, ~300°C, ~65 atm (high pressure).
A. The compound contains C=C or C≡C unsaturation. Decolourisation does NOT confirm it is specifically an alkene — a cycloalkane (also C5H10) would NOT decolourise. An alkyne (also unsaturated) would. Additional confirmation: reaction with H2/Ni to give a saturated product, or hydration to give an alcohol.
B. Hydrogenation. CH3CH=CH2 + H2 → CH3CH2CH3 (propane). The product is bromine-inactive because the C=C double bond has been consumed — propane is saturated and does not react with bromine water.
C. 2-methylpropene: CH2=C(CH3)2. Apply Markovnikov: C1 (=CH2) has 2H → H adds here. C2 (=C(CH3)2) has 0H → Br adds here (tertiary carbon). Major product: (CH3)3CBr = 2-bromo-2-methylpropane. The tertiary carbocation at C2 is very stable (three alkyl groups), making this the strongly favoured pathway.
D. Hydration. CH2=CH2 + H2O → CH3CH2OH (ethanol). High temperature (300°C) provides activation energy for the acid-catalysed reaction. High pressure (65 atm) drives the equilibrium toward the alcohol product (Le Chatelier — fewer gas moles on the product side).
1. B — Apply Markovnikov's rule to 2-methylpropene: C1 (=CH2) has 2H → H here; C2 (=C(CH3)2) has 0H → Br here (tertiary carbon). Product: 2-bromo-2-methylpropane. Option A is the anti-Markovnikov product; option D would be halogenation (two Br atoms).
2. B — Decolourisation confirms unsaturation (C=C or C≡C). Alkynes also decolourise; some aldehydes can too. Not decolourising means no accessible unsaturation — consistent with alkane, cycloalkane, ketone, or alcohol. Option A is too specific; option D incorrectly identifies Q as definitely a cycloalkane.
3. B — Industrial hydration: H2O (steam), H3PO4 catalyst, ~300°C, high pressure (~65 atm). Option A is halogenation. Option C has wrong catalyst (Ni is for hydrogenation) and wrong conditions. Option D is hydrohalogenation.
4. C — Addition: two reactants → one product; the pi bond breaks and one atom of the reagent bonds to each C of the former double bond. The sigma bond remains intact. Option B has this backwards — it is the pi bond, not the sigma bond, that breaks.
5. A — Propene: CH3CH=CH2. Markovnikov: H to C3 (=CH2, 2H); OH to C2 (=CH-, 1H). Product: CH3CH(OH)CH3 = propan-2-ol. Option B (propan-1-ol) would be the anti-Markovnikov product.
Q1 (3 marks): Alkenes contain a C=C double bond consisting of a sigma bond and a pi bond [1]. The pi bond is weaker, electron-rich, and accessible — it can be broken by approaching reagents, allowing two reactants to add across the double bond (addition reaction) [1]. Alkanes have only sigma bonds, which are stronger and less accessible — they require radical initiation (UV light) for halogenation, and the reaction is substitution (one H replaced by one halogen) rather than addition, because there is no multiple bond to open [1].
Q2 (4 marks): But-1-ene + HBr: CH3CH2CH=CH2 + HBr → CH3CH2CHBrCH3 [1]. Major product: 2-bromobutane [1]. Markovnikov's rule: H adds to C1 (more H atoms, 2H); Br adds to C2 (fewer H atoms, more substituted) [1]. The carbocation intermediate at C2 is secondary — more stable than a primary carbocation at C1 due to electron donation from the adjacent alkyl groups — so this pathway is faster and the Markovnikov product predominates [1].
Q3 (5 marks): The bromine water test involves adding the compound to orange/brown Br2(aq) — decolourisation is a positive result [1]. The test confirms that the compound contains unsaturation: a C=C or C≡C bond, which reacts with Br2 by addition to form a colourless dihalo-product [1]. However, the test does NOT specifically confirm an alkene — alkynes also decolourise bromine water (addition occurs to the triple bond) [1], and cycloalkanes (which have the same molecular formula CnH2n as alkenes) do NOT react with bromine water [1]. To specifically identify an alkene, additional evidence is needed: for example, reaction with H2/Ni catalyst to give a saturated product (confirming C=C), followed by hydration with H2O/H3PO4 to give an alcohol; or IR/NMR data to confirm the presence of C=C but not C≡C [1].
Return to your original response about margarine production. You should now be able to give a precise HSC-style explanation:
State the four types of alkene addition reactions, and the reagent that opens the C=C in each.
State Markovnikov's rule in plain English and apply it to: but-1-ene + HCl. What is the major product?
What does the bromine water test confirm? What is it unable to confirm, and why?
List all three conditions required for the industrial hydration of ethene to produce ethanol.
Why does the Ni catalyst in hydrogenation not appear in the balanced equation?
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