Year 12 Chemistry Module 7 · Organic Chemistry ⏱ ~45 min 5 MC · 3 Short Answer Lesson 6 of 23 Core Core

Reactions of Alkenes, Hydrogenation, Halogenation, Hydrohalogenation & Hydration

Master the four ways alkenes react: discover why the pi bond is the reactive hotspot, how to write balanced equations for hydrogenation, halogenation, hydrohalogenation and hydration, and why making industrial ethanol needs heat, pressure, and an acid catalyst all at once.

Today's hook: Margarine is made from liquid vegetable oil, the only chemical difference between the oil and the solid spread is a few missing double bonds. How does adding hydrogen gas turn a liquid into a solid?
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You’re here

Connect: map the four alkene additions

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions
Before You Read

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.

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Addition Reactions of Alkenes, All Four Types
Alkene + H2 → Alkane
Hydrogenation Ni/Pd/Pt catalyst, ~150–200°C
Alkene + X2 → Dihaloalkane
Halogenation (X = Cl or Br) No catalyst, room temperature
Alkene + HX → Monohalogenated alkane
Hydrohalogenation Unsymmetrical alkenes give two possible products
Alkene + H2O → Alcohol
Hydration H3PO4 catalyst, ~300°C, high pressure
In every addition reaction: two reactants → one product; the C=C pi bond opens and the reagent adds across it, nothing is lost.
Learning Intentions
goals
Know
  • The reagents, conditions, and products for all four alkene addition reactions
  • That bromine water decolourises as a test for unsaturation (C=C or C≡C)
  • Markovnikov's rule (enrichment, not required for the NSW exam): H adds to the carbon with more H atoms
Understand
  • Why the pi bond, not the sigma bond, is the reactive part of the C=C bond
  • Why unsymmetrical alkenes can form two possible products
  • Why industrial ethanol production requires high temperature, pressure, and acid catalyst
Can Do
  • Write balanced equations for all four addition reactions with full conditions
  • Write the addition equation for HX or H₂O with an unsymmetrical alkene (using Markovnikov's rule to name the major product is enrichment, not required for the NSW exam)
  • Evaluate the bromine water test and identify its limitations
Scan these before reading
vocab
addition reactionTwo reactants combine to form one product. The C=C pi bond opens and each reagent atom bonds to one carbon of the former double bond.
pi bondThe weaker, more reactive component of a C=C bond, the weaker of the two bonds in a C=C double bond. It restricts rotation around the bond axis and is more reactive than the single bond and is broken in addition reactions.
hydrogenationAddition of H2 across a C=C bond to produce a saturated alkane. Requires a Ni, Pd, or Pt catalyst.
halogenationAddition of X2 (Br2 or Cl2) across a C=C bond to produce a vicinal dihaloalkane.
Markovnikov's ruleWhen HX adds to an unsymmetrical alkene, H goes to the carbon with more H atoms (the less-substituted carbon) and X goes to the more-substituted carbon.
hydrationAddition of H2O across a C=C bond to produce an alcohol. Requires an acid catalyst, high temperature, and high pressure.
📚 Core Content
2

Explain: the pi bond is the reactive site

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Why Alkenes Undergo Addition Reactions

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:

  • Sigma bond the framework bond along the C-C axis; strong, not easily broken
  • Pi bond the weaker bond in a C=C double bond; more reactive, restricts rotation around the C=C axis, more easily broken by approaching reagents

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.

ReactionReagent addedProduct typeWhat opens the C=C
HydrogenationH, HAlkane (saturated)H2 across C=C
HalogenationX, X (Br2/Cl2)DihaloalkaneX2 across C=C
HydrohalogenationH, XMonohalogenated alkaneHX across C=C
HydrationH, OHAlcoholH2O across C=C
Exam Rule: In every alkene addition question, identify the C=C first, then show what adds to each carbon. The fact that all four reactions add ACROSS the double bond, one atom or group to each carbon, is the unifying principle. Once you understand this, you can predict the product of any alkene addition reaction from the identity of the reagent.
Common Error: Students confuse addition reactions with substitution reactions. In addition: two reactants → one product, nothing is lost, the double bond disappears. In substitution: two reactants → two products, one group is replaced by another (HBr or H₂O is also produced). The key test, if the product has MORE atoms than either reactant alone, it is addition.
Quick check: In an alkene addition reaction, which bond breaks?
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Represent: hydrogenation and halogenation

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Hydrogenation and Halogenation

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.

Hydrogenation

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.

Conditions, Hydrogenation

REAGENT
H2 gas
CATALYST
Ni (or Pd or Pt), heterogeneous, solid metal surface
CONDITIONS
~150–200°C; high pressure (industrial scale)
EQUIPMENT
Pressure vessel (industrial); sealed apparatus (lab)

Examples:

  • CH2=CH2 + H2 → CH3-CH3 (ethene → ethane)
  • CH3CH=CHCH3 + H2 → CH3CH2CH2CH3 (but-2-ene → butane)

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.

Halogenation & the Bromine Water Test

Reactants: alkene + X2 (Br2 or Cl2). Product: dihaloalkane, one halogen on each carbon of the former double bond (vicinal dihalide).

Conditions, Halogenation

REAGENT
Br2 or Cl2 (as gas, pure liquid, or in solution)
CATALYST
None required
CONDITIONS
Room temperature
EQUIPMENT
Standard glassware; fume cupboard for Cl2

Examples:

  • CH2=CH2 + Br2 → CH2BrCH2Br (1,2-dibromoethane)
  • CH3CH=CH2 + Cl2 → CH3CHClCH2Cl (1,2-dichloropropane)

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.

Bromine water unsaturation test as an ordered reaction-and-interpretation sequence
Bromine water test: orange/brown → colourless confirms unsaturation. Not specific to alkenes alone.
Exam Rule: For hydrogenation, state the catalyst as Ni, Pd, OR Pt, any is acceptable. The catalyst is heterogeneous (solid), the reactants are gases or liquids. You must state both the catalyst AND the temperature, "hydrogen gas alone" does not react with alkenes at room temperature.
Common Error: The bromine water test confirms unsaturation (C=C or C≡C) it does NOT specifically confirm an alkene. Alkynes also decolourise bromine water. Saying "the compound is an alkene because it decolourises bromine water" without further qualification is incomplete and loses marks.
Safety: Hydrogenation uses hydrogen gas, which is highly flammable and forms explosive mixtures with air, keep all naked flames and sparks away and work in a well-ventilated area; the finely divided nickel (or Pd/Pt) catalyst can be pyrophoric and an irritant, so it is handled with care and this reaction is normally a teacher demonstration or industrial process, not a student bench practical. Bromine water is corrosive and toxic and its vapour irritates the eyes and lungs, use small volumes of dilute bromine water in a fume cupboard with goggles and gloves; the halogenated organic products (e.g. 1,2-dibromoethane) are harmful, so collect them as halogenated organic waste rather than pouring them down the sink.
True or False: The bromine water test specifically confirms the presence of an alkene (C=C) and no other functional group.
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Respond: predict hydrohalogenation products

Beyond the syllabus. Addition of hydrogen, halogens, hydrogen halides and water, with conditions, products and the bromine-water test, is core. Markovnikov orientation is fine as a quick prediction aid, but anti-Markovnikov chemistry and mechanistic explanations are extension — the exam requires the reaction families, not the mechanism.
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Hydrohalogenation and Markovnikov's Rule

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. The syllabus asks you to write the addition equation; Markovnikov's rule, covered below as enrichment, describes which product forms predominantly.

Reactants: alkene + HX (HCl, HBr, or HI). Product: monohalogenated alkane. Conditions: HX gas or HX dissolved in organic solvent; no catalyst; room temperature.

Conditions, Hydrohalogenation

REAGENT
HCl, HBr, or HI (gas or in organic solvent)
CATALYST
None required
CONDITIONS
Room temperature
EQUIPMENT
Standard glassware; fume cupboard (HX gases are corrosive)

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.

Markovnikov's Rule

The Rule in Plain English

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):

Major product (Markovnikov)

CH3CH=CH2 + HBr → CH3CHBrCH3

H → C1 (has 2H, more H). Br → C2 (has 1H, more substituted). Product: 2-bromopropane, the major product.

Minor product (anti-Markovnikov)

CH3CH=CH2 + HBr → CH3CH2CH2Br

H → C2. Br → C1 (primary carbon). Product: 1-bromopropane, the minor product.

Step-by-step method for applying Markovnikov's rule:

  1. Draw or identify the C=C and label both carbons
  2. Count H atoms on each carbon of the double bond
  3. H adds to the carbon with MORE H atoms
  4. X adds to the carbon with FEWER H atoms (more substituted)
  5. Write and name the major product
Markovnikov addition of HBr to propene, comparing C1 and C2
Markovnikov's rule: H adds to C1 (more H atoms), Br adds to C2 (fewer H, more substituted) → major product is 2-bromopropane.
Beyond the syllabus (extension): Predicting which product dominates (Markovnikov's rule / regiochemistry) is not required by NSW HSC Chemistry Module 7. The syllabus asks only that you write the addition reaction of an unsaturated hydrocarbon with H2, a halogen, a hydrogen halide or water. Markovnikov's rule is included here as enrichment for interested students; you will not lose marks in the NSW exam for omitting it. If you do use it, identify which C=C carbon has more H atoms first.
Common Error: Students apply Markovnikov's rule to symmetrical alkenes like but-2-ene (CH3CH=CHCH3), where both C=C carbons are identical. For symmetrical alkenes, there is only ONE possible product, applying Markovnikov's rule is unnecessary and can produce a wrong answer if applied carelessly.
Propene + HBr: which is the MAJOR product according to Markovnikov's rule?
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Feedback: hydration, conditions and misconceptions

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Hydration, Making Alcohols from Alkenes

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.

Conditions, Hydration

REAGENT
H2O (steam)
CATALYST
Dilute H3PO4 or H2SO4, a homogeneous acid catalyst
CONDITIONS
~300°C, high pressure (60–70 atm for industrial scale)
EQUIPMENT
High-pressure reactor (industrial); reflux condenser (lab scale)

Examples:

  • CH2=CH2 + H2O → CH3CH2OH (ethanol, industrial production)
  • CH3CH=CH2 + H2O → CH3CH(OH)CH3 (propan-2-ol, Markovnikov)

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).

Hydration vs Fermentation for Ethanol Production

Hydration (industrial)
• Ethene (from crude oil)
• 300°C, 65 atm, H3PO4 catalyst
• CH2=CH2 + H2O → CH3CH2OH
• Fast, continuous process
• High purity (~95% ethanol)
• Not renewable (crude oil derived)
Fermentation (biological)
• Glucose/starch (from plants)
• ~35°C, anaerobic, yeast enzyme
• C6H12O6 → 2CH3CH2OH + 2CO2
• Slow, batch process
• Low purity (~15%, must distil)
• Renewable (plant-based glucose)
Exam Rule: Hydration requires ALL THREE conditions: catalyst (H3PO4 or H2SO4) AND high temperature (~300°C) AND high pressure (~65 atm). Stating only one or two earns partial marks. The most commonly omitted condition is pressure.
Common Error: Students write "hydration uses HCl as the catalyst." HCl is the reagent in hydrohalogenation, it adds HCl across the double bond to give a haloalkane. Hydration uses H2O as the REAGENT and H3PO4 or H2SO4 as the CATALYST. Confusing these is a mark-losing error.
Match the reaction type to its correct catalyst/conditions. Which combination describes HYDRATION of an alkene?
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High-Frequency Misconceptions
  • ✕

    "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.

Complete the sentence: Industrial hydration of ethene requires H₃PO₄ as a ______, a temperature of approximately ______°C, and ______ pressure, all three conditions must be stated.
🧠 Worked Examples

Worked Example 1

Writing Equations with Conditions

Problem: Write balanced equations and state all conditions for: (a) propene + hydrogen; (b) ethene + bromine; (c) but-1-ene + water.

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(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

2

(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

3

(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)

Worked Example 2

Markovnikov's Rule (enrichment)

Problem: (a) Write the major product of the reaction between but-1-ene and HBr, and name it. (b) Write the minor product. Markovnikov's rule is enrichment beyond the NSW syllabus; the required skill is writing the addition equation.

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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).

2

Major product: H adds to C1 (more H). Br adds to C2 (fewer H, more substituted). Product: CH3CH2CHBrCH3 = 2-bromobutane.

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Why it is the major product: Markovnikov's rule predicts that H adds to the double-bond carbon that already has more H atoms (C1), placing Br on the more substituted carbon (C2). This gives 2-bromobutane as the major product. (The underlying reaction mechanism is beyond the NSW syllabus.)

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Minor product: The reverse addition (H to C2, Br to C1, called anti-Markovnikov) gives 1-bromobutane (CH3CH2CH2CH2Br), which forms in smaller amounts.

Answer: (a) Major: 2-bromobutane (CH3CH2CHBrCH3), H to C1 (more H), Br to C2 (more substituted). (b) Minor: 1-bromobutane (CH3CH2CH2CH2Br). The required skill is writing the addition equation; identifying the major product via Markovnikov's rule is enrichment.

Worked Example 3

Extended Response, Bromine Water Test & Evaluation

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?

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(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.

2

(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).

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(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.

Interactive Tool, Hydrocarbon Reactions Predictor Open fullscreen ↗
True or false? Bromine water is decolourised by alkenes and alkynes (Br₂ adds across the multiple bond) but not by alkanes in the dark.
🔀 Sort the Steps +7 XP
Arrange these steps in the correct order for writing the product when HBr adds to an unsymmetrical alkene such as propene (naming which product dominates uses Markovnikov's rule, enrichment beyond the syllabus):
Open the pi bond so each carbon can form one new single bond
Add H and Br across the former double bond (major product: H to the carbon with more H, by Markovnikov's rule)
Write and name the resulting monohalogenated alkane
Identify the C=C double bond and label the two carbons
Count the hydrogen atoms already on each double-bond carbon
6

Apply: evidence, equations and product prediction

Complete the Learn phase to unlock Practice.

✏️ Activities
A1
Activity 1, Guided Practice: Addition Reactions

For each reaction below, write the balanced equation, identify the reaction type, and state all conditions (reagent, catalyst, temperature, equipment).

Reactants
1. Ethene + Cl2
2. Propene + H2
3. But-1-ene + HCl (apply Markovnikov)
4. Ethene + H2O (industrial)
Your equation, reaction type, and conditions
A2
Activity 2, Test Evidence and Reaction Classification

A chemist carries out four experiments. For each observation, identify the reaction type, write a possible equation, and evaluate any conclusion made.

Observation
A. C₅H₁₀ is added to bromine water, rapid decolourisation.
B. Propene + H₂/Ni/150°C, reaction product is bromine-inactive.
C. 2-methylpropene + HBr, major product: tertiary haloalkane.
D. Ethene + H₂O/H₃PO₄/300°C, alcohol product confirmed.
Your analysis
📝 Check Your Understanding
MC
Randomised Question Bank

Complete five questions drawn from the ten aligned alkene-addition items.

SA
Exam-Style Practice

1. Explain why alkenes undergo addition reactions while alkanes typically undergo substitution. Refer to bond types in your answer. (3 marks)

2. Write balanced equations, with conditions, for the reaction of but-1-ene with (i) hydrogen, (ii) bromine, and (iii) steam (water) with an acid catalyst. For the reaction with HBr, identify the major product. (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)

Show All Answers

Activity 1, Guided Practice

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).

Activity 2, Test Evidence

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. (Predicting the major product uses Markovnikov's rule, which is enrichment beyond the syllabus.)

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).

Multiple Choice Answers

1. Correct response: 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. Correct response: 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. Correct response: 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. Correct response: 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. Correct response: 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.

Short Answer Model Answers

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): (i) Hydrogenation: CH3CH2CH=CH2 + H2 → CH3CH2CH2CH3 (butane); Ni catalyst, ~150–200°C [1]. (ii) Halogenation: CH3CH2CH=CH2 + Br2 → CH3CH2CHBrCH2Br (1,2-dibromobutane); no catalyst, room temperature [1]. (iii) Hydration: CH3CH2CH=CH2 + H2O → CH3CH2CH(OH)CH3 (butan-2-ol); H3PO4 catalyst, ~300°C, high pressure [1]. HBr major product: 2-bromobutane, CH3CH2CHBrCH3 (Markovnikov, enrichment) [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].

How did your thinking change?

Return to your original response about margarine production. You should now be able to give a precise HSC-style explanation:

  • The hydrogen gas undergoes hydrogenation H2 adds across the C=C double bonds in the vegetable oil's fatty acid chains
  • Each H atom bonds to one carbon of the former double bond, the pi bond breaks, the carbons become saturated
  • The nickel catalyst provides a surface on which both the oil's double bonds and H2 adsorb, lowering activation energy
  • Saturated fatty acids have higher melting points (stronger London dispersion forces between straight chains) → the oil becomes solid/semi-solid margarine
7

Review: retrieve the addition patterns

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🏆 Review

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?