Ch 2 · Alkenes · AS & A Level Chemistry 9701

Topic 2 of 2 in Hydrocarbons — AS & A Level Chemistry 9701 — 77 lessons.

Introduction

Alkenes are far more reactive than alkanes because the exposed, electron-rich pi bond of the C=C double bond readily attracts electrophiles. This single structural feature explains every reaction in this lesson: addition of hydrogen, steam and halogens, oxidation by manganate(VII), polymerisation, and the bromine water test, and the electrophilic addition mechanism must be reproduced accurately with curly arrows and carbocation intermediates.

Key ideas — Producing alkenes: elimination and dehydration

Alkenes can be produced from a halogenoalkane by elimination: heating with a concentrated solution of NaOH or KOH dissolved in ethanol removes HX across adjacent carbons, forming a C=C bond, e.g. CH_3CH_2Br + KOH xrightarrowethanol, heat CH_2=CH_2 + KBr + H_2O. Alkenes can also be produced from an alcohol by dehydration: passing the alcohol vapour over a heated catalyst such as Al_2O_3, or heating with a concentrated acid catalyst such as concentrated H_3PO_4 or H_2SO_4, eliminates water, e.g. CH_3CH_2OH xrightarrowAl_2O_3, heat CH_2=CH_2 + H_2O.

🧑‍🏫 Teacher's explanation

Think of making an alkene like taking a piece of double-sided tape and turning it into a single-sided piece of tape by removing a strip. In organic chemistry, we often want to create a double bond (a C=C bond) to make a molecule more reactive. To do this, we have to 'remove' two small atoms or groups from adjacent carbon atoms.

📖 Definitions

🔍 Worked example — Easy

Calculate the molar mass of the alkene produced by the dehydration of ethanol, CH_3 extCH_2 extOH, and state its formula.

🔍 Worked example — Standard

A sample of 5.00 g of bromoethane, CH_3 extCH_2 extBr, undergoes elimination with KOH in ethanol. Calculate the mass of ethene, CH_2=CH_2, produced. (A_r: H=1, C=12, Br=57)

🔍 Worked example — Exam-style

Predict the major product when 2-chlorobutane undergoes elimination with concentrated KOH in ethanol and explain why using Zaitsev's rule (the most substituted alkene is preferred).

⚠️ Common errors

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🎯 Exam question

Key ideas — Electrophilic addition: hydrogen, steam, hydrogen halides and halogens

Alkenes undergo electrophilic addition across the C=C bond with several reagents:

🧑‍🏫 Teacher's explanation

Think of the double bond in an alkene like a pair of magnets that are very close together, creating a dense cloud of negative charge. This cloud is 'electron-rich'. Because it is so negative, it attracts 'electrophiles'—which are molecules or ions that are 'electron-loving' and looking for a positive charge to neutralize themselves.

📖 Definitions

🔍 Worked example — Easy

Calculate the molar mass of the product formed when ethene (C_2H_4) reacts with HBr.

🔍 Worked example — Standard

Explain why the reaction of propene (CH_3CH=CH_2) with HBr yields 2-bromopropane as the major product rather than 1-bromopropane.

🔍 Worked example — Exam-style

A sample of ethene (C_2H_4) is reacted with Br_2 in an organic solvent. If 0.50 mol of ethene reacts completely, what volume of Br_2 is required at RTP? (Assume Br_2 is a gas for calculation purposes, V_m = 24 dm^3mol^-1)

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🎯 Exam question

Key ideas — Mechanism of electrophilic addition

Using the addition of hydrogen bromide to propene as an example: the electron-rich pi bond of propene attacks the delta^+ hydrogen atom of the polarised H-Br molecule, and a pair of electrons from the pi bond forms a new C-H bond. This heterolytically breaks the H-Br bond, releasing a bromide ion, Br^-, and leaving a carbocation (an ion with a positively charged carbon) on the other carbon that was part of the double bond. The bromide ion then acts as a nucleophile, donating a lone pair to the positive carbon to form the C-Br bond and complete the addition.

🧑‍🏫 Teacher's explanation

Think of the double bond in an alkene as a dense cloud of electrons sitting above and below the carbon atoms. Because these electrons are 'exposed' (not tucked between nuclei like a single bond), they are highly attractive to anything looking for electrons, such as an electrophile.

📖 Definitions

🔍 Worked example — Easy

Draw the mechanism for the reaction between ethene (C_2H_4) and HBr.

🔍 Worked example — Standard

Predict the major product when propene (CH_3CH=CH_2) reacts with HBr.

🔍 Worked example — Exam-style

Calculate the mass of 2-bromopropane produced from the reaction of 0.50 moles of propene with excess HBr.

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🎯 Exam question

(a) Draw the mechanism for the formation of the major product. [4] (b) Explain, in terms of carbocation stability, why the major product is formed. [2] [6 marks]

Key ideas — Carbocation stability and Markovnikov's rule

Alkyl groups are electron-releasing through the positive inductive effect (+I effect): they push electron density towards the positively charged carbon of a carbocation, spreading out (delocalising) the charge and making the ion more stable. A secondary carbocation, with two alkyl groups attached to the positive carbon, is therefore more stable than a primary carbocation, with only one. Since the more stable carbocation forms preferentially and reacts faster, the major product of HBr adding to propene is 2-bromopropane (via the secondary carbocation), with 1-bromopropane (via the less stable primary carbocation) as only a minor product. This is summarised by Markovnikov's rule: the hydrogen atom adds preferentially to the carbon of the double bond that already carries the greater number of hydrogen atoms.

🧑‍🏫 Teacher's explanation

Imagine a group of people trying to hold a heavy, heavy weight. If one person is holding it alone, they are under immense pressure (high charge density). If two or three people help hold it, the weight is shared, and the pressure on any single person is much lower (lower charge density). In chemistry, a carbocation is like that person under pressure. An alkyl group (like -CH_3) acts like a helper, pushing electron density toward the positive carbon to 'share the load'. This makes the positive charge less intense and the ion more stable.

📖 Definitions

🔍 Worked example — Easy

Identify the type of carbocation formed when propene reacts with H^+: CH_3CH=CH_2 + H^+ arrow [CH_3CH^+CH_3] or [CH_3CH_2CH_2^+]?

🔍 Worked example — Standard

Predict the major organic product when propene reacts with HBr.

🔍 Worked example — Exam-style

Calculate the mass of 2-bromopropane formed if 0.500 mol of propene reacts completely with HBr (assume 100% yield of major product). M_r of C_3H_7Br = 123.

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🎯 Exam question

Key ideas — Oxidation by potassium manganate(VII)

Cold, dilute, acidified potassium manganate(VII), KMnO_4, oxidises an alkene to a diol (two -OH groups on adjacent carbons), decolourising the purple manganate(VII) as it is reduced, e.g. ethene forms ethane-1,2-diol. Hot, concentrated, acidified KMnO_4 oxidises the double bond more vigorously, breaking the C=C bond completely: each carbon is oxidised to a ketone (if it carries two alkyl/aryl groups) or all the way to a carboxylic acid or carbon dioxide (if it carries at least one hydrogen). This oxidative cleavage is used to help deduce the position of a double bond in an unknown alkene from its oxidation products.

🧑‍🏫 Teacher's explanation

Think of the carbon-carbon double bond (C=C) as a high-energy, reactive 'zipper' in the molecule. When you use a mild reagent, it's like carefully unzipping just the two strands and adding a small attachment to each side. This is the 'cold, dilute' method which produces a diol.

📖 Definitions

🔍 Worked example — Easy

Ethene reacts with cold, dilute, acidified KMnO_4. Write the balanced equation for the formation of the organic product.

🔍 Worked example — Standard

Propene (CH_3CH=CH_2) is treated with hot, concentrated, acidified KMnO_4. Predict the two organic products formed.

🔍 Worked example — Exam-style

An unknown alkene, C_4H_8, undergoes vigorous oxidation with hot, concentrated KMnO_4 to produce only one organic product: CH_3 extCH_2 extCOOH. Identify the structure of the original alkene.

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🎯 Exam question

(a) Identify the alkene. (b) Write the balanced equation for the reaction (you may use [O] to represent the oxygen from the oxidizing agent). [3 marks]

Key ideas — Addition polymerisation

Under high pressure with a catalyst, many alkene monomer molecules join together by repeated addition across their C=C bonds, forming a long-chain addition polymer with no other by-product. Ethene monomers form poly(ethene): nCH_2=CH_2 arrow (-CH_2CH_2-)_n. Propene monomers similarly form poly(propene): nCH_3CH=CH_2 arrow (-CH(CH_3)CH_2-)_n. The repeat unit is drawn with the two open bonds shown extending from a bracket, and n denotes a very large number of repeat units.

🧑‍🏫 Teacher's explanation

Imagine a long chain of paperclips. Each individual paperclip is like a monomer—a small, independent molecule. When you hook them all together, you get a long, continuous chain called a polymer. In addition polymerisation, the 'hooks' are the double bonds (C=C) in the alkene molecules. When these bonds break, they allow the molecules to link up end-to-end without losing any atoms.

📖 Definitions

🔍 Worked example — Easy

Calculate the molar mass of the repeat unit of poly(ethene) formed from ethene, CH_2=CH_2.

🔍 Worked example — Exam-style

Ethene is used to produce poly(ethene). If 1.00 mol of ethene is reacted to form a polymer, what is the theoretical mass of the poly(ethene) produced, assuming 100% yield and no by-products?

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🎯 Exam question

Key ideas — The bromine water test for C=C

Shaking an alkene with aqueous bromine (bromine water) causes rapid electrophilic addition across the double bond, decolourising the orange-brown bromine solution as it is converted to a colourless dibromo product, e.g. CH_2=CH_2 + Br_2 arrow CH_2BrCH_2Br. This rapid decolourisation at room temperature, without UV light, is used as a simple chemical test to distinguish an unsaturated alkene from a saturated alkane, which does not decolourise bromine water under these conditions.

🧑‍🏫 Teacher's explanation

Imagine you are trying to dye a piece of fabric. If the fabric is smooth and tightly woven (like an alkane), the dye just sits on the surface. But if the fabric has open loops and gaps (like an alkene's double bond), the dye can dive right into the structure and become part of the weave. In chemistry, the double bond is that 'open loop' that is hungry for new atoms.

Ch 2 · Alkenes · AS & A Level Chemistry 9701

All chapters in Hydrocarbons