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    Topic 6: Organic Chemistry I — Edexcel A-Level Chemistry

    Test yourself on Topic 6: Organic Chemistry I with PEARSON EDEXCEL A-Level practice questions.

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    Topic 6: Organic Chemistry I explained

    This topic introduces the concept of oxidation numbers as a systematic method for classifying redox reactions, including disproportionation.

    Read the full explanation

    Students learn to define oxidation and reduction in terms of electron transfer and changes in oxidation number, and apply these principles to write and balance ionic half-equations.

    What to demonstrate

    1. Correct calculation of oxidation numbers in compounds and ions, including peroxides and metal hydrides.
    2. Correct identification of oxidation and reduction based on electron transfer and oxidation number changes.
    3. Correct identification of oxidising and reducing agents.
    Show all 6 objectives
    1. Correct identification of disproportionation reactions.
    2. Correct use of Roman numerals to indicate oxidation numbers.
    3. Correct construction of full ionic equations from ionic half-equations.

    Topic 6: Organic Chemistry I exam tips

    Topic Overview

    Topic 6: Organic Chemistry I introduces the fundamental principles of organic chemistry, focusing on the chemistry of alkanes, alkenes, and haloalkanes. You will explore the concept of homologous series, functional groups, and the nomenclature of organic compounds. This topic also covers the mechanisms of key reactions, including free-radical substitution, electrophilic addition, and nucleophilic substitution, which are essential for understanding how organic molecules transform.

    Understanding organic chemistry is crucial because it forms the basis for more advanced topics such as organic synthesis, polymers, and biochemistry. The skills you develop here—such as drawing mechanisms, predicting products, and explaining reaction conditions—are directly applicable to later topics and to real-world applications like pharmaceuticals and materials science. Mastery of this topic will also strengthen your ability to analyse and interpret organic reactions in exams.

    This topic builds on your GCSE knowledge of hydrocarbons and introduces the systematic approach required at A-Level. You will learn to name compounds using IUPAC rules, identify isomers, and understand how the structure of a molecule influences its reactivity. By the end of this topic, you should be able to write balanced equations, draw reaction mechanisms, and explain the factors that affect reaction rates and yields.

    Key Concepts
    • →Nomenclature: Understand how to name alkanes, alkenes, and haloalkanes using IUPAC rules, including the use of prefixes, suffixes, and locants.
    • →Isomerism: Recognise structural isomers (chain, position, functional group) and stereoisomers (E/Z isomerism in alkenes).
    • →Reaction mechanisms: Be able to draw and explain the mechanisms for free-radical substitution (alkanes), electrophilic addition (alkenes), and nucleophilic substitution (haloalkanes).
    • →Reactivity trends: Understand how bond polarity and bond enthalpy affect the reactivity of alkanes, alkenes, and haloalkanes.
    • →Practical techniques: Know how to carry out tests for unsaturation (bromine water) and distinguish between primary, secondary, and tertiary haloalkanes using silver nitrate.
    Marking Points
    • Correct calculation of oxidation numbers in compounds and ions, including peroxides and metal hydrides.
    • Correct identification of oxidation and reduction based on electron transfer and oxidation number changes.
    • Correct identification of oxidising and reducing agents.
    • Correct identification of disproportionation reactions.
    • Correct use of Roman numerals to indicate oxidation numbers.
    • Correct construction of full ionic equations from ionic half-equations.
    Examiner Tips
    • 💡Always check that the sum of oxidation numbers in a neutral compound equals zero and in an ion equals the charge of the ion.
    • 💡Remember that oxidising agents are reduced (gain electrons) and reducing agents are oxidised (lose electrons).
    • 💡When balancing half-equations, ensure the total charge on both sides is equal.
    • 💡Practice identifying oxidation numbers in various contexts, especially for s- and p-block elements.
    • 💡Always show the movement of electrons using curly arrows in mechanisms. Ensure arrows start from a bond or lone pair and point to the atom where the electrons are going. Missing or incorrect arrows lose marks.
    • 💡When naming compounds, remember to number the carbon chain to give the lowest possible locants to functional groups and substituents. Check for alphabetical order when listing substituents.
    • 💡For practical questions on haloalkane reactivity, remember that silver nitrate test results depend on the halogen (AgCl white, AgBr cream, AgI yellow) and the rate of precipitation (tertiary > secondary > primary).
    Common Mistakes
    • Confusing the direction of electron transfer in oxidation and reduction.
    • Incorrectly assigning oxidation numbers in complex ions or species.
    • Failing to balance both atoms and charges when constructing ionic half-equations.
    • Misidentifying the species being oxidised or reduced in a disproportionation reaction.
    • Misconception: Alkanes are unreactive because they are saturated. Correction: Alkanes do undergo reactions, such as combustion and free-radical substitution, but they require specific conditions (e.g., UV light for substitution).
    • Misconception: In electrophilic addition, the electrophile is always H⁺. Correction: The electrophile can be a positive ion (e.g., H⁺ from HBr) or a molecule with a partial positive charge (e.g., Br₂ in the presence of a catalyst).
    • Misconception: All haloalkanes undergo nucleophilic substitution at the same rate. Correction: The rate depends on the halogen (I > Br > Cl > F) and the class of haloalkane (tertiary > secondary > primary for SN1, but primary > secondary > tertiary for SN2).
    Frequently Asked Questions
    What is the difference between structural isomerism and stereoisomerism?
    Structural isomers have the same molecular formula but different arrangements of atoms, such as chain, position, or functional group isomers. Stereoisomers have the same structural formula but different spatial arrangements, like E/Z isomers in alkenes due to restricted rotation around the double bond.
    How do I know which mechanism (SN1 or SN2) a haloalkane will undergo?
    The mechanism depends on the structure of the haloalkane. Primary haloalkanes typically undergo SN2 (one step, backside attack), while tertiary haloalkanes undergo SN1 (two steps, carbocation intermediate). Secondary haloalkanes can undergo either, depending on conditions. Also, SN1 is favoured in polar protic solvents, while SN2 is favoured in polar aprotic solvents.
    Why do alkenes undergo addition reactions but alkanes do not?
    Alkenes have a carbon-carbon double bond, which is electron-rich and can act as a nucleophile, attracting electrophiles. The double bond is also weaker than a single bond, making it easier to break. Alkanes have only strong C-C and C-H single bonds, so they require more energy (e.g., UV light) to react via substitution.
    What is the test for unsaturation and how does it work?
    The test uses bromine water (orange). If the compound is unsaturated (contains a double bond), the bromine adds across the bond, decolourising the solution from orange to colourless. This is an electrophilic addition reaction. Saturated compounds do not decolourise bromine water under normal conditions.
    How do I name a compound with both a double bond and a halogen?
    The double bond takes priority in numbering, so the carbon chain is numbered to give the double bond the lowest locant. The halogen is named as a substituent (e.g., fluoro, chloro, bromo, iodo) and placed in alphabetical order with other substituents. For example, 3-bromoprop-1-ene.
    What factors affect the rate of nucleophilic substitution in haloalkanes?
    The rate depends on the halogen (C-I bonds break fastest due to low bond enthalpy), the class of haloalkane (tertiary fastest for SN1, primary fastest for SN2), and the solvent (polar protic solvents favour SN1, polar aprotic favour SN2). Also, the nucleophile strength influences the rate in SN2 reactions.