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    Topic 17: Organic Chemistry II — Edexcel A-Level Chemistry

    Test yourself on Topic 17: Organic Chemistry II with PEARSON EDEXCEL A-Level practice questions.

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    Topic 17: Organic Chemistry II explained

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

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    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 17: Organic Chemistry II exam tips

    Topic Overview

    Organic Chemistry II builds on the foundations of AS organic chemistry, delving deeper into the mechanisms, reactions, and properties of organic compounds. This topic covers key functional groups including aldehydes, ketones, carboxylic acids, esters, amines, and amides, as well as the chemistry of aromatic compounds like benzene. You will explore how these compounds react through mechanisms such as nucleophilic addition, nucleophilic substitution, and electrophilic substitution, and learn to predict products and conditions. Understanding these reactions is crucial for synthesising complex molecules and for appreciating the role of organic chemistry in pharmaceuticals, polymers, and biochemistry.

    This topic is central to the Edexcel A-Level Chemistry specification because it connects fundamental principles of bonding and reactivity to real-world applications. You will develop skills in drawing reaction mechanisms, interpreting spectroscopic data (IR and NMR) to identify organic structures, and planning multi-step syntheses. Mastery of Organic Chemistry II is essential for tackling exam questions that require you to apply knowledge to unfamiliar compounds, and it forms the basis for further study in chemistry or related fields such as medicine, materials science, and environmental science.

    In the wider context, Organic Chemistry II demonstrates how the structure of molecules dictates their reactivity and properties. For example, the difference in reactivity between aldehydes and ketones in nucleophilic addition reactions highlights the influence of steric and electronic effects. Similarly, the stability of benzene due to delocalisation explains why it undergoes substitution rather than addition reactions. By the end of this topic, you should be able to rationalise reaction outcomes, design synthetic routes, and analyse organic compounds using spectroscopic techniques.

    Key Concepts
    • →Nucleophilic addition reactions of aldehydes and ketones with HCN, NaBH4, and 2,4-DNPH, including mechanisms and stereochemistry (racemic mixtures).
    • →Electrophilic substitution reactions of benzene: nitration, halogenation, Friedel-Crafts alkylation and acylation, and the directing effects of substituents (activating vs deactivating groups).
    • →Carboxylic acids and their derivatives: formation of esters, acyl chlorides, and amides; nucleophilic addition-elimination mechanisms; relative reactivity of derivatives.
    • →Amines as bases and nucleophiles: preparation from halogenoalkanes and nitriles, reactions with acyl chlorides to form amides, and the formation of azo dyes via diazotisation.
    • →Spectroscopic identification of organic compounds using infrared (IR) spectroscopy for functional groups and nuclear magnetic resonance (NMR) spectroscopy for carbon-hydrogen environments, including integration and splitting patterns.
    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 draw curly arrows accurately: they must start from a lone pair or a bond, and point to where the electrons are going. In mechanisms, show all lone pairs and charges clearly. Examiners look for precision in arrow pushing.
    • 💡When asked to suggest a synthesis route, work backwards from the target molecule. Identify functional group interconversions and consider protecting groups if necessary. State reagents and conditions explicitly, and include equations.
    • 💡For NMR questions, remember that the number of signals indicates the number of different hydrogen environments. Use the integration (peak area) to determine the ratio of hydrogens, and splitting patterns (n+1 rule) to deduce neighbouring hydrogens. Don't forget to account for exchangeable protons (e.g., OH, NH) which may not show splitting.
    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: Benzene undergoes addition reactions like alkenes. Correction: Benzene is unusually stable due to delocalisation and undergoes electrophilic substitution, not addition, to maintain its aromatic ring.
    • Misconception: Aldehydes and ketones react identically with nucleophiles. Correction: Aldehydes are more reactive than ketones due to less steric hindrance and greater partial positive charge on the carbonyl carbon; ketones require more forcing conditions.
    • Misconception: In nucleophilic addition-elimination reactions of acyl chlorides, the nucleophile attacks the carbonyl carbon first. Correction: The mechanism involves addition of the nucleophile to form a tetrahedral intermediate, followed by elimination of the leaving group (Cl-).
    Frequently Asked Questions
    Why does benzene undergo substitution rather than addition reactions?
    Benzene has a delocalised ring of electrons that makes it unusually stable. Addition reactions would break this delocalisation, requiring a large input of energy. Instead, benzene undergoes electrophilic substitution, where an electrophile replaces a hydrogen atom, preserving the aromatic ring and its stability. This is why benzene does not decolourise bromine water like alkenes do.
    How do I distinguish between an aldehyde and a ketone using chemical tests?
    Aldehydes can be oxidised to carboxylic acids, while ketones are resistant to oxidation. Use Tollens' reagent (ammoniacal silver nitrate): aldehydes produce a silver mirror, ketones do not. Alternatively, Fehling's solution (blue) gives a brick-red precipitate with aldehydes but not ketones. Also, aldehydes give a positive result with Schiff's reagent (magenta colour), whereas ketones do not.
    What is the difference between nucleophilic addition and nucleophilic addition-elimination?
    Nucleophilic addition occurs with aldehydes and ketones: the nucleophile attacks the carbonyl carbon, forming an alkoxide intermediate that is then protonated. No leaving group is lost. Nucleophilic addition-elimination occurs with carboxylic acid derivatives (e.g., acyl chlorides): the nucleophile adds to the carbonyl, forming a tetrahedral intermediate, which then eliminates a leaving group (e.g., Cl-) to restore the carbonyl. The key difference is the presence of a good leaving group in the latter.
    How do I determine the structure of an unknown organic compound from IR and NMR spectra?
    Start with IR: identify functional groups by characteristic absorptions (e.g., C=O at ~1700 cm⁻¹, O-H broad at ~3300 cm⁻¹ for alcohols/carboxylic acids). Then use ¹H NMR: count the number of signals (different H environments), use integration to find the ratio of H atoms, and apply the n+1 rule to deduce neighbouring H atoms. For example, a triplet indicates a CH₂ next to a CH₃. Combine this information to propose a structure, and check consistency with the molecular formula if given.
    Why are acyl chlorides more reactive than carboxylic acids?
    Acyl chlorides are more reactive because the chlorine atom is a good leaving group (weak base) and is highly electronegative, increasing the partial positive charge on the carbonyl carbon. In contrast, the -OH group in carboxylic acids is a poor leaving group, and the lone pair on oxygen can donate into the carbonyl, reducing its electrophilicity. Additionally, acyl chlorides undergo nucleophilic addition-elimination readily, while carboxylic acids often require activation (e.g., using SOCl₂) to react.
    What are the directing effects of substituents on benzene in electrophilic substitution?
    Substituents on benzene can be activating (electron-donating) or deactivating (electron-withdrawing). Activating groups (e.g., -OH, -NH₂, -CH₃) direct incoming electrophiles to the ortho and para positions because they donate electrons into the ring, stabilising the intermediate. Deactivating groups (e.g., -NO₂, -CN, -COOH) are electron-withdrawing and direct to the meta position, as they destabilise the ortho/para intermediates. Halogens are deactivating but ortho/para directing due to a combination of inductive and resonance effects.