Cambridge AS & A Level9701

Arenes

Chemistry 9701 Chapter Notes

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1. Benzene: Structure and Bonding

Arenes are aromatic hydrocarbons containing one or more benzene rings. The simplest arene is benzene, C₆H₆. Its structure was a long-standing puzzle. The historical Kekulé structure proposed a flat ring of six carbons with alternating single and double bonds. However, this model is incorrect. Evidence shows that benzene is more stable and has a different geometry.

  1. Bond Lengths: X-ray diffraction shows all carbon-carbon bonds in benzene are identical in length (0.139 nm). This is intermediate between a typical C-C single bond (0.154 nm) and a C=C double bond (0.134 nm). The Kekulé structure would have alternating long and short bonds.
  1. Enthalpy of Hydrogenation: Hydrogenating one C=C bond in cyclohexene releases -120 kJ/mol. For Kekulé's structure with three C=C bonds, we would expect an enthalpy of hydrogenation of 3 x (-120) = -360 kJ/mol. The actual experimental value for benzene is only -208 kJ/mol. Benzene is 152 kJ/mol more stable than the Kekulé structure would suggest. This extra stability is called delocalisation energy.
  1. Reactivity: Alkenes readily undergo electrophilic addition reactions, which break the C=C double bond. Benzene resists addition reactions because this would destroy the stable delocalised system. Instead, it undergoes substitution reactions.

The modern model describes benzene as a planar, hexagonal molecule where each carbon atom is sp² hybridised. Each carbon forms three σ-bonds (one to a hydrogen atom, two to adjacent carbon atoms), creating a 120° bond angle. The remaining p-orbital on each carbon atom overlaps sideways with its neighbours, forming a continuous ring of electron density above and below the plane of the molecule. This forms a delocalised π-system containing 6 electrons.

Key term

Delocalisation: The spreading of π (pi) electrons over three or more atoms in a molecule, rather than being confined between two atoms in a localised bond.

Examiner insight

Examiners look for a clear link between a piece of evidence (e.g., bond length) and how it refutes the Kekulé model while supporting the delocalised model.

Common pitfall

Describing benzene as having alternating single and double bonds. This is the incorrect Kekulé structure; you must refer to the delocalised π-system.

Fun fact

The symbol for benzene, a hexagon with a circle inside, was suggested by British chemist Sir Robert Robinson in 1925, although the hexagon alone was already in use.

Worked example 14 marks

Explain, with reference to two distinct types of evidence, why the delocalised model for benzene is preferred over the Kekulé structure. [4]

  1. 1

    Step 1: State the first piece of evidence. Evidence from bond lengths shows that all C-C bonds in benzene are 0.139 nm, which is an intermediate length between a C-C single bond (0.154 nm) and a C=C double bond (0.134 nm). [1 mark]

  2. 2

    Step 2: Explain the evidence. This contradicts the Kekulé structure which would have alternating short (C=C) and long (C-C) bonds. The delocalised model, with electrons spread evenly, explains the uniform bond length. [1 mark]

  3. 3

    Step 3: State the second piece of evidence. Thermochemical evidence from enthalpy of hydrogenation shows benzene is more stable than predicted by the Kekulé model. The expected value for hydrogenating three C=C bonds is -360 kJ/mol, but the actual value is -208 kJ/mol. [1 mark]

  4. 4

    Step 4: Explain the evidence. This difference of 152 kJ/mol, known as the delocalisation energy, shows the delocalised structure is significantly more stable. Benzene's tendency to undergo substitution rather than addition also supports this, as substitution preserves the stable delocalised system. [1 mark]

Recap

  • Benzene's formula is C₆H₆.
  • It is a planar hexagonal molecule with bond angles of 120°.
  • All C-C bonds have an intermediate length of 0.139 nm.
  • The structure is stabilised by a delocalised π-system of 6 electrons.
  • Evidence for this model comes from bond lengths, enthalpy data, and its chemical reactivity.

Quick check

  1. What is the C-C-C bond angle in a benzene molecule?1 mark
  2. Why is the enthalpy of hydrogenation of benzene less exothermic than expected for cyclohexatriene?1 mark

2. The Mechanism of Electrophilic Substitution

The delocalised π-system in benzene is an area of high electron density, making it susceptible to attack by electrophiles (electron-loving species). However, unlike alkenes which undergo addition, benzene undergoes substitution. This is because substitution allows the highly stable delocalised ring to be preserved in the final product. The general mechanism for electrophilic substitution occurs in two main steps:

Step 1: Electrophilic Attack A curly arrow is drawn from the delocalised π-ring of benzene to the electrophile (E⁺). This shows the π-system providing a pair of electrons to form a new C-E bond. This step is slow (rate-determining) as it disrupts the stable aromatic system. It forms a highly unstable, positively charged intermediate called a carbocation or Wheland intermediate. In this intermediate, the positive charge is delocalised over five of the six carbon atoms, which is often represented by an incomplete circle or 'horseshoe' within the hexagon.

Step 2: Restoration of Aromaticity The C-H bond on the same carbon atom that the electrophile attacked then breaks. A curly arrow is drawn from this C-H bond back into the ring. The pair of electrons from the C-H bond reforms the delocalised π-system, restoring the aromatic stability. The hydrogen leaves as a proton (H⁺), which is typically removed by the conjugate base of the catalyst used to generate the electrophile. This second step is very fast.

General Mechanism Step 1: C₆H₆ + E⁺ → [C₆H₆E]⁺

General Mechanism Step 2: [C₆H₆E]⁺ → C₆H₅E + H⁺

Key term

Electrophile: An electron-deficient species, such as an ion or molecule, that is attracted to electron-rich regions and accepts a pair of electrons to form a new covalent bond.

Examiner insight

Marks are consistently awarded for the correct use of curly arrows. The arrow must start from the π-ring for the initial attack and from the C-H bond for the final step.

Common pitfall

Drawing the first curly arrow from a specific C=C double bond (Kekulé style) instead of from the delocalised ring. Also, drawing the horseshoe of the intermediate incorrectly, it must not include the carbon atom that has been attacked.

Worked example 13 marks

Draw the mechanism for the reaction of benzene with a general electrophile, E⁺. Use curly arrows to show the movement of electron pairs. [3]

  1. 1

    Step 1: Draw the benzene ring and the electrophile E⁺. Draw a curly arrow originating from the centre of the delocalised π-ring and pointing to the E⁺. [1 mark]

  2. 2

    Step 2: Draw the unstable carbocation intermediate. This should be a hexagon with a bond to E and a bond to H on one carbon, and a positive charge inside an incomplete ring covering the other five carbons. [1 mark]

  3. 3

    Step 3: Draw a curly arrow from the C-H bond (on the carbon bonded to E) pointing back into the ring. Show the final products: the substituted benzene ring and H⁺. [1 mark]

Recap

  • Benzene's main reaction type is electrophilic substitution.
  • The reaction preserves the stable delocalised π-system.
  • The two-step mechanism involves attack by an electrophile and loss of an H⁺ ion.
  • The intermediate is an unstable carbocation where the aromaticity is temporarily broken.
  • Always draw curly arrows from an area of high electron density (the π-ring or a bond) to an area of low electron density (the electrophile or positive charge).

Quick check

  1. What is the name of the unstable intermediate formed during electrophilic substitution of benzene?1 mark
  2. Why does benzene undergo substitution rather than addition reactions?1 mark

3. Halogenation and Nitration of Benzene

Halogenation and nitration are classic examples of electrophilic substitution. Both require specific conditions to generate a strong enough electrophile to attack the stable benzene ring.

Halogenation (e.g., Bromination) Benzene does not react with bromine water. To brominate benzene, you need pure bromine and a halogen carrier catalyst, such as iron(III) bromide (FeBr₃) or iron filings (which react with bromine to form FeBr₃ in situ). The catalyst's job is to polarise the Br-Br bond, making one bromine atom sufficiently positive to act as an electrophile (Br⁺). Reagents: Br₂ Catalyst: FeBr₃ (or Fe) Conditions: Room temperature, in the dark (to prevent free-radical reactions).

Nitration Nitration introduces a nitro group (-NO₂) onto the ring. This requires a 'nitrating mixture' of concentrated nitric acid and concentrated sulfuric acid. The sulfuric acid is the catalyst; it is a stronger acid than nitric acid and protonates it, causing it to lose a water molecule and generate the powerful nitronium ion electrophile (NO₂⁺). Reagents: Concentrated HNO₃ and concentrated H₂SO₄ Conditions: Maintain temperature at 50°C. If the temperature rises above this, multiple substitutions can occur, forming dinitrobenzene.

Generation of Bromonium ion: Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻

Overall Halogenation: C₆H₆ + Br₂ --(FeBr₃)--> C₆H₅Br + HBr

Generation of Nitronium ion: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻

Overall Nitration: C₆H₆ + HNO₃ --(conc. H₂SO₄, 50°C)--> C₆H₅NO₂ + H₂O

Key term

Halogen Carrier: A Lewis acid catalyst, such as AlCl₃ or FeBr₃, that polarises a halogen molecule to generate a more powerful electrophile for aromatic substitution.

Common pitfall

Forgetting the halogen carrier for halogenation or using 'UV light' as a condition, which promotes a different mechanism (free-radical substitution).

Fun fact

Nitrobenzene has a characteristic smell of almonds or shoe polish. It is highly toxic and is primarily used to produce aniline, a precursor for many dyes and pharmaceuticals.

Worked example 14 marks

State the reagents and conditions needed to convert benzene into nitrobenzene. Write a balanced equation for the formation of the electrophile. [4]

  1. 1

    Step 1: State the reagents. The reagents are concentrated nitric acid (HNO₃) and concentrated sulfuric acid (H₂SO₄). [1 mark]

  2. 2

    Step 2: State the conditions. The reaction is carried out at a constant temperature of 50°C. [1 mark]

  3. 3

    Step 3: Write the equation for the formation of the electrophile. The electrophile is the nitronium ion, NO₂⁺. [1 mark]

  4. 4

    Step 4: The balanced equation is: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻. (An alternative simplified equation HNO₃ + H₂SO₄ → NO₂⁺ + H₂O + HSO₄⁻ is often accepted). [1 mark]

Recap

  • Halogenation of benzene requires a halogen (Cl₂ or Br₂) and a halogen carrier catalyst (e.g., AlCl₃, FeBr₃).
  • Nitration of benzene requires a nitrating mixture of concentrated HNO₃ and H₂SO₄.
  • The temperature for mononitration should be kept at 50°C to avoid further substitution.
  • In both reactions, the catalyst's role is to generate a strong electrophile (e.g., Br⁺ or NO₂⁺).
  • These reactions are examples of electrophilic substitution.

Quick check

  1. What is the electrophile in the nitration of benzene?1 mark
  2. Why is a catalyst needed to react chlorine with benzene?1 mark

4. Friedel-Crafts Alkylation and Acylation

Friedel-Crafts reactions are important methods for forming new carbon-carbon bonds to a benzene ring, allowing for the synthesis of more complex molecules.

Friedel-Crafts Alkylation This reaction attaches an alkyl group (like -CH₃ or -CH₂CH₃) to the benzene ring. It is achieved by reacting benzene with a haloalkane (R-X) in the presence of a Lewis acid catalyst, typically anhydrous aluminium chloride (AlCl₃). The catalyst helps generate a carbocation electrophile (R⁺) from the haloalkane. Example: Benzene reacts with chloroethane to form ethylbenzene. `C₆H₆ + CH₃CH₂Cl --(AlCl₃)--> C₆H₅CH₂CH₃ + HCl` Alkylation has some drawbacks: the alkyl group product is more reactive than benzene, leading to polysubstitution (multiple alkyl groups adding). Also, the carbocation intermediate can rearrange to form a more stable carbocation, leading to unexpected products.

Friedel-Crafts Acylation This reaction attaches an acyl group (R-C=O) to the benzene ring, forming a phenylketone. It is achieved by reacting benzene with an acyl chloride (RCOCl) in the presence of anhydrous AlCl₃. The catalyst helps generate a resonance-stabilised acylium ion (R-C≡O⁺), which is the electrophile. Example: Benzene reacts with ethanoyl chloride to form phenylethanone. `C₆H₆ + CH₃COCl --(AlCl₃)--> C₆H₅COCH₃ + HCl` Acylation is often preferred over alkylation because the acylium ion does not rearrange, and the ketone product is less reactive than benzene, preventing polysubstitution. The resulting ketone can then be reduced to an alkyl group if needed, providing a more controlled way to make alkylbenzenes.

Alkylation: C₆H₆ + R-X --(AlCl₃)--> C₆H₅-R + HX

Acylation: C₆H₆ + RCOCl --(AlCl₃)--> C₆H₅COR + HCl

Generation of Acylium Ion: RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻

Key term

Acylium ion: A cation with the formula R-C≡O⁺, which acts as the electrophile in Friedel-Crafts acylation reactions.

Examiner insight

Be precise with naming. For the product of benzene and ethanoyl chloride, 'phenylethanone' is correct; 'acetophenone' is a common name but may not be accepted unless specified. 'Phenyl ethanone' with a space is incorrect.

Fun fact

Friedel-Crafts reactions, discovered in 1877, are widely used in industry to synthesise products ranging from high-octane fuels to detergents and plastics.

Worked example 12 marks

Benzene reacts with propanoyl chloride (CH₃CH₂COCl) in the presence of a catalyst to form a ketone. Name the ketone product and the catalyst required. [2]

  1. 1

    Step 1: Identify the product. The propanoyl group (CH₃CH₂CO-) will substitute a hydrogen on the benzene ring. The product is a phenylketone with a total of 9 carbons. The name is 1-phenylpropan-1-one. [1 mark]

  2. 2

    Step 2: Identify the catalyst. This is a Friedel-Crafts acylation, which requires a Lewis acid catalyst. The standard catalyst is anhydrous aluminium chloride (AlCl₃). [1 mark]

Worked example 22 marks

Give two reasons why Friedel-Crafts acylation is often preferred to alkylation in organic synthesis. [2]

  1. 1

    Reason 1: Acylation does not lead to polysubstitution because the ketone product is deactivated towards further substitution. [1 mark]

  2. 2

    Reason 2: The acylium ion electrophile does not undergo rearrangement, whereas the carbocation in alkylation can rearrange, leading to a mixture of products. [1 mark]

Recap

  • Friedel-Crafts reactions form C-C bonds with a benzene ring.
  • Alkylation adds an alkyl group using a haloalkane and AlCl₃ catalyst.
  • Acylation adds an acyl group (RCO-) using an acyl chloride and AlCl₃ catalyst.
  • Acylation is generally more useful as it avoids polysubstitution and carbocation rearrangement.
  • All Friedel-Crafts reactions require anhydrous conditions as the catalyst reacts with water.

Quick check

  1. What functional group is formed in the Friedel-Crafts acylation of benzene?1 mark
  2. What is the formula of the catalyst used in Friedel-Crafts reactions?1 mark

5. Reactions of Alkylbenzenes

Alkylbenzenes, such as methylbenzene (toluene), have two regions of reactivity: the aromatic ring and the alkyl side-chain. The reaction that occurs depends entirely on the conditions used.

1. Reactions of the Aromatic Ring The alkyl group is an 'activating' group, meaning it donates electron density into the benzene ring. This makes the ring more electron-rich and therefore more reactive towards electrophiles than benzene itself. For example, methylbenzene reacts faster than benzene in nitration and halogenation. The reaction conditions are often milder. The alkyl group directs incoming electrophiles to the 2- (ortho) and 4- (para) positions. Example: Halogenation of methylbenzene with Br₂ and an FeBr₃ catalyst gives a mixture of 2-bromomethylbenzene and 4-bromomethylbenzene.

2. Reactions of the Side-Chain Under different conditions, the alkyl side-chain reacts just like an alkane.

a) Free-Radical Substitution: In the presence of ultraviolet (UV) light, halogens react with the side-chain via a free-radical mechanism. No catalyst is needed. The aromatic ring is unaffected. Example: Methylbenzene reacts with chlorine in UV light to form (chloromethyl)benzene. `C₆H₅CH₃ + Cl₂ --(UV light)--> C₆H₅CH₂Cl + HCl`

b) Oxidation: Alkyl side-chains are readily oxidised to a carboxylic acid group (-COOH) by strong oxidising agents. The standard reagent is hot, alkaline potassium manganate(VII) (KMnO₄), followed by acidification with a dilute acid (e.g., H₂SO₄). A key point is that any alkyl side-chain, regardless of its length, is oxidised back to a single -COOH group attached to the ring. For example, both methylbenzene and ethylbenzene are oxidised to benzoic acid. `C₆H₅CH₂CH₃ + 6[O] → C₆H₅COOH + CO₂ + 2H₂O`

Ring Halogenation: C₆H₅CH₃ + Br₂ --(FeBr₃)--> CH₃C₆H₄Br + HBr (mixture of 2- and 4-isomers)

Side-Chain Halogenation: C₆H₅CH₃ + Cl₂ --(UV light)--> C₆H₅CH₂Cl + HCl

Side-Chain Oxidation: C₆H₅CH₃ + 3[O] --(Hot alk. KMnO₄, then H⁺)--> C₆H₅COOH + H₂O

Key term

Side-Chain: An alkyl group or other substituent that is attached to a main molecular chain or, in this context, a benzene ring.

Examiner insight

Examiners frequently test the distinction between ring and side-chain reactivity. Be prepared to state the specific reagents and conditions for each and draw the resulting products.

Common pitfall

Confusing the conditions for ring versus side-chain halogenation. Remember: Catalyst = Ring, UV Light = Side-chain.

Worked example 13 marks

Predict the structure of the major organic product formed in each of the following reactions of ethylbenzene (C₆H₅CH₂CH₃).a) Reaction with Cl₂ in the presence of AlCl₃.b) Reaction with hot, alkaline potassium manganate(VII), followed by acidification. [3]

  1. 1

    Step 1 (a): Identify the reaction type. Cl₂ with AlCl₃ catalyst is electrophilic substitution on the aromatic ring. The ethyl group is 2,4-directing. This will produce a mixture of 1-chloro-2-ethylbenzene and 1-chloro-4-ethylbenzene. [1 mark for either correct structure]

  2. 2

    Step 2 (b): Identify the reaction type. Hot, alkaline KMnO₄ is a strong oxidising agent that attacks the alkyl side-chain. [1 mark]

  3. 3

    Step 3 (b): Determine the product. Any alkyl side-chain, regardless of length, is oxidised to a carboxylic acid group (-COOH) attached to the ring. The product is benzoic acid (C₆H₅COOH). [1 mark]

Recap

  • Alkylbenzenes can react at the ring or the side-chain depending on conditions.
  • Electrophilic substitution occurs on the ring with a catalyst (e.g., AlCl₃).
  • Free-radical substitution occurs on the side-chain with UV light.
  • Strong oxidation (hot KMnO₄) converts any alkyl side-chain to a -COOH group.
  • Alkyl groups are activating and 2,4-directing for ring substitution.

Quick check

  1. What conditions are needed to substitute a chlorine atom onto the side-chain of methylbenzene?1 mark
  2. What is the product when propylbenzene is heated with alkaline KMnO₄ and then acidified?1 mark

6. Directing Effects and Reactivity

When a benzene ring already has a substituent, that group influences where a second substituent will add. It also affects the overall rate of reaction. This is known as the directing effect.

Activating Groups (2,4-directing) These groups donate electron density to the benzene ring, making it more nucleophilic and thus more reactive towards electrophiles. The reaction is faster than with benzene itself. They direct the incoming electrophile to the 2- (ortho) and 4- (para) positions. The 4-isomer is often the major product due to less steric hindrance. Examples of activating groups: `-OH`, `-NH₂`, `-OR` (e.g., -OCH₃), `-R` (alkyl groups, e.g., -CH₃). Phenol (-OH) and aniline (-NH₂) are so strongly activated that they react with bromine water at room temperature without a catalyst, often substituting at all available 2, 4, and 6 positions.

Deactivating Groups (3-directing) These groups withdraw electron density from the benzene ring, making it less nucleophilic and less reactive towards electrophiles. The reaction is slower than with benzene. They direct the incoming electrophile to the 3- (meta) position. Examples of deactivating groups: `-NO₂`, `-CN`, `-COOH`, `-CHO`, `-COR`. For example, nitrating nitrobenzene requires harsher conditions (higher temperature) than nitrating benzene, and the product is 1,3-dinitrobenzene.

Halogens: The Exception Halogens (`-F`, `-Cl`, `-Br`, `-I`) are an anomaly. They are deactivating because their high electronegativity withdraws electron density from the ring through the sigma bond (the inductive effect). This makes the ring less reactive than benzene. However, they have lone pairs of electrons which can be donated into the ring through resonance. This donation increases electron density at the 2- and 4- positions, so they are 2,4-directing. So, chlorobenzene reacts more slowly than benzene, but it forms a mixture of 1,4-dichlorobenzene and 1,2-dichlorobenzene.

Key term

Activating Group: A substituent on a benzene ring that increases the rate of electrophilic substitution compared to benzene itself and directs incoming groups to the 2- and 4- positions.

Examiner insight

Questions on multi-step synthesis rely heavily on directing effects. You must choose the correct order of reactions to place substituents in the desired positions.

Fun fact

The directing effects of substituents are crucial in the pharmaceutical industry for synthesising complex drug molecules with specific structures and activities.

Worked example 13 marks

Predict the major organic product(s) when methylbenzene undergoes nitration with concentrated nitric and sulfuric acids. Explain your reasoning. [3]

  1. 1

    Step 1: Identify the directing effect. The methyl group (-CH₃) is an alkyl group, which is an activating, 2,4-directing group. [1 mark]

  2. 2

    Step 2: Determine the products. The incoming nitro group (-NO₂) will substitute at the 2- and 4- positions. This will form a mixture of 2-nitromethylbenzene and 4-nitromethylbenzene. [1 mark for drawing/naming both]

  3. 3

    Step 3: Justify the major product. The 4-isomer (4-nitromethylbenzene) is generally the major product due to reduced steric hindrance compared to the 2-isomer where the groups are adjacent. [1 mark]

Worked example 23 marks

Arrange benzene, phenol (C₆H₅OH), and nitrobenzene (C₆H₅NO₂) in order of increasing reactivity towards electrophilic substitution. Explain your order. [3]

  1. 1

    Step 1: State the order. The order of increasing reactivity is: Nitrobenzene < Benzene < Phenol. [1 mark]

  2. 2

    Step 2: Explain the least reactive. The -NO₂ group in nitrobenzene is strongly electron-withdrawing, deactivating the ring and making it least reactive. [1 mark]

  3. 3

    Step 3: Explain the most reactive. The -OH group in phenol is strongly electron-donating, activating the ring by increasing its electron density, making it the most reactive. Benzene is the baseline for comparison. [1 mark]

Recap

  • Substituents on a benzene ring influence the position and rate of further substitution.
  • Activating groups (-OH, -NH₂, -R) are 2,4-directing and increase reactivity.
  • Deactivating groups (-NO₂, -COOH) are 3-directing and decrease reactivity.
  • Halogens are an exception: they are deactivating but 2,4-directing.
  • The 4- (para) isomer is often the major product over the 2- (ortho) isomer due to less steric hindrance.

Quick check

  1. Is the -COOH group activating or deactivating?1 mark
  2. To which positions does a chlorine atom direct incoming electrophiles?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. An aromatic hydrocarbon was found to contain 90.6% carbon and 9.4% hydrogen by mass. The relative molecular mass was determined to be 106.0. Determine the empirical and molecular formulas of the hydrocarbon. [4]4 marks
  2. Draw the full mechanism, including the generation of the electrophile, for the mononitration of benzene using a nitrating mixture. Use curly arrows to show the movement of electron pairs. [5]5 marks
  3. Ethylbenzene, C₆H₅CH₂CH₃, can undergo chlorination under two different sets of conditions to give two different monochlorinated products, A and B. Give the reagents and conditions to form A and B, and draw the structure of each product. [4]4 marks
  4. Explain why phenol (C₆H₅OH) is more reactive towards electrophiles than benzene, but chlorobenzene (C₆H₅Cl) is less reactive than benzene. [4]4 marks
  5. Propose a two-step synthesis to prepare 3-bromobenzoic acid starting from benzene. For each step, state the reagents and conditions required. [5]5 marks
  6. Write an equation for the reaction of benzene with propanoyl chloride in the presence of an AlCl₃ catalyst. Name the organic product and the type of reaction. [3]3 marks
  7. Explain, by comparing the expected and actual enthalpy changes of hydrogenation, how thermochemical data provides evidence for the stability of benzene. Use numerical values in your answer. [3]3 marks
  8. Draw the displayed formula for the three structural isomers of dimethylbenzene. Give the systematic name for each isomer. [3]3 marks
  9. What would be observed when hot, alkaline potassium manganate(VII) is heated under reflux with butylbenzene (C₆H₅(CH₂)₃CH₃), followed by acidification? Name the organic product formed. [3]3 marks
  10. Compound D is an isomer of C₈H₁₀. When D is treated with chlorine in the presence of AlCl₃, only one possible aromatic monochloro-substituted product is formed. Deduce the structure of D and the structure of the product. [3]3 marks

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