i. The IUPAC name of the following compound is (Diagram of a benzene ring with Cl atoms at positions 2, 4, and 6, and an OH group at position 1)
Answer: b) 2,6-dichlorophenol
Explanation: The -OH group is the principal functional group, so the carbon it is attached to is C1. The two chlorine atoms are at positions 2 and 6 relative to the -OH group.
ii. The type of intermolecular forces in benzene are
Answer: b) London dispersion forces
Explanation: Benzene is a nonpolar molecule. The only intermolecular forces present are weak London dispersion forces (a type of van der Waals force).
iii. Which one of the following carbon atom in monosubstituted benzene ring is called meta carbon?
Answer: c) carbon 3
Explanation: In a monosubstituted benzene ring, the carbon attached to the substituent is C1. The adjacent carbons are ortho (C2 and C6), the next ones are meta (C3 and C5), and the one opposite is para (C4).
iv. The unsaturated nature of benzene can be justified by
Answer: a) Hydrogenation
Explanation: Benzene can add hydrogen (hydrogenation) in the presence of a catalyst to form cyclohexane. This addition reaction demonstrates its unsaturated character.
v. Which one of the following challenges was faced by Kekulé's structure of benzene?
Answer: b) benzene is less reactive than alkenes
Explanation: Kekulé's structure proposed alternating single and double bonds. If this were true, benzene should undergo addition reactions readily like alkenes. However, benzene is much less reactive than alkenes and prefers substitution reactions. This was a major challenge to the Kekulé structure.
vi. The C-C-C bond angle in benzene molecule is
Answer: b) 120°
Explanation: Benzene is a planar molecule with sp² hybridized carbons. The bond angles in a regular hexagon are 120°.
vii. Delocalization of pi (π) electrons in benzene molecule is because
Answer: b) it is hexagonal
Explanation: The hexagonal, planar structure allows the unhybridized 2p orbitals on each carbon to overlap sideways continuously, allowing the π electrons to delocalize around the ring.
viii. The electrophile used in bromination of benzene molecule is
Answer: b) bromonium ion (Note: The actual electrophile is Br⁺, often represented as a bromonium ion or a polarized Brδ⁺-Brδ⁻ complex. Option b is the closest.)
ix. The electrophile used in Friedel-Craft acylation is
Answer: b) acylium ion
Explanation: In Friedel-Craft acylation, the electrophile is the acylium ion (R-C⁺=O), generated by the reaction of an acyl chloride with a Lewis acid catalyst like AlCl₃.
x. Aromatic hydrocarbons are found in the extracts of many plants. Their common characteristic is that they have
Answer: b) benzene unit
Explanation: Aromatic hydrocarbons are defined by the presence of a benzene ring or a related aromatic system in their structure.
xi. Which one of the following compounds can easily be nitrated?
Answer: b) phenol
Explanation: The -OH group in phenol is strongly activating and ortho/para directing. Phenol is much more reactive than benzene towards electrophilic substitution and can be nitrated easily.
xii. The most reactive compound, in terms of addition reactions, among the following is:
Answer: a) propene (Note: Alkynes are generally more reactive than alkenes, but among the given options, propene (an alkene) is much more reactive than benzene (aromatic) and propane (alkane). If ethyne were the intended comparison, it would be more reactive than propene.)
i. Cyclohexene decolourises bromine water whereas benzene cannot, why?
Answer: Cyclohexene contains a carbon-carbon double bond (C=C). This double bond can undergo an addition reaction with bromine, breaking the π bond and adding Br atoms across the double bond. This reaction consumes bromine, causing the
brown colour of bromine water to disappear. Benzene, on the other hand, does not have a true C=C double bond. Its π electrons are delocalized over the entire ring, giving it aromatic stability. Benzene does not undergo addition reactions with bromine under normal conditions; it requires a catalyst (like AlCl₃ or FeBr₃) and undergoes substitution instead.
ii. Consider the synthesis of 3-chlorobenzoic acid from methylbenzene given: (Diagram: Methylbenzene [source symbol unavailable] reaction 1 [source symbol unavailable] A [source symbol unavailable] reaction 2 [source symbol unavailable] 3-chlorobenzoic acid)
Answer: The intermediate A is benzoic acid.
Answer:
Reaction 1: Oxidation of methylbenzene to benzoic acid. Reagents: Hot alkaline KMnO₄ followed by acidification (H⁺). Reaction 2: Chlorination of benzoic acid. Reagents: Cl₂ in the presence of a Lewis acid catalyst (e.g., FeCl₃ or AlCl₃).
iii. Predict the products and their structural formulae of the following chemical reactions of benzene. Benzene + Cl₂ + AlCl₃ [source symbol unavailable] A A + HNO₃ + H₂SO₄ (55°C) [source symbol unavailable] B
Answer:
A: Chlorobenzene (C₆H₅Cl). The reaction is electrophilic chlorination. B: A mixture of ortho-chloronitrobenzene and para-chloronitrobenzene. The -Cl group is ortho/para directing.
iv. In Friedel-Craft acylation of benzene, the aluminium chloride (AlCl₃) acts as Lewis acid. Justify the statement.
Answer: In Friedel-Craft acylation, AlCl₃ acts as a Lewis acid because it accepts a lone pair of electrons from the acyl chloride (RCOCl). This interaction polarizes the acyl chloride, generating the electrophilic acylium ion (RCO⁺) and forming a complex with the chloride ion (AlCl₄⁻). Thus, AlCl₃ facilitates the reaction by accepting electron pairs.
v. Explain the difference between halogenation of benzene when reacted with
Answer: This is an electrophilic substitution reaction. AlCl₃ acts as a Lewis acid catalyst, polarizing Cl₂ to generate the electrophile Cl⁺. The product is chlorobenzene and HCl.
Answer: This is a free radical substitution reaction. UV light provides energy to homolytically cleave Cl₂ into chlorine free radicals. The product is benzene hexachloride (BHC), an addition product, not a substitution product.
vi. Explain how are arenium ion formed in electrophilic substitution reactions of benzene?
Answer: An arenium ion (also called a sigma complex) is formed when an electrophile (E⁺) attacks the delocalized π electron cloud of benzene. The electrophile forms a sigma bond with one carbon atom. This disrupts the aromaticity of the ring, localizing the positive charge on the adjacent carbons. The arenium ion is a resonance-stabilized carbocation intermediate. It then loses a proton to restore aromaticity, forming the substitution product.
vii. State the difference between substitution reactions of alkanes and benzene.
Answer:
| Feature | Alkanes | Benzene |
| Mechanism | Free radical substitution | Electrophilic substitution |
| Conditions | UV light or high temperature | Lewis acid catalyst (e.g., AlCl₃) |
| Electrophile | Free radicals (e.g., Cl•) | Electrophiles (e.g., Cl⁺, NO₂⁺) |
| Intermediate | Free radical | Arenium ion |
| Example | CH₄ + Cl₂ [source symbol unavailable] CH₃Cl + HCl | C₆H₆ + Cl₂ [source symbol unavailable] C₆H₅Cl + HCl |
viii. Explain the structure of benzene with the help of atomic orbital treatment.
Answer: Each carbon in benzene is sp² hybridized. The three sp² hybrid orbitals form sigma bonds with two adjacent carbons and one hydrogen atom. This creates a planar, hexagonal ring with bond angles of 120°. Each carbon also has one unhybridized 2p orbital perpendicular to the plane of the ring. These 2p orbitals overlap sideways with each other, creating a continuous π electron cloud above and below the plane of the ring. This delocalization of the six π electrons gives benzene its exceptional
stability.
ix. How can unhybrid 2p-orbitals of benzene adopt a doughnut shape? Show diagram.
Answer: (Diagram Description) The six unhybridized 2p orbitals on the carbon atoms of benzene are oriented perpendicular to the plane of the ring. They overlap laterally with each other. This lateral overlap creates two continuous rings (or lobes) of π electron density, one above and one below the plane of the carbon ring. Together, these two lobes resemble a doughnut shape enveloping the benzene ring.
i. Benzene reacts with bromine.
Answer: C₆H₆ + Br₂ [source symbol unavailable] C₆H₅Br + HBr (in the presence of FeBr₃ catalyst)
Answer: Iron(III) bromide (FeBr₃) or anhydrous aluminium bromide (AlBr₃).
Answer: The reddish-brown colour of bromine disappears, and hydrogen bromide (HBr) gas is evolved (white fumes in moist air).
ii. Benzene also reacts haloalkanes.
Answer: Anhydrous aluminium chloride (AlCl₃).
Answer: The electrophile formed is the isopropyl cation: (CH₃)₂CH⁺.
iii. The exceptional stability of benzene molecule is due to resonance in it.
Answer: Benzene has delocalized carbon-carbon double bonds because the six unhybridized p-orbitals overlap continuously around the ring, allowing the six π electrons to move freely throughout the entire ring rather than being localized between specific pairs of carbon atoms.
Answer: The delocalization of π electrons in benzene lowers its energy compared to the hypothetical cyclohexa-1,3,5-triene, which has three isolated double bonds. This energy difference is called the resonance energy (about 150 kJ/mol), making benzene more stable.
Answer: The carbon-carbon bonds in benzene are neither pure single bonds nor pure double bonds. The delocalization of π electrons gives each C-C bond a bond order of 1.5, meaning it has partial double bond character. This is why all C-C bond lengths in benzene are equal (139 pm), intermediate between a single bond (154 pm) and a double bond (134 pm).
iv. Benzene is more stable compounds than expected by chemists.
Answer: Benzene's stability is quantified by its resonance energy. When benzene is hydrogenated, it releases less energy than expected for a hypothetical molecule with three isolated double bonds. The difference (about 150 kJ/mol) is the resonance energy, which is the extra stability gained from delocalization of π electrons.
Answer: (Diagram Description) A graph showing the energy levels of cyclohexene (one double bond), cyclohexa-1,4-diene (two double bonds), hypothetical cyclohexa-1,3,5-triene (three isolated double bonds), and benzene. Benzene is shown at a lower energy level than the hypothetical triene by an amount equal to the resonance energy.
v. The electrophilic attack on a mono-substituted benzene is governed by the substituent already attached to benzene ring.
Answer: Meta-directing substituents are groups that direct incoming electrophiles primarily to the meta position (position 3 and
5) on the benzene ring. Examples: -NO₂ (nitro), -COOH (carboxyl), -CHO (aldehyde), -CN (cyano).
Answer:
Ortho/Para Directors (Activating): Groups like -OH, -NH₂, -CH₃ donate electrons to the ring (via resonance or inductive effect). This stabilizes the arenium ion intermediate when the electrophile attacks at ortho or para positions, lowering the activation energy for these pathways. Meta Directors (Deactivating): Groups like -NO₂, -COOH withdraw electrons from the ring. They destabilize the arenium ion. However, the meta attack places the positive charge on carbons that are not directly bonded to the electron-withdrawing group, resulting in a relatively more stable intermediate compared to ortho/para attacks.
vi. A nitrobenzene undergoes nitration much more slowly than benzene. Explain why this happens and predict how the rate would change if a methoxy group were present instead.
Answer: Nitrobenzene undergoes nitration much more slowly than benzene because the -NO₂ group is strongly electron- withdrawing (deactivating). It withdraws electron density from the ring, making it less nucleophilic and less able to attack the
electrophile (NO₂⁺). If a methoxy group (-OCH₃) were present instead (anisole), the rate would be much faster. The -OCH₃ group is strongly electron-donating (activating) through resonance, increasing the electron density on the ring and making it more reactive towards electrophiles.
vii. Toluene undergoes bromination in the presence of FeBr₃. Predict the major product(s) and justify the position(s) of substitution.
Answer: The major products are ortho-bromotoluene and para-bromotoluene. The methyl group (-CH₃) in toluene is an activating group and an ortho/para director. It donates electron density to the ring through inductive and hyperconjugative effects, stabilizing the arenium ion intermediates for ortho and para attack. The para isomer is usually the major product due to less steric hindrance. Meta attack is less favored.
viii. Chlorobenzene is treated with nitric acid and sulfuric acid. Explain which position the NO₂ group is most likely to occupy and why, considering both resonance and inductive effects of Cl.
Answer: The NO₂ group will most likely occupy the ortho and para positions. The chlorine atom in chlorobenzene is an ortho/ para director. It has two opposing effects: a strong electron-withdrawing inductive effect (-I) that deactivates the ring, and a strong electron-donating resonance effect (+R) that activates the ortho and para positions. The resonance effect dominates in stabilizing the arenium ion intermediate for ortho/para attack. Therefore, the major products are ortho-chloronitrobenzene and para-chloronitrobenzene.
ix. You are asked to synthesize 4-nitroaniline from benzene in the fewest steps possible. What steps would you take, and in what order? Justify your sequence based on directing effects.
Answer:
1. Nitration: Benzene [source symbol unavailable] Nitrobenzene (using HNO₃/H₂SO₄). The -NO₂ group is a meta director.
2. Reduction: Nitrobenzene [source symbol unavailable] Aniline (using Sn/HCl or Fe/HCl). The -NH₂ group is an ortho/para director.
3. Acetylation (Protection): Aniline [source symbol unavailable] Acetanilide (using CH₃COCl). This protects the -NH₂ group from oxidation and reduces its activating power to prevent polysubstitution.
4. Nitration: Acetanilide [source symbol unavailable] p-Nitroacetanilide (using HNO₃/H₂SO₄). The acetamido group is an ortho/para director, and the para product is favored.
5. Hydrolysis: p-Nitroacetanilide [source symbol unavailable] 4-Nitroaniline (using H₂O/H⁺).
Justification: Direct nitration of aniline is not done because aniline is very reactive and can be oxidized. Also, the -NH₂ group would be protonated in the strong acid medium, becoming a meta director (-NH₃⁺). Protecting the amine group as an amide ensures it remains an ortho/para director and prevents oxidation.
x. Design a synthesis for m-bromonitrobenzene starting from benzene.
Answer:
1. Nitration: Benzene [source symbol unavailable] Nitrobenzene (using HNO₃/H₂SO₄). The -NO₂ group is a meta director.
2. Bromination: Nitrobenzene [source symbol unavailable] m-Bromonitrobenzene (using Br₂/FeBr₃). The -NO₂ group directs the incoming bromine to the meta position.
Justification: The -NO₂ group is a meta director, so bromination occurs at the meta position. If bromination were done first (to give bromobenzene), the -Br group (ortho/para director) would direct the nitro group to ortho/para positions, not meta. Therefore, nitration must be done first.
Name the following aromatic compounds. (Diagrams of four substituted benzene rings)
i. Benzene with Br at position 1 and Cl at position 2.
Answer: 1-bromo-2-chlorobenzene (or o-bromochlorobenzene)
ii. Benzene with CHO at position 1 and I at position 2.
Answer: 2-iodobenzaldehyde
iii. Benzene with OH at position 1 and NO₂ at position 3.
Answer: 3-nitrophenol (or m-nitrophenol)
iv. Benzene with CH₃ at position 1 and Br at position 2.
Answer: 1-bromo-2-methylbenzene (or o-bromotoluene)
1. Draw the two Kekulé structures of benzene. How did Kekulé switch to closed structure of benzene molecule, although other chemists tried open structure for it?
Answer: (Diagram Description)
Kekulé Structure 1: A six-membered ring with alternating single and double bonds. Kekulé Structure 2: The same ring, but with the double bonds shifted to the alternate positions.
Kekulé proposed a closed, cyclic structure for benzene (a six-membered ring) based on the fact that benzene has only one monosubstitution product (C₆H₅X). If benzene had an open-chain structure with multiple double bonds, it would have given multiple monosubstitution products. The closed ring structure explained this observation. Additionally, the cyclic structure accounted for the molecular formula C₆H₆ and the equal number of carbon and hydrogen atoms.
2. Define resonance in benzene. Why is benzene exceptionally stable than alkenes?
Answer: Resonance in benzene refers to the delocalization of the six π electrons over the entire ring. The actual structure of benzene is a hybrid of the two Kekulé structures. The π electrons are not localized in three double bonds but are spread evenly around the ring. This delocalization lowers the energy of the molecule.
Benzene is exceptionally stable than alkenes because of this resonance stabilization (resonance energy of about 150 kJ/mol). In alkenes, the π electrons are localized in a single double bond. In benzene, the delocalization of π electrons over the entire ring provides extra stability, making benzene much less reactive than alkenes towards addition reactions.
