Chemistry 2007 Jamb Examination

A. Ionic bonding is correct because metals usually transfer electrons to halogens, forming electrovalent compounds. Bromine, being a halogen, accepts electrons from the metal atom, creating strong electrostatic forces between the oppositely charged ions.

This ionic character dominates in metal halides such as XBr₂ and XBr₃. Metallic bonding cannot occur since bromine is non-metallic, while covalent or dative bonding is not expected because the electronegativity difference between the metal and bromine favors ionic bond formation.

C. The correct method is the use of separating funnel because petrol and water are immiscible liquids. A separating funnel allows them to be separated based on differences in density, with water settling below petrol.

Other methods like centrifugation, filtration, or distillation are not suitable because they are either for solid-liquid separation or miscible liquids. Here, immiscibility makes the separating funnel the most effective option.

A. C₆H₅OH, phenol, is aromatic because it contains a benzene ring. Aromatic compounds must satisfy Huckel’s rule and have a delocalized π-electron system within a cyclic planar structure.

The other options do not contain such a ring structure and therefore are aliphatic. The aromatic nature of phenol makes it chemically distinct, especially in substitution reactions compared to typical alcohols.

C. Cyanide ion (CN⁻) is nucleophilic because it donates a lone pair of electrons to the carbon atom bonded to chlorine. The chlorine atom is more electronegative, making the carbon electron-deficient and susceptible to attack.

This is a classic nucleophilic substitution reaction (Sᴺ2). The cyanide attacks the carbon, replacing chlorine and forming ethyl cyanide. Such reactions highlight the role of CN⁻ as a strong nucleophile.

D. The best anti-knock agent is (CH₃)₃C–CH₂CH₂CH₃, a highly branched hydrocarbon. The greater the branching in hydrocarbons, the higher their resistance to knocking during combustion in engines.

Straight-chain alkanes burn less smoothly, leading to premature ignition and knocking. Branching increases stability of the fuel-air mixture, thus improving octane number and making branched hydrocarbons ideal anti-knock agents in petrol.

Styrene (C6H5CH=CH2) is produced industrially by dehydrogenation of ethylbenzene, which itself is obtained by alkylation of benzene with ethene:

Hence, the hydrocarbon used is ethene (option D).

Butane has two carbon skeletons: n-butane and isobutane.

Chlorination can occur at different positions:

• n-butane → 1-chlorobutane, 2-chlorobutane
• isobutane → 1-chloro-2-methylpropane, 2-chloro-2-methylpropane

Total structural isomers = 4. Correct answer: 4 (option A).

Reforming is the process in which straight-chain alkanes are converted into branched-chain, cyclic, or aromatic hydrocarbons:

It improves octane number of fuels and produces aromatics.

Hence correct answer: reforming (option B).

Boiling point increases with increasing molecular mass and surface area due to stronger van der Waals forces:

• Methane (CH₄) < propane (C₃H₈) < ethane (C₂H₆) < pentane (C₅H₁₂)

Hence, pentane has the highest boiling point (option A).

Waxes are esters formed from long-chain fatty acids (alkanoic acids) and long-chain alcohols (alkanols). They are mainly composed of alkanoates (option D).