Chemistry 2011 Jamb Examination

A primary amine is defined as an amine in which the nitrogen atom is bonded to one alkyl (or aryl) group and two hydrogen atoms (general formula RNH₂). Let's inspect each option carefully:

• Option A: (CH₃)₃N — this is tertiary amine. Nitrogen is bonded to three alkyl groups and no hydrogen (R₃N). Not primary.

• Option B: CH₃NH₂ — here nitrogen is bonded to one methyl group (CH₃) and two hydrogens: structure is RNH₂. This matches the definition of a primary amine.

• Option C: CH₃C—NH₂ (the structure shown is a carbonyl directly adjacent to the NH₂ group if read with the double-bonded O above the carbon) — this is actually an amide (–CONH₂) rather than a simple amine; amides are not classified as primary amines in the usual sense because the nitrogen is bonded to the carbonyl carbon (and its properties differ).

• Option D: (CH₃)₂NH — this is a secondary amine (R₂NH), nitrogen bonded to two alkyl groups and one hydrogen.

Conclusion: the structure that meets the primary amine definition is CH₃NH₂ (Option B).

Fehling's test is a classical qualitative test for the presence of an aldehyde group (specifically reducing aldehydes). The reagent contains Cu(II) complex ions which are reduced to insoluble Cu(I) oxide (Cu₂O), a brick-red precipitate, when an aldehyde is present.

1. Fehling's solution reacts with aldehydes (R–CHO) but not with most ketones (R–CO–R).
2. Alkanol (alcohol) typically does not give Fehling's test (unless it is a reducing sugar or has an aldehyde functional group).
3. Alkanes are saturated hydrocarbons and are inert to Fehling's reagent.
4. Alkanone (a ketone) generally does not reduce Fehling's solution under normal conditions.

Since K gives a brick-red precipitate, it contains an aldehyde functional group (an alkanal). Therefore option C is correct.

Let's compare the two molecules systematically:

• 2-Methylpropane (also called isobutane) has the formula C₄H₁₀ and is a branched isomer of butane.
• Butane has the formula C₄H₁₀ and is the straight-chain isomer.

1. Homologous series: members differ by a CH₂ unit (same functional group). Isomers (like butane and 2-methylpropane) belong to the same homologous series because they have the same functional group (alkanes) and differ only in structure.
2. Boiling points: isomers can have different boiling points. For example, branched isomers often have slightly lower boiling points due to decreased surface area and weaker London dispersion forces compared to straight-chain isomers.
3. Number of carbon atoms: both compounds have the same number (4 carbons).
4. Chemical properties: many chemical properties are similar because both are alkanes (e.g., combustion, substitution under radical conditions), but physical properties (like boiling point) and some reactivity nuances differ.

Therefore the true statement is that they are members of the same homologous series (both are C_nH_2n+2 alkanes). Option A is correct.

This reaction is the classical esterification (Fischer esterification) where a carboxylic acid (ethanoic acid) reacts with an alcohol (ethanol) in the presence of a strong acid catalyst (conc. H₂SO₄) to give an ester (ethyl ethanoate) and water.

1. Mechanism overview (conceptual): the acid protonates the carbonyl oxygen of the carboxylic acid making the carbonyl carbon more electrophilic; the alcohol attacks, followed by proton transfers and loss of water to form the ester.
2. Role of conc. H₂SO₄: acts as both a catalyst and a dehydrating agent, shifting equilibrium toward ester formation by removing water.
3. Why not the others? 'Condensation' is a general term for reactions that join two molecules with loss of a small molecule (water); esterification is a specific type of condensation but the conventional name for this reaction is esterification. 'Saponification' is the base-catalysed hydrolysis of an ester to give alcohol and carboxylate salt. 'Neutralization' is acid + base → salt + water, which is not what's happening here.

Therefore the best, most specific answer is A: esterification.

Polymerization typically involves monomers that contain double bonds (unsaturated units) which can open up and link together to form long chains. Let's consider each option:

• A: C₂H₄ (ethene) — contains a C=C double bond and is the classic monomer used to make polyethylene via addition polymerization.
• B: C₂H₅COOH (propanoic/ethanoic-like acid depending on notation) — carboxylic acids can undergo condensation polymerization in special cases (with di-functional monomers), but a single monofunctional carboxylic acid is not a typical polymerizable monomer on its own.
• C: C₂H₆ (ethane) — saturated hydrocarbon, not readily polymerizable under normal polymerization mechanisms.
• D: C₂H₅OH (ethanol) — an alcohol; alone it does not undergo addition polymerization (it could be involved in condensation polymerizations if bifunctional but not here).

Therefore ethene (C₂H₄) is the monomer that will undergo polymerization (option A).

Definition — Vapour density (VD) is commonly defined relative to hydrogen as:\[ \text{Vapour density} = \dfrac{\text{Molecular mass}}{2} \]So the molecular mass (M) is:\[ M = 2 \times \text{VD} \]Substitute the given vapour density:\[ M = 2 \times 45 = 90 \; \text{g mol}^{-1} \] Compute the molar mass of the empirical formula CH₂O:\[ M_{\text{empirical}} = M_C + 2M_H + M_O \]\[= 12 + 2(1) + 16 = 30 \; \text{g mol}^{-1} \]Determine the integer multiple (n) relating empirical formula mass to molecular mass:\[ n = \dfrac{M}{M_{\text{empirical}}} = \dfrac{90}{30} = 3 \]Multiply the empirical formula by this integer to get the molecular formula:\[ \text{Molecular formula} = (\text{CH}_2\text{O})_3 = \text{C}_3\text{H}_6\text{O}_3 \]  

Concentrated nitric acid (HNO₃) is a strong oxidizing agent when heated. If an initially colourless solution turns brown on boiling with HNO₃, it usually means an oxidation has occurred producing Fe³⁺ (which gives yellow→brown colours).

steps:
1. Identify the oxidant: HNO₃ (on heating) can oxidize Fe²⁺ to Fe³⁺, and in some cases produces brown NO₂ fumes as a side product.
2. If the solution originally contained Fe²⁺, oxidation converts it to Fe³⁺, and the solution then becomes yellow/brown.
3. Write the relevant half-reaction to show oxidation of iron:\[ \mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-} \] 4. A typical reduction half-reaction for nitrate under these conditions (simplified) is:\[ \mathrm{NO_3^- + 2H^+ + e^- \rightarrow NO_2 + H_2O} \] 5. Combining these shows nitrate oxidizes Fe²⁺ to Fe³⁺, producing brown colour (Fe³⁺) and brown NO₂ fumes sometimes.
Conclusion: the original cation present (which is oxidized to give the brown result) is Fe²⁺.

In a typical Daniell-type galvanic cell, zinc metal is oxidized at the anode. That oxidation produces electrons which then travel through the external circuit to the cathode where reduction occurs.
1. Oxidation at the anode (zinc electrode):\[ \mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-} \] 2. The 2e^- produced flow through the external circuit to the cathode where they are consumed in a reduction reaction (for example, Cu²⁺ + 2e^- → Cu(s)).
3. Therefore the zinc electrode is the source (producer) of electrons — it is the anode and undergoes oxidation, not reduction.

This reaction is the radical substitution of methane by chlorine to give chloromethane (and hydrogen chloride). It is a photochemical reaction initiated by light (especially UV).

Mechanism summary (why light matters):
1. Initiation: Cl–Cl bonds are broken homolytically by photons (hv) to give chlorine radicals:\[ \mathrm{Cl_2 \xrightarrow{hv} 2Cl^{\bullet}} \]2. Propagation: chlorine radicals abstract H from CH₄, producing CH₃^• and HCl, then CH₃^• reacts with Cl₂ to give CH₃Cl and regenerate Cl^• radicals.
3. Termination: radicals combine to end the chain.

Because the initiation step depends directly on light (photon energy to form radicals), the presence/intensity of light is the principal factor controlling the rate. Heat or concentration can influence rates, but without light the chain reaction does not start efficiently.

Therefore 'light' (option D) is the major factor.

Corrosion of iron (rusting) is an electrochemical process that requires an electrolyte (usually water) and an oxidizing agent (oxygen). Carbon(IV) oxide (CO₂) can accelerate corrosion by forming carbonic acid in water, but the essential requirements are water and oxygen.

Electrochemical corrosion simplified:
• Anodic reaction (iron oxidised):\[ \mathrm{Fe \rightarrow Fe^{2+} + 2e^-} \]• Cathodic reaction (oxygen reduced in presence of water):\[ \mathrm{O_2 + 4H^+ + 4e^- \rightarrow 2H_2O} \]\or in neutral/basic water:\[ \mathrm{O_2 + 2H_2O + 4e^- \rightarrow 4OH^-} \]
Water provides the medium (electrolyte) for ionic movement; oxygen accepts electrons at the cathode. Without either, the electrochemical cell cannot operate effectively and rusting is greatly slowed.

Hence the required condition (essential pair) is water and oxygen — option D.