Test: Principles Related to Practical Chemistry - JEE MCQ

In Lassaigne’s test for nitrogen, the organic compound is fused with sodium to form sodium cyanide (NaCN) if nitrogen is present. During the test, the sodium fusion extract is treated with ferrous sulfate (FeSO₄) and ferric chloride (FeCl₃), followed by acidification. This leads to the formation of a complex compound, ferric ferrocyanide, which has the formula Fe₄[Fe(CN)₆]₃ and gives the characteristic Prussian blue color.

• B is incorrect as it does not represent the Prussian blue complex.
• C is not a stable compound.
• D is related to the test for sulfur and nitrogen together, not just nitrogen.

Lassaigne’s test for nitrogen detects nitrogen in organic compounds by forming sodium cyanide (NaCN) upon fusion with sodium. For this to happen, the compound must contain carbon and nitrogen in its structure, as the carbon is necessary to form the cyanide ion.

• Urea (NH₂CONH₂): Contains carbon and nitrogen, gives a positive test.
• Hydrazine (N₂H₄): Although it lacks carbon, hydrazine can form NaCN in the presence of a carbon source (e.g., sodium carbonate) during fusion, so it gives a positive test.
• Azobenzene (C₆H₅N=NC₆H₅): Contains a nitrogen-nitrogen double bond, which is stable and does not form NaCN during fusion, so it does not give a positive test.
• Phenyl hydrazine (C₆H₅NHNH₂): Contains carbon and nitrogen, gives a positive test.

Thus, azobenzene is the correct answer.

• Vapour density (VD) = 30.
• Molecular weight = 2 × Vapour density = 2 × 30 = 60 g/mol.
• Empirical formula = CH₂O.
• Empirical formula weight = 12 (C) + 2×1 (H) + 16 (O) = 30 g/mol.
• Molecular formula = n × Empirical formula, where n = Molecular weight / Empirical weight = 60 / 30 = 2.
• Molecular formula = 2 × (CH₂O) = C₂H₄O₂.

However, among the options, C₂H₄O has a molecular weight of 12×2 + 1×4 + 16 = 44 g/mol, which does not match the calculated molecular weight. This suggests a possible error in the options or question. Based on standard calculations, C₂H₄O₂ (acetic acid) is expected, but since C₂H₄O is listed and commonly tested, it may be a contextual fit (e.g., acetaldehyde).

• Empirical formula = CH₂.
• Empirical formula weight = 12 (C) + 2×1 (H) = 14 g/mol.
• Molecular weight = 42 g/mol (given).
• n = Molecular weight / Empirical weight = 42 / 14 = 3.
• Molecular formula = 3 × (CH₂) = C₃H₆.

Verification:

• Molecular weight of C₃H₆ = 3×12 + 6×1 = 42 g/mol, which matches.
• C₃H₆ (cyclopropane or propene) is a valid hydrocarbon.

Thus, the correct answer is C. C₃H₆.

• Sulphur content = 3.4% = 3.4 g of sulphur per 100 g of insulin.
• Atomic mass of sulphur = 32 g/mol.
• To find the minimum molecular mass, assume the molecule contains at least one sulphur atom.
• If 1 sulphur atom (32 g) constitutes 3.4% of the molecular mass (M), then:
  (32 / M) × 100 = 3.4
  M = 32 × 100 / 3.4 ≈ 941.18 g/mol.
• The closest option to 941.18 is 940.

Thus, the minimum molecular mass is A. 940.

• Empirical formula = CH₂O.
• Empirical formula weight = 12 + 2×1 + 16 = 30 g/mol.
• Let the molecular formula be (CH₂O)_n = C_nH_2nO_n.
• Moles of carbohydrate = 0.0833 mol.
• Mass of hydrogen in 0.0833 mol = 1.00 g.
• Molecular formula = C_nH_2nO_n, so 2n hydrogen atoms per molecule.
• Atomic mass of hydrogen = 1 g/mol, so mass of hydrogen in 1 mole of carbohydrate = 2n g.
• For 0.0833 moles, mass of hydrogen = 0.0833 × 2n = 1.00 g.
• Solving: 2n × 0.0833 = 1.00
  n = 1.00 / (2 × 0.0833) ≈ 6.
• Molecular formula = C₆H₁₂O₆.

Verification:

• Molecular weight of C₆H₁₂O₆ = 6×12 + 12×1 + 6×16 = 180 g/mol.
• Hydrogen mass in 1 mole = 12 g.
• In 0.0833 moles, hydrogen mass = 12 × 0.0833 ≈ 1.00 g, which matches.

Thus, the correct answer is D. C₆H₁₂O₆.

Organic compounds are primarily derived from:

• Coal tar: A byproduct of coal processing, rich in aromatic compounds like benzene, toluene, and naphthalene.
• Petroleum: A major source of hydrocarbons (alkanes, alkenes, aromatics) used to synthesize organic compounds.
• Natural gas: Contains methane and other hydrocarbons, serving as a source for organic chemicals.

Since all three are significant sources, the correct answer is D. All of the above.

Bond energy depends on the hybridization of the carbon atoms, which affects the bond length and strength:

• C(sp³)-C(sp³) (e.g., in ethane, single bond): Longer bond, lower bond energy (~348 kJ/mol).
• C(sp²)-C(sp²) (e.g., in ethene, double bond): Shorter bond, higher bond energy (~614 kJ/mol).
• C(sp)-C(sp) (e.g., in acetylene, triple bond): Shortest bond, highest bond energy (~839 kJ/mol).

Thus, the order of bond energies is:
C(sp³)-C(sp³) < C(sp²)-C(sp²) < C(sp)-C(sp), which corresponds to A. I < II < III.

• C₂H₅OC₂H₅ (diethyl ether): Formula C₄H₁₀O, with the ether linkage between two ethyl groups (C₂H₅-O-C₂H₅).
• CH₃OCH₂CH₂CH₃ (methyl propyl ether): Formula C₄H₁₀O, with the ether linkage between a methyl group and a propyl group (CH₃-O-CH₂CH₂CH₃).
• Metamerism: A type of isomerism where compounds have the same molecular formula and functional group but differ in the alkyl groups attached to the functional group.
• Both compounds are ethers (same functional group) with the same molecular formula (C₄H₁₀O) but different alkyl groups around the oxygen.

Thus, they are metamers, and the correct answer is C. metamers.

• Enantiomers require chirality.
• Geometrical isomers involve restricted rotation (e.g., in alkenes).
• Conformational isomers involve rotation around single bonds.

A branched structure in a hydrocarbon means the carbon chain has at least one carbon atom attached to three or more other carbon atoms (a tertiary carbon) or a side chain.

• 1 carbon (CH₄): No branching possible.
• 2 carbons (C₂H₆): Only a straight chain (ethane).
• 3 carbons (C₃H₈): Only a straight chain (propane).
• 4 carbons (C₄H₁₀): Can form a branched structure, e.g., isobutane (2-methylpropane), where one carbon is attached to three others.

Thus, the minimum number of carbon atoms required is 4, and the correct answer is D. four.