Question 1Level 2 · ExamSingle correct
The three isomeric pentanes are pentane (CH3CH2CH2CH2CH3), 2-methylbutane ((CH3)2CHCH2CH3) and 2,2-dimethylpropane ((CH3)4C). The correct order of their boiling points and the correct order of their melting points, respectively, is ______. Exactly one option is correct.
Ab.p.: pentane > 2-methylbutane > 2,2-dimethylpropane; m.p.: pentane > 2-methylbutane > 2,2-dimethylpropane
Bb.p.: 2,2-dimethylpropane > 2-methylbutane > pentane; m.p.: 2,2-dimethylpropane > pentane > 2-methylbutane
Cb.p.: pentane > 2-methylbutane > 2,2-dimethylpropane; m.p.: 2,2-dimethylpropane > pentane > 2-methylbutane
Db.p.: pentane > 2-methylbutane > 2,2-dimethylpropane; m.p.: 2-methylbutane > pentane > 2,2-dimethylpropane
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Answer: C
Boiling point depends on the strength of London (dispersion) forces, which depends on the surface area of contact. Branching makes the molecule more compact (nearly spherical for 2,2-dimethylpropane), lowering the contact area. b.p.: pentane (309 K) > 2-methylbutane (301 K) > 2,2-dimethylpropane (283 K).
Melting point depends also on how well the molecules pack in the crystal. The highly symmetrical, nearly spherical (CH3)4C packs very well and has the highest m.p. (about 257 K); the unsymmetrical 2-methylbutane packs worst and has the lowest m.p. (about 113 K); pentane is in between (about 143 K). m.p.: 2,2-dimethylpropane > pentane > 2-methylbutane.
So branching lowers the b.p. steadily, but the m.p. order is set by symmetry and packing.
Answer: (C).
Difficulty: HardQuestion 2Level 2 · ExamSingle correct
An equimolar mixture of bromomethane, CH3Br, and 2-bromopropane, (CH3)2CHBr, is treated with sodium in dry ether. Ignoring side reactions such as elimination and disproportionation, the alkanes formed are ______. Exactly one option is correct.
Aethane, 2-methylpropane and 2,3-dimethylbutane
Bethane, butane and 2,3-dimethylbutane
C2-methylpropane only
Dethane, propane and hexane
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Answer: A
Wurtz reaction: 2R–X + 2Na → R–R + 2NaX. With two different halides RX and R'X, all three couplings occur: R–R, R–R' and R'–R'.
R = CH3, R' = (CH3)2CH:
CH3–CH3 = ethane;
CH3–CH(CH3)2 = 2-methylpropane;
(CH3)2CH–CH(CH3)2 = 2,3-dimethylbutane.
Butane (B) would need two ethyl or a methyl + propyl group; propane and hexane (D) would need ethyl/propyl groups. Option (C) ignores the symmetric couplings; this mixture is why the Wurtz reaction is not used to make unsymmetrical alkanes.
Answer: (A).
Difficulty: HardQuestion 3Level 3 · AdvancedOne or more correct
Methane reacts with chlorine in diffused sunlight to give chloromethane and HCl. Which of the following statements about this reaction are correct? One or more options may be correct.
AIn the propagation steps, the radical used up in one step is regenerated in the next, so the absorption of one photon can lead to the formation of many molecules of CH3Cl.
BIn the first propagation step Cl∙ abstracts a hydrogen atom from CH4 to give HCl and CH3∙, rather than forming CH3Cl and H∙; the alternative step would be strongly endothermic.
CThe formation of a small amount of ethane as a by-product supports the free-radical mechanism.
DMethane and chlorine react rapidly in the dark at room temperature, because the reaction is exothermic overall.
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Answer: A, B, C
Mechanism. Initiation: Cl2 → 2Cl∙ (light). Propagation: Cl∙ + CH4 → CH3∙ + HCl; CH3∙ + Cl2 → CH3Cl + Cl∙. Termination: radical + radical.
(A) Correct. Cl∙ consumed in the first propagation step is regenerated in the second, so it is a chain reaction: one initiation event gives many product molecules.
(B) Correct. Forming H–Cl (about 431 kJ mol−1) almost compensates for breaking C–H (about 435 kJ mol−1). Forming C–Cl (about 350 kJ mol−1) while breaking C–H, and releasing H∙, would be endothermic by about 85 kJ mol−1, so this path is not followed.
(C) Correct. Ethane arises from the termination step CH3∙ + CH3∙ → C2H6, which is evidence that methyl radicals are present.
(D) Incorrect. A negative overall ΔH does not make a reaction fast. Initiation needs light (or a high temperature) to break Cl–Cl; in the dark at room temperature the mixture does not react.
Answer: (A), (B), (C).
Difficulty: Very hard