which of the following organic compounds has the lowest boiling point?

Analyze the question: We are asked which compound has the lowest boiling point among the options given.

Concept: For alkanes, boiling point depends on the strength of the van der Waals (London dispersion) forces between molecules. These forces get stronger as the molecular size (chain length and molar mass) increases. So the smaller the molecule, the lower the boiling point.

Looking at the options:

• A. C₆H₁₄ (hexane): b.p. about 69 °C
• B. C₅H₁₂ (pentane): b.p. about 36 °C
• C. C₄H₁₀ (butane): b.p. about 0 °C
• D. (CH₃)₃, which is written incompletely. Taken literally, it would be three CH₃ groups, which is not a stable compound. The likely intended compound is (CH₃)₃CH (2-methylpropane, isobutane), or possibly (CH₃)₃C-type notation, which is also a 4-carbon species. Most likely it is isobutane, C₄H₁₀.

Solution:

• Hexane and pentane have higher boiling points because they have more carbon atoms.
• Butane and 2-methylpropane are both C₄H₁₀, but 2-methylpropane is branched. Branching makes the molecule more compact and spherical, with less surface area for contact. This gives weaker dispersion forces and a lower boiling point (about -12 °C, compared with about 0 °C for butane).

So the order of boiling points is: (CH₃)₃CH (-12 °C) < C₄H₁₀ (≈0 °C) < C₅H₁₂ (36 °C) < C₆H₁₄ (69 °C)

Note: The formula in option D appears to be typed incompletely. My working agrees with the key on that reading.

Answer: D

Tip: For alkanes, remember two rules: more carbons means a higher boiling point, and for the same number of carbons, more branching means a lower boiling point.