Which compound will have the lowest boiling point?

Analyze the question: We need the compound with the weakest intermolecular forces, which gives the lowest boiling point.

Concept:

• Alcohols (–OH) and carboxylic acids (–COOH) form hydrogen bonds, so their boiling points are high. Carboxylic acids are highest because they form dimers.
• Alkanes are non-polar and have only weak van der Waals (London dispersion) forces.
• Among alkanes of the same molecular formula, branching lowers the boiling point. Branched molecules are more compact, have less surface area for contact, and so have weaker dispersion forces.

Going through the options:

• A. CH3OH (methanol): hydrogen bonding, b.p. about 65 °C
• B. CH3COOH (ethanoic acid): strong hydrogen bonding and dimers, b.p. about 118 °C
• C. CH3CHOH (this is probably a typo for CH3CH2OH, ethanol): hydrogen bonding, b.p. about 78 °C
• D. C4H10: if this is n-butane, the b.p. is about 0 °C (−0.5 °C)
• E. (CH3)3CH (2-methylpropane, isobutane): also C4H10, but branched, b.p. about −12 °C

Comparing D and E: Both have the formula C4H10, so they are isomers. D is the straight-chain n-butane. E is the branched isomer, isobutane. Because of its branching, E has weaker van der Waals forces and the lower boiling point (−12 °C compared with −0.5 °C).

Answer: E.

Tip: For boiling points, remember the order: hydrogen bonding (acids > alcohols) is much greater than van der Waals forces in alkanes. For isomers, more branching means a lower boiling point.