Arrange the following compounds in the decreasing order of their boiling points:
(A) Butyraldehyde
(B) n-Butyl alcohol
(C) Sec-butyl alcohol
(D) n-butane
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B), (C), (A), (D)
For molecules of similar molar mass, the boiling point depends on the strength of the intermolecular forces: hydrogen bonding > dipole-dipole attraction > van der Waals forces alone. Among isomers, branching lowers the boiling point, because a more compact molecule has a smaller surface area for contact.
(B) n-butyl alcohol (butan-1-ol): intermolecular hydrogen bonding through –OH, straight chain (about 391 K)
(C) sec-butyl alcohol (butan-2-ol): also hydrogen bonded, but branched, so slightly lower (about 373 K)
(A) butyraldehyde (butanal): polar C=O gives dipole-dipole attraction, but no O–H, so no hydrogen bonding between its molecules (about 348 K)
(D) n-butane: non-polar, only weak van der Waals forces (about 273 K)
\( \displaystyle \text{butan-1-ol} \gt \text{butan-2-ol} \gt \text{butanal} \gt \text{butane} \)
Shortcut: The straight chain alcohol boils highest and the alkane lowest, so the order starts with (B) and ends with (D). Only Option (3) does this.