Question:medium

Identify the correct increasing order of boiling points of the given compounds :

Show Hint

Boiling Point \(\propto\) Molecular Mass.
Boiling Point \(\propto \frac{1}{\text{Branching}}\).
Straight chains always have higher boiling points than their branched isomers.
Updated On: Jul 22, 2026
  • Propan–1–ol < butan–1–ol < butan–2–ol < pentan–1–ol
  • Pentan–1–ol < butan–1–ol < butan–2–ol < Propan–1–ol
  • Propan–1–ol < butan–2–ol < butan–1–ol < pentan–1–ol
  • Butan–1–ol < Butan–2–ol < Propan –1–ol < Pentan–1–ol
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: What controls the boiling point of an alcohol.
Alcohols boil the way they do mostly because of two things pulling in opposite directions: a longer, heavier carbon chain gives stronger van der Waals attraction between molecules and raises the boiling point, while branching near the $-OH$ makes the molecule more compact and rounded, which weakens those same attractions and lowers the boiling point.
Step 2: Sorting the compounds by chain length first.
Propan-1-ol has three carbons, butan-1-ol and butan-2-ol both have four, and pentan-1-ol has five, so purely by mass we already expect propan-1-ol to boil lowest and pentan-1-ol to boil highest, with the two butanols sitting in between.
Step 3: Breaking the tie between the two four-carbon alcohols.
Butan-1-ol is a straight primary alcohol, while butan-2-ol carries its $-OH$ on the second carbon, making the molecule slightly more compact in shape. That extra compactness lowers the intermolecular attraction a bit, so butan-2-ol boils just below butan-1-ol.
Step 4: Putting the full order together.
Stringing all of this together gives propan-1-ol lowest, then butan-2-ol, then butan-1-ol, then pentan-1-ol highest, which matches the increasing order given as the correct choice. \[ \boxed{\text{Propan-1-ol} < \text{Butan-2-ol} < \text{Butan-1-ol} < \text{Pentan-1-ol}} \]
Was this answer helpful?
0