Question

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

• A. Propan–1–ol < butan–1–ol < butan–2–ol < pentan–1–ol
• B. Pentan–1–ol < butan–1–ol < butan–2–ol < Propan–1–ol
• C. Propan–1–ol < butan–2–ol < butan–1–ol < pentan–1–ol
• D. Butan–1–ol < Butan–2–ol < Propan –1–ol < Pentan–1–ol

Hint:

Boiling Point \(\propto\) Molecular Mass.
Boiling Point \(\propto \frac{1}{\text{Branching}}\).
Straight chains always have higher boiling points than their branched isomers.

Answer 1

The Correct Option is C

Concept: The boiling point of organic compounds like alcohols is influenced by two main factors:

• Molecular Weight/Mass: As the number of carbon atoms increases, the magnitude of Van der Waals forces (London dispersion forces) increases, leading to a higher boiling point.

• Branching: For isomeric compounds (same molecular formula), increased branching results in a more spherical shape. This reduces the surface area in contact with other molecules, thereby weakening the Van der Waals forces and lowering the boiling point.
Step 1: Ranking by carbon chain length.
The given compounds are:

• Propan-1-ol (\(3 \text{ carbons}\))

• Butan-1-ol (\(4 \text{ carbons}\))

• Butan-2-ol (\(4 \text{ carbons}\))

• Pentan-1-ol (\(5 \text{ carbons}\))
By mass: \(\text{Propan-1-ol} \lt \text{Butanols} \lt \text{Pentan-1-ol}\).

Step 2: Comparing the isomers (Butan-1-ol vs Butan-2-ol).
Both have 4 carbons.

• Butan-1-ol is a primary alcohol with a straight chain.

• Butan-2-ol has the hydroxyl group on the second carbon, which effectively creates a "branch" in terms of molecular surface area.
Because Butan-2-ol is more branched than the straight-chain Butan-1-ol, its boiling point is lower. Thus: \(\text{Butan-2-ol} \lt \text{Butan-1-ol}\).

Step 3: Final Assembly.
Combining the observations: Propan-1-ol (\(lowest\)) < Butan-2-ol < Butan-1-ol < Pentan-1-ol (\(highest\)).