Question

Arrange in increasing order of boiling points: (A) 2,2-dimethylpropane, (B) 2-methylbutane, (C) n-pentane, (D) n-butane.

• A. A $<$ B $<$ C $<$ D
• B. C $<$ A $<$ B $<$ D
• C. D $<$ A $<$ B $<$ C
• D. D $<$ C $<$ B $<$ A

Hint:

More branching decreases boiling point because branching reduces molecular surface area and weakens van der Waals attraction.

Answer 1

The Correct Option is C

Concept: Boiling point of alkanes depends mainly upon:
• Molecular mass
• Surface area
• Strength of van der Waals forces More branching causes:
• More compact structure
• Smaller surface area
• Weaker intermolecular attraction
• Lower boiling point

Step 1: Compare Carbon Number n-Butane has: \[ 4\ carbon\ atoms \] Others contain: \[ 5\ carbon\ atoms \] Thus n-butane has lowest boiling point. Hence: \[ D \] comes first.

Step 2: Compare Pentane Isomers Among pentane isomers: \[ \mathrm{n\text{-}Pentane} \] is straight chain and has maximum surface area. \[ \mathrm{2\text{-}Methylbutane} \] is branched. \[ \mathrm{2,2\text{-}Dimethylpropane} \] is highly branched and almost spherical. Therefore boiling point order becomes: \[ \mathrm{2,2\text{-}Dimethylpropane} < \mathrm{2\text{-}Methylbutane} < \mathrm{n\text{-}Pentane} \] Combining all compounds: \[ \mathrm{n\text{-}Butane} < \mathrm{2,2\text{-}Dimethylpropane} < \mathrm{2\text{-}Methylbutane} < \mathrm{n\text{-}Pentane} \] Thus: \[ D < A < B < C \] Hence, the correct answer is: \[ \boxed{(C)} \]