Question

Arrange the compounds in order of their increasing boiling points. (A) 1-bromoethane (B) 1-bromobutane (C) 2-bromo-2-methylpropane (D) 2-bromobutane

• A. (A), (C), (D), (B)
• B. (A), (D), (C), (B)
• C. (C), (D), (A), (B)
• D. (C), (A), (D), (B)

Hint:

For alkyl halides of comparable molecular mass, greater branching means lower boiling point because of reduced surface area available for intermolecular attractions.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Boiling points of haloalkanes are influenced by: 1. Molecular Weight: Higher molar mass increases London dispersion forces (boiling point increases). 2. Branching: Increased branching reduces the surface area available for intermolecular interactions (boiling point decreases).

Step 2: Detailed Explanation:
• (A) 1-bromoethane: $C_2H_5Br$ (Lowest mass, lowest boiling point).
• (B) 1-bromobutane: $C_4H_9Br$ (Straight chain, highest surface area, highest boiling point).
• (C) 2-bromo-2-methylpropane: $C_4H_9Br$ (Highly branched, lowest surface area, lower than D).
• (D) 2-bromobutane: $C_4H_9Br$ (Branched, but less than C, higher boiling point than C).

Step 3: Comparing the four:
Order of mass: A < (B, C, D are isomers). Order of boiling point for isomers: Straight chain (B) > Less branched (D) > More branched (C). Combining these: A < C < D < B.

Step 4: Final Answer:
The increasing order is (A), (C), (D), (B).