Question

Which is the correct decreasing order of boiling points for different compounds from following ?

• A. \(\text{CH}_3\text{Cl} > \text{CH}_3\text{Br} > \text{CH}_2\text{Br}_2 > \text{CHBr}_3\)
• B. \(\text{CH}_3\text{Br} > \text{CH}_2\text{Br}_2 > \text{CHBr}_3 > \text{CH}_3\text{Cl}\)
• C. \(\text{CH}_2\text{Br}_2 > \text{CHBr}_3 > \text{CH}_3\text{Br} > \text{CH}_3\text{Cl}\)
• D. \(\text{CHBr}_3 > \text{CH}_2\text{Br}_2 > \text{CH}_3\text{Br} > \text{CH}_3\text{Cl}\)

Hint:

Boiling point of similar haloalkanes rises with molecular mass and surface area, so more bromine atoms means a higher boiling point.

Answer 1

The Correct Option is D

Step 1: Understand the concept
All four compounds are polar molecules. Their boiling points depend mainly on the strength of van der Waals forces. These forces grow with molecular mass, size of the electron cloud and surface area.

Step 2: Compare the molecules
Molar masses: \(\text{CH}_3\text{Cl}\) = 50.5, \(\text{CH}_3\text{Br}\) = 95, \(\text{CH}_2\text{Br}_2\) = 174, \(\text{CHBr}_3\) = 253 g/mol. Bromine is larger and more polarisable than chlorine, so each added Br raises the dispersion forces strongly.

Step 3: Known boiling points
\(\text{CH}_3\text{Cl}\) about 249 K, \(\text{CH}_3\text{Br}\) about 277 K, \(\text{CH}_2\text{Br}_2\) about 370 K and \(\text{CHBr}_3\) about 422 K. The trend rises steadily with mass.

Step 4: Why the other orders fail
Option (A) puts the lightest compounds at the top and the heaviest at the bottom, which is exactly reversed. Options (B) and (C) place \(\text{CH}_3\text{Cl}\) or \(\text{CH}_3\text{Br}\) above heavier bromides, which breaks the mass trend. Only option (D) lists the compounds from heaviest to lightest.

Final Answer:
Boiling point falls from CHBr3 to CH3Cl as molar mass decreases. This is option (D). \[ \boxed{\text{(D) }\text{CHBr}_3 > \text{CH}_2\text{Br}_2 > \text{CH}_3\text{Br} > \text{CH}_3\text{Cl}} \]