Question

Arrange the following in decreasing order of their boiling points.

$A$: n-butane

$B$: 2-methylbutane

$C$: n-pentane

$D$: 2,2-dimethylpropane

• A. $d>c>b>a$
  
  
• B. $a>b>c>d$
  
  
• C. $b>c>d>a$
  
  
• D. $c>b>d>a$

Answer 1

The correct option is D $c>b>d>a$

Boiling point $\propto$ molecular mass

For same molecular mass,

$\text{Boiling point}\propto \frac{1}{\text{Branching}}$

Thus, as molecular mass increases, boiling point increases. When comparing molecules of equal molecular mass, as branching increases, boiling point decreases.

The given molecules are:

• Among the given molecules, n-pentane has maximum molecular mass while n-butane has minimum molecular mass. So, n-butane has the lowest boiling point.
• Moreover 2,2-Dimethylpropane has maximum branching.

Thus, order of decreasing boiling points:
$\text{n-pentane}\left(c\right)>2-\text{Methyl butane}\left(b\right)>2,2-\text{Dimethylpropane}\left(d\right)>n-\text{butane}\left(a\right)$.
Hence, option (D) is correct.