Arrange the following in decreasing order of their boiling points.

(A) n–butane

(B) 2–methylbutane

(C) n-pentane

(D) 2,2–dimethylpropane

Following are the structures of the given hydrocarbons-

Boiling point depends directly on molar mass and surface area. More the molar mass and surface area more will be the boiling point.

Out of all 4 given hydrocarbons, n- butane (A) has less number of carbon and hydrogen atoms which means its molar mass will be less than other given hydrocarbons. Therefore, its boiling point will be the least out of all the others.

Now, (B), (C) and (D) can be compared on the basis of surface area. Surface area depends on branching. More will be the branching less will be the surface area. From their structures it can be seen that (D) is the most branched hydrocarbon, then comes (B) and (C) is the least branched. Therefore, the surface area and thus boiling point will be in the following order- (D) < (B) < (C)

Thus, the order of all the given four hydrocarbons will be (iv) C > B > D > A