London dispersion forces are the primary intermolecular forces...

London dispersion forces depend on the molecular mass and the surface area of the molecules. The magnitude of these forces increases with an increase in molecular mass due to higher polarizability.

The molecular formulas of the given hydrocarbons are:
(a) 2,2-Dimethylbutane: $C_6H_{14}$
(b) n-Butane: $C_4H_{10}$
(c) n-Hexane: $C_6H_{14}$
(d) 2-Methylpropane: $C_4H_{10}$
(e) n-Pentane: $C_5H_{12}$

Based on molecular mass, the general order of dispersion forces is $C_4H_{10} < C_5H_{12} < C_6H_{14}$.

For isomers having the same molecular mass, branching decreases the surface area of the molecule, which weakens the London dispersion forces.

Comparing the $C_4H_{10}$ isomers, 2-methylpropane is branched while n-butane is a straight chain. Thus, the forces are weaker in 2-methylpropane: (d) $<$ (b).

Comparing the $C_6H_{14}$ isomers, 2,2-dimethylbutane is highly branched while n-hexane is a straight chain. Thus, the forces are weaker in 2,2-dimethylbutane: (a) $<$ (c).

Combining these observations, the increasing order of the magnitude of London dispersion forces is (d) $<$ (b) $<$ (e) $<$ (a) $<$ (c).

Answer: (d) $<$ (b) $<$ (e) $<$ (a) $<$ (c)