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(A) The boiling point of alkanes depends on the surface area of the molecule. As the number of carbon atoms increases,the boiling point increases. For

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$D > C > A > B$

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(A) The boiling point of alkanes depends on the surface area of the molecule. As the number of carbon atoms increases,the boiling point increases. For isomers,the boiling point decreases with an increase in branching because branching reduces the surface area,leading to weaker van der Waals forces. The given compounds are: $(A)$ $2-$Methylbutane ($C_5H_{12}$,branched) $(B)$ $2,2-$Dimethylpropane ($C_5H_{12}$,highly branched) $(C)$ Pentane ($C_5H_{12}$,straight chain) $(D)$ Hexane ($C_6H_{14}$,straight chain) Comparing the $C_5$ isomers ($A$,$B$,$C$): The order of boiling points is $Pentane > 2-$Methylbutane $> 2,2-$Dimethylpropane $(C > A > B)$. Hexane $(D)$ has the highest boiling point due to the highest number of carbon atoms. Therefore,the decreasing order is $D > C > A > B$.

Arrange the following in decreasing order of their boiling points:
$(A)$ $2-$Methylbutane
$(B)$ $2,2-$Dimethylpropane
$(C)$ Pentane
$(D)$ Hexane

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Arrange the following in decreasing order of their boiling points: $(A)$ $2-$Methylbutane $(B)$ $2,2-$Dimethylpropane $(C)$ Pentane $(D)$ Hexane