Arrange the following alkanes in order of the | vedclass

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(A) The boiling point of alkanes depends on the surface area of the molecule.As the branching in the alkane increases,the surface area decreases,which

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$A > B > C$

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(A) The boiling point of alkanes depends on the surface area of the molecule.As the branching in the alkane increases,the surface area decreases,which leads to weaker van der Waals forces of attraction.Pentane $(A)$ is a straight-chain alkane with the largest surface area.$2-$Methylbutane $(B)$ has one branch,reducing its surface area compared to pentane.$2,2-$Dimethylpropane $(C)$ is the most branched isomer,resulting in the smallest surface area and the lowest boiling point.Therefore,the order of boiling points is $A > B > C$.

Arrange the following alkanes in order of their boiling points:
$A$. Pentane
$B$. $2-$Methylbutane
$C$. $2,2-$Dimethylpropane

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Arrange the following alkanes in order of their boiling points:$A$. Pentane$B$. $2-$Methylbutane$C$. $2,2-$Dimethylpropane