Question

Rank the following in order of increasing surface tension at a given temperature, and explain your ranking: (a) ${\mathbf{CH}}_3{\mathbf{CH}}_2{\mathbf{CH}}_2\mathbf{OH}$(b)role="math" localid="1658943316217" ${\mathbf{OHCH}}_2\mathbf{CH}\mathbf{\left(}\mathbf{OH}\mathbf{\right)}\mathbf{C}{\mathbf{H}}_2\mathbf{OH}$ (c) ${\mathbf{OHCH}}_2{\mathbf{CH}}_2\mathbf{OH}$

Short answer

On the basis of surface tension, at a given temperature the increasing order of the given compounds is${\mathbf{CH}}_3{\mathbf{CH}}_2{\mathbf{CH}}_2\mathbf{OH}\mathbf{<}{\mathbf{OHCH}}_2{\mathbf{CH}}_2\mathbf{OH}\mathbf{<}{\mathbf{OHCH}}_2\mathbf{CH}\mathbf{\left(}\mathbf{OH}\mathbf{\right)}\mathbf{C}{\mathbf{H}}_2\mathbf{OH}$.

Step-by-step solution

Surface Tension

Surface tension is the property of liquids where the liquid experiences force on the liquid surface due to the unbalanced intermolecular forces of attraction on the surface in comparison to bulk.

H-bonding interactions

For the alcoholic compounds, the functional group –OH experiences strong H-bonding interactions. Considering the above compounds, there is a difference in these H-bonding interactions.

${\mathbf{CH}}_3{\mathbf{CH}}_2{\mathbf{CH}}_2\mathbf{OH}$ has only one –OH bond, ${\mathbf{OHCH}}_2{\mathbf{CH}}_2\mathbf{OH}$ has two –OH bonds, whereas ${\mathbf{OHCH}}_2\mathbf{CH}\mathbf{\left(}\mathbf{OH}\mathbf{\right)}\mathbf{C}{\mathbf{H}}_2\mathbf{OH}$has three –OH bonds. Hence, the H- bonding interactions are stronger in${\mathbf{OHCH}}_2\mathbf{CH}\mathbf{\left(}\mathbf{OH}\mathbf{\right)}\mathbf{C}{\mathbf{H}}_2\mathbf{OH}$ than ${\mathbf{OHCH}}_2{\mathbf{CH}}_2\mathbf{OH}$, thus the surface tension of the former is more than the latter. Propanol will have the lowest value of surface tension.