Q12.63P - Chemistry: Molecular Nature Of Matter And Change

Question

Rank the compounds in Problem 12.61 in order of decreasing viscosity at a given temperature; explain your ranking.

Step-by-Step Solution

Answer

On the basis of viscosity, at a given temperature the decreasing order of the mentioned compounds is ${\mathbf{OHCH}}_2\mathbf{CH}\mathbf{\left(}\mathbf{OH}\mathbf{\right)}\mathbf{C}{\mathbf{H}}_2\mathbf{OH}\mathbf{>}{\mathbf{OHCH}}_2{\mathbf{CH}}_2\mathbf{OH}\mathbf{>}{\mathbf{CH}}_3{\mathbf{CH}}_2{\mathbf{CH}}_2\mathbf{OH}$.

Viscosity

The viscosity is the shear resistance of any liquid for deformation under external pressure.

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 former will have more viscosity than the latter, and in this line, propanol will have the least viscosity.