Question

Pentanol ( ${\mathbf{C}}_5{\mathbf{H}}_{11}\mathbf{OH}$,$\mathbf{\mu }\mathbf{=}\mathbf{88}\mathbf{.}\mathbf{15}\mathbf{g}\mathbf{/}\mathbf{mol}$ ) has nearly the same molar mass as hexane (${\mathbf{C}}_6{\mathbf{H}}_{14}$,$\mathbf{\mu }\mathbf{=}\mathbf{86}\mathbf{.}\mathbf{17}\mathbf{g}\mathbf{/}\mathbf{mol}$ ) but is more than 12 times as viscous at $\mathbf{20}\mathbf{°}\mathbf{C}$. Explain.

Short answer

Pentanol $\left({\mathbf{C}}_5{\mathbf{H}}_{11}\mathbf{OH}\mathbf{,}\mathbf{\mu }\mathbf{=}\mathbf{88}\mathbf{.}\mathbf{15}\mathbf{g}\mathbf{/}\mathbf{mol}\right)$ is 12 times more viscous than hexane $\left({\mathbf{C}}_6{\mathbf{H}}_{14}\mathbf{,}\mathbf{\mu }\mathbf{=}\mathbf{86}\mathbf{.}\mathbf{17}\mathbf{g}\mathbf{/}\mathbf{mol}\right)$ at $\mathbf{20}\mathbf{°}\mathbf{C}$ because of the strong intermolecular H-bonding interactions.

Step-by-step solution

Bonding interactions

Pentanol $\left({\mathbf{C}}_5{\mathbf{H}}_{11}\mathbf{OH}\mathbf{,}\mathbf{\mu }\mathbf{=}\mathbf{88}\mathbf{.}\mathbf{15}\mathbf{g}\mathbf{/}\mathbf{mol}\right)$ consists of –OH functional group which is involved in the strong H-bonding interactions. Hexane $\left({\mathbf{C}}_6{\mathbf{H}}_{14}\mathbf{,}\mathbf{\mu }\mathbf{=}\mathbf{86}\mathbf{.}\mathbf{17}\mathbf{g}\mathbf{/}\mathbf{mol}\right)$ being a non-polar hydrocarbon involves only weak London dispersion forces.

Effect on viscosity

Stronger intermolecular H-bonding in the case of pentanol makes the bonding in the former case stronger than hexane. Thus, pentanol has high viscosity than hexane.