Question: Explain the observed trend in the melting points of the following three isomeric alcohols:

${\left({\mathrm{CH}}_3\right)}_{\mathbf{2}}{\mathbf{CHCH}}_{\mathbf{2}}\mathbf{OH}\mathbf{}\left(-108°C\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OH}\mathbf{}\left(-90°C\right)\mathbf{}\mathbf{,}{\left({\mathrm{CH}}_3\right)}_{\mathbf{3}}\mathbf{COH}\mathbf{}\left(26°C\right)$

The point at which a particular substance transforms from a solid state to a liquid state is the melting point.Numerous factors influence the melting point, and one among them is the pressure.

Isomeric alcohols are alcohols comprising equivalent molecular formulas but differ in their spatial arrangement. An example of isomeric alcohol is 1-butanol and 2-butanol.

The melting point of a particular compound is directly proportional to the surface area and symmetry. An increase in surface area or symmetry increases the melting point of the particular compound.

A branched polymer has a lesser surface area in comparison to unbranched primary alcohol.${\left({\mathrm{CH}}_3\right)}_2{\mathrm{CHCH}}_2\mathrm{OH}$ is branched primary alcohol and ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$is an unbranched primary alcohol.

Hence, the melting point of ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$is higher than${\left({\mathrm{CH}}_3\right)}_2{\mathrm{CHCH}}_2\mathrm{OH}$ .

Because of the greater symmetry in ${\left({\mathrm{CH}}_3\right)}_3\mathrm{COH}$, it has a larger melting point than${\left({\mathrm{CH}}_3\right)}_2{\mathrm{CHCH}}_2\mathrm{OH}$ and ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$.