Question

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

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

Short answer

Answer

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

Step-by-step solution

Step-by-Step SolutionStep 1: Melting point

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.

 Step 2: Isomeric alcohols

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.

Observed trend in the melting points of isomeric alcohols

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 ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}$.