Use the ${\mathbf{pK}}_{\mathbf{a}}$table in Appendix B to determine in which direction the

Equilibrium is favored.

(b)${\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OH}\mathbf{+}\dot{\mathbf{N}}{\mathbf{H}}_{\mathbf{2}}\mathbf{\rightleftarrows }{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}\dot{\mathbf{O}}\mathbf{+}{\mathbf{NH}}_{\mathbf{3}}$

IUPAC name of propanol is propan-1-ol. As alcohol is attached to the first carbon atom, it is a primary alcohol.

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

${\mathrm{pK}}_{\mathrm{a}}$Value of Propanol = 16.85

In this compound, the nitrogen atom is attached to three hydrogen atoms so as to complete its valency. In ammonia, the nitrogen atom is ${\mathrm{Sp}}^3$ hybridized, and it has a pyramidal shape. ${\mathrm{NH}}_3$

${\mathrm{pK}}_{\mathrm{a}}$Value of Ammonia = 38

The ${\mathrm{pK}}_{\mathrm{a}}$Value of ammonia is more than the ${\mathrm{pK}}_{\mathrm{a}}$Value of propanol. Therefore, the acidic strength of propanol is more than the acidic strength of ammonia. And as we know, equilibrium always favors in that direction in which acidic strength is less; we can say that basic strength is more. Hence, the equilibrium will favor the product side of the reaction.

${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH}+\dot{\mathrm{N}}{\mathrm{H}}_2\rightleftarrows {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\dot{\mathrm{O}}+{\mathrm{NH}}_3$