Q. 2-2-64E-b - Organic Chemistry (Mcmurry)

Question

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

Equilibrium is favored.

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

Step-by-Step Solution

Answer

A product is favored in this reaction.

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

pK a values of Propanol and Ammonia

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

Identify in which direction equilibrium will be favored

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.

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