Q14P - Organic Chemistry (Mcmurry)

Question

Will either of the following reactions take place to a significant extent as written, according to the data in Table 2-3?

a)  $\mathbf{HCN}\mathbf{+}{\mathbf{CH}}_{\mathbf{3}}\mathbf{C}{{\mathbf{O}}_{\mathbf{2}}}^{\mathbf{-}}{\mathbf{Na}}^{\mathbf{+}}\stackrel{\mathbf{?}}{\mathbf{\to }\mathbf{N}{{\mathbf{a}}^{\mathbf{+}}}^{\mathbf{-}}\mathbf{CN}\mathbf{+}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CO}}_{\mathbf{2}}\mathbf{H}}$                                                        b)  ${\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{OH}\mathbf{+}{{\mathbf{Na}}^{\mathbf{+}}}^{\mathbf{-}}\mathbf{CN}\stackrel{\mathbf{?}}{\mathbf{\to }{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{O}}^{\mathbf{-}}{\mathbf{Na}}^{\mathbf{+}}\mathbf{+}\mathbf{HCN}}$

Step-by-Step Solution

Answer

a) Reaction will not take place to a significant extent in the direction written.

 $\mathrm{H}—\mathrm{CN}+{\mathrm{CH}}_3\mathrm{C}{{\mathrm{O}}_2}^{-}{\mathrm{Na}}^{+}\stackrel{?}{\to \mathrm{N}{{\mathrm{a}}^{+}}^{-}\mathrm{CN}+{\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H}}$     

$\text{p}{K}_a$=9.3$\text{p}{K}_a$=4.7

                                       Weaker acidStronger acid

b) Reaction will not take place to a significant extent in the direction written.

 ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{O}—\mathrm{H} +{\mathrm{Na}}^{+}{}^{-}\mathrm{CN}\stackrel{?}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{O}}^{-}{\mathrm{Na}}^{+}+\mathrm{HCN}$

$\text{p}{K}_a$=16 $\text{p}{K}_a$= 9.3

Weaker acid              Stronger acid

Step 1: Acid and base strength

Acids differ in their ability to donate H⁺. Stronger acids such as HCl, react almost completely with water, whereas weaker acids, such as acetic acid (CH₃COOH), react only slightly. The exact strength of the given acid HA in water solution is described using the acidity constant ($K_a$ ) for the acid-dissociation equilibrium.

Example:

 $$\begin{gathered} {\text{HA + H}}_2\text{O }\rightleftarrows {\text{ A}}^{-}{\text{ + H3O}}^{-} \\ {K}_a\text{ = }\frac{{\text{[H3O}}^{+}{\text{][A}}^{-}\text{]}}{\left[HA\right]} \end{gathered}$$

Acid strengths normally use $\text{p}{K}_a$ values rather than $K_a$ values, where the $\text{p}{K}_a$ is the negative common logarithm of $K_a$:

 $\text{p}K\text{a =-log }{K}_a$

The stronger acid has a smaller  $\text{p}{K}_a$ and a weaker acid has a larger $\text{p}{K}_a$ .

Step 2(a) : Determine the significant extent of reaction (a)

A reaction takes place if the stronger acid is the reactant.

(a)        $\mathrm{H}—\mathrm{CN}+{\mathrm{CH}}_3\mathrm{C}{{\mathrm{O}}_2}^{-}{\mathrm{Na}}^{+}\stackrel{?}{\to \mathrm{N}{{\mathrm{a}}^{+}}^{-}\mathrm{CN}+{\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H}}$                                                               

 $\text{p}{K}_a$=9.3                           $\text{p}{K}_a$=4.7

Weaker acid Stronger acid

 

                                                                              

As we know from step 1 that the lower the value, the stronger the acid. Thus ${\mathrm{CH}}_3{\mathrm{CO}}_2\mathrm{H}$ , not HCN, is the stronger acid. So the above reaction will not take place to a significant extent in the direction written.

Step 3(b): Determine the significant extent of reaction (b)

A reaction takes place if the stronger acid is the reactant.

(b) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{O}—\mathrm{H}+{\mathrm{Na}}^{+} {}^{-}\mathrm{CN}\stackrel{\mathrm{heat}}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{O}}^{-}{\mathrm{Na}}^{+}+\mathrm{HCN}$

 $\text{p}{K}_a$=16 $\text{p}{K}_a$= 9.3

Weaker acid              Stronger acid

 

We know from step 1 that the lower the value, the stronger the acid. Thus HCN, not ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}$, is the stronger acid. So, the above reaction will not take place to a significant extent in the direction written.