Q153CP - Chemistry: Molecular Nature Of Matter And Change

Question

Chlorobenzene, ${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}\mathbf{\text{Cl}}$, is a key intermediate in the manufacture of dyes and pesticides. It is made by the chlorination of benzene, catalyzed by ${\mathbf{\text{FeCl}}}_{\mathbf{\text{3}}}$, in this series of steps: 

 (1) $\mathbf{ }{\mathbf{\text{Cl}}}_{\mathbf{\text{2}}}{\mathbf{\text{ + FeCl}}}_{\mathbf{\text{3}}}\mathbf{\rightleftharpoons }{\mathbf{\text{FeCl}}}_{\mathbf{\text{5}}}{\mathbf{\text{(orCl}}}^{\mathbf{\text{ + }}}\mathbf{\text{FeC}}{{\mathbf{\text{l}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{ - }}}\mathbf{\text{)}}$

 (2) ${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{6}}}{\mathbf{\text{ + Cl}}}^{\mathbf{\text{ + }}}{\mathbf{\text{FeCl}}}_{\mathbf{\text{4}}}^{\mathbf{\text{ - }}}\mathbf{\rightleftharpoons }{\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{6}}}{\mathbf{\text{Cl}}}^{\mathbf{\text{ + }}}{\mathbf{\text{ + FeCl}}}_{\mathbf{\text{4}}}^{\mathbf{\text{ - }}}$

 (3) ${\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{6}}}{\mathbf{\text{Cl}}}^{\mathbf{\text{ + }}}\mathbf{\rightleftharpoons }{\mathbf{\text{C}}}_{\mathbf{\text{6}}}{\mathbf{\text{H}}}_{\mathbf{\text{5}}}{\mathbf{\text{Cl + H}}}^{\mathbf{\text{ + }}}$

 (4) ${\mathbf{\text{H}}}^{\mathbf{\text{ + }}}{\mathbf{\text{ + FeCl}}}_{\mathbf{\text{4}}}^{\mathbf{\text{ - }}}\mathbf{\rightleftharpoons }{\mathbf{\text{HCl + FeCl}}}_{\mathbf{\text{3}}}$

(a) Which of the step(s) is (are) Lewis acid-base reactions?

 (b) Identify the Lewis acids and bases in each of those steps.

Step-by-Step Solution

Answer

(a) Reactions (1), (2), and (3) are Lewis acid-base reactions.

(b) ${\text{Cl}}_2$ is a Lewis base and  ${\text{FeCl}}_3$is a Lewis acid.

      ${\text{C}}_6{\text{H}}_6$  is a Lewis base and ${\text{Cl}}^{+} {\text{FeCl}}_4^{-}$ is a Lewis acid.

       ${\text{FeCl}}_4^{-}$ is a Lewis base and ${\text{H}}^{+}$ is a Lewis acid.

Step 1(a): Find the steps of lewis acid-base reactions

Evaluate each reaction step.

${\text{Cl}}_{\text{2}}{\text{ + FeCl}}_{\text{3}}\rightleftharpoons {\text{FeCl}}_{\text{5}}$

${\text{Cl}}_2$ reaction is electron rich. An electron pair will attack the ${\text{FeCl}}_3$  to form an adduct ${\text{FeCl}}_5$ 

Therefore, reaction (1) is a Lewis acid-base reaction.

${\text{C}}_6{\text{H}}_6+{\text{Cl}}^{+}{\text{FeCl}}_4^{-}\rightleftharpoons {\text{C}}_6{\text{H}}_6{\text{Cl}}^{+}+{\text{FeCl}}_4^{-}$

This reaction is electron rich. It will donate one of its electron pair from the double bond to the partially positive chlorine of ${\text{Cl}}^{+}{\text{FeCl}}^{-}$ to form the products above.

Therefore, reaction (2) is a Lewis acid-base reaction.

${\text{C}}_6{\text{H}}_6{\text{Cl}}^{+}\rightleftharpoons {\text{C}}_6{\text{H}}_5\text{Cl}+{\text{H}}^{+}$

Reaction (3) is not an acid-base reaction but more like deprotonation. It releases a proton because it is more stable that way.

 ${\text{H}}^{+}+{\text{FeCl}}_4^{-}\rightleftharpoons \text{HCl}+{\text{FeCl}}_3$

${\text{FeCl}}_4^{-}$ donates its pair of electrons to ${\text{H}}^{+}$.

Therefore, reaction (4) is a Lewis acid-base reaction.

Step 2(b): Identify the Lewis acids and bases

(1) ${\text{Cl}}_2+{\text{FeCl}}_3\rightleftharpoons {\text{FeCl}}_5$

The Lewis acid is an electron acceptor and the Lewis base is an electron donor. Recall that in this reaction, ${\text{Cl}}_2$ donated its lone pair to ${\text{FeCl}}_3$.

Therefore, ${\text{Cl}}_2$  is a Lewis base and ${\text{FeCl}}_3$is a Lewis acid.

(2)  _{${\text{C}}_6{\text{H}}_6+{\text{Cl}}^{+}{\text{FeCl}}_4^{-}\rightleftharpoons {\text{C}}_6{\text{H}}_6{\text{Cl}}^{+}+{\text{FeCl}}_4^{-}$}

The Lewis acid is an electron acceptor, and the Lewis base is an electron donor. Recall that in this reaction$,{\text{C}}_6{\text{H}}_6$ is electron-rich and donated its lone pair to ${\text{Cl}}^{+}{\text{FeCl}}_4^{-}$Therefore,$,{\text{C}}_6{\text{H}}_6$ is a Lewis base and ${\text{Cl}}^{+}{\text{FeCl}}_4^{-}$ is a Lewis acid.

(4) ${\text{H}}^{+}+{\text{FeCl}}_4^{-}\rightleftharpoons \text{HCl}+{\text{FeCl}}_3$

The Lewis acid is an electron acceptor, and the Lewis base is an electron donor. Recall that in this reaction, ${\text{FeCl}}_4^{-}$ donates its lone pair to ${\text{H}}^{+}$.

Therefore, ${\text{FeCl}}_4^{-}$ is a Lewis base and ${\text{H}}^{+}$ is a Lewis acid.