Question

Write an equation for the reaction between each Lewis acid-base pair, showing electron flow by means of curved arrows. (a) \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~B}+\mathrm{OH}^{-} \longrightarrow\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{AlCl}_{3} \longrightarrow\)

Short answer

Question: Identify the Lewis acid and Lewis base in each pair, and using curved arrows, depict the electron flow in the reaction between them. (a) \((\mathrm{CH}_{3}\mathrm{CH}_{2})_{3}\mathrm{B}\) and \(\mathrm{OH}^{-}\) Lewis acid: \((\mathrm{CH}_{3}\mathrm{CH}_{2})_{3}\mathrm{B}\) Lewis base: \(\mathrm{OH}^{-}\) Electron flow: $$ (\mathrm{CH}_{3} \mathrm{CH}_{2})_{3} \mathrm{~B}+\mathrm{OH}^{-} \longrightarrow (\mathrm{CH}_{3} \mathrm{CH}_{2})_{3}\mathrm{B}\text{---}\mathrm{O}^{-}\mathrm{H} $$ (b) \(\mathrm{CH}_{3} \mathrm{Cl}\) and \(\mathrm{AlCl}_{3}\) Lewis acid: \(\mathrm{AlCl}_{3}\) Lewis base: \(\mathrm{CH}_{3} \mathrm{Cl}\) Electron flow: $$ \mathrm{CH}_{3}\mathrm{Cl}+\mathrm{AlCl}_{3} \longrightarrow \mathrm{CH}_{3}\text{---}\mathrm{Cl}\text{---}\mathrm{AlCl}_{3} $$

Step-by-step solution

(a) Identify the Lewis acid and Lewis base

In the given equation, \((\mathrm{CH}_{3}\mathrm{CH}_{2})_{3}\mathrm{B}\) is the Lewis acid and \(\mathrm{OH}^{-}\) is the Lewis base. The Lewis acid has an empty electron orbital that can accept an electron pair, while the Lewis base is a species containing an electron pair that can be donated.

(a) Show the reaction using curved arrows

The electron pair in the oxygen atom of the hydroxide ion, \(\mathrm{OH}^{-}\), will be donated to the electron-deficient boron atom in \((\mathrm{CH}_{3}\mathrm{CH}_{2})_{3}\mathrm{B}\). To represent this, a curved arrow will be drawn from the oxygen atom to the boron atom, indicating the flow of electrons: $$ (\mathrm{CH}_{3} \mathrm{CH}_{2})_{3} \mathrm{~B}+\mathrm{OH}^{-} \longrightarrow (\mathrm{CH}_{3} \mathrm{CH}_{2})_{3}\mathrm{B}\text{---}\mathrm{O}^{-}\mathrm{H} $$

(b) Identify the Lewis acid and Lewis base

In the given equation, \(\mathrm{CH}_{3} \mathrm{Cl}\) is the Lewis base and \(\mathrm{AlCl}_{3}\) is the Lewis acid. The Lewis base has a lone pair of electrons on the chlorine atom that can be donated, while the Lewis acid has an empty electron orbital in the aluminum atom that can accept an electron pair.

(b) Show the reaction using curved arrows

The electron pair in the chlorine atom of the methyl chloride molecule, \(\mathrm{CH}_{3} \mathrm{Cl}\), will be donated to the electron-deficient aluminum atom in \(\mathrm{AlCl}_{3}\). To show this reaction, a curved arrow will be drawn from the chlorine atom to the aluminum atom, indicating the flow of electrons: $$ \mathrm{CH}_{3}\mathrm{Cl}+\mathrm{AlCl}_{3} \longrightarrow \mathrm{CH}_{3}\text{---}\mathrm{Cl}\text{---}\mathrm{AlCl}_{3} $$