Question

For each equation, label the Lewis acid and the Lewis base. In addition, show all unshared pairs of electrons on the reacting atoms and use curved arrows to show the flow of electrons in each reaction. (a) \(\mathrm{F}^{-}+\mathrm{BF}_{3} \longrightarrow \mathrm{BF}_{4}^{-}\) (b) C=C[C+]C=[OH+]

Short answer

Question: Identify the Lewis acid and Lewis base in each reaction, and explain the electron flow in each process. a) F⁻ + BF₃ → BF₄⁻ b) CH₃C=CCH₂C⁺=OH⁺ Answer: a) In the reaction F⁻ + BF₃ → BF₄⁻, F⁻ is the Lewis base as it donates an electron pair, and BF₃ is the Lewis acid as it accepts an electron pair. The electron flow is from the lone pair of electrons on the fluorine atom of F⁻ towards the boron atom of BF₃. b) In the reaction CH₃C=CCH₂C⁺=OH⁺, the double-bonded carbons act as the Lewis base by donating a pi electron pair, and the OH⁺ group acts as the Lewis acid by accepting the electron pair. The electron flow is from the pi bond between the carbon atoms towards the oxygen atom in the OH⁺ group.

Step-by-step solution

Identify the Lewis acid and Lewis base

In this reaction, \(\mathrm{F}^{-}\) has a lone pair of electrons and it can donate an electron pair, making it a Lewis base. On the other hand, \(\mathrm{BF}_{3}\) can accept a pair of electrons, making it a Lewis acid.

Show the unshared pairs of electrons

We will represent unshared pairs of electrons on the reacting atoms using dots. For the Lewis base, \(\mathrm{F}^{-}\), it has 3 lone pairs on the fluorine atom in addition to the negatively charged lone pair that is donated.

Use curved arrows to show the electron flow

We will draw a curved arrow starting from the lone pair of electrons on the fluorine atom of \(\mathrm{F}^{-}\) (Lewis base) and point it towards the boron atom of \(\mathrm{BF}_{3}\) (Lewis acid). This indicates that the electron pair is being donated by the Lewis base to the Lewis acid. Combining all the information above, the reaction can be represented as follows: \(\underset{\text{Lewis base}}{\mathrm{F}^{-}:\overset{..}}+\underset{\text{Lewis acid}}{\mathrm{BF}_{3}} \longrightarrow \mathrm{BF}_{4}^{-}\) (b) C=C[C+]C=[OH+]

Convert SMILES notation to a chemical equation

Converting the given SMILES notation to its equivalent chemical equation, we get: \(\mathrm{CH}_{3}\mathrm{C}=\mathrm{C}\mathrm{CH}_{2}\mathrm{C}^{+}={\mathrm{OH}}^{+}\)

Identify the Lewis acid and Lewis base

In this reaction, the negatively charged oxygen atom in \({\mathrm{OH}}^{+}\) can accept a pair of electrons, making it a Lewis acid. The pi bond formed by two electrons from the double bond between the carbon atoms can be donated as an electron pair, making the double-bonded carbons a Lewis base.

Show the unshared pairs of electrons

We will represent unshared pairs of electrons on the reacting atoms using dots. For the Lewis base, the double-bonded carbons, we will show the pi electrons as a unshared pair between the carbons. For the Lewis acid, \({\mathrm{OH}}^{+}\), the oxygen atom has two lone pairs.

Use curved arrows to show the electron flow

We will draw a curved arrow starting from the pi bond between the carbon atoms (Lewis base) and point it towards the oxygen atom in \({\mathrm{OH}}^{+}\) (Lewis acid). This indicates that the electron pair is being donated by the Lewis base to the Lewis acid. Combining all the information above, the reaction can be represented as follows: \(\underset{\text{Lewis base}}{\mathrm{CH}_{3}\mathrm{C}=\mathrm{C}\mathrm{CH}_{2}\mathrm{C}}+\underset{\text{Lewis acid}}{{\overset{..}{\mathrm{O}}}\mathrm{H}^{+}} \longrightarrow \mathrm{CH}_{3}\mathrm{C}\mathrm{CH}_{2}\mathrm{COH}\)