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Chapter 4: Problem 8

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

Expert verified
1. Lewis acid: \((\mathrm{CH}_{3} \mathrm{CH}_{2})_{3} \mathrm{~B}\) Lewis base: \(\mathrm{OH}^{-}\) 2. Lewis acid: \(\mathrm{AlCl}_{3}\) Lewis base: \(\mathrm{CH}_{3} \mathrm{Cl}\) Answer: 1. \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~B}+\mathrm{OH}^{-} \longrightarrow (\mathrm{CH}_{3} \mathrm{CH}_{2})_{3} \mathrm{BOH}\). The electron flow is from the oxygen atom in \(\mathrm{OH}^{-}\) to the boron atom in \((\mathrm{CH}_{3} \mathrm{CH}_{2})_{3} \mathrm{~B}\). 2. \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{AlCl}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{AlCl}_{4}\). The electron flow is from the chlorine atom in \(\mathrm{CH}_{3} \mathrm{Cl}\) to the aluminum atom in \(\mathrm{AlCl}_{3}\).

Step by step solution

01

Identify Lewis acids and bases in each reaction

In the first reaction: (a) \((\mathrm{CH}_{3} \mathrm{CH}_{2})_{3} \mathrm{~B}\) is the Lewis acid, as boron has an empty orbital available to accept a pair of electrons. \(\mathrm{OH}^{-}\) is the Lewis base, as it can donate a lone pair of electrons from its oxygen atom. In the second reaction: (b) \(\mathrm{CH}_{3} \mathrm{Cl}\) is the Lewis base, as the chlorine atom can donate a lone pair of electrons. \(\mathrm{AlCl}_{3}\) is the Lewis acid, as aluminum has an empty orbital available to accept a pair of electrons.
02

Write a balanced equation for each reaction, showing the electron flow using curved arrows

For the first reaction (a): \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~B}+\mathrm{OH}^{-} \longrightarrow (\mathrm{CH}_{3} \mathrm{CH}_{2})_{3} \mathrm{BOH} \) The electron flow in this reaction is from the oxygen atom bonding pair in \(\mathrm{OH}^{-}\) (Lewis base) to the boron atom in \((\mathrm{CH}_{3} \mathrm{CH}_{2})_{3} \mathrm{~B}\) (Lewis acid). So, we will show this flow using a curved arrow. For the second reaction (b): \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{AlCl}_{3} \longrightarrow \mathrm{CH}_{3} \mathrm{AlCl}_{4}\) The electron flow in this reaction is from the chlorine atom bonding pair in \(\mathrm{CH}_{3} \mathrm{Cl}\) (Lewis base) to the aluminum atom in \(\mathrm{AlCl}_{3}\) (Lewis acid). So, we will show this flow using a curved arrow.

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Most popular questions from this chapter

An ester is a derivative of a carboxylic acid in which the hydrogen of the carboxyl group is replaced by an alkyl group (Section 1.3E). Draw a structural formula of methyl acetate, which is derived from acetic acid by replacement of the \(\mathrm{H}\) of its \(-\mathrm{OH}\) group by a methyl group. Determine if proton transfer to this compound from \(\mathrm{HCl}\) occurs preferentially on the oxygen of the \(\mathrm{C}=\mathrm{O}\) group or the oxygen of the \(\mathrm{OCH}_{3}\) group.

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