Problem 82 - Chemistry The Molecular Science

Question

Trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}:\), reacts readily with diborane, \(\mathrm{B}_{2} \mathrm{H}_{6}\). The diborane dissociates to two \(\mathrm{BH}_{3}\) fragments, each of which can react with trimethylamine to form a complex, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}: \mathrm{B} \mathrm{H}_{3}\). Write an equation for this reaction and interpret it in terms of Lewis acid-base theory.

Step-by-Step Solution

Answer

The reaction is \((\mathrm{CH}_{3})_{3}\mathrm{N} + \mathrm{B}_{2}\mathrm{H}_{6} \rightarrow 2((\mathrm{CH}_{3})_{3}\mathrm{N}:\mathrm{BH}_{3}).\) Trimethylamine acts as a Lewis base, and \(\mathrm{BH}_{3}\) acts as a Lewis acid.

Step 1: Write the Reactants

Identify the two substances participating in the reaction. Here, the reactants are trimethylamine, \((\mathrm{CH}_{3})_{3}\mathrm{N}:\), and diborane, \(\mathrm{B}_{2}\mathrm{H}_{6}\).

Step 2: Recognize Diborane Dissociation

According to the problem, diborane \((\mathrm{B}_{2}\mathrm{H}_{6})\) dissociates into two \(\mathrm{BH}_{3}\) units. This gives us two molecules of \(\mathrm{BH}_{3}\) available to react with the trimethylamine.

Step 3: Form the Complex

Each \(\mathrm{BH}_{3}\) unit reacts with a trimethylamine molecule \((\mathrm{CH}_{3})_{3}\mathrm{N}:\) to form a complex \((\mathrm{CH}_{3})_{3}\mathrm{N}:\mathrm{BH}_{3}\). This means we will have two complexes, as each unit of \(\mathrm{BH}_{3}\) participates in the reaction.

Step 4: Write the Overall Equation

Combine the reactants and products of the reaction in a single chemical equation: \[ (\mathrm{CH}_{3})_{3}\mathrm{N} + \mathrm{B}_{2}\mathrm{H}_{6} \rightarrow 2\ ((\mathrm{CH}_{3})_{3}\mathrm{N}:\mathrm{BH}_{3}). \]

Step 5: Lewis Acid-Base Interpretation

In Lewis acid-base theory, a Lewis acid is an electron-pair acceptor (\(\mathrm{BH}_{3}\)), and a Lewis base is an electron-pair donor (\((\mathrm{CH}_{3})_{3}\mathrm{N}:\)). Here, \((\mathrm{CH}_{3})_{3}\mathrm{N}:\) donates an electron pair to \(\mathrm{BH}_{3}\), forming a product complex.

Key Concepts

Trimethylamine

Trimethylamine is an organic compound that acts as a Lewis base. It is composed of a nitrogen atom surrounded by three methyl groups, giving it the chemical formula \((\mathrm{CH}_3)_3\mathrm{N}:\). This configuration allows the nitrogen atom to have a lone pair of electrons.
This lone pair is essential in its role as a Lewis base. In chemical terms, a Lewis base is defined by its ability to donate an electron pair to another species, typically a Lewis acid.
In reactions, such as with diborane, the nitrogen's lone pair on trimethylamine becomes available to form a bond with electrophilic centers like the boron atom which are found in Lewis acids.

Diborane Reaction

Diborane \((\mathrm{B}_2\mathrm{H}_6)\) is a molecule consisting of boron and hydrogen. In its reaction with trimethylamine, diborane plays a significant role. It undergoes dissociation to form two \(\mathrm{BH}_3\) fragments.
This process is crucial because these \(\mathrm{BH}_3\) units are the active participants in further reactions with trimethylamine. In the described reaction, each \(\mathrm{BH}_3\) fragment functions as a Lewis acid. This means that they have vacant p orbitals available for accepting an electron pair.
Through this acceptance, a stable complex is formed when the \((\mathrm{CH}_3)_3\mathrm{N}:\) trimethylamine molecules donate their lone pair to the \(\mathrm{BH}_3\) fragments, resulting in \((\mathrm{CH}_3)_3\mathrm{N}:\mathrm{BH}_3\) complexes.

Chemical Equations

Chemical reactions can be expressed through balanced chemical equations, which help to visualize the conversion of reactants to products. The reaction between trimethylamine and diborane can be represented by the equation: \[ (\mathrm{CH}_3)_3\mathrm{N} + \mathrm{B}_2\mathrm{H}_6 \rightarrow 2 ((\mathrm{CH}_3)_3\mathrm{N}:\mathrm{BH}_3) \]This equation shows the stoichiometry of the reaction.
It implies that one molecule of \((\mathrm{CH}_3)_3\mathrm{N}\) reacts with one molecule of \(\mathrm{B}_2\mathrm{H}_6\) to produce two molecules of the complex \((\mathrm{CH}_3)_3\mathrm{N}:\mathrm{BH}_3)\).

This visual representation helps in applying the Lewis acid-base theory, where the transfer of the electron pair is clearly shown. Equations like these provide a convenient shorthand for chemical transformations, allowing chemists to predict the outcomes of reactions and the quantities of substances required or produced.