Question

Draw the product formed when ${\left({\mathrm{CH}}_3{\mathrm{CH}}_2\right)}_{\mathbf{3}}\mathbf{N}$, a Lewis base, reacts with each Lewis acid:

a. role="math" localid="1648902828598" $\mathbf{B}{\left({\mathrm{CH}}_3\right)}_{\mathbf{3}}$

b. role="math" localid="1648902808294" ${\left({\mathrm{CH}}_3\right)}_{\mathbf{3}}{\mathbf{C}}^{\mathbf{+}}$

c.${\mathbf{AlCl}}_{\mathbf{3}}$

Short answer

a.

b.

c.

Step-by-step solution

Lewis acid and Lewis base

The Lewis acids and Lewis bases can be differentiated based on the presence of lone pair of electrons.

• The Lewis acids are electron-deficient species that can accept at least one lone pair of electrons.
• The Lewis bases have filled orbitals that can donate at least one lone pair of electrons.

Therefore, the electrons are transferred from Lewis bases to the Lewis acid and form an adduct.

Formation of the product or adduct 

a.The $\mathrm{B}{\left({\mathrm{CH}}_3\right)}_3$molecule is a Lewis acid or electrophile and localid="1648903802240" ${\left({\mathrm{CH}}_3{\mathrm{CH}}_2\right)}_3\mathrm{N}:$ is a Lewis base.

Therefore, the lone pair of electrons are transferred from localid="1648904001847" ${\left({\mathrm{CH}}_3{\mathrm{CH}}_2\right)}_3\mathrm{N}:$ to $\mathrm{B}{\left({\mathrm{CH}}_3\right)}_3$and form an adduct by forming a covalent bond between nitrogen and boron.

Formation of an adduct

b.The ${\left({\mathrm{CH}}_3\right)}_3{\mathrm{C}}^{+}$ molecule is a Lewis acid or electrophile and ${\left({\mathrm{CH}}_3{\mathrm{CH}}_2\right)}_3\mathrm{N}:$is a Lewis base. Therefore, the lone pair of electrons are transferred from ${\left({\mathrm{CH}}_3{\mathrm{CH}}_2\right)}_3\mathrm{N}:$ to ${\left({\mathrm{CH}}_3\right)}_3{\mathrm{C}}^{+}$and form an adduct by forming a covalent bond between nitrogen and carbon.

Formation of an adduct

c. The ${\mathrm{AlCl}}_3$ molecule is a Lewis acid or electrophile and ${\left({\mathrm{CH}}_3{\mathrm{CH}}_2\right)}_3\mathrm{N}:$ is a Lewis base. Therefore, the lone pair of electrons are transferred from ${\left({\mathrm{CH}}_3{\mathrm{CH}}_2\right)}_3\mathrm{N}:$ to ${\mathrm{AlCl}}_3$and form an adduct by forming a covalent bond between nitrogen and aluminum.

Formation of an adduct