Question

Which of the following compounds undergo Friedel–Crafts alkylation with ${\mathbf{CH}}_{\mathbf{3}}\mathbf{Cl}$and ${\mathbf{AlCl}}_{\mathbf{3}}$? Draw the products formed when a reaction occurs.

Short answer

Friedel-Craft alkylation substitutes the alkyl group at the ortho and para positions of the substrate having an electron-donating group.

Step-by-step solution

Friedel-Craft reaction

Friedel-Craft alkylation substitutes the alkyl group at the ortho and para positions of the substrate having an electron-donating group.

Effect of substituents

Friedel-Crafts reactions do not occur in the presence of strongly deactivating substituents such as ${\text{NO}}_{\text{2}}$,${\text{NH}}_{\text{2}}$ ,${\text{NR}}_{\text{2}}$ , NHR groups.

Explanation

a. The ${\text{SO}}_3\text{H}$compound is highly deactivating. Therefore, Friedel-Craft alkylation does not occur.

b. The Cl compound is also electron-donating. Therefore, Friedel-Craft alkylation gives ortho and para products.

c. The $\text{N}{\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}$compound is also highly deactivating. Therefore, Friedel-Craft alkylation does not occur.

d. The ${\text{NHCOCH}}_3$ compound is electron-withdrawing, but the ring is sufficient for Friedel-Craft alkylation. Therefore, Friedel-Craft alkylation gives ortho and para products.