Question

Draw the products formed when each compound is treated with ${\mathbf{HNO}}_{\mathbf{3}}$ and ${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$.

Short answer

To draw the product of reaction with these disubstituted benzene derivatives,${\text{HNO}}_3$ and , ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$,remember the following points:

• If two directing effects reinforce each other, the new substituent will be in the position reinforced by both.
• If the directing effects oppose each other, the stronger activator wins.
• No substitution occurs between two meta substituents.

Step-by-step solution

Friedel-Craft reaction

To draw the product of reaction with these disubstituted benzene derivatives,${\text{HNO}}_3$ and , ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$,remember the following points:

• If two directing effects reinforce each other, the new substituent will be in the position reinforced by both.
• If the directing effects oppose each other, the stronger activator wins.
• No substitution occurs between two meta substituents.

Ortho-para and meta directing groups

The ortho and para directing groups are -OH, OR, R (alkyl group), Cl, and groups containing lone pair of electrons.

The meta-directing groups are ${\text{NO}}_{\text{2}}$, -COCl, -CO, -CHO, etc.

Explanation

a. The nitration of the given compound results in the substitution of the nitro group at ortho with respect to the methoxy group.

b. The nitration of the given compound results in the substitution of the nitro group at ortho with respect to the Br group.

c. The nitration of the given compound results in the substitution of the nitro group at ortho with respect to the methyl group.

d. The nitration of the given compound results in the substitution of the nitro group at ortho with respect to the Cl and para with respect to the Br group.