Question

Draw the products formed when each compound is treated with ${\mathbf{HNO}}_{\mathbf{3}}$ and ${\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$. State whether the reaction occurs faster or slower than a similar reaction with benzene.

Short answer

Nitration is the process of substituting ${\text{NO}}_{\text{2}}$ group. The substitution depends upon the substrate’s nature. The ${\text{NO}}_{\text{2}}$ is an electron-withdrawing group.

Step-by-step solution

Nitration

Nitration is the process of substituting ${\text{NO}}_{\text{2}}$ group. The substitution depends upon the substrate’s nature. The ${\text{NO}}_{\text{2}}$ is an electron-withdrawing group.

Reactivity towards nitration

The electron-withdrawing substituents on the benzene ring slow the rate of nitration reaction. The substituents attract the electron towards themselves.

The electron-donating substituents on the benzene ring increase the rate of nitration reaction. The substituents donate the electron towards ${\text{NO}}_{\text{2}}$.

Explanation

a. The group on benzene is an electron-withdrawing substituent; therefore, the reaction occurs slower than the nitration reaction.

b. The CN group on benzene is an electron-withdrawing substituent; therefore, the reaction occurs slower than the nitration reaction.

c. The OH group on benzene is an electron-donating substituent; therefore, the reaction occurs faster than the nitration reaction.

d. The Cl group on benzene is an electron-withdrawing substituent; therefore, the reaction occurs slower than the nitration reaction.

e. The ${\text{- CH}}_2{\text{CH}}_3$group on benzene is an electron-donating substituent; therefore, the reaction occurs faster than the nitration reaction.