Friedel Crafts Alkylation Reaction Statements Analysis

To determine the correct statements regarding the given Friedel-Crafts alkylation reaction of benzene with 1-chloropropane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}$) in the presence of anhydrous $\text{AlCl}_3$:

$\text{C}_6\text{H}_6 + \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} \xrightarrow{\text{Anhydrous } \text{AlCl}_3} \text{Alkylbenzene}$

Let me analyze each statement step-by-step:

1. Analysis of Intermediate and Major Product (Statements A & B):

   • The reaction initiates when 1-chloropropane reacts with Lewis acid $\text{AlCl}_3$ to form an initial primary carbocation complex: $\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{AlCl}_3 \longrightarrow \text{CH}_3\text{CH}_2\stackrel{\oplus}{\text{C}}\text{H}_2 + \text{AlCl}_4^-$
   • The primary carbocation ($\text{1}^\circ$) undergoes a rapid 1,2-hydride shift to generate a more stable secondary carbocation ($\text{2}^\circ$), known as the isopropyl carbocation: $\text{CH}_3-\text{CH}_2-\stackrel{\oplus}{\text{C}}\text{H}_2 \xrightarrow{1,2-\text{H}^- \text{ shift}} \text{CH}_3-\stackrel{\oplus}{\text{C}}\text{H}-\text{CH}_3$
   • Attack of benzene on the isopropyl carbocation yields isopropylbenzene (cumene) as the major product, while n-propylbenzene is formed only as a minor product.
   • Therefore, Statement A is incorrect and Statement B is correct.

2. Multiple Substitution / Polyalkylation (Statement C):

   • Alkyl groups (such as $-\text{CH}(\text{CH}_3)_2$) are electron-donating due to inductive effects ($+I$) and hyperconjugation.
   • Consequently, the product (alkylbenzene) is more reactive towards electrophilic aromatic substitution than the starting material (benzene).
   • This causes polyalkylation (multiple substitution) to occur readily and makes it an inevitable limitation of Friedel-Crafts alkylation under standard conditions.
   • Therefore, Statement C is correct.

3. Effect of Electron-Donating Substituents (Statement D):

   • Electron-donating substituents activate the benzene ring towards electrophilic attack, making Friedel-Crafts alkylation proceed even faster to give alkylbenzenes. (It is electron-withdrawing groups like $-\text{NO}_2$ that retard or prevent the reaction).
   • Therefore, Statement D is incorrect.

Thus, the correct statements are B and C only.

Correct Option: B