Chemistry Hydrocarbons MDCAT 2017
PMDC Verified Question 47 of 65
For halogenation of benzene, which reagent is used:
A
\( \text{H}_2\text{SO}_4 \)
B
\( \text{HNO}_3 \)
C
\( \text{AlCl}_3 \)
D
\( \text{HCl} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: \( \text{AlCl}_3 \)
Concept:

Halogens alone are not electrophilic enough to break the aromaticity of benzene. A Lewis acid catalyst is strictly required.

Solution:

  • To substitute a halogen onto a benzene ring, a catalyst like \( \text{AlCl}_3 \), \( \text{FeCl}_3 \), or \( \text{FeBr}_3 \) is used.


  • These Lewis acids possess empty d-orbitals that accept a lone pair from the halogen molecule (e.g., \( \text{Cl}_2 \)), polarizing the bond and forming the strongly electrophilic halonium ion (\( \text{Cl}^+ \)).


Why other options are incorrect:

  • \( \text{H}_2\text{SO}_4 \) and \( \text{HNO}_3 \) are used for nitration and sulfonation, not halogenation.


  • HCl is a Brønsted acid and cannot effectively polarize molecular halogens for this reaction.

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