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 ${C_6}{H_6} + C{l_2}\xrightarrow[{{\text{dark, cold}}}]{{{\text{Anhydrous AlC}}{{\text{l}}_3}}}P$

Last updated date: 20th Jun 2024
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Hint: We are given the question, the reaction between benzene and chlorine gas in the presence of anhydrous aluminium chloride under dark and cold conditions. Aluminium chloride is known to be a very good Lewis acid and readily accepts a lone pair of electrons.

Complete answer:
Anhydrous aluminium chloride is an electron deficient species and a Lewis acid. In presence of Lewis acids, benzene undergoes electrophilic substitution reaction. In the reaction given in the question, aluminium chloride acts as a catalyst. It has empty orbitals which accept a lone pair from chlorine forming a complex $AlCl_4^ - $ with anion formed as a result of the heterolytic cleavage of the chlorine molecule. The other species obtained is $C{l^ + }$ which is highly electrophilic. It attacks the benzene ring and substitutes a hydrogen of the benzene resulting into the formation of chlorobenzene. The hydrogen ion released reacts with $AlCl_4^ - $ to produce $HCl$ and recover $AlC{l_3}$.
Under the given conditions and in presence of excess amounts of chlorine, the benzene ring undergoes repeated electrophilic substitution reactions till all the hydrogens of the ring are replaced by chlorine atoms. Hence hexa- chloro benzene $\left( {{C_6}C{l_6}} \right)$ is produced as the product.

Thus, the correct answer is D.

Benzene is a completely conjugated, planar aromatic species with delocalized $\pi $ electron density spread both above and below the ring. As a result, it is rich in electrons. This electron density attracts electron deficient atoms or ions towards it, thus, making the benzene ring susceptible to electrophilic attack. Hence, benzene undergoes electrophilic substitution reactions.