
Chlorobenzene is formed by reaction of chlorine with benzene in the presence of $AlC{l_3}$. Which of the following species attacks the benzene ring in this reaction$?$
(i) $C{l^ - }$
(ii) $C{l^ + }$
(iii) $AlC{l_3}$
(iv) ${\left[ {AlC{l_4}} \right]^ - }$
Answer
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Hint: The starting compound for this reaction is benzene. It is reacting with the reagent chlorine in presence of $AlC{l_3}$ as a catalyst to give chlorobenzene as the desired product. $AlC{l_3}$ is electron deficient and reacts with electron rich \[C{l_2}\] molecules to undergo heterolytic cleavage and one of the cleaved products reacts with benzene.
Complete step-by-step answer:For getting chlorobenzene as the end product, benzene is reacted with a \[C{l_2}\] molecule in the presence of $AlC{l_3}$ as a catalyst.
Now, $AlC{l_3}$ being a catalyst activates the chlorine molecule and reduces the activation energy of the reaction. Also being an electron deficient molecule it reacts with electron rich \[C{l_2}\] molecule to undergo heterolytic cleavage leading to the formation of two species: ${\left[ {AlC{l_4}} \right]^ - }$ and $C{l^ + }$, where $C{l^ + }$ is an electrophile and hence prefer to get attacked by electron rich centers while ${\left[ {AlC{l_4}} \right]^ - }$is a nucleophile and hence attacks electron deficient centers.
$AlC{l_3} + C{l_2} \to {\left[ {AlC{l_4}} \right]^ - } + C{l^ + }$
Now we know benzene is electron rich in nature due to the presence of delocalized $\pi $ electrons. Hence the electrophile $C{l^ + }$is attacked by the electron rich benzene ring to form chlorobenzene.
${C_6}{H_6}\, + \,C{l^ + }\, \to \,{C_6}{H_5}Cl$
Additional Information: The intermediate formed during the reaction of benzene and chloronium ion is known as sigma complex and in the complex the carbon atom to which the chlorine atom gets attached is $s{p^3}$ hybridized.
Hence the correct answer is (ii) $C{l^ + }$.
Note: Do take proper note of the starting compound, the reagents used and the end product of the reaction. Also do properly notice which one of the reacting species is a nucleophile and hence attacks electron deficient centers and vice versa.
Complete step-by-step answer:For getting chlorobenzene as the end product, benzene is reacted with a \[C{l_2}\] molecule in the presence of $AlC{l_3}$ as a catalyst.
Now, $AlC{l_3}$ being a catalyst activates the chlorine molecule and reduces the activation energy of the reaction. Also being an electron deficient molecule it reacts with electron rich \[C{l_2}\] molecule to undergo heterolytic cleavage leading to the formation of two species: ${\left[ {AlC{l_4}} \right]^ - }$ and $C{l^ + }$, where $C{l^ + }$ is an electrophile and hence prefer to get attacked by electron rich centers while ${\left[ {AlC{l_4}} \right]^ - }$is a nucleophile and hence attacks electron deficient centers.
$AlC{l_3} + C{l_2} \to {\left[ {AlC{l_4}} \right]^ - } + C{l^ + }$
Now we know benzene is electron rich in nature due to the presence of delocalized $\pi $ electrons. Hence the electrophile $C{l^ + }$is attacked by the electron rich benzene ring to form chlorobenzene.
${C_6}{H_6}\, + \,C{l^ + }\, \to \,{C_6}{H_5}Cl$
Additional Information: The intermediate formed during the reaction of benzene and chloronium ion is known as sigma complex and in the complex the carbon atom to which the chlorine atom gets attached is $s{p^3}$ hybridized.
Hence the correct answer is (ii) $C{l^ + }$.
Note: Do take proper note of the starting compound, the reagents used and the end product of the reaction. Also do properly notice which one of the reacting species is a nucleophile and hence attacks electron deficient centers and vice versa.
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