Chemistry - Aromatic Hydrocarbon Question with Solution | TestHub

ChemistryAromatic HydrocarbonAROMATIC NUCLEPHIC SUBSTITUTION BY SNAE and SNEAMedium2 minPYQ_2019
ChemistryMediumsingle choice

The major product obtained in the given reaction is:

Options:

Answer:
D
Solution:

We know that AlCl3 is a lewis acid. It can give Friedel Crafts reactions. Here, we can see that AlCl3 acts as a lewis acid and accepts the electron pair from the chlorine atom. This formsAlCl4. Alongside this, a carbocation is also formed.

Now, this carbocation is an electrophilic carbon. So, benzene rings can give electrophilic substitution reactions here. So, as a result, the double bond of benzene will attack on the carbocation and a C-C bond will be formed there. Then after losing a proton, the carbon ring becomes aromatic. This, we obtain a compound which has two ring structures. One ring is aromatic and the other is not aromatic.

Friedel-Craft alkylation takes place to give a six membered ring.

Stream:JEESubject:ChemistryTopic:Aromatic HydrocarbonSubtopic:AROMATIC NUCLEPHIC SUBSTITUTION BY SNAE and SNEA
2mℹ️ Source: PYQ_2019

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