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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Hydrocarbons

JEE Main 2019 · 10 April · Evening Shift · Chemistry · Hydrocarbons

The major product obtained in the given reaction is:

The major product obtained in the given reaction is:

The major product obtained in the given reaction is:

Options

  1. ALet E _ 1 (r), E _ 2 (r) and E _ 3 (r) be the respective electric fields at a distance r from a point charge Q, an
  2. BLet E _ 1 (r), E _ 2 (r) and E _ 3 (r) be the respective electric fields at a distance r from a point charge Q, an
  3. CLet E _ 1 (r), E _ 2 (r) and E _ 3 (r) be the respective electric fields at a distance r from a point charge Q, an
  4. DLet E _ 1 (r), E _ 2 (r) and E _ 3 (r) be the respective electric fields at a distance r from a point charge Q, an

Correct answer

D. Let E _ 1 (r), E _ 2 (r) and E _ 3 (r) be the respective electric fields at a distance r from a point charge Q, an

Step-by-step solution

We know that AlCl3{AlCl}_{3} is a lewis acid. It can give Friedel Crafts reactions. Here, we can see that AlCl3{AlCl}_{3} acts as a lewis acid and accepts the electron pair from the chlorine atom. This forms−AlCl4.- {AlCl}_{4} . 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.

The major product obtained in the given reaction is:

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