Join WhatsApp Icon JEE WhatsApp Group
Question 41

image


'X' is:

1. Formation of the Carbocation (Protonation)

The reaction uses hydrofluoric acid ($$\text{HF}$$) as an acid catalyst. The exocyclic double bond of methylenecyclohexane undergoes protonation according to Markovnikov's rule:

  • The $$H^+$$ from $$\text{HF}$$ adds to the terminal carbon ($$=\text{CH}_2$$), converting it into a methyl group ($$-\text{CH}_3$$).
  • This creates a highly stable tertiary ($$3^\circ$$) carbocation at the ring carbon (the 1-methylcyclohexyl cation).

2. Electrophilic Aromatic Substitution (Friedel-Crafts Alkylation)

Benzene acts as the nucleophile and attacks this stable tertiary carbocation:

  • The $$\pi$$-electrons of the benzene ring attack the positively charged carbon atom of the 1-methylcyclohexyl cation.
  • Subsequent deprotonation restores the aromaticity of the benzene ring.
image

Get AI Help

Video Solution

video

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests
Ask AI