Verified Solution • 2025–26 Session moderate

QJEEM19Cha010: JEE Main 2019 Chemistry - Class 12 JEEM19Cha010 Solution: The major product in the following conversion is –

The major product in the following conversion is –

JEE Main 2019 (Online) 12th January Evening Slot Chemistry - Haloalkanes and Haloarenes Question 127 English

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/d5bfbe8155b992f39731dff1bba22735d920d0237204bdd3d6eab8eb4d2401ad.webp
  • B https://ik.imagekit.io/novatra/photos/6bcd4e5e83cd250df53748ddffbeb49e2adf2fccb4147d1695986f888531bc7f.webp
  • C https://ik.imagekit.io/novatra/photos/00b0afd4352f77debc609845a660a53e53866281d6764714ad5570b37eb52e21.webp
  • D https://ik.imagekit.io/novatra/photos/c53d4aa3ad38330cba1a864f3bcd4680a2147a4df34b45a58d3b741d88555d78.webp

Step-by-Step Verified Solution

JEE Main 2019 (Online) 12th January Evening Slot Chemistry - Haloalkanes and Haloarenes Question 127 English Explanation 1
As here HBr is present in excess, so HBr will react to alkene and ether part sinultaneously.

According to Markovnikov rule. Carbocation close to benzene ring is more stable.

In the given reagent, carbocation is →

JEE Main 2019 (Online) 12th January Evening Slot Chemistry - Haloalkanes and Haloarenes Question 127 English Explanation 2
So in HBr, Br− adds to the carbocation.

And in the ether part because of lone pair of oxygen, it will take part in resonance a double bond will create between O atom and benzene ring. So, this bond will be stronger. That is why CH3 − O bond will break.

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