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QJEEM23Chd788: JEE Main 2023 Chemistry - Class 12 JEEM23Chd788 Solution: Choose the halogen which is most reactive towards S N 1…

Choose the halogen which is most reactive towards SN1 reaction in the given compounds (A, B, C & D)

JEE Main 2023 (Online) 8th April Morning Shift Chemistry - Haloalkanes and Haloarenes Question 38 English

Multiple Choice Options

  • A A−Br(b);B−I(a);C−Br(a);D−Br(a)
  • B A−Br(a);B−I(a);C−Br(a);D−Br(a)
  • C A−Br(b);B−I(b);C−Br(b);D−Br(b)
  • D A−Br(a);B−I(a);C−Br(b);D−Br(a)

Step-by-Step Verified Solution

Organic compounds are more reactive towards SN 1 , if the carbocation is stable compared to others. Reactivity in reactions depends upon the stability of the carbocation intermediate.

JEE Main 2023 (Online) 8th April Morning Shift Chemistry - Haloalkanes and Haloarenes Question 38 English Explanation 1
Br⁡(a)\operatorname{Br}(\mathrm{a}) , Becuase formed intermediate carbocation get stable by conjugation with phenyl ring

JEE Main 2023 (Online) 8th April Morning Shift Chemistry - Haloalkanes and Haloarenes Question 38 English Explanation 2
I(a)\mathrm{I}(\mathrm{a}) , Becuase formed intermediate carbocation become more stable by conjugation

JEE Main 2023 (Online) 8th April Morning Shift Chemistry - Haloalkanes and Haloarenes Question 38 English Explanation 3
Br⁡(b)\operatorname{Br}(\mathrm{b}) , Becuase we can't remove Br(a)B r(a) from bridge head carbon (Bredt's rule)

JEE Main 2023 (Online) 8th April Morning Shift Chemistry - Haloalkanes and Haloarenes Question 38 English Explanation 4
Br⁡(a)\operatorname{Br}(\mathrm{a}) , because formed intermediate 3∘3^{\circ} carbocation is more stable (stability of carbocation 3^{\circ}>2^{\circ} \left.>1^{\circ}\right)

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