Chemistry-
General
Easy

Question

Give the decreasing order of the reaction rates of the following benzyl with HBr

  1. I greater than I I greater than I I I greater than I V    
  2. I V greater than I I I greater than I I greater than I    
  3. I I greater than I greater than I I I greater than I V    
  4. I V greater than I I I greater than I greater than I I    

The correct answer is: I greater than I I greater than I I I greater than I V


    The positive charge that develops on benzylic C in this S N to the power of 1 end exponentreaction is most effectively delocalised by (OMe) group (+R and – I right parenthesis. The left parenthesis p minus N O subscript 2 end subscript right parenthesis group withdraws stack e with minus on top density from the ring by – I and – R left parenthesis p-Cl) group withdraws stack e with minus on tops’ from the ring by – I only. More EDG left parenthesis e. g minus O M e right parenthesisstabilies positive chargeon the benzyl C atom, whereas EWG (e. g comma negative N O subscript 2 end subscript comma negative C l right parenthesis destabilises the positive charge. Hence, the reactivity order is:I>II>III>IV

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