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Question

The decreasing order of rate of reaction with HBr for the following compounds is :

  1. (a)>(c)>(b)
  2. (c)>(a)>(b)
  3. (a)>(b)>(c)
  4. (c)>(b)>(a)


Solution

The correct option is B (c)>(a)>(b)
Reaction with HBr takes place through carbocation intermediate formation.
More the stability of intermediate formed, more is the rate of reaction.
Intermediates formed are :

 Intermediates (a) and (c) are stabilized by +M effect , whereas (b) is only stabilized by hyperconjugation.
So (b) is the least stable
Since electronegativity of oxygen is more than that of nitrogen, N will have a higher tendency to donate electrons as compared to O.
So, carbocation is more stable in case of (c) as comapred to (a)
Order of intermediate stability : 
(c)>(a)>(b)
Rate of reaction order :
(c)>(a)>(b)

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