Question

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

• A. $\left(a\right)>\left(c\right)>\left(b\right)$
• B. $\left(c\right)>\left(a\right)>\left(b\right)$
• C. $\left(c\right)>\left(b\right)>\left(a\right)$
• D. $\left(a\right)>\left(b\right)>\left(c\right)$

Answer 1

The correct option is B $\left(c\right)>\left(a\right)>\left(b\right)$

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 :
$\left(c\right)>\left(a\right)>\left(b\right)$
Rate of reaction order :
$\left(c\right)>\left(a\right)>\left(b\right)$