Question

Give the decreasing order of reactivity of the following compounds with HBr.

(I)

(II)

(III)

(IV)

• A. III > IV > II > I
• B. III > II > IV > I
• C. III > II > I > IV
• D. II > III > IV > I

Answer 1

Answer: (C)

Electron Releasing Group favours ${\text{S}}_{\text{N}}\text{1}$ reactivity with $\text{HBr}$ . $\text{EWG}$ retards ${\text{S}}_{\text{N}}\text{1}$ reactivity with $\text{HBr}$ .

$\text{ERG}$ effect of $\left(-OMe\right)(+R,-I)$ is greater than $(Me)$ group $\left(+IandH.C.\right).$ So, reactivity of $III>II.$

In $\left(IV\right),\left(-OMe\right)$ group is at m-position and exerts only $-I$ effect and so reactivity of $\left(IV\right)$ is less than $\left(I\right).$

So the reactivity order is: $III>II>I>IV$