Question

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

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

Answer 1

The correct option is B $III>II>I>IV$

Electron donating groups (EDG) favours $S_{N}1$ reactivity with $HBr$. Electron withdrawing groups (EWG) retards $S_{N}1$ reactivity with $HBr$.
Electron donating effect of $(-OMe)$ group $(+R,-I)$ is greater than $(-Me)$ group $(+I\text{and}H.C.)$. So, reactivity of $III>II$.
In $\left(IV\right),(-OMe)$, 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: $III>II>I>IV$.