Question

The correct order of rates of solvolysis of the following compounds is:

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

Answer 1

The correct option is D $II>III>I$

Rate of solvolysis can be explained on the basis of ease of carbocation formation.

In structure $\left(I\right)$, the carbocation formed after the removal of $B{r}^{-}$ is an antiaromatic compound, which will be the least stable. So, the rate of solvolysis of the bromo-compound will be the least.


In case of structure $\left(II\right)$, the carbocation formed after the removal of $B{r}^{-}$ is an aromatic compound. So, it will be the most stable and the rate of solvolysis will be more.


In structure $\left(III\right)$, the carbocation formed after the removal of $B{r}^{-}$ is in resonance with an aromatic ring. So, it will be more stable than the carbocation formed in structure$-I$. So, the rate of solvolysis of structure $\left(III\right)$ will be more than the bromo-compound in structure $\left(I\right)$.

Therefore, the order of rates of solvolysis is $II>III>I$.