The decreasing order of reactivity of the com | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Solvolysis proceeds via the formation of a carbocation intermediate. The stability of the carbocation determines the rate of reaction.In all three

Vedclass · Accepted Answer

$II > I > III$

Vedclass · Answer

(A) Solvolysis proceeds via the formation of a carbocation intermediate. The stability of the carbocation determines the rate of reaction.In all three cases,a tertiary benzylic carbocation is formed upon the loss of the chloride ion.Compound $II$ has an additional electron-donating methyl group $(-CH_3)$ at the ortho position relative to the carbocation center,which stabilizes the carbocation through the inductive effect.Compound $I$ has a cyclopropyl group at the para position. The cyclopropyl group is electron-donating due to $\sigma$-bond resonance (bent bonds),which stabilizes the carbocation.Compound $III$ is a simple tertiary benzylic carbocation without any additional electron-donating groups on the ring.Comparing the stabilization: The methyl group in $II$ provides significant inductive stabilization. The cyclopropyl group in $I$ also provides stabilization,but the inductive effect of the methyl group in $II$ is more effective in this specific benzylic system.Thus,the order of stability of the carbocations,and hence the rate of solvolysis,is $II > I > III$.

The decreasing order of reactivity of the compounds given below towards solvolysis under identical conditions is:

Similar Questions

The period of the function $f(\theta) = \sin \frac{\theta}{3} + \cos \frac{\theta}{2}$ is (in $\pi$)

In the figure shown,the velocities of different blocks are given. The velocity of $C$ is $.........\,m/s$.

The set of solutions of the equation $(\sqrt{3}-1) \sin \theta+(\sqrt{3}+1) \cos \theta=2$ is

Which of the following statements is correct for the peroxide ion $(O_2^{2-})$?
$A.$ It has completely filled antibonding molecular orbitals.
$B.$ It is diamagnetic.
$C.$ It has a bond order of $1$.
$D.$ It is isoelectronic with Neon.

In the following reaction,$A$ and $B$ respectively are:
$A$ $\xrightarrow{HBr} C_2H_5Br$ $\xrightarrow{B} A$

Vedclass Test Series

Exam Paper Generator

Online Exam Module