Question

The rate of solvolysis of I–IV follows 

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

Hint:

In solvolysis, more substituted carbocations lead to faster reactions due to greater stability.

Answer 1

The Correct Option is C

The problem involves determining the rate of solvolysis of the given compounds (I–IV). Solvolysis is a type of nucleophilic substitution where the solvent acts as the nucleophile. The rate of solvolysis is influenced by the stability of the carbocation formed during the reaction. More stable carbocations lead to faster solvolysis rates.

Let's analyze each compound:

1. Compound III: This compound can form a secondary carbocation, which is stabilized by hyperconjugation and possible resonance effects. It is the most likely candidate to solvolyze rapidly.
2. Compound II: This compound forms a less stable secondary carbocation compared to III but is more stable than the carbocation in compound I due to lesser steric strain.
3. Compound I: The carbocation here is less stable due to steric factors and limited hyperconjugation, leading to a slower rate compared to III and II.
4. Compound IV: This compound forms the least stable carbocation due to significant ring strain and steric hindrance, resulting in the slowest solvolysis rate.

Therefore, the order of solvolysis rate from fastest to slowest is:

III > II > I > IV