Question

The reactivity of the following compounds on the basis of the \( \mathrm{S_N1} \) mechanism is

(I), (II), (III), (IV) as shown in the figure.

• A. \( \text{IV}>\text{III}>\text{I}>\text{II} \)
• B. \( \text{II}>\text{IV}>\text{III}>\text{I} \)
• C. \( \text{III}>\text{IV}>\text{I}>\text{II} \)
• D. \( \text{IV}>\text{III}>\text{II}>\text{I} \)

Hint:

For \( \mathrm{S_N1} \) reactions, always rank compounds by:
Carbocation stability
Presence of resonance (benzylic>allylic>alkyl)
Degree of substitution Leaving group is the same, so it does not affect comparison here.

Answer 1

The Correct Option is A

Concept: In an \( \mathrm{S_N1} \) reaction, the rate-determining step is the formation of a carbocation. Hence, the reactivity order depends on the stability of the carbocation formed after the leaving group departs. Carbocation stability increases due to:
Greater degree of substitution (\(3^\circ>2^\circ>1^\circ\))
Resonance stabilization (benzylic, allylic)
Hyperconjugation
Step 1: Analyze compound (IV). Compound (IV) forms a tertiary benzylic carbocation with two phenyl rings. The positive charge is extensively stabilized by:
Resonance with both phenyl rings
Hyperconjugation Thus, (IV) forms the most stable carbocation and reacts fastest.
Step 2: Analyze compound (III). Compound (III) is a tertiary alkyl bromide. It forms a tertiary carbocation stabilized by hyperconjugation, but no resonance stabilization is present. Hence, its reactivity is less than (IV) but greater than secondary carbocations.
Step 3: Analyze compounds (I) and (II). Both (I) and (II) are secondary cyclohexyl bromides. However:
In (I), the carbocation formed is relatively more stable due to better structural accommodation.
In (II), the carbocation is less stable due to poorer overlap and higher ring strain. Thus, \[ \text{Stability: } \text{I}>\text{II} \]
Step 4: Write the overall order based on carbocation stability. \[ \boxed{\text{IV}>\text{III}>\text{I}>\text{II}} \] This matches Option (1).