Q.Which of the following compounds would undergo $\mathrm{S_N1}$ reaction faster and why?

Concept: SN1 reactivity depends on the stability of the carbocation intermediate formed after the leaving group departs. The more stable the carbocation, the faster the SN1 reaction.

Reasoning:

1. In (A), the leaving group is on a primary carbon attached to a cyclohexane ring. The carbocation formed would be a primary carbocation, which is highly unstable and not stabilized by resonance.
2. In (B), the leaving group is on a carbon directly attached to a benzene ring. The carbocation formed is a benzylic carbocation, which is strongly stabilized by resonance delocalization of the positive charge into the aromatic ring. …

Why This Approach Works

The $\mathrm{S_N1}$ mechanism is a two-step process: first, the leaving group (chloride) departs, forming a carbocation intermediate; second, the nucleophile attacks this carbocation. The rate-determining step is the first step — carbocation formation. So, the faster the carbocation forms and the more stable it is, the faster the $\mathrm{S_N1}$ reaction proceeds.

This means we don't need to think about nucleophile strength or steric hindrance at the transition state (which matters in $\mathrm{S_N2}$). Instead, we compare the stability of the carbocations that would form from each compound.

Step-by-Step Reasoning

1. Identify the leaving group and the carbon it's attached to.

   Both compounds have a chlorine atom as the leaving group. In (A), the chlorine is on a primary carbon (a $\mathrm{-CH_2Cl}$ group attached to a cyclohexane ring). In (B), the chlorine is on a benzylic carbon (the carbon directly attached to the benzene ring).

2. Consider the carbocation formed after $\mathrm{Cl^-}$ leaves.

   For (A): The carbocation is $\mathrm{C_6H_{11}-CH_2^+}$ — a primary carbocation. Primary carbocations are highly unstable because they have only one alkyl group donating electron density via hyperconjugation and the inductive effect.

   For (B): The carbocation is $\mathrm{C_6H_5-CH_2^+}$ — a benzylic carbocation. Here, the positive charge can be delocalized into the benzene ring's $\pi$ system.

3. Analyze the stabilization in the benzylic carbocation.

   The benzene ring has three resonance structures that spread the positive charge onto the ortho and para positions of the ring. This delocalization dramatically lowers the energy of the carbocation, making it much more stable than any simple primary carbocation.

   Resonance stabilization of benzyl carbocation:

   $\mathrm{C_6H_5-CH_2^+ \leftrightarrow \overset{+}{C}_6H_5=CH_2 \text{ (and other resonance forms)}}$

4. Compare the rates directly. …

Method: Carbocation Stability Analysis for SN1 Reactivity

SN1 reactions proceed via a carbocation intermediate. The rate-determining step is the formation of this carbocation. Therefore, the more stable the carbocation, the faster the SN1 reaction.

Step-by-Step Reasoning

Step 1: Identify the carbocation formed by each compound

• (A) (chloromethyl)cyclohexane loses Cl⁻ to form a primary carbocation on the CH₂ group attached to the cyclohexane ring:

  $\text{C}_6\text{H}_{11} \text{—} \overset{+}{\text{C}}\text{H}_2$

• (B) Benzyl chloride loses Cl⁻ to form a benzyl carbocation:

  $\text{C}_6\text{H}_5 \text{—} \overset{+}{\text{C}}\text{H}_2$

Step 2: Compare carbocation stability

• The benzyl carbocation is resonance-stabilized — the positive charge is delocalised into the aromatic ring. This resonance makes it roughly as stable as a tertiary alkyl carbocation (much like an allylic carbocation), even though the CH₂ centre is formally primary. …

This is a classic trap in SN1 reactivity. Let’s break down the common mistakes and how to avoid them.

🧠 The Core Concept First

SN1 reactions depend on carbocation stability. The rate-determining step is the formation of a carbocation after the leaving group departs.

More stable carbocation → faster SN1 reaction.

✗ Mistake #1: Confusing SN1 with SN2

What students do:

They think “less steric hindrance = faster reaction” and pick (A) because the $\ce{-CH2Cl}$ group is on a primary carbon that looks “less crowded.”

Why it’s wrong:

SN1 does not depend on steric hindrance — that’s SN2. SN1 depends only on carbocation stability.

How to avoid:

• If the question says SN1, immediately switch your thinking to carbocation stability.
• Draw the carbocation that would form after $\ce{Cl-}$ leaves.

✗ Mistake #2: Forgetting to draw the carbocation

What students do:

They compare the starting molecules directly without imagining the intermediate.

Why it’s wrong:

You cannot judge SN1 reactivity from the starting compound alone. You must see what cation forms.

How to avoid:

Always sketch the carbocation:

• (A) $\ce{C6H11-CH2+}$ → a primary carbocation (very unstable).
• (B) $\ce{C6H5-CH2+}$ → a benzyl carbocation (highly stabilized by resonance with the benzene ring).

✗ Mistake #3: Ignoring resonance stabilization

What students do:

They think “cyclohexane is bigger, so it might stabilize better” or “benzene is electron-withdrawing, so it slows reaction.”

Why it’s wrong:

The benzyl carbocation is resonance-stabilized — the positive charge is delocalized into the ring. This makes it much more stable than a simple primary carbocation.

How to avoid:

• Memorize the carbocation stability order: Benzyl ≈ Allyl ≈ 3° > 2° > 1° > Methyl
• Draw the resonance structures for benzyl carbocation — you’ll see the charge spread over multiple atoms. …