Q.Out of $C_6H_5CH_2Cl$ and $C_6H_5CHClC_6H_5$, which is more easily hydrolysed by aqueous KOH.

The key idea is SN1 reactivity: the rate depends on the stability of the carbocation intermediate formed after the leaving group departs.

1. Both compounds are benzylic chlorides, so both can form resonance-stabilised carbocations.
2. $C_6H_5CH_2Cl$ gives a primary benzylic carbocation ($C_6H_5CH_2^+$), stabilised by resonance with one phenyl ring. …

This question is about SN1 hydrolysis — a reaction where the leaving group (Cl) departs first, forming a carbocation intermediate, which is then attacked by water (or OH⁻ from KOH). The rate depends entirely on how stable that carbocation is. Aqueous KOH provides a polar, protic environment that favours SN1 over SN2 for these benzylic halides.

Let’s compare the two compounds.

1. Identify the leaving group and the potential carbocation.

   Both are benzylic chlorides. In $C_6H_5CH_2Cl$, the carbon bearing Cl is attached to one phenyl ring and two hydrogens. In $C_6H_5CHClC_6H_5$, it is attached to two phenyl rings and one hydrogen.

2. Think about carbocation stability.

   A carbocation is stabilised by resonance with adjacent π-systems (like phenyl rings) and by hyperconjugation from alkyl groups. The more resonance contributors you can draw, the more the positive charge is delocalised, and the more stable the carbocation.

   • For $C_6H_5CH_2Cl$: The carbocation formed is $C_6H_5CH_2^+$ (benzyl carbocation). The positive charge can be delocalised into the phenyl ring — you can draw resonance structures where the charge moves to the ortho and para positions. This is moderately stable.
   • For $C_6H_5CHClC_6H_5$: The carbocation formed is $C_6H_5CH^+C_6H_5$ (diphenylmethyl carbocation, also called benzhydryl carbocation). Here, the positive charge is delocalised into two phenyl rings simultaneously. That means twice as many resonance structures and much greater charge dispersal.

   Carbocation stability order:

   $CH_3^+ < \text{primary} < \text{secondary} < \text{tertiary} < \text{allylic/benzylic} < \text{diphenylmethyl} < \text{triphenylmethyl}$

3. Relate stability to SN1 rate.

   In SN1, the rate-determining step is the formation of the carbocation. The more stable the carbocation, the lower the activation energy for its formation, and the faster the reaction. So the compound that gives the more stable carbocation will hydrolyse more easily. …

Method: Carbocation Stability Analysis for SN1 Reactivity

Concept: SN1 reactions proceed via a carbocation intermediate. The rate depends on the stability of that carbocation — more stable carbocation → faster reaction.

Steps

1. Identify the leaving group

   Both compounds have Cl as the leaving group. The nucleophile (OH⁻ from KOH) is the same. So the difference lies in the carbocation formed.

2. Draw the carbocation formed after Cl⁻ leaves

   • $C_6H_5CH_2Cl$ → benzylic carbocation: $C_6H_5CH_2^+$
   • $C_6H_5CHClC_6H_5$ → diphenylmethyl carbocation: $(C_6H_5)_2CH^+$

3. Compare carbocation stability

   • Both are benzylic (resonance-stabilized by the phenyl ring).
   • The diphenylmethyl carbocation has two phenyl rings donating electron density via resonance → more delocalization of positive charge. …

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

🧠 The Core Concept First

Both compounds are benzylic halides, so both can form a resonance-stabilized carbocation. The question is: which carbocation is more stable?

• $C_6H_5CH_2Cl$ (benzyl chloride) → forms $C_6H_5CH_2^+$ (primary benzylic carbocation)
• $C_6H_5CHClC_6H_5$ (diphenylmethyl chloride / benzhydryl chloride) → forms $(C_6H_5)_2CH^+$ (secondary benzylic carbocation, stabilized by two phenyl rings)

More phenyl rings = more resonance stabilization → more stable carbocation → faster SN1 reaction.

✓ Correct answer: $C_6H_5CHClC_6H_5$ is more easily hydrolysed.

✗ Common Mistake #1: Confusing “more easily hydrolysed” with “more reactive toward nucleophile”

The error: Students think “easily hydrolysed” means the compound reacts faster with OH⁻ directly (SN2). They compare steric hindrance and say the less hindered one (benzyl chloride) reacts faster.

Why it’s wrong: Aqueous KOH is a weak base in water (due to solvation). SN1 is favoured over SN2 for benzylic halides because:

• The carbocation is highly stabilized.
• The nucleophile (OH⁻) is weak and solvated.

How to avoid: Always check the mechanism first. For benzylic/allylic/tertiary halides, SN1 dominates in protic solvents like water/ethanol.

✗ Common Mistake #2: Forgetting that both are benzylic — then guessing based on “primary vs secondary” alone

The error: Students say “primary halide reacts faster in SN2, so benzyl chloride is more reactive.” They forget that both are benzylic and both favour SN1.

Why it’s wrong: In SN1, the rate depends only on carbocation stability, not on the leaving group’s steric environment. The diphenylmethyl carbocation is more stable because the positive charge is delocalized over two aromatic rings.

How to avoid: Draw the resonance structures of both carbocations. Count the number of resonance contributors — more = more stable.

✗ Common Mistake #3: Ignoring the role of the solvent (aqueous KOH)

The error: Students treat this as a dry reaction and assume SN2 is always faster for primary halides. …