Q.Which of the compounds will react faster in $\mathrm{S_N1}$ reaction with the $\mathrm{{}^-OH}$ ion?
$\mathrm{CH_3-CH_2-Cl}$ or $\mathrm{C_6H_5-CH_2-Cl}$

The key idea is that SN1 reactivity depends on carbocation stability — the rate-determining step forms a carbocation, so the compound that gives a more stable carbocation reacts faster.

Reasoning:

1. In SN1, the leaving group (Cl) departs first, generating a carbocation intermediate.
2. $\mathrm{CH_3-CH_2^+}$ is a primary carbocation — highly unstable, with no resonance or hyperconjugative stabilisation beyond a few alkyl groups.
3. $\mathrm{C_6H_5-CH_2^+}$ (benzyl carbocation) is stabilised by resonance: the positive charge is delocalised into the aromatic ring, making it far more stable than a primary alkyl carbocation. …

The $\mathrm{S_N1}$ mechanism is a two-step nucleophilic substitution where the rate-determining step is the formation of a carbocation intermediate. The nucleophile ($\mathrm{{}^-OH}$ here) attacks only after the leaving group ($\mathrm{Cl}^-$) has left. So the entire reaction rate depends on how easily the C–Cl bond breaks — and that depends entirely on the stability of the carbocation that forms.

Let’s compare the two candidates.

1. Ethyl chloride ($\mathrm{CH_3-CH_2-Cl}$)

   If the chloride ion leaves, we get an ethyl carbocation: $\mathrm{CH_3-CH_2^+}$. This is a primary carbocation — the positive carbon is attached to only one alkyl group. Primary carbocations are highly unstable because there is very little electron-donating hyperconjugation or inductive stabilization. They are so high in energy that $\mathrm{S_N1}$ reactions with primary substrates are essentially impossible under normal conditions; the reaction would instead follow an $\mathrm{S_N2}$ pathway if a good nucleophile is present. Here, with $\mathrm{{}^-OH}$ (a strong nucleophile), ethyl chloride would react via $\mathrm{S_N2}$, not $\mathrm{S_N1}$.

2. Benzyl chloride ($\mathrm{C_6H_5-CH_2-Cl}$)

   Loss of chloride gives a benzyl carbocation: $\mathrm{C_6H_5-CH_2^+}$. This is not a simple primary carbocation — the positive charge is delocalized into the aromatic ring. The benzene ring’s $\pi$ electrons can overlap with the empty p-orbital on the benzylic carbon, spreading the charge over several atoms. This resonance stabilization dramatically lowers the energy of the carbocation, making it far more stable than any simple alkyl primary carbocation. In fact, the benzyl carbocation is comparable in stability to a tertiary carbocation. …

Method: Carbocation Stability Analysis

This is the Carbocation Stability Method — the rate-determining step of an $\mathrm{S_N1}$ reaction is the formation of a carbocation. The more stable the carbocation intermediate, the faster the reaction.

Steps

1. Identify the leaving group

   Both compounds have $\mathrm{Cl}$ as the leaving group — identical. So the difference lies in the carbon skeleton.

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

   • For $\mathrm{CH_3CH_2Cl}$: $\mathrm{CH_3CH_2^+}$ (a primary carbocation)
   • For $\mathrm{C_6H_5CH_2Cl}$: $\mathrm{C_6H_5CH_2^+}$ (a benzyl carbocation)

3. Compare carbocation stability

   • Benzyl carbocation is highly stabilized by resonance — the positive charge is delocalized into the benzene ring. …

This is a classic trap in SN1 reactivity. Let's break it down.

The Correct Answer

$\mathrm{C_6H_5-CH_2-Cl}$ (benzyl chloride) reacts faster in SN1 with $\mathrm{{}^-OH}$.

The reason: SN1 proceeds via a carbocation intermediate. Benzyl chloride forms a benzyl carbocation ($\mathrm{C_6H_5-CH_2^+}$), which is stabilized by resonance with the benzene ring. The ethyl carbocation from $\mathrm{CH_3-CH_2-Cl}$ is only weakly stabilized by hyperconjugation and induction.

Common Mistakes & How to Avoid Them

Mistake 1: Confusing SN1 and SN2 requirements

• The error: Students think "less steric hindrance = faster" and pick $\mathrm{CH_3-CH_2-Cl}$.
• Why it's wrong: SN1 rate depends on carbocation stability, not steric hindrance. SN2 is the one that cares about sterics.
• How to avoid: Always ask: "Which step is rate-determining?" For SN1, it's carbocation formation — so stability of the carbocation decides the rate.

Mistake 2: Forgetting resonance stabilization of benzyl carbocation

• The error: Treating $\mathrm{C_6H_5-CH_2^+}$ like a simple primary carbocation.
• Why it's wrong: The positive charge is delocalized into the benzene ring — this makes it much more stable than a typical primary carbocation.
• How to avoid: Draw the resonance structures. The benzyl carbocation has at least 4 resonance forms — the charge is spread over multiple atoms.

Mistake 3: Thinking "primary vs primary" means equal reactivity

• The error: Both substrates are primary alkyl halides, so students assume similar SN1 rates.
• Why it's wrong: SN1 rarely happens on simple primary carbocations (they are too unstable). Benzyl and allyl halides are exceptions because of resonance.
• How to avoid: Memorize the special stability order: Allyl ≈ Benzyl > 3° > 2° > 1° > Methyl (for carbocation stability)

Mistake 4: Ignoring the leaving group and nucleophile

• The error: Thinking $\mathrm{{}^-OH}$ (a strong base) forces SN2. …