Q.Benzylamine may be alkylated as shown in the following equation:
$C_6H_5CH_2NH_2 + R{-}X \rightarrow C_6H_5CH_2NHR$
Which of the following alkyl halides is best suited for this reaction through $S_N1$ mechanism?

SN1 reactivity depends on carbocation stability - the rate-determining step forms a carbocation, so the alkyl halide that gives the most stable carbocation reacts fastest.

Reasoning:

1. CH3Br gives an unstable primary methyl carbocation; C2H5Br gives an unstable primary ethyl carbocation.
2. C6H5Br would require a phenyl carbocation, which never forms - aryl halides do not undergo SN1. …

The key to this problem lies in understanding what controls the rate of an SN1 reaction. Unlike SN2, where steric hindrance and nucleophile strength dominate, SN1 is all about the stability of the intermediate carbocation. The rate-determining step is the ionization of the alkyl halide to form a carbocation and a halide ion. The more stable that carbocation, the faster the reaction proceeds.

Benzylamine is a good nucleophile, but in SN1, the nucleophile attacks after the carbocation has formed, so its identity matters less than how easily the alkyl halide ionizes.

Let's examine each option.

1. (i) CH3Br (methyl bromide) - methyl carbocations are the least stable of all; this substrate reacts almost exclusively via SN2, not SN1.
2. (ii) C6H5Br (bromobenzene) - the bromine sits directly on an aromatic ring. The resulting phenyl cation is extremely unstable (the empty orbital is orthogonal to the ring's pi system, so there is no resonance stabilisation). Aryl halides do not undergo SN1 or SN2 under ordinary conditions. …

Method: Carbocation Stability Analysis for SN1 Reactivity

Concept First

In SN1 reactions, the rate-determining step is the formation of a carbocation intermediate.

The more stable the carbocation, the faster the SN1 reaction proceeds.

Steps

1. Identify the carbocation formed by each alkyl halide after loss of the leaving group ($Br^-$):

   • (A) $CH_3Br$ → $\overset{+}{C}H_3$ (methyl carbocation)
   • (B) $C_6H_5Br$ → $C_6H_5^+$ (phenyl carbocation)
   • (C) $C_6H_5CH_2Br$ → $C_6H_5\overset{+}{C}H_2$ (benzyl carbocation)
   • (D) $C_2H_5Br$ → $CH_3\overset{+}{C}H_2$ (ethyl carbocation)

2. Rank carbocation stability (most stable → least stable):

   • Benzyl carbocation ($C_6H_5CH_2^+$) — resonance-stabilized by the benzene ring
   • Ethyl carbocation ($CH_3CH_2^+$) — hyperconjugation from 3 alpha C–H bonds
   • Methyl carbocation ($CH_3^+$) — no hyperconjugation, no resonance …

✗ Mistake 1: Confusing SN1 with SN2 requirements

What students do:

They see a primary alkyl halide like $CH_3Br$ or $C_2H_5Br$ and think “smaller = more reactive” — which is true for SN2, not SN1.

Why it’s wrong:

SN1 needs a stable carbocation intermediate. Primary carbocations are extremely unstable, so primary halides almost never react via SN1.

How to avoid:

Always ask: “Can this form a stable carbocation?”

• SN1 favours tertiary > secondary > benzylic/allylic > primary > methyl.
• Methyl and primary halides are SN2 specialists, not SN1.

✗ Mistake 2: Forgetting that benzyl carbocation is exceptionally stable

What students do:

They pick $C_6H_5Br$ (bromobenzene) because it has a benzene ring, thinking “benzene = stable”.

Why it’s wrong:

In bromobenzene, the bromine is directly attached to the sp² carbon of the ring. The carbocation that would form (phenyl cation) is highly unstable — it’s an antiaromatic species. SN1 does not occur on aryl halides under normal conditions.

How to avoid:

• Aryl halides ($Ar-X$) do not undergo SN1 or SN2 easily.
• Benzylic halides ($ArCH_2-X$) are the ones that form resonance-stabilised carbocations.

✗ Mistake 3: Ignoring the role of the leaving group

What students do:

They focus only on the carbocation stability and forget that the reaction also requires a good leaving group.

Why it’s wrong:

All four options have Br⁻ as the leaving group — so here it’s not a differentiating factor. But in other problems, a poor leaving group (like F⁻ or OH⁻) can kill SN1 even if the carbocation would be stable.

How to avoid:

Always check both:

1. Carbocation stability (rate-determining step)
2. Leaving group ability (if given different halogens)

✓ Correct approach for this question

Step 1: Identify the mechanism — SN1.

Step 2: Rate depends on carbocation stability.

Step 3: Compare the carbocations formed:

• (A) $CH_3Br$ → $CH_3^+$ (methyl, very unstable) ✗ …