Q.tert-Butylbromide reacts with aq. NaOH by $\mathrm{S_N1}$ mechanism while n-butylbromide reacts by $\mathrm{S_N2}$ mechanism. Why?

The key idea is that SN1 reactivity depends on carbocation stability, while SN2 reactivity depends on steric hindrance.

1. For tert-butyl bromide: The carbon attached to bromine is tertiary. In an SN1 mechanism, the C–Br bond breaks first to form a carbocation. A tertiary carbocation is highly stabilised by three alkyl groups (hyperconjugation and inductive effect), so this step is fast and favourable. The bulky tert-butyl group also strongly hinders any backside attack, making SN2 impossible. …

The key to understanding this lies in the stability of the intermediate formed in each mechanism. $\mathrm{S_N1}$ reactions proceed through a carbocation intermediate, while $\mathrm{S_N2}$ reactions occur in a single step with no intermediate. The structure of the alkyl halide dictates which pathway is feasible.

1. Carbocation stability determines $\mathrm{S_N1}$ feasibility. For an $\mathrm{S_N1}$ reaction, the rate-determining step is the departure of the leaving group (bromide) to form a carbocation. The more stable the carbocation, the faster this step occurs. Carbocation stability follows the order:

This is because alkyl groups are electron-donating via hyperconjugation and inductive effects, which delocalize the positive charge.

2. tert-Butyl bromide forms a stable tertiary carbocation.

   The structure is $(\text{CH}_3)_3\text{C-Br}$. When bromide leaves, the resulting carbocation is $(\text{CH}_3)_3\text{C}^+$, a tertiary carbocation. This is highly stabilized by three methyl groups donating electron density. The activation energy for this step is low, making $\mathrm{S_N1}$ the dominant mechanism.

   Tip

   A quick way to remember: tertiary alkyl halides almost always react via $\mathrm{S_N1}$ (or $\mathrm{E1}$) because the carbocation is stable enough to form.

3. n-Butyl bromide cannot form a stable carbocation.

   The structure is $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{-Br}$. If bromide were to leave, the resulting carbocation would be primary ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2^+$). Primary carbocations are extremely unstable — they lack sufficient alkyl groups to stabilize the positive charge. The energy required to form such a high-energy intermediate is prohibitively high.

   Watch out

   A common mistake is to think that n-butyl bromide could undergo a hydride shift to form a more stable carbocation. In practice, such rearrangements are possible only under strongly acidic conditions or with good leaving groups in polar solvents — but here, with aqueous NaOH, the $\mathrm{S_N2}$ pathway is much faster than any rearrangement.

4. Steric hindrance also plays a role. …

Concept: Steric Hindrance in Nucleophilic Substitution

The key concept is steric hindrance — the physical bulk of alkyl groups around the reaction centre (the carbon attached to the leaving group). This determines whether the mechanism is $\mathrm{S_N1}$ or $\mathrm{S_N2}$.

Method: Steric Accessibility Analysis

Steps

1. Identify the substrate structure
   • tert-Butylbromide: The central carbon is bonded to three methyl groups and one bromine. Structure:

• n-Butylbromide: The bromine is on a primary carbon (end of a straight chain). Structure:

2. Analyse steric hindrance at the reaction centre

   • In $\mathrm{S_N2}$, the nucleophile must attack from the backside of the carbon–bromine bond.
   • tert-Butylbromide: Three bulky methyl groups block the backside approach — high steric hindrance.
   • n-Butylbromide: Only a hydrogen and an alkyl chain are behind the carbon — low steric hindrance.

3. Match mechanism to steric environment

   • High steric hindrance → $\mathrm{S_N2}$ is impossible → reaction proceeds via $\mathrm{S_N1}$ (two-step, carbocation intermediate).
   • Low steric hindrance → $\mathrm{S_N2}$ is favoured (one-step, direct attack).

4. Check carbocation stability (for $\mathrm{S_N1}$)

   • tert-Butyl carbocation $(\text{CH}_3)_3\text{C}^+$ is tertiary — highly stabilised by hyperconjugation and inductive effects. …

This is a classic comparison that tests your understanding of substrate structure and reaction mechanisms in organic chemistry. Let's break down the common mistakes and how to avoid them.

🧠 The Core Concept

The key difference lies in the stability of the carbocation intermediate and steric hindrance around the carbon bearing the leaving group.

• tert-Butyl bromide (3° alkyl halide) → forms a stable tertiary carbocation → favours $\mathrm{S_N1}$ (two-step, via carbocation).
• n-Butyl bromide (1° alkyl halide) → forms a highly unstable primary carbocation → cannot use $\mathrm{S_N1}$ → instead uses $\mathrm{S_N2}$ (one-step, backside attack).

✗ Common Mistake #1: "Both react by the same mechanism because both have Br as leaving group"

Why it's wrong:

The leaving group is the same, but the alkyl group structure determines the mechanism. The rate-determining step in $\mathrm{S_N1}$ depends only on the substrate (not the nucleophile), while $\mathrm{S_N2}$ depends on both.

How to avoid:

Always check the degree of substitution at the carbon attached to the leaving group:

• 1° → almost always $\mathrm{S_N2}$ (unless special conditions)
• 3° → almost always $\mathrm{S_N1}$ (or E1)
• 2° → can go either way depending on solvent, nucleophile, etc.

✗ Common Mistake #2: "tert-Butyl bromide reacts faster because it has more alkyl groups"

Why it's wrong:

More alkyl groups slow down $\mathrm{S_N2}$ (steric hindrance) but speed up $\mathrm{S_N1}$ (carbocation stabilisation). So the reason for the mechanism choice is opposite for the two pathways.

How to avoid:

Remember the steric vs electronic trade-off:

• $\mathrm{S_N2}$: more alkyl groups = more steric hindrance → slower
• $\mathrm{S_N1}$: more alkyl groups = more hyperconjugation/inductive effect → more stable carbocation → faster

✗ Common Mistake #3: "n-Butyl bromide reacts by $\mathrm{S_N1}$ because it has a primary carbocation"

Why it's wrong:

A primary carbocation is extremely unstable — it does not form under normal conditions. The activation energy is too high.

How to avoid:

Memorise the carbocation stability order:

Only 3° and some 2° carbocations are stable enough for $\mathrm{S_N1}$ in typical exam problems. If you see a primary alkyl halide, assume $\mathrm{S_N2}$ unless told otherwise (e.g., with a very good leaving group and weak nucleophile in a polar protic solvent — but that's rare).

✗ Common Mistake #4: "The nucleophile (OH⁻) is strong, so both should be $\mathrm{S_N2}$"

Why it's wrong: …