Q.Arrange the compounds of each set in order of reactivity towards $S_N2$ displacement:

Concept: Steric Hindrance in $S_N2$ Reactions. Reactivity decreases as the alkyl halide becomes more substituted because the backside attack is blocked.

Reasoning:

1. $S_N2$ reactivity order: methyl > 1° > 2° > 3°. More alkyl groups on the carbon bearing the leaving group increase steric hindrance.
2. For (i): 1-Bromopentane (1°) > 2-Bromopentane (2°) > 2-Bromo-2-methylbutane (3°).
3. For (ii): 1-Bromo-3-methylbutane (1°) > 2-Bromo-3-methylbutane (2°) > 2-Bromo-2-methylbutane (3°). …

The Concept: Why Steric Hindrance Rules $S_N2$

In an $S_N2$ reaction, the nucleophile attacks the carbon from the back, pushing the leaving group out from the front. This requires the nucleophile to get close to the carbon — any bulky groups near that carbon physically block the approach. The more substituted the carbon (primary < secondary < tertiary), the more crowded it is, and the slower the reaction.

For alkyl halides, the order of $S_N2$ reactivity is:

But within the same class (e.g., all primary), branching on nearby carbons also matters — a neopentyl group is much slower than a simple primary because the bulky tert-butyl group blocks the backside.

Let’s apply this to each set.

(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

1. Identify the carbon type:

   • 2-Bromo-2-methylbutane: The bromine is on a carbon bonded to three other carbons — tertiary.
   • 2-Bromopentane: Bromine on a carbon bonded to two other carbons — secondary.
   • 1-Bromopentane: Bromine on a carbon bonded to only one other carbon — primary.

2. Apply the reactivity order: Tertiary is the slowest, primary is the fastest. So:

   • Fastest: 1-Bromopentane (primary)
   • Middle: 2-Bromopentane (secondary)
   • Slowest: 2-Bromo-2-methylbutane (tertiary)

Order for (i): 1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane

(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane

1. Identify the carbon type:

   • 1-Bromo-3-methylbutane: The bromine is on a primary carbon (CH$_2$Br at the end of a chain).
   • 2-Bromo-3-methylbutane: The bromine is on a secondary carbon (CHBr in the middle).
   • 2-Bromo-2-methylbutane: Again, tertiary.

2. Check for extra hindrance: The secondary one (2-bromo-3-methylbutane) has a methyl branch on the adjacent carbon (C3). That adds some steric bulk near the reaction center, but it’s still secondary — faster than tertiary, slower than primary.

3. Order: Primary > Secondary > Tertiary.

Order for (ii): 1-Bromo-3-methylbutane > 2-Bromo-3-methylbutane > 2-Bromo-2-methylbutane

(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane

All four are primary bromides — the bromine is on a CH$_2$ group at the end of a chain. So we must compare steric hindrance from nearby branching.

1. Draw the structures:
   • 1-Bromobutane: CH$_3$CH$_2$CH$_2$CH$_2$Br — a straight chain, no branching near the reactive carbon. …

Method: Steric Hindrance Analysis in $S_N2$ Reactions

The Steric Hindrance Approach is used here. In $S_N2$ reactions, the nucleophile attacks the electrophilic carbon from the back side. Bulkier groups around the reaction centre slow the reaction because they physically block the approach.

Steps

1. Identify the electrophilic carbon (the carbon bonded to the leaving group, Br).
2. Classify the carbon as primary (1°), secondary (2°), or tertiary (3°).
3. Compare steric hindrance — more alkyl groups on the carbon = more hindrance = slower $S_N2$.
4. Rank in decreasing order of reactivity: 1° > 2° > 3°.

(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

Order of reactivity:

1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane

(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane

Order of reactivity:

1-Bromo-3-methylbutane > 2-Bromo-3-methylbutane > 2-Bromo-2-methylbutane

(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane …

This is a classic test of $S_N2$ reactivity and the effect of steric hindrance. Let’s break down the common mistakes and how to avoid them.

🧠 Core Concept: $S_N2$ Reactivity Order

For $S_N2$ reactions, the rate depends on how easily the nucleophile can access the carbon bearing the leaving group.

Less steric hindrance → faster reaction.

Reactivity order:

Methyl > 1° > 2° > 3° (primary > secondary > tertiary)

✗ Common Mistake #1: Confusing Substrate Class with Carbon Count

Example: Thinking 2-bromopentane (a 2° halide) is more reactive than 1-bromopentane (a 1° halide) because “pentane has more carbons.”

Why it’s wrong:

$S_N2$ reactivity depends on substitution at the reacting carbon, not total chain length.

• 1-Bromopentane: primary (1°) — least hindered, fastest.
• 2-Bromopentane: secondary (2°) — slower.
• 2-Bromo-2-methylbutane: tertiary (3°) — extremely slow (often negligible).

✓ How to avoid:

Always identify the class of the carbon attached to the leaving group (1°, 2°, or 3°) before comparing.

✗ Common Mistake #2: Ignoring Branching Near the Reaction Site

Example: In set (iii), thinking 1-bromo-2,2-dimethylpropane (neopentyl bromide) is as reactive as other 1° bromides.

Why it’s wrong:

Neopentyl bromide is primary but has three methyl groups on the β-carbon. This creates massive steric hindrance — the nucleophile cannot easily approach the backside. Its $S_N2$ rate is much slower than a normal 1° halide.

✓ How to avoid:

For $S_N2$, β-branching matters as much as α-substitution. Draw the structure and check for bulky groups near the reaction centre.

✗ Common Mistake #3: Forgetting That 3° Halides Are Essentially Unreactive in $S_N2$

Example: Placing 2-bromo-2-methylbutane (3°) anywhere except last in the reactivity order.

Why it’s wrong:

Tertiary halides are so sterically hindered that $S_N2$ is practically impossible. They react via $S_N1$ instead.

✓ How to avoid:

In any $S_N2$ ranking, all 3° halides go at the bottom — no exceptions.

✓ Correct Answers (with reasoning)

Set (i): 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

Order:

1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylbutane

Set (ii): 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane

| Compound | Class | Reactivity |

|----------|-------|------------| …