Q.Predict the order of reactivity of the following compounds in $S_N1$ and $S_N2$ reactions:

Concept: $S_N1$ vs $S_N2$ Reactivity. The key is carbocation stability for $S_N1$ and steric hindrance for $S_N2$.

(i) Four isomeric bromobutanes

• $S_N1$ depends on carbocation stability: tertiary > secondary > primary. Of the two primary bromides, the carbocation from 1-bromo-2-methylpropane ($(CH_3)_2CHCH_2Br$) is more stable than from 1-bromobutane, because of the stronger electron-donating inductive effect of the $(CH_3)_2CH-$ group -- so 1-bromo-2-methylpropane is more reactive than 1-bromobutane in $S_N1$.

  $S_N1$: tert-butyl bromide > sec-butyl bromide > 1-bromo-2-methylpropane > 1-bromobutane.

• $S_N2$ depends on steric hindrance: methyl > primary > secondary > tertiary.

  $S_N2$: n-butyl > isobutyl > sec-butyl > tert-butyl.

(ii) Benzylic/allylic bromides

• $S_N1$ follows carbocation stability: $C_6H_5CH(C_6H_5)Br$ gives a diphenylmethyl carbocation (very stable, resonance with two phenyls). $C_6H_5C(CH_3)(C_6H_5)Br$ gives a tertiary benzylic carbocation (even more stable due to +I of methyl). $C_6H_5CH(CH_3)Br$ gives a secondary benzylic carbocation. $C_6H_5CH_2Br$ gives a primary benzylic carbocation (least stable among these). …

Part (i): the four isomeric bromobutanes

$CH_3CH_2CH_2CH_2Br$ (1-bromobutane, $1^\circ$), $CH_3CH_2CHBrCH_3$ (2-bromobutane, $2^\circ$), $(CH_3)_2CHCH_2Br$ (1-bromo-2-methylpropane, $1^\circ$), $(CH_3)_3CBr$ (2-bromo-2-methylpropane, $3^\circ$).

$S_N1$ (carbocation stability): among the two primary bromides, the carbocation formed from $(CH_3)_2CHCH_2Br$ is more stable than the one from $CH_3CH_2CH_2CH_2Br$, because of the greater electron-donating inductive effect of the $(CH_3)_2CH-$ group -- so 1-bromo-2-methylpropane is more reactive than 1-bromobutane in $S_N1$:

$S_N2$ (least steric hindrance fastest -- the reverse order):

Part (ii): benzylic bromides

(A) $C_6H_5CH_2Br$ ($1^\circ$ benzylic), (B) $C_6H_5CH(C_6H_5)Br$ ($2^\circ$, two Ph), (C) $C_6H_5CH(CH_3)Br$ ($2^\circ$, one Ph), (D) $C_6H_5C(CH_3)(C_6H_5)Br$ ($3^\circ$, two Ph). …

Method: Carbocation Stability & Steric Hindrance Analysis

This method uses two fundamental principles:

• $S_N1$ depends on carbocation stability (more stable carbocation → faster reaction)
• $S_N2$ depends on steric hindrance (less hindered substrate → faster reaction)

(i) The four isomeric bromobutanes

Compounds:

• 1-bromobutane (primary)
• 2-bromobutane (secondary)
• 1-bromo-2-methylpropane (primary, branched)
• 2-bromo-2-methylpropane (tertiary)

$S_N1$ Reactivity Order

Step 1: Identify the carbocation formed after Br⁻ leaves.

Step 2: Order by carbocation stability. Of the two primary bromides, the carbocation from 1-bromo-2-methylpropane is more stable than from 1-bromobutane, because of the greater electron-donating inductive effect of the $(CH_3)_2CH-$ group.

$S_N1$ order:

2-bromo-2-methylpropane > 2-bromobutane > 1-bromo-2-methylpropane > 1-bromobutane

$S_N2$ Reactivity Order

Step 1: Count the number of alkyl groups on the carbon bearing Br (steric hindrance).

Step 2: Order by increasing steric hindrance.

$S_N2$ order:

1-bromobutane > 1-bromo-2-methylpropane > 2-bromobutane > 2-bromo-2-methylpropane

(ii) Benzylic/allylic bromides

Compounds:

• $C_6H_5CH_2Br$ (primary benzylic)
• $C_6H_5CH(CH_3)Br$ (secondary benzylic)
• $C_6H_5CH(C_6H_5)Br$ (secondary benzylic, diphenyl)
• $C_6H_5C(CH_3)(C_6H_5)Br$ (tertiary benzylic)

$S_N1$ Reactivity Order

Step 1: Evaluate carbocation stability — benzylic carbocations are stabilized by resonance with the phenyl ring. More phenyl groups = more resonance stabilization.

Common Mistakes: $S_N1$ vs $S_N2$ Substitution Reactivity

This question tests your understanding of carbocation stability (for $S_N1$) and steric hindrance (for $S_N2$). Here are the most frequent errors students make:

Mistake 1: Confusing $S_N1$ and $S_N2$ Reactivity Trends

The error: Students often apply the same reasoning to both mechanisms — e.g., assuming that what makes a compound reactive in $S_N1$ also makes it reactive in $S_N2$.

Why it happens: Both reactions involve breaking the C–Br bond, but the rate-determining steps are completely different.

How to avoid:

• $S_N1$ → Rate depends on carbocation stability (more substituted = more stable = faster)
• $S_N2$ → Rate depends on steric hindrance (less hindered = faster)

Key rule: For $S_N1$, think tertiary > secondary > primary > methyl. For $S_N2$, think methyl > primary > secondary > tertiary.

Mistake 2: Forgetting Resonance Stabilisation in Benzylic Systems

The error: Treating $C_6H_5CH_2Br$ (benzyl bromide) as a simple primary alkyl halide.

Why it happens: Students memorise "primary = fast $S_N2$, slow $S_N1$" without considering special cases.

How to avoid:

• The benzyl carbocation ($C_6H_5CH_2^+$) is resonance-stabilised — the positive charge is delocalised into the benzene ring.
• This makes benzyl halides very reactive in $S_N1$ despite being "primary" in a formal sense.
• Similarly, the benzylic position is less hindered than it looks — the flat benzene ring doesn't block backside attack as much as an alkyl group would.

Mistake 3: Misordering the Benzylic Series

The error: Assuming $C_6H_5CH(C_6H_5)Br$ (diphenylmethyl bromide) is less reactive than $C_6H_5CH(CH_3)Br$ in $S_N1$.

Why it happens: Students count alkyl groups but forget that phenyl rings stabilise carbocations even more than alkyl groups.

How to avoid:

• Compare carbocation stability: more phenyl groups = more resonance stabilisation
• Order for $S_N1$: $C_6H_5C(CH_3)(C_6H_5)Br$ (tertiary + 2 phenyl) > $C_6H_5CH(C_6H_5)Br$ (secondary + 2 phenyl) > $C_6H_5CH(CH_3)Br$ (secondary + 1 phenyl) > $C_6H_5CH_2Br$ (primary + 1 phenyl)

Mistake 4: Ignoring Steric Effects in $S_N2$ for Benzylic Halides

The error: Assuming all benzylic halides are equally fast in $S_N2$ because they're "benzylic."

Why it happens: Students over-focus on the benzylic stabilisation and forget that bulky groups still block backside attack.

How to avoid:

• For $S_N2$, steric hindrance dominates:
  • $C_6H_5CH_2Br$ (least hindered) → fastest
  • $C_6H_5CH(CH_3)Br$ (one methyl group) → slower
  • $C_6H_5CH(C_6H_5)Br$ (one phenyl group) → even slower
  • $C_6H_5C(CH_3)(C_6H_5)Br$ (tertiary, two bulky groups) → very slow or no reaction

Mistake 5: Mixing Up the Isomeric Bromobutanes …