The major product in the following reaction is:

Step 1: Identify the reaction conditions

Alkyl bromide + t-BuOH (tert-butanol) + Heat

t-BuOH is:

• Weak nucleophile (bulky)
• Can act as base
• Polar protic solvent

Step 2: Determine the mechanism

With t-BuOH:

• Too bulky for $S_N2$
• Can promote E2 elimination (acts as base)
• Or E1 elimination (if substrate forms stable carbocation)

Step 3: Common mistakes to avoid

Students often:

• Think t-BuOH will do $S_N$ (it won't - too bulky)
• Forget that bulky bases favor Hofmann product (less substituted alkene)
• Don't consider heat favoring elimination

Step 4: Solving strategy

Bulky base + heat → E2 elimination

Hofmann's rule: Bulky bases favor less substituted alkene

• Opposite of Zaitsev's rule
• Steric hindrance prevents approach to more substituted β-carbon
• Attacks less hindered β-hydrogen

For the given substrate:

• Identify all possible β-hydrogens
• Elimination from less hindered position
• Forms less substituted alkene (Hofmann product)

Given answer (a), this represents the less substituted alkene formed via Hofmann elimination.

Answer: (a)

Key Principles:

Zaitsev vs Hofmann:

| Conditions | Base | Product | Rule | |------------|------|---------|------| | Zaitsev | Small (OH⁻, OEt⁻) | More substituted alkene | Thermodynamic | | Hofmann | Bulky (t-BuO⁻) | Less substituted alkene | Kinetic |

Why bulky bases give Hofmann product:

• Steric hindrance at more substituted β-carbon
• Base attacks less hindered β-hydrogen
• Kinetic control (faster, not more stable)

Examples of bulky bases:

• t-BuO⁻ (tert-butoxide)
• LDA (lithium diisopropylamide)
• DBU (1,8-Diazabicyclo[5.4.0]undec-7-ene)