Question

The major product \(P\) from the following reaction is:

• A. Option 1
• B. Option 2
• C. Option 3
• D. Option 4

Hint:

Bulky bases like \(t\)-BuO⁻ favor E2 elimination and often give the more stable conjugated alkene as the major product.

Answer 1

The Correct Option is C

Step 1: Understanding the nature of substrate.
The given substrate is a substituted benzyl bromide system containing a highly substituted (benzylic tertiary) carbon bearing a bromine atom. Such substrates are highly prone to elimination reactions under strong basic conditions due to the stability of the resulting alkene.

Step 2: Nature of reagent (Me\(_3\)CONa).
Sodium tert-butoxide (Me\(_3\)CONa) is a strong, bulky base. Bulky bases favor elimination (E2) over substitution because steric hindrance prevents nucleophilic attack but allows proton abstraction.

Step 3: Identifying reaction pathway.
The reaction proceeds via E2 elimination mechanism where the base abstracts a \(\beta\)-hydrogen anti to the leaving group (Br), leading to removal of HBr and formation of a double bond.

Step 4: Formation of alkene product.
Elimination leads to the formation of the most stable conjugated alkene system. In this case, the product is a substituted styrene-type alkene where the double bond is formed between the benzylic carbon and adjacent carbon, giving maximum stability due to conjugation with the aromatic ring.

Step 5: Stability consideration.
The product is stabilized by resonance with the benzene ring (conjugation), making it the major product over any substitution product. Hence, elimination dominates and gives the alkene shown in option (3).

Step 6: Final verification.
Among the given options, only option (3) represents the conjugated alkene formed via E2 elimination with a bulky base.
Final Answer: \[ \boxed{\text{Option (3)}} \]