Question

Which type of elimination takes place when 2-phenylethylbromide heated with NaOEt ?

• A. El elimination
• B. E2 elimination
• C. ElcB elimination
• D. E2 or E1cB

Answer 1

The correct option is B E2 elimination

A strong base favours E2 reaction as it readily attacks the proton.
Since, strong base and good leaving are involved in this reaction it wont go by $E1cB$ reaction mechanism.

Substrate has only one $\beta$-carbons so only one product is formed as shown below.