Q.The major product formed when 2-bromo-2-methylbutane is refluxed with ethanolic KOH is

Step 1. 2-Bromo-2-methylbutane, CH3-CH2-C(Br)(CH3)-CH3, is a TERTIARY alkyl halide; refluxing it with ethanolic (alcoholic) KOH is a classic base-promoted ELIMINATION (E2/E1) condition, not substitution, for a tertiary substrate.

Step 2. The C-Br carbon (C2) has beta-hydrogens available on TWO different neighbouring carbons: the C1 methyl group and the C3 methylene (CH2) group, so elimination can go in two directions.

Step 3. Eliminating toward C3 (removing a C3-H) gives 2-methylbut-2-ene, CH3-C(CH3)=CH-CH3 -- a TRISUBSTITUTED alkene (three alkyl groups total on the two double-bond carbons).

Step 4. Eliminating toward C1 (removing a C1-H) gives 2-methylbut-1-ene, CH2=C(CH3)-CH2-CH3 -- a DISUBSTITUTED alkene. …

Identify all beta-hydrogens available for E2/E1 elimination from the halide carbon, draw the alkene each direction would give, and apply Zaitsev's rule …

• Forgetting that ethanolic/alcoholic KOH specifically favours ELIMINATION over substitution -- picking an alcohol product (options b or d) would be the mistake of applying aqueous-substitution logic to an elimination-favouring solvent/base combination. …