Question

What happens when 2-Bromobutane is made to react with alcoholic potassium hydroxide?

Hint:

Alcoholic KOH = Elimination (Alkene). Aqueous KOH = Substitution (Alcohol). Remember Zaitsev's rule for the major alkene.

Answer 1

Step 1: Reaction type.
Alcoholic KOH acts as a strong base and promotes beta-elimination (dehydrohalogenation). A hydrogen from the beta-carbon and the bromine from the alpha-carbon are removed, forming an alkene.
Step 2: Available beta-carbons in 2-bromobutane.
2-Bromobutane ($CH_3CH_2CH(Br)CH_3$) has two sets of beta-hydrogens: at C1 (giving but-1-ene) and at C3 (giving but-2-ene).
Step 3: Zaitsev's rule.
According to Zaitsev's rule, the major product is the more substituted (thermodynamically more stable) alkene. But-2-ene ($CH_3CH=CHCH_3$, disubstituted) is more stable than but-1-ene ($CH_3CH_2CH=CH_2$, monosubstituted).
Step 4: Products.
\[ CH_3CH_2CH(Br)CH_3 \xrightarrow{\text{alc. KOH}} CH_3CH=CHCH_3~(\text{major}) + CH_3CH_2CH=CH_2~(\text{minor}) + KBr + H_2O \] But-2-ene is the major product; but-1-ene is the minor product.