Single correct · +4 / −1
The product of dehydrohalogenation of 2-bromobutane with alc. KOH is mainly:
- (a) 1-butene
- (b) 2-butene
- (c) Butane
- (d) But-1-ene
Answer and solution
Answer (b) 2-butene
Solution
Chemistry ma’am · 4 steps
Two kinds of β-hydrogen here, and Saytzeff says the more substituted alkene wins.
- What alc. KOH does. Alcoholic KOH removes HBr: the Br leaves from C2 and a hydrogen leaves from a neighbouring (β) carbon, so a C=C forms between them.
- Find the β-carbons. In CH₃-CHBr-CH₂-CH₃ the carbon with Br is C2. Both C1 and C3 carry hydrogens, so two alkenes are possible.
H removed from Product Alkyl groups on the double bond Share C1 But-1-ene, CH₂=CH-CH₂CH₃ 1 (monosubstituted) minor C3 But-2-ene, CH₃-CH=CH-CH₃ 2 (disubstituted) major - Saytzeff rule. When more than one alkene can form, the major product is the more substituted one. More alkyl groups on the double bond mean more hyperconjugation and a more stable alkene.
- So. The main product is 2-butene, option (b).
The idea to keep. In dehydrohalogenation, list every β-carbon that has a hydrogen. The alkene with more alkyl groups on the double bond is the major product (Saytzeff).
Answer from the NTA final answer key. Solution by PrepWiser.