dehydrohalogenation

Question

Dehydrohalogenation of 2-bromobutane with alcoholic KOH gives:

RankGuru · Accepted Answer

**Dehydrohalogenation** is a $\beta$-elimination reaction. 
- Reagent: **Alcoholic $KOH$** favors elimination over substitution.
- Reaction: $CH_3-CH(Br)-CH_2-CH_3 + KOH(alc) 	o 	ext{Alkene} + KBr + H_2O$
- There are two types of $\beta$-hydrogens available (on C1 and C3).
- **Saytzeff's Rule**: The major product is the more highly substituted alkene (the one with the greater number of alkyl groups attached to the double-bonded carbons).
- Product: **2-Butene** ($CH_3-CH=CH-CH_3$) is more stable and forms the major product (~80%), while 1-Butene is the minor product.

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dehydrohalogenation

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