Question

Elimination of bromine from 2-bromobutane results in the formation of

• A. predominantly 2-butyne
• B. predominantly 1-butene
• C. predominantly 2-butene
• D. equimolar mixture of 1 and 2-butene

Answer 1

The correct option is C predominantly 2-butene

$E_1$ reaction is a two step process.
In step 1, leaving group leaves and form a carbocation.
In step 2, the base will attack the proton and proton abstraction takes place to form alkenes.
The major product will be the alkene which has more substitution by Saytzeff rule. Since, it gives the stable alkene.
In given reaction, proton abstraction from carbocation can lead to 2 different product (b) and (c).

Comparing (b) and (c), we can say that (b) is the more substituted alkene. Thus, by Saytzeff rule, compound (b) is predominantly formed in the given reaction.