Question

Formation of 2-butene from 2-bromobutane is according to
A. Markownikoff’s rule
B. Bayer
C. Saytzeff
D. Wurtz

Answer 1

Hint: For the formation of alkene mostly an elimination reaction takes place which is also known as dehydrohalogenation. It is a one-step mechanism where two sigma bonds are broken to form a pi bond. No intermediate is formed in this reaction.

Complete Step by Step Answer:
For the formation of 2-butene from 2-bromobutane the reaction will be as follows-
$CH_3CHBrCH_2CH_3 + \ \ \ Alcoholic\ KOH\ \ \ \rightarrow CH_3CH=CHCH_3 + HCl$
This reaction proceeds through E2 elimination. In E2 elimination the major alkene that is stable alkene formed is a Saytzeff alkene and the alkene which is unstable is called Hoffmann alkene. For the formation of saytzeff alkene, the β-hydrogen is required because for the formation of stable alkene removal of hydrogen from β-carbon is required. This reaction follows regioselectivity.

In elimination reaction base abstracts the proton from the carbon but the proton present in an alkene is not so acidic in nature so to abstract that proton from the alkene a strong base is required and alcoholic KOH acts as a strong base.
$KOH+EtOH \rightarrow EtOK+H_2O$

This reaction establishes equilibrium at a very early state just after the 1% formation of the product so it acts as a strong alkali. Thus alcoholic KOH proceeds through elimination reaction. But aqueous KOH is not used for elimination reaction because KOH is highly soluble in water due to its solvation in water its alkalinity decreases.

Thus, Option (A) is correct

Note: Regioselectivity is the selection or the preference of the bond, the selection of which bond will break and the position where the reagent will attack. Since it selects the bond so the product formed is a selective product thus it is called regioselectivity.