Question

How do we convert \[But-1-ene\] and \[But-2-ene.\]

Answer 1

Hint: We need to understand the conversions of alcohols and halides to alkenes. We know that the general formula of alkenes is where n is the number of atoms in the molecule. \[But-2-ene\] and \[But-2-ene\] are structural isomers. They differ in the position of double bonds where but\[1-ene\] has the double bond present between the first and second carbon and \[But-2-ene\] has the double bond between the \[~2nd\] and the \[3rd\] carbon atom. We will discuss the preparation of these two isomers from alcohols and halides.

Complete step by step answer:
We can prepare \[But-1-ene\] and \[But-2-ene\] as following methods, From Alcohols: Alkenes can be synthesized from alcohols by dehydration of alcohols. Alcohols lose water to form a double bond. Secondary and tertiary alcohols dehydrate through the $E1$ mechanism when heated in the presence of (an acid).
The ion leaves first and forms a carbocation as the reaction intermediate and the water then abstracts a proton from an adjacent carbon, forming a double bond. The alkene formed depends on which proton is removed. If proton is removed from a terminal carbon of butanol, \[But-2-ene\] is formed and if removed from a \[~2nd\] carbon of the same butanol, \[But-2-ene\] is formed. The reactions are as follows:
$C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}\xrightarrow{HCl}C{{H}_{3}}-C{{H}_{2}}-CH(Cl)-C{{H}_{3}}$
$CH3-CH2-CH(Cl)-CH3\xrightarrow{alc.KOH}C{{H}_{3}}-CH=CH-C{{H}_{3}}$

Note: It must be noted that the reaction of alcohol to produce alkenes produces both \[But-1-ene\] and \[But-2-ene\] together. Here the Zaitsev rule comes into play. The Zaitsev rule predicts that the major product is \[But-2-ene\] or but\[2-ene.\] Also, the strong base in the production of alkenes from alkyl halides removes the slightly acidic hydrogen proton from the alkyl halide.