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

How do we convert But1eneBut-1-ene and But2ene.But-2-ene.

Explanation

Solution

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. But2eneBut-2-ene and But2eneBut-2-ene are structural isomers. They differ in the position of double bonds where but1ene1-ene has the double bond present between the first and second carbon and But2eneBut-2-ene has the double bond between the  2nd~2nd and the 3rd3rd carbon atom. We will discuss the preparation of these two isomers from alcohols and halides.

Complete step by step answer:
We can prepare But1eneBut-1-ene and But2eneBut-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 E1E1 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, But2eneBut-2-ene is formed and if removed from a  2nd~2nd carbon of the same butanol, But2eneBut-2-ene is formed. The reactions are as follows:
CH3CH2CH=CH2HClCH3CH2CH(Cl)CH3C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}\xrightarrow{HCl}C{{H}_{3}}-C{{H}_{2}}-CH(Cl)-C{{H}_{3}}
CH3CH2CH(Cl)CH3alc.KOHCH3CH=CHCH3CH3-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 But1eneBut-1-ene and But2eneBut-2-ene together. Here the Zaitsev rule comes into play. The Zaitsev rule predicts that the major product is But2eneBut-2-ene or but2ene.2-ene. Also, the strong base in the production of alkenes from alkyl halides removes the slightly acidic hydrogen proton from the alkyl halide.