Question

Write the chemical equation for preparation of but-1-ene and but-2-ene from (a) alcohols (b) alkyl halides (c) vicinal dihalides.

Answer 1

Hint: We need to understand the conversions of alcohols and halides to alkenes. We know that the general formula of alkenes is \[{C_n}{H_{2n}}\] where n is the number of atoms in the molecule. But-1-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 $H_2SO_4$(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-1-ene is formed and if removed from a 2nd carbon of the same butanol, but-2-ene is formed. The reactions are as follows:
\[C{H_3}C{H_2}C{H_2}C{H_2}OH\xrightarrow[{{H_2}S{O_4}}]{\Delta }C{H_3}C{H_2}CH = C{H_2}\]
$C{H_3}C{H_2}CH\left( {OH} \right)C{H_3}\xrightarrow[{{H_2}S{O_4}}]{\Delta }C{H_3}C{H_2} = CHC{H_3}$
From alkyl halides: The dehydrohalogenation of alkyl halides produces alkenes by a β elimination reaction which involves the loss of a hydrogen and a halide from an alkyl halide (RX) in the presence of a strong base such as potassium hydroxide or sodium ethoxide. For example, 1-bromobutane will give but-1-ene and 2-bromobutane will give but-2-ene.
The reactions are as follows:
\[C{H_3}C{H_2}C{H_2}C{H_2}Br + KOH\xrightarrow[{}]{{}}C{H_3}C{H_2}CH = C{H_2} + KBr\]
$C{H_3}C{H_2}CH\left( {Br} \right)C{H_3} + KOH \to C{H_3}C{H_2} = CHC{H_3} + KBr$
From vicinal dihalides: Vicinal dihalides are hydrocarbons which have 2 halogens on the same side of the molecule on adjacent carbon atoms. These molecules on reaction with zinc form alkenes. This is also a dehalogenation reaction. 1,2-dibromobutane will give but-1-ene and 2,2-dibromobutane will give but-2-ene. The reactions are as follows:
$C{H_3}C{H_2}CH\left( {Br} \right)C{H_2}Br\xrightarrow{{Zn}}C{H_3}C{H_2}CH = C{H_2} + ZnBr$
$C{H_3}CH\left( {Br} \right)CH\left( {Br} \right)C{H_3}\xrightarrow{{Zn}}C{H_3}CH = CHC{H_3} + KBr$

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 2‐butene 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 via an acid‐base reaction.