Question
Question: When (E)-3-bromo-3-hexene treated with \({\text{C}}{{\text{H}}_3}{{\text{O}}^ - }\) in \({\text{C}}{...
When (E)-3-bromo-3-hexene treated with CH3O− in CH3OH. The product formed is:
(A) 3-hexyne
(B) 2-hexyne
(C) 2,3-hexadiene
(D) 2,4-hexadiene
Solution
To solve this we can use the Saytzeff’s rule. Saytzeff’s rule is used in the analysis of elimination reactions. The elimination reactions of alcohols and halides produce alkenes. In elimination reactions, the major product is obtained by the Saytzeff’s rule.
Complete step by step solution: We are given (E)-3-bromo-3-hexene which is treated with CH3O− in CH3OH. The product of this reaction is obtained by Saytzeff's rule.
According to Saytzeff’s rule, the major alkene is formed by the removal of hydrogen from a β-carbon which has a low number of substituents.
When (E)-3-bromo-3-hexene is treated with CH3O− in CH3OH it undergoes dehydrobromination reaction. in the dehydrobromination reaction, a hydrogen bromide molecule is eliminated.
The reaction of (E)-3-bromo-3-hexene with CH3O− in CH3OH is as follows:
According to the Saytzeff’s rule, (E)-3-bromo-3-hexene undergoes dehydrobromination and hydrogen bromide gets eliminated fro (E)-3-bromo-3-hexene and 3-hexyne is formed as the major product.
Thus, when (E)-3-bromo-3-hexene is treated with CH3O− in CH3OH, the product formed is 3-hexyne.
Thus, the correct option is (A) 3-hexyne.
Note: We have solved the given reaction of (E)-3-bromo-3-hexene with CH3O− in CH3OH, with the help of Saytzeff’s rule. We know that Saytzeff’s rule favours the alkene with less number of hydrogen on double bonded carbon atoms. In the reaction dehydrobromination of (E)-3-bromo-3-hexene occurs. Thus, Saytzeff’s rule is applicable in this case. During an elimination reaction, a proton is removed from the carbon atom having less number of substituents. Thus when (E)-3-bromo-3-hexene is treated with CH3O− in CH3OH, we get 3-hexyne as a major product.