
The isomerization of 1−butyne to 2−butyne can be achieved by treatment with:
A. Hydrochloric acid
B. Ammoniacal silver nitrate
C. Ammoniacal cuprous chloride
D. Ethanolic potassium hydroxide
Answer
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Hint: The isomerization of 1−butyne to 2−butyne can be achieved by treatment of 1-butyne with
ethanolic potassium hydroxide. In the presence of ethanolic potassium hydroxide, there will be a pi-cloud shift in the structure.
Complete step by step answer:
The isomerization of 1- butyne to 2- butyne can be done by ethanolic \[KOH\] , as it is a strong base. In 1- butyne the triple bond is between 1st and 2nd carbon. When it reacts with ethanolic \[KOH\], the OH- group of \[KOH\] attacks on the 3rd carbon of 1-butyne.
Now 3rd C has -ve charge on it. By tautomerism, the negative charge will shift on 2nd carbon. So, a double bond will form between 2nd and 3rd carbon and the triple bond between 1st and 2nd carbon will convert into a double bond with addition of H on 1st carbon. This is a very unstable species.
Now, OH- again attacks on 3rd carbon and the same process occurs again and a triple bond will form between 2nd and 3rd carbon and the double bond between 1st and 2nd carbon will convert into a single bond and 2-butyne will form. These types of reactions are stereotypic in nature and the attack of nucleophiles takes place from the opposite side such that the conformation of the compound changes.
The rest of the three reagents are used for a substitution reaction and not any kind of elimination process.
So, option D. Ethanolic potassium hydroxide is correct.
Note:
The isomerization of 2−butyne to 1−butyne can be done by treating it with\[NaN\mathop H\nolimits_2 /liq{\text{ }}N\mathop H\nolimits_2 \]. In the case of isomers, the molecular formula remains the same while the structural formula keeps on varying.
ethanolic potassium hydroxide. In the presence of ethanolic potassium hydroxide, there will be a pi-cloud shift in the structure.
Complete step by step answer:
The isomerization of 1- butyne to 2- butyne can be done by ethanolic \[KOH\] , as it is a strong base. In 1- butyne the triple bond is between 1st and 2nd carbon. When it reacts with ethanolic \[KOH\], the OH- group of \[KOH\] attacks on the 3rd carbon of 1-butyne.
Now 3rd C has -ve charge on it. By tautomerism, the negative charge will shift on 2nd carbon. So, a double bond will form between 2nd and 3rd carbon and the triple bond between 1st and 2nd carbon will convert into a double bond with addition of H on 1st carbon. This is a very unstable species.
Now, OH- again attacks on 3rd carbon and the same process occurs again and a triple bond will form between 2nd and 3rd carbon and the double bond between 1st and 2nd carbon will convert into a single bond and 2-butyne will form. These types of reactions are stereotypic in nature and the attack of nucleophiles takes place from the opposite side such that the conformation of the compound changes.
The rest of the three reagents are used for a substitution reaction and not any kind of elimination process.
So, option D. Ethanolic potassium hydroxide is correct.
Note:
The isomerization of 2−butyne to 1−butyne can be done by treating it with\[NaN\mathop H\nolimits_2 /liq{\text{ }}N\mathop H\nolimits_2 \]. In the case of isomers, the molecular formula remains the same while the structural formula keeps on varying.
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