
When 1-butyne undergoes oxymercuration with the help of $ HgS{O_4} + {H_2}S{O_4} $ the product(s) formed is/are:
(A) $ C{H_3}C{H_2}COOH + HCOOH $
(B) $ C{H_3}C{H_2}COC{H_3} $
(C) $ C{H_3}C{H_2}C{H_2}COOH $
(D) $ C{H_3}C{H_2}C{H_2}CHO $
Answer
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Hint : At first we will write what is given in the question. We will explain the Markownikoff’s rule according to which the reaction is going to occur. In the end we will choose the correct option.
Complete Step By Step Answer:
Step1. The reactant that is going under the reaction is 1-butyne. It undergoes oxymercuration with the help of $ HgS{O_4} + {H_2}S{O_4} $ .
Step2. Oxymercuration is a type of electrophilic addition reaction. IN this reaction it transforms an alkene into a neutral alcohol.
Step3. The Markownikoff’s rule is an outcome of chemical addition reaction. When a protic acid is mixed with the asymmetric alkene, the acidic hydrogen attaches it to the carbon having a greater number of hydrogen substituents. The halide group attaches itself to the carbon atom, the one which has the most amount of alkyl substituents.
Step4. So the reaction happening is
$ C{H_3}C{H_2} - C \equiv CH\xrightarrow[{HgS{O_4} + {H_2}S{O_4}}]{{2{H_2}O}}C{H_3}C{H_2} - {C_2}{H_5}{O_2} $
From here we will reduce the water from it and the reaction will down below
$ C{H_3}C{H_2}COH = C{H_2} $
Now from here the double bond will go from carbon to the oxygen and the hydrogen will move to the carbon.
So the new product will be $ C{H_3}C{H_2}COH = C{H_2} \rightleftarrows C{H_3}C{H_2}COC{H_3} $
Hence the correct option will be B.
Note :
1-butyne is an organic compound whose formula is $ {C_4}{H_6} $ . It is flammable and it is often stored as compressed gas. It has a molecular weight of fifty six grams. It is also often known as ethyl acetylene. It is reactive with alkyne. It occurs as colorless gas.
Complete Step By Step Answer:
Step1. The reactant that is going under the reaction is 1-butyne. It undergoes oxymercuration with the help of $ HgS{O_4} + {H_2}S{O_4} $ .
Step2. Oxymercuration is a type of electrophilic addition reaction. IN this reaction it transforms an alkene into a neutral alcohol.
Step3. The Markownikoff’s rule is an outcome of chemical addition reaction. When a protic acid is mixed with the asymmetric alkene, the acidic hydrogen attaches it to the carbon having a greater number of hydrogen substituents. The halide group attaches itself to the carbon atom, the one which has the most amount of alkyl substituents.
Step4. So the reaction happening is
$ C{H_3}C{H_2} - C \equiv CH\xrightarrow[{HgS{O_4} + {H_2}S{O_4}}]{{2{H_2}O}}C{H_3}C{H_2} - {C_2}{H_5}{O_2} $
From here we will reduce the water from it and the reaction will down below
$ C{H_3}C{H_2}COH = C{H_2} $
Now from here the double bond will go from carbon to the oxygen and the hydrogen will move to the carbon.
So the new product will be $ C{H_3}C{H_2}COH = C{H_2} \rightleftarrows C{H_3}C{H_2}COC{H_3} $
Hence the correct option will be B.
Note :
1-butyne is an organic compound whose formula is $ {C_4}{H_6} $ . It is flammable and it is often stored as compressed gas. It has a molecular weight of fifty six grams. It is also often known as ethyl acetylene. It is reactive with alkyne. It occurs as colorless gas.
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