All of the following contribute to the large, negative, free-energy change upon hydrolysis of "high-energy" compounds except: a.
1 answer:
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Answer:
See explanation below
Explanation:
In this case we have the starting reactant which is the ethine, In the first step reacts with NaNH₂, a strong base. This base will substract the hydrogen from one of the carbon of the ethine, and form a carbanion. This will react with the propane bromide, displacing the bromine and forming a 5 carbon chain with the triple bond on the carbon 1 and 2.
In the second step, reacts with the lindlar catalyst to do a reduction, and form a double bond between carbon 1 and 2. In essence, compound A is similar to compound B.
Finally B reacts with water in acid and makes a addition reaction, and form an alcohol.
The whole process can be seen in the picture below.
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