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Magazine Chapter 23: Q66. (page 924)
Question: Synthesize (Z)-hept-5-en-2-one from ethyl acetoacetate (CH3COCH2CO2Et ) and the given starting material. You may also use any other organic compounds or required inorganic reagents.
Short Answer
Answer
Step by step solution
Lindlar reagent
It is a type of heterogeneous catalyst that is composed of palladium, which is deposited on barium sulfate.
Steps involved in the reaction
In the first step of the reaction, ethyl 3-oxobutanoate reacts with NaOEt and 3-bromoprop-1-yne to form ethyl 2-acetylpent-4-ynoate.
In the second step, it reacts with ethane-1,2-diol in acidic condition to form ethyl 2-(2-methyl-1,3-dioxolan-2-yl)pent-4-ynoate.
In the third step, it reacts with n-BuLi and Me-I to form ethyl 2-(2-methyl-1,3-dioxolan-2-yl)hex-4-ynoate.
In the fourth step, it undergoes hydrolysis to form 2-acetylhex-4-ynoic acid. In the fifth step, it forms hept-5-yn-2-one under heating.
In the last step, hept-5-yn-2-one reacts with the Lindar reagent to form (Z)-hept-5-en-2-one.
The mechanism is shown below:
Mechanism of the reaction
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