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lukilgm
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Hello "organic". I have a question about the reagents that should be used to make the transformation that I attach below.
The first reagent I see is potassium permanganate to oxidize the aldehyde group to carboxylic acid and not alter the alcohol. But then, I don't see how to remove the water to close the ring.
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lukilgm
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I don't understand stage 3 and stage 4 of the link you gave me.
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German Fernandez
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Stage 3 involves the loss of a proton from the oxygen that belonged to the alcohol and the protonation of one of the hydroxyls to transform it into a good leaving group (water). Stage 4
involves the elimination of the water molecule, aided by the transfer of the pair hydroxyl free.
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lukilgm
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In stage 3: How does the proton lose oxygen from the alcohol and then become protonated to the hydroxyl (which then forms water)?
To what extent does it depend on the medium being acidic/basic/neutral for the equilibrium to shift to one of the sides? In stage 4: Why does the hydroxyl lone pair move to form the double bond? (it is that since it does not have any reagent and it does not indicate the medium in which it is found, I do not understand why there is an elimination) To what extent does it depend on the medium being acidic/basic/neutral for it to move to one of the sides of the balance?
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German Fernandez
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Hello, the esterification reaction is carried out in an acid medium (catalytic sulfuric).
In this acid medium, the basic atoms (oxygen) are continuously protonated and depotated. When the mechanism is written, the appropriate acid-base balances are drawn. Thus, to remove a hydroxyl group we protonate it to transform it into a good leaving group. Acid derivatives always follow a reaction mechanism called addition-elimination.
In a first stage, they add the nucleophile (after protonation if the medium is acid), to later eliminate a group, a stage favored by the strong steric congestion that carbon presents. Greetings
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