Use the narrative of the mechanism, steps 1. through 6. below, to diagram/draw the mechanism of the following reaction. Show each step described in the narrative. N(CH3)2 carbonyl compound--ketone 1. Nucleophilic addition of HN(CH3)2 to form a tetrahedral intermediate that contains both an ammonium ion and an oxide. enamine 2. the oxide removes one of the hydrogens from the ammonium to form an aminol intermediate. This compound is not stable--compounds with sp³ hybridized carbons bound to two or more non-halogen heteroatom substituents tend to be unstable. 3. The hydroxyl oxygen takes a proton from the trace amount of acid H₂O* in the reaction mixture to form an oxonium intermediate. 4. The amine lone pair forms a pi bond with the former carbonyl carbon and forces water to leave; giving rise to an imminum ion/ammoniumion. 5. A free molecule of the secondary amine (or a molecule of the solvent, or water from the trace acid solution) removes a hydrogen from a carbon that was alpha to the former carbonyl carbon. 6. The carbanion lonepair becomes a pi bond between the former calphha carbon and the former carbonyl carbon; concurrently, the pi bond between the nitrogen and the former carbonyl carbon becomes a nitrogen lonepair. accessibility: Investigate
Use the narrative of the mechanism, steps 1. through 6. below, to diagram/draw the mechanism of the following reaction. Show each step described in the narrative. N(CH3)2 carbonyl compound--ketone 1. Nucleophilic addition of HN(CH3)2 to form a tetrahedral intermediate that contains both an ammonium ion and an oxide. enamine 2. the oxide removes one of the hydrogens from the ammonium to form an aminol intermediate. This compound is not stable--compounds with sp³ hybridized carbons bound to two or more non-halogen heteroatom substituents tend to be unstable. 3. The hydroxyl oxygen takes a proton from the trace amount of acid H₂O* in the reaction mixture to form an oxonium intermediate. 4. The amine lone pair forms a pi bond with the former carbonyl carbon and forces water to leave; giving rise to an imminum ion/ammoniumion. 5. A free molecule of the secondary amine (or a molecule of the solvent, or water from the trace acid solution) removes a hydrogen from a carbon that was alpha to the former carbonyl carbon. 6. The carbanion lonepair becomes a pi bond between the former calphha carbon and the former carbonyl carbon; concurrently, the pi bond between the nitrogen and the former carbonyl carbon becomes a nitrogen lonepair. accessibility: Investigate
Organic Chemistry: A Guided Inquiry
2nd Edition
ISBN:9780618974122
Author:Andrei Straumanis
Publisher:Andrei Straumanis
Chapter20: Acidity And Pka Of Phenols
Section: Chapter Questions
Problem 11E
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Reactions of Ethers
Ethers (R-O-R’) are compounds formed by replacing hydrogen atoms of an alcohol (R-OH compound) or a phenol (C6H5OH) by an aryl/ acyl group (functional group after removing single hydrogen from an aromatic ring). In this section, reaction, preparation and behavior of ethers are discussed in the context of organic chemistry.
Epoxides
Epoxides are a special class of cyclic ethers which are an important functional group in organic chemistry and generate reactive centers due to their unusual high reactivity. Due to their high reactivity, epoxides are considered to be toxic and mutagenic.
Williamson Ether Synthesis
An organic reaction in which an organohalide and a deprotonated alcohol forms ether is known as Williamson ether synthesis. Alexander Williamson developed the Williamson ether synthesis in 1850. The formation of ether in this synthesis is an SN2 reaction.
Question
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Transcribed Image Text:Use the narrative of the mechanism, steps 1. through 6. below, to diagram/draw the
mechanism of the following reaction. Show each step described in the narrative.
N(CH3)2
carbonyl compound--ketone
1. Nucleophilic addition of HN(CH3)2 to form a tetrahedral intermediate that contains both an
ammonium ion and an oxide.
enamine
2. the oxide removes one of the hydrogens from the ammonium to form an aminol intermediate.
This compound is not stable--compounds with sp³ hybridized carbons bound to two or more
non-halogen heteroatom substituents tend to be unstable.
3. The hydroxyl oxygen takes a proton from the trace amount of acid H₂O* in the reaction mixture
to form an oxonium intermediate.
4. The amine lone pair forms a pi bond with the former carbonyl carbon and forces water to leave;
giving rise to an imminum ion/ammoniumion.
5. A free molecule of the secondary amine (or a molecule of the solvent, or water from the trace
acid solution) removes a hydrogen from a carbon that was alpha to the former carbonyl carbon.
6. The carbanion lonepair becomes a pi bond between the former calphha carbon and the former
carbonyl carbon; concurrently, the pi bond between the nitrogen and the former carbonyl carbon
becomes a nitrogen lonepair.
Accessibility: Investigate
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