The following reaction proceeds in two parts. The first part produces an intermediate compound A and...
The following reaction proceeds in two parts. The first part produces an intermediate compound A and byproduct HCOO–. The product of the second part is a bicyclic ?,?-unsaturated carbonyl compound.
Homework Answers
The given reaction involves between an aldehyde and an - unsaturated carbonyl compound. This reaction is known as Robinson’s annulation and results in the formation of a ring.
Robinson’s annulation is a reaction which involves two reactions that is a Michael addition followed by aldol condensation.
In Michael addition, firstly formation of an enolate takes place by deprotonation by a base. Then the enolate formed attacks the carbonyl moiety via 1,4 conjugate addition. The final step of Michael addition involves protonation.
Then the aldol condensation takes place to give the final product.
The intermediate A is as follows:
The mechanism of the first step is follows:
The mechanism of the second step is follows:
The mechanism of the third step is follows:
The mechanism of the fourth step is follows:
The mechanism of the fifth step is follows:
The mechanism is as follows:
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The following reaction proceeds in two parts. The first part produces an intermediate compound A and...
Please answer the last two parts. The first part is correct This question has multiple parts....
Please answer the last two parts. The first part is
correct
This question has multiple parts. Work all the parts to get the most points The Stork reaction is a condensation reaction between an enamine donor and an aB-unsaturated carbonyl acceptor. The overall reaction consists of a three-step sequence of 1 formation of an enamine from a ketone, 2 Michael adition to an a.p-unsaturated carbonyl compound, and 3. hydrolysis of the enamine in dilute acid to regenerate the ketone. Consider...
answer all parts Predict the two structures of the major products obtained when the following compound...
answer all parts
Predict the two structures of the major products obtained when the following compound undergoes acid hydrolysis. CH, HO Hint: This compound is an enamine, formed from a carbonyl compound (ketone or aldehyde) and a secondary amine. Predict the two structures of the major products obtained when the following compound undergoes acid hydrolysis. Hint: This compound is an enamine, formed from a carbonyl compound (ketone or aldehyde) and a secondary amine. Predict the two structures of the major...
Draw the structures of the intermediate compound and final organic product in the following two-step reaction....
Draw the structures of the intermediate compound and final
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formula of the final product is given.
Why does the following reaction use 2 equivalents of amine? 2@g CHANH. Products??? The first equivalent...
Why does the following reaction use 2 equivalents of amine? 2@g CHANH. Products??? The first equivalent undergoes nucleophilie attack with the carbonyl carbon, the second equivalent neutralizes the acid byproduct The first equivalent undergoes electrophilie attack with the carbonyl carbon, the second equivalent undergoes nucleophilic attack with the carbonyl oxygen The first equivalent undergoes nucleophili tak with the carbonyl carbon, the second equivalent undergoes nucleophilic attack with the carbonyl oxygen The first equivalent undergoes nucleophilic attack with the carbonyl carbon,...
Part A Choose the correct reaction coordinate diagram for a two-step reaction in which the first...
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A 3-ethylcyclohexan-1-one reacts with both propMgBr, and water. In the first part of the reaction, the propane part of propMgBr takes the bonding electrions from MgBr, becoming a carbanion. The carbanion attacks the slightly positive carbon of the carbonyl group. The pi obnd between the oxygen and carbon break, and the electron pair moves toward the oxygen atom. Now the intermediate reacts with water. First a hydrogen atom loses it's bonding pair to the oxygen in the water molecule. Now...
For a reaction that proceeds by the following two-step mechanism A2 ⇄ 2A (fast) A +...
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Part A Draw the intermediate (in the first box) that forms in the reaction of 1-methylcyclopentene...
Part A Draw the intermediate (in the first box) that forms in the reaction of 1-methylcyclopentene with Br2 in CCl4, then draw the expected product (in the second box). Do not provide stereochemistry. Interactive 3D display mode CH3 Draw the species on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars, including charges where needed. The single bond is active by default. Marvin JS Cl Br
nothing else was given. Part A Draw the intermediate (in the first box) that immediately precedes...
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Please answer the last two parts. The first part is
correct
This question has multiple parts. Work all the parts to get the most points The Stork reaction is a condensation reaction between an enamine donor and an aB-unsaturated carbonyl acceptor. The overall reaction consists of a three-step sequence of 1 formation of an enamine from a ketone, 2 Michael adition to an a.p-unsaturated carbonyl compound, and 3. hydrolysis of the enamine in dilute acid to regenerate the ketone. Consider...
answer all parts
Predict the two structures of the major products obtained when the following compound undergoes acid hydrolysis. CH, HO Hint: This compound is an enamine, formed from a carbonyl compound (ketone or aldehyde) and a secondary amine. Predict the two structures of the major products obtained when the following compound undergoes acid hydrolysis. Hint: This compound is an enamine, formed from a carbonyl compound (ketone or aldehyde) and a secondary amine. Predict the two structures of the major...
Draw the structures of the intermediate compound and final
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Why does the following reaction use 2 equivalents of amine? 2@g CHANH. Products??? The first equivalent undergoes nucleophilie attack with the carbonyl carbon, the second equivalent neutralizes the acid byproduct The first equivalent undergoes electrophilie attack with the carbonyl carbon, the second equivalent undergoes nucleophilic attack with the carbonyl oxygen The first equivalent undergoes nucleophili tak with the carbonyl carbon, the second equivalent undergoes nucleophilic attack with the carbonyl oxygen The first equivalent undergoes nucleophilic attack with the carbonyl carbon,...
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Part A Draw the intermediate (in the first box) that forms in the reaction of 1-methylcyclopentene with Br2 in CCl4, then draw the expected product (in the second box). Do not provide stereochemistry. Interactive 3D display mode CH3 Draw the species on the canvas by choosing buttons from the Tools (for bonds), Atoms, and Advanced Template toolbars, including charges where needed. The single bond is active by default. Marvin JS Cl Br
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