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Chapter 24: Q.13. (page 962)

Question: Draw a stepwise mechanism for the conversion of heptane-2,6-dione to 3-methylcyclohex-2-enone with NaOEt, EtOH.

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Step by step solution

01

Aldol reaction

Aldol reaction in the presence of base is termed as base-catalyzed aldol reaction. It is performed under NaOEt, EtOH and is a reversible reaction.

02

Explanation for the mechanism

In the first step of the reaction, EtO-from NaOEt abstracts an alpha hydrogen to create a nucleophilic carbon center. Then, the intermolecular attack takes place to form a six-membered ring.

Negative charge on oxygen abstracts proton from EtOH to form hydroxyl group in the compound. Further,EtO-abstracts proton to create negative charge. The rearrangement takes place to form 3-methylcyclohex-2-enone.

The mechanism is shown below:

Mechanism of the reaction

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Most popular questions from this chapter

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Question: Devise a synthesis of 2-methylcyclopentanone from cyclohexene. You may also use any required reagents

Question: Answer the following questions about 2-acetylcyclopentanone.

a. What starting materials are needed to form 2-acetylcyclopentanone by a Claisen reaction that forms bond (a)?

b. What starting materials are needed to form 2-acetylcyclopentanone by a Claisen reaction that forms bond (b)?

c. What product is formed when 2-acetylcyclopentanone is treated with NaOCH2CH3, followed by CH3I?

d. Draw the Robinson annulation product(s) formed by reaction of 2-acetylcyclopentanone with methyl vinyl ketone (CH2=CHCOCH3).

e. Draw the structure of the most stable enol tautomer(s).
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