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Chapter 22: Q18. (page 892)

Draw a stepwise mechanism for the following reaction.

Short Answer

Expert verified

Answer

Formation of the final product

Step by step solution

01

Step-by-Step SolutionStep 1: The intramolecular esterification

The intramolecular esterification takes place from the and hydroxy carboxylic acids. It occurs in the presence of a strong acid and forms five or six-membered lactones.

During this reaction, the acid protonates the carbonyl group and causes the intramolecular ring closure resulting in the formation of lactones.

02

The stepwise mechanism of the given reaction

  • The first step involves the protonation of the carboxylic acid group followed by the removal of a proton.

First step

  • The second step is the ring closure by the nucleophilic attack of the -OH group on the carbonyl carbon.

Second step

  • The next step is the protonation of the alcohol group.

Third step

  • Finally, a proton is removed by the group:

Formation of the final product

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Most popular questions from this chapter

Question:What reagents are needed to convert phenylacetonitrile (C6H5CH2CN) to each compound:

a.C6H5CH2CO(CH3)b.C6H5CH2COC(CH3)3c.C6H5CH2CHOd.C6H5CH2COOH

Draw all stereoisomers formed in each reaction.

What organocuprate reagent is needed to convert CH3CH2COCIto each ketone?

a.

b.

c.

Identify the lettered compounds in the following reaction scheme. Compounds F, G, and K are isomers of molecular formula C13H18O. How could 1HNMR spectroscopy distinguish these three compounds from each other?

Question: Give the IUPAC or common name for each compound.

a.

b.

c.

d.

e.

f.
See all solutions

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