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Chapter 22: Q16. (page 892)

Question: Draw the products of each reaction.

a.

b.

c.

d.

Short Answer

Expert verified

Answer

a.

b.

c.

d.

Step by step solution

01

Step-by-Step Solution  Step 1: Fischer-esterification

Carboxylic acids react with alcohol in the presence of acids to form the corresponding ester. This reaction is known as the Fischer-esterification reaction.

It’s a two-step reaction involving the addition of nucleophiles and the elimination of leaving groups.

02

Reaction of carboxylic acid

a. Carboxylic acid reacts with an alcohol in an acidic medium to form an ester.

Reaction with alcohol

b. Carboxylic acid reacts with an alcohol in an acidic medium to form an ester.

Reaction with alcohol

c. NaOCH3 acts as a base and abstracts the acidic proton to form the corresponding sodium salt.

Reaction with base

d. Acid and alcohol are present in the same molecule. Therefore, intramolecular cyclization will occur in the acidic medium to form cyclic ester (lactone).


Formation of cyclic ester

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Most popular questions from this chapter

What ester and Grignard reagent are needed to prepare each alcohol?

a.

b.

c.

Draw the structure of both an acid chloride and an ester that can be used to prepare each compound by reduction.

Review the oxidation reactions using Cr+6 reagents in Section 12.12. Then, draw the product formed when compound B is treated with each reagent.

a. NaBH4, CH3OH

b. [1] LiAlH4 ; [2]H2O

c. PCC

d.Ag2O, NH4OH

e. CrO3, H2SO4, H2O

Draw the product when each compound is treated with either (CH3)2CuLi , followed by H2O , HC≡CLi , followed by H2O .

a.

b.

c.

What alcohol is formed when each compound is treated with NaBH4in CH3OH?
See all solutions

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