
${\text{RCOOR'}}$ can be prepared by:
(a) Esterification of RCOOH
(b) Esterification of $(RCO)_2O$
(c) Baeyer - Villiger oxidation of ${\text{RCOR'}}$ with peroxy acid
(d) Reaction of RCOCl with ${\text{R'OH}}$
Answer
222.6k+ views
Hint: In order to answer the question, we have to have knowledge about the structure of carboxylic acid. Esterification process can be explained in an easier manner then.
Complete step-by-step answer:
On esterification of carboxylic acid (RCOOH), or an acid anhydride (RCOOCOR) with an alcohol (${\text{R'OH}}$), an ester (${\text{RCOOR'}}$) is formed. This reaction can be backed by the following chemical equation.
$\left. \text{RCOOH + {R}'OH }\overset{{}}\leftrightarrows{\text{ RCOO{R}' + }}{{\text{H}}_{\text{2}}}\text{O} \right\}\text{Esterification of Carboxylic Acid with Alcohol}$
\[\left. \text{RCOOCOR + {R}'OH }\overset{{}}\leftrightarrows{\text{ RCOO{R}' + RCOOH}} \right\}\text{Esterification of Acid Anhydrides with Alcohol}\]
An example of the same is shown by the following chemical equation.
\[\begin{align}
& \text{ C}{{\text{H}}_{\text{3}}}\text{COOH + }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH }\overset{{}}\leftrightarrows{\text{ C}}{{\text{H}}_{\text{3}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{ + }{{\text{H}}_{2}}\text{O} \\
& \text{ Acid Chloride Alcohol Ester Water} \\
\end{align}\]
\[\begin{align}
& \text{ }{{\left( \text{C}{{\text{H}}_{\text{3}}}\text{CO} \right)}_{\text{2}}}\text{O + }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH }\overset{{}}{\leftrightarrows}\text{ C}{{\text{H}}_{\text{3}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{ + C}{{\text{H}}_{\text{3}}}\text{COOH} \\
& \text{Acid Anhydrade Alcohol Ester Carboxylic Acid} \\
\end{align}\]
An ester (${\text{RCOOR'}}$) is obtained when a ketone (${\text{RCOR'}}$) is treated with peroxy acid (${\text{R''COOOR}}$), also known as the Baeyer Villiger oxidation. This reaction can be backed by the following chemical equation.
$\left. \text{RCO{R}' + {R}''COOOH }\overset{-\text{{R}''COOH}}{\leftrightarrows}\text{ RCOO{R}' } \right\}\text{Baeyer Villiger Oxydation}$
An example of the same is shown by the following chemical equation.
\[\begin{align}
& \text{C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{C}}_{2}}{{\text{H}}_{5}}\text{ + }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{COOOH }\overset{-{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{COOH}}{\leftrightarrows}\text{ C}{{\text{H}}_{\text{3}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}} \\
& \text{ Ketone Peroxy Acid Ester} \\
\end{align}\]
Also an ester (${\text{RCOOR'}}$) is formed when an acid chloride (RCOCl) reacts with an alcohol (${\text{R'OH}}$). This reaction can be backed by the following chemical equation.
$\left. \text{RCOCl + {R}'OH }\overset{{}}{\leftrightarrows}\text{ RCOO{R}' + HCl} \right\}\text{Reaction of Acid Chloride with Alcohol}$
An example of the same is shown by the following chemical equation.
\[\begin{align}
& \text{ C}{{\text{H}}_{\text{3}}}\text{COCl + }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH }\overset{{}}{\leftrightarrows}\text{ C}{{\text{H}}_{\text{3}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{ + HCl} \\
& \text{ Acid Chloride Alcohol Ester Hydrochloric Acid} \\
\end{align}\]
So we can conclude that ${\text{RCOOR'}}$can be prepared by:
Esterification of RCOOH
Baeyer - Villiger oxidation of ${\text{RCOR'}}$ with peroxy acid
Reaction of RCOCl with ${\text{R'OH}}$
Therefore, the correct options are Option A, Option B, Option C and Option D.
Note: We should be familiar with the concept of esterification. So here is the definition. Esterification is a general name for a chemical reaction in which two reactants, which are typically alcohols and an acid, form an ester as a reaction product. Esters are common in organic chemistry and biological materials and often have a pleasant characteristic fruity odor.
Let us now know the definition of Baeyer - Villiger oxidation. Baeyer - Villiger oxidation is an organic reaction that forms an ester from a ketone or a lactone from acyclic ketones using peroxyacids or peroxides as the oxidant.
Complete step-by-step answer:
On esterification of carboxylic acid (RCOOH), or an acid anhydride (RCOOCOR) with an alcohol (${\text{R'OH}}$), an ester (${\text{RCOOR'}}$) is formed. This reaction can be backed by the following chemical equation.
$\left. \text{RCOOH + {R}'OH }\overset{{}}\leftrightarrows{\text{ RCOO{R}' + }}{{\text{H}}_{\text{2}}}\text{O} \right\}\text{Esterification of Carboxylic Acid with Alcohol}$
\[\left. \text{RCOOCOR + {R}'OH }\overset{{}}\leftrightarrows{\text{ RCOO{R}' + RCOOH}} \right\}\text{Esterification of Acid Anhydrides with Alcohol}\]
An example of the same is shown by the following chemical equation.
\[\begin{align}
& \text{ C}{{\text{H}}_{\text{3}}}\text{COOH + }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH }\overset{{}}\leftrightarrows{\text{ C}}{{\text{H}}_{\text{3}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{ + }{{\text{H}}_{2}}\text{O} \\
& \text{ Acid Chloride Alcohol Ester Water} \\
\end{align}\]
\[\begin{align}
& \text{ }{{\left( \text{C}{{\text{H}}_{\text{3}}}\text{CO} \right)}_{\text{2}}}\text{O + }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH }\overset{{}}{\leftrightarrows}\text{ C}{{\text{H}}_{\text{3}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{ + C}{{\text{H}}_{\text{3}}}\text{COOH} \\
& \text{Acid Anhydrade Alcohol Ester Carboxylic Acid} \\
\end{align}\]
An ester (${\text{RCOOR'}}$) is obtained when a ketone (${\text{RCOR'}}$) is treated with peroxy acid (${\text{R''COOOR}}$), also known as the Baeyer Villiger oxidation. This reaction can be backed by the following chemical equation.
$\left. \text{RCO{R}' + {R}''COOOH }\overset{-\text{{R}''COOH}}{\leftrightarrows}\text{ RCOO{R}' } \right\}\text{Baeyer Villiger Oxydation}$
An example of the same is shown by the following chemical equation.
\[\begin{align}
& \text{C}{{\text{H}}_{\text{3}}}\text{CO}{{\text{C}}_{2}}{{\text{H}}_{5}}\text{ + }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{COOOH }\overset{-{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{COOH}}{\leftrightarrows}\text{ C}{{\text{H}}_{\text{3}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}} \\
& \text{ Ketone Peroxy Acid Ester} \\
\end{align}\]
Also an ester (${\text{RCOOR'}}$) is formed when an acid chloride (RCOCl) reacts with an alcohol (${\text{R'OH}}$). This reaction can be backed by the following chemical equation.
$\left. \text{RCOCl + {R}'OH }\overset{{}}{\leftrightarrows}\text{ RCOO{R}' + HCl} \right\}\text{Reaction of Acid Chloride with Alcohol}$
An example of the same is shown by the following chemical equation.
\[\begin{align}
& \text{ C}{{\text{H}}_{\text{3}}}\text{COCl + }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{OH }\overset{{}}{\leftrightarrows}\text{ C}{{\text{H}}_{\text{3}}}\text{COO}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{ + HCl} \\
& \text{ Acid Chloride Alcohol Ester Hydrochloric Acid} \\
\end{align}\]
So we can conclude that ${\text{RCOOR'}}$can be prepared by:
Esterification of RCOOH
Baeyer - Villiger oxidation of ${\text{RCOR'}}$ with peroxy acid
Reaction of RCOCl with ${\text{R'OH}}$
Therefore, the correct options are Option A, Option B, Option C and Option D.
Note: We should be familiar with the concept of esterification. So here is the definition. Esterification is a general name for a chemical reaction in which two reactants, which are typically alcohols and an acid, form an ester as a reaction product. Esters are common in organic chemistry and biological materials and often have a pleasant characteristic fruity odor.
Let us now know the definition of Baeyer - Villiger oxidation. Baeyer - Villiger oxidation is an organic reaction that forms an ester from a ketone or a lactone from acyclic ketones using peroxyacids or peroxides as the oxidant.
Recently Updated Pages
JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Other Pages
Solutions Class 12 Chemistry Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 12 Chemistry Chapter 1 Solutions - 2025-26

The D and F Block Elements Class 12 Chemistry Chapter 4 CBSE Notes - 2025-26

NCERT Solutions for Class 12 Chemistry Chapter Chapter 7 Alcohol Phenol and Ether

NCERT Solutions ForClass 12 Chemistry Chapter Chapter 8 Aldehydes Ketones And Carboxylic Acids

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

