${\text{RCOOR'}}$ 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
249.3k+ 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 Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Other Pages
CBSE Class 12 Chemistry Question Paper 2026 PDF Download (All Sets) with Answer Key

NCERT Solutions For Class 12 Chemistry Chapter 10 Biomolecules - 2025-26

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

NCERT Solutions For Class 12 Chemistry Chapter 2 Electrochemistry - 2025-26

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

NCERT Solutions For Class 12 Chemistry Chapter 3 Chemical Kinetics - 2025-26

