The general formula of an ester (where R represents an alkyl group) is:
(A) ${\text{R - OH}}$
(B) ${\text{R - COOH}}$
(C) ${\text{R - COOR}}$
(D) ${\text{R - H}}$
Answer
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Hint: The naturally occurring chemical substances that are formed by combination of an acid and an alcohol are known as esters. Esters are organic in nature.
Step by step answer: The functional group of an alcohol is ${\text{ - OH}}$. Thus, ${\text{R - OH}}$ is a general formula of an alcohol.
Thus, option (A) is not correct.
The functional group of an acid is ${\text{ - COOH}}$. Thus, ${\text{R - COOH}}$ is a general formula of an acid.
Thus, option (B) is not correct.
Esters are formed by the combination of an acid and an alcohol. Thus, the general formula of an ester is ${\text{R - COOR}}$.
Thus, option (C) is correct.
Adding a hydrogen atom to an alkyl group gives the parent alkane. Thus, the general formula ${\text{R - H}}$ is for an alkane.
Thus, option (D) is not correct.
Thus, the correct option is (C) ${\text{R - COOR}}$.
Additional Information: The general reaction for formation of an ester by combination of an acid and an alcohol is as follows:
${\text{R - OH + R - COOH}} \to {\text{R - COOR + }}{{\text{H}}_2}{\text{O}}$
In the reaction, a water molecule is eliminated.
The reaction of propanol and methanoic acid produces an ester named propyl methanoate. The reaction is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - OH + H - COOH }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - COOH + }}{{\text{H}}_2}{\text{O}}$
Note: When esters are named, name the alcoholic group first, then name the acid group with –oate ending.
In ${\text{C}}{{\text{H}}_{\text{3}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - COOH}}$, the alcoholic group is propyl and the acid group is methane. In the acid group, replace –e with –oate. Thus, the name of the ester is propyl methanoate.
Other example of ester is ethyl propanoate ${\text{C}}{{\text{H}}_{\text{3}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - COO - C}}{{\text{H}}_2}{\text{ - C}}{{\text{H}}_3}$. Ethyl propanoate is formed by the reaction of ethanol $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{ - OH}}} \right)$ with propanoic acid $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{ - COOH}}} \right)$.
Step by step answer: The functional group of an alcohol is ${\text{ - OH}}$. Thus, ${\text{R - OH}}$ is a general formula of an alcohol.
Thus, option (A) is not correct.
The functional group of an acid is ${\text{ - COOH}}$. Thus, ${\text{R - COOH}}$ is a general formula of an acid.
Thus, option (B) is not correct.
Esters are formed by the combination of an acid and an alcohol. Thus, the general formula of an ester is ${\text{R - COOR}}$.
Thus, option (C) is correct.
Adding a hydrogen atom to an alkyl group gives the parent alkane. Thus, the general formula ${\text{R - H}}$ is for an alkane.
Thus, option (D) is not correct.
Thus, the correct option is (C) ${\text{R - COOR}}$.
Additional Information: The general reaction for formation of an ester by combination of an acid and an alcohol is as follows:
${\text{R - OH + R - COOH}} \to {\text{R - COOR + }}{{\text{H}}_2}{\text{O}}$
In the reaction, a water molecule is eliminated.
The reaction of propanol and methanoic acid produces an ester named propyl methanoate. The reaction is as follows:
${\text{C}}{{\text{H}}_{\text{3}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - OH + H - COOH }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - COOH + }}{{\text{H}}_2}{\text{O}}$
Note: When esters are named, name the alcoholic group first, then name the acid group with –oate ending.
In ${\text{C}}{{\text{H}}_{\text{3}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - COOH}}$, the alcoholic group is propyl and the acid group is methane. In the acid group, replace –e with –oate. Thus, the name of the ester is propyl methanoate.
Other example of ester is ethyl propanoate ${\text{C}}{{\text{H}}_{\text{3}}}{\text{ - C}}{{\text{H}}_{\text{2}}}{\text{ - COO - C}}{{\text{H}}_2}{\text{ - C}}{{\text{H}}_3}$. Ethyl propanoate is formed by the reaction of ethanol $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{ - OH}}} \right)$ with propanoic acid $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{ - COOH}}} \right)$.
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