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Chemistry Question on Aldehydes, Ketones and Carboxylic Acids

Methyl benzoate can be prepared by

A

C6H5COOH+CH3OH→H+C _{6} H _{5} COOH + CH _{3} OH \xrightarrow{ H ^{+}}

B

C6H5COOCl+CH3OH→ Pyridine C _{6} H _{5} COOCl + CH _{3} OH \xrightarrow{\text { Pyridine }}

C

C6H5COOH+CH2N2→C _{6} H _{5} COOH + CH _{2} N _{2} \rightarrow

D

All of the above method

Answer

All of the above method

Explanation

Solution

C6H5COOH+CH3OH→H+C6H5COOHCH3methyl benzoate+H2O{C_6}{H_5}COOH + C{H_3}OH\xrightarrow{{H^+ }} \underset{\text{methyl benzoate}}{{C_6}{H_5}COOHC{H_3}} + {H_2}O (The reaction is methyl benzoate known as Fischer - Sprier esterification reaction.) C6H5COCl+CH3OH→Pyridine C6H5COOCH3+HCl{C_6}{H_5}COCl + C{H_3}OH\xrightarrow{\text{Pyridine }}{C_6}{H_5}COOC{H_3} + HCl C6H5COOH+CH2 diazomethane →N2C6H5COOCH3+N2C _{6} H _{5} COOH +\underset{\text { diazomethane }}{ CH _{2}} \xrightarrow{ N _{2}} C _{6} H _{5} COOCH _{3}+ N _{2} (Methyl easters can be prepared by treating carboxylic acid with ethereal solution of diazomethane. Hence, methyl benzoate can be prepared by all these above methods.