Question

Propan-2-ol + ethanoic acid $ \xrightarrow{{}} $

• A. $\left( CH _{3}\right)_{2} CHCOOCH _{3}$
• B. $CH _{3} COOCH \left( CH _{3}\right)_{2}$
• C. $CH _{3} COOCH _{2} CH _{3}$
• D. $\left( CH _{3}\right)_{2} CHCOOCH _{2} CH _{3}$

Answer 1

Answer: (B)

$\underset{\text{ethanoic acid}}{CH _{3} COOH} + \underset{\text{propan -2-ol}}{HOCH \left( CH _{3}\right)_{2}}$
$ \xrightarrow{H_2SO_4} \underset{\text{iso-Propyl ethanoate}}{CH_3COOCH(CH_3)_2}$
This reaction is called esterification as ester is the final product.