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Question

Chemistry Question on Amines

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

A

(CH3)2CHCOOCH3\left( CH _{3}\right)_{2} CHCOOCH _{3}

B

CH3COOCH(CH3)2CH _{3} COOCH \left( CH _{3}\right)_{2}

C

CH3COOCH2CH3CH _{3} COOCH _{2} CH _{3}

D

(CH3)2CHCOOCH2CH3\left( CH _{3}\right)_{2} CHCOOCH _{2} CH _{3}

Answer

CH3COOCH(CH3)2CH _{3} COOCH \left( CH _{3}\right)_{2}

Explanation

Solution

CH3COOHethanoic acid+HOCH(CH3)2propan -2-ol\underset{\text{ethanoic acid}}{CH _{3} COOH} + \underset{\text{propan -2-ol}}{HOCH \left( CH _{3}\right)_{2}}
>[H2SO4]CH3COOCH(CH3)2isoPropylethanoate{ ->[H_2SO_4] \underset{iso-Propyl ethanoate}{CH3COOCH(CH_3)_2}}
This reaction is called esterification as ester is the final product.