Question

Consider the following reaction.
${CH}_3-\stackrel{\stackrel{O^{16}}{\parallel }}{C}-O^{18}-C_2{H}_5+H_2{O}^{16}\underset{H^{+}}{\overset{Heat}{\to }}$
The products formed in the reaction are:

• A. ${CH}_3-\stackrel{\stackrel{O^{16}}{\parallel }}{C}-O^{16}H$ and $C_2{H}_5{O}^{18}H$
• B. ${CH}_3-\stackrel{\stackrel{O^{16}}{\parallel }}{C}-O^{18}H$ and $C_2{H}_5{O}^{16}H$
• C. ${CH}_3-\stackrel{\stackrel{O^{16}}{\parallel }}{C}-O^{16}H$ and $C_2{H}_5{O}^{16}H$
• D. ${CH}_3-\stackrel{\stackrel{O^{18}}{\parallel }}{C}-O^{18}H$ and $C_2{H}_5{O}^{18}H$

Answer 1

The correct option is A ${CH}_3-\stackrel{\stackrel{O^{16}}{\parallel }}{C}-O^{16}H$ and $C_2{H}_5{O}^{18}H$

From mechanism, it can be seen that, in esterification, alcohol acts as an acid and gives hydrogen ion.