In the following reaction, trace the position of isotopic $^{18} O$ :

H_{2} O^{18}

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C H_{3} - C ∣ ∣^{18} O - O C H_{2} C H_{3}

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C H_{3} - C ∣ ∣ O -^{18} O C H_{2} C H_{3}

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Can be in all three above

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Solution

The correct option is C $C H_{3} - C ∣ ∣ O -^{18} O C H_{2} C H_{3}$
This esterification reaction between carboxylic acid and ethanol takes place in the presence of concentrated sulfuric acid. The ethanoic acid takes a proton (a hydrogen ion) from the concentrated sulphuric acid. Then the positive charge on the carbon atom is attacked by one of the lone pairs on the oxygen of the ethanol molecule. Hence the isotopic $O^{18}$ is in the ester bonded between $C O$ and $C_{2} H_{5}$ groups.

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