The scheme shows the formation of ethanoic anhydride from ethanoic acid.
Which molecule is removed when the two molecules of ethanoic acid join to form ethanoic anhydride?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Water
In the scheme the two acid molecules are joined through one oxygen atom: the –OH of one molecule and the H of the –OH of the other are lost together as water. The formulas show the same:
\( \displaystyle \ce{2CH3COOH -> (CH3CO)2O + H2O} \)
2 \( \ce{C2H4O2} \) − \( \ce{C4H6O3} \) = \( \ce{H2O} \)
H+ (a mineral acid such as H2SO4) or P2O5, with heat, brings about this loss of water.
Why the other options are incorrect
Option (1) Carbon dioxide — Incorrect. The anhydride keeps all four carbon atoms of the two acid molecules, so no carbon is lost.
Option (2) Methane — Incorrect. Both CH3 groups are still present in the anhydride.
Option (4) Hydrogen — Incorrect. One oxygen atom is lost along with two hydrogen atoms; losing only H2 would leave an extra oxygen atom in the product.
NCERT (2026-27 edition):
8.9.2 Reactions Involving Cleavage of C–OH Bond – “1. Formation of anhydride
Carboxylic acids on heating with mineral acids such as H2SO4 or with P2O5 give corresponding anhydride.”
Formation of anhydride (ethanoic acid to ethanoic anhydride)