Aldehydes react with an excess of a monohydric alcohol in the presence of dry hydrogen chloride. What is the final product?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → An acetal
Dry HCl protonates the carbonyl oxygen and makes the carbonyl carbon more open to attack by the alcohol. The addition takes place in two stages:
One molecule of alcohol adds to give an alkoxyalcohol, the hemiacetal.
With alcohol in excess, a second molecule reacts with the hemiacetal: its –OH is replaced by another –OR′ group and water is lost. The gem-dialkoxy compound formed is the acetal.
Both steps take place in the presence of dry HCl:
\( \displaystyle \ce{RCHO + R'OH} \rightleftharpoons \ce{RCH(OH)OR'} \) (hemiacetal)
\( \displaystyle \ce{RCH(OH)OR' + R'OH} \rightleftharpoons \ce{RCH(OR')2 + H2O} \) (acetal)
Why the other options are incorrect
Option (1) A hemiacetal — Incorrect. The hemiacetal is only the intermediate; with excess alcohol it reacts further to give the acetal.
Option (3) A gem-diol — Incorrect. A gem-diol has two –OH groups on one carbon; it would come from adding water, not an alcohol. The acetal has two –OR′ groups on that carbon.
Option (4) A vicinal diol — Incorrect. A vicinal diol has –OH groups on two adjacent carbons; addition of an alcohol to one C=O group cannot put groups on two carbons.
NCERT (2026-27 edition):
8.4 Chemical Reactions – “(d) Addition of alcohols: Aldehydes react with one equivalent of monohydric alcohol in the presence of dry hydrogen chloride to yield alkoxyalcohol intermediate, known as hemiacetals, which further react with one more molecule of alcohol to give a gem-dialkoxy compound known as acetal as shown in the reaction.”