Aldehydes form gem-dialkoxy compound when they are reacted with :

The correct answer is Option (2) → two equivalent of monohydric alcohol in the presence of dry hydrogen chloride

The carbonyl carbon of an aldehyde is electrophilic and adds alcohols. Dry HCl protonates the carbonyl oxygen and makes the carbon more electrophilic. The addition goes in two stages:

    \( \displaystyle \ce{RCHO + R'OH} \xrightarrow{\text{dry HCl}} \ce{RCH(OH)OR'}\ \ (\text{hemiacetal}) \)

    \( \displaystyle \ce{RCH(OH)OR' + R'OH} \xrightarrow{\text{dry HCl}} \ce{RCH(OR')2 + H2O}\ \ (\text{acetal}) \)

The acetal has two alkoxy groups on the same carbon, so it is the gem-dialkoxy compound. Its formation uses two molecules of a monohydric alcohol per aldehyde molecule.

Why the other options are incorrect

• Option (1) one equivalent of monohydric alcohol in the presence of dry hydrogen chloride — Incorrect. One molecule of alcohol stops at the hemiacetal, which has one −OH and only one alkoxy group on that carbon.

• Option (3) ethylene glycol in the presence of dry hydrogen chloride — Incorrect. Ethylene glycol is a dihydric alcohol; its two −OH groups close a five-membered ring, giving a cyclic acetal (with ketones, the ethylene glycol ketal), not the open-chain dialkoxy compound formed from a monohydric alcohol.

• Option (4) hydrogen cyanide in the presence of sodium hydroxide — Incorrect. Base-catalysed addition of HCN gives a cyanohydrin, RCH(OH)CN, not an alkoxy compound.

NCERT (2026-27 edition):
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.”