Question:

In the following reaction :
\(\text{Aldehyde} + \text{Alcohol} \xrightarrow{\ \mathrm{HCl}\ } \text{Acetal}\)
AldehydeAlcohol
HCHOtBuOH
\(\mathrm{CH_3CHO}\)MeOH
The best combination is

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Acetal formation is favoured by a more electrophilic, less hindered carbonyl and a less hindered, more nucleophilic alcohol.
Updated On: Nov 5, 2026
  • \(\mathrm{CH_3CHO}\) and MeOH
  • HCHO and MeOH
  • \(\mathrm{CH_3CHO}\) and tBuOH
  • HCHO and tBuOH
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The Correct Option is B

Solution and Explanation

Acetal formation: the protonated carbonyl is attacked by ROH (twice).
HCHO has no electron-donating alkyl group and no steric hindrance, so it is the most reactive carbonyl compound (more than \(\mathrm{CH_3CHO}\)).
tBuOH is very bulky (tertiary), which hinders attack and the formation of a crowded acetal; MeOH is small and a good nucleophile.
Best combination: HCHO and MeOH, option (2).
Answer: (2) HCHO and MeOH
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