Which of the following is used in the acetylation of alcohols with acetyl chloride to remove the HCl formed?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Pyridine
Acetylation puts an acetyl (\( \ce{CH3CO} \)) group on the oxygen of an alcohol. With acetyl chloride, HCl is formed as the second product:
\( \displaystyle \ce{R-OH + CH3COCl} \xrightarrow{\text{pyridine}} \ce{R-OCOCH3 + HCl} \)
Pyridine is a base: it neutralises the HCl as it forms, which shifts the equilibrium to the right and gives a better yield of the ester.
Why the other options are incorrect
Option (1) Palladium — Incorrect. Finely divided palladium is a hydrogenation catalyst (for example, reducing aldehydes and ketones to alcohols); it does not take up HCl.
Option (2) Anhydrous zinc chloride — Incorrect. It is a Lewis acid, used with conc. HCl in the Lucas reagent and as a dehydration catalyst; it is not a base.
Option (4) Anhydrous aluminium chloride — Incorrect. It is the Lewis acid catalyst of Friedel-Crafts reactions; it cannot neutralise HCl.
NCERT (2026-27 edition):
7.4.4 Chemical Reactions – “… The reaction with acid chloride is carried out in the presence of a base (pyridine) so as to neutralise HCl which is formed during the reaction. It shifts the equilibrium to the right hand side. The introduction of acetyl (CH3CO) group in alcohols or phenols is known as acetylation.”