Glucose is acetylated with acetic anhydride to give glucose pentaacetate. Which other product is formed in this reaction, and how many moles of it are formed from one mole of glucose?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Acetic acid; 5 mol
Acetylation is the introduction of an acetyl (CH3CO) group in place of the hydrogen of an –OH group. With an acid anhydride, each –OH group that reacts gives one ester group and sets free one molecule of the acid:
\( \displaystyle \ce{R-OH + (CH3CO)2O <=>[H+] R-OCOCH3 + CH3COOH} \)
Glucose has five –OH groups, so five acetyl groups go in and five molecules of acetic acid come out:
\( \displaystyle \ce{C6H12O6 + 5(CH3CO)2O -> C16H22O11 + 5CH3COOH} \)
Here \( \ce{C16H22O11} \) is glucose pentaacetate. It is a stable compound, which shows that the five –OH groups are attached to different carbon atoms.
NCERT (2026-27 edition):
10.1.2.1 Glucose – “5. Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms.”
7.4.4 Chemical Reactions – “Alcohols and phenols react with carboxylic acids, acid chlorides and acid anhydrides to form esters.”
\( \displaystyle \ce{Ar/ROH + R'-COOH <=>[H+] Ar/ROCOR' + H2O} \)
\( \displaystyle \ce{Ar/R-OH + (R'CO)2O <=>[H+] Ar/ROCOR' + R'COOH} \)
“…The introduction of acetyl (CH3CO) group in alcohols or phenols is known as acetylation.”