Question

How do you explain the presence of five – OH groups in glucose molecule?

Answer 1

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.

\[\begin{array}{cc}
\phantom{....}\ce{CHO}\phantom{...................}\ce{CHO}\phantom{......}\ce{O}\phantom{...........}\\
\phantom{....}|\phantom{........................}|\phantom{..........}||\phantom{...........}\\
\phantom{}\ce{(CHOH)4 ->[Acetic anhydride] \phantom{.}(CH - O - C - CH3)4}\phantom{}\\
\phantom{..................}|\phantom{..........}\ce{O}\\
\phantom{....}|\phantom{........................}|\phantom{..........}||\phantom{...........}\\
\phantom{.....}\ce{CH2OH}\phantom{.................}\ce{CH2 - O - C - CH3}\phantom{.....}
\end{array}\]