Question

How do you explain the presence of an aldehydic group in a glucose molecule?

Answer 1

Glucose reacts with hydroxylamine to form a monoxime, and adds one molecule of hydrogen cyanide to give cyanohydrin.

\[\begin{array}{cc}
\phantom{.......}\ce{CHO}\phantom{....................}\ce{CH = N - OH}\phantom{..}\\
\phantom{......}|\phantom{.........................}|\phantom{..............}\\
\ce{(CH2OH)4 + NH2OH -> (CHOH)4}\\
\phantom{......}|\phantom{.........................}|\phantom{..............}\\
\phantom{}\ce{CH2OH}\phantom{.................}\ce{\underset{Glucose oxime}{CH2OH}\phantom{.}}
\end{array}\]

\[\begin{array}{cc}
\phantom{.....}\ce{CHO}\phantom{.....................}\ce{CH(OH)CN}\phantom{..}\\
\phantom{.......}|\phantom{.........................}|\phantom{...............}\\
\phantom{.....}\ce{(CH2OH)4 + HCN -> \phantom{...}(CHOH)4}\phantom{.....}\\
\phantom{......}|\phantom{.........................}|\phantom{..............}\\
\phantom{.....}\ce{CH2OH}\phantom{...............}\ce{\underset{GlucoseCyanohydrin}{CH2OH}\phantom{...}}
\end{array}\]

Therefore, it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a carboxylic acid containing six carbon atoms.

\[\begin{array}{cc}
\phantom{}\ce{CHO}\phantom{....................}\ce{COOH}\phantom{...}\\
\phantom{...}|\phantom{.........................}|\phantom{...........}\\
\phantom{...}\ce{(CHOH)4 + O ->[Br2, H2O] \phantom{...}(CHOH)4}\phantom{...}\\
\phantom{......}|\phantom{.........................}|\phantom{..............}\\
\phantom{}\ce{CH2OH}\phantom{..................}\ce{\underset{Glucose acid}{CH2OH}\phantom{.}}
\end{array}\]

This indicates that carbonyl group present in glucose is an aldehydic group.