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Question
Account for the following:
There are 5 OH groups in glucose
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Solution
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.
\[\begin{array}{cc}
\phantom{...}\ce{CHO}\phantom{...................}\ce{CHO}\phantom{....}\phantom{}\ce{O}\phantom{.........}\\
\phantom{..}|\phantom{.......................}|\phantom{}\phantom{........}||\phantom{........}\\
\phantom{}\ce{(CHOH)4 ->[Acetic anhydride] (CH-O-C-CH3)4}\\
\phantom{..........................}\ce{O}\\
\phantom{..}|\phantom{.......................}|\phantom{.}\phantom{........}||\phantom{........}\\
\phantom{}\ce{CH2OH}\phantom{................}\ce{CH2-O-C-CH3}\phantom{}
\end{array}\]
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RELATED QUESTIONS
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(A)
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Reason: ‘D’ represents its dextrorotatory nature.
Consider the following reactions:
(i) \[\ce{Glucose + R-OH ->[Conc. HNO3] [A] ->[X eq of][(CH3CO)2O] Acetyl derivative}\]
(ii) \[\ce{Glucose ->[Ni/H2] [A] ->[Y eq of][(CH3CO)2O] Acetyl derivative}\]
(iii) \[\ce{Glucose ->[Z eq of][(CH3CO)2O] Acetyl derivative}\]
'X, 'Y' and 'Z' in these reactions are respectively:
